106598053 Resolucao Mecanica Vetorial para Engenheiros 9ed cap 2 5 estatica

PROBLEM 2.1 Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P = 75 N and Q = 125 N, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION (a)

Parallelogram law:

(b)

Triangle rule:

We measure:

R = 179 N, α = 75.1°

R = 179 N

75.1°

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 3

PROBLEM 2.2 Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P = 60 lb and Q = 25 lb, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION (a)

Parallelogram law:

(b)

Triangle rule:

We measure:

R = 77.1 lb, α = 85.4°

R = 77.1 lb

85.4°

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 4

PROBLEM 2.3 The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

We measure: (a)

Parallelogram law:

(b)

Triangle rule:

We measure:

α = 51.3° β = 59.0°

γ = 67.0°

R = 139.1 lb,

R = 139.1 lb

67.0°

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 5

PROBLEM 2.4 Two forces are applied at Point B of beam AB. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION (a)

Parallelogram law:

(b)

Triangle rule:

We measure:

R = 3.30 kN, α = 66.6°

R = 3.30 kN

66.6°

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 6

PROBLEM 2.5 The 300-lb force is to be resolved into components along lines a-a′ and b-b′. (a) Determine the angle α by trigonometry knowing that the component along line a-a′ is to be 240 lb. (b) What is the corresponding value of the component along b-b′?

SOLUTION (a)

Using the triangle rule and law of sines: sin β sin 60° = 240 lb 300 lb sin β = 0.69282 β = 43.854° α + β + 60° = 180°

α = 180° − 60° − 43.854° = 76.146°

(b)

Law of sines:

Fbb′ 300 lb = sin 76.146° sin 60°

α = 76.1° Fbb′ = 336 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 7

PROBLEM 2.6 The 300-lb force is to be resolved into components along lines a-a′ and b-b′. (a) Determine the angle α by trigonometry knowing that the component along line b-b′ is to be 120 lb. (b) What is the corresponding value of the component along a-a′?

SOLUTION Using the triangle rule and law of sines: (a)

sin α sin 60° = 120 lb 300 lb

sin α = 0.34641 α = 20.268°

(b)

α = 20.3°

α + β + 60° = 180° β = 180° − 60° − 20.268° = 99.732° Faa′ 300 lb = sin 99.732° sin 60°

Faa′ = 341 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 8

PROBLEM 2.7 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R.

SOLUTION Using the triangle rule and law of sines: (a)

sin α sin 25° = 50 N 35 N sin α = 0.60374

α = 37.138° (b)

α = 37.1°

α + β + 25° = 180° β = 180° − 25° − 37.138° = 117.86° R 35 N = sin117.86 sin 25°

R = 73.2 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 9

PROBLEM 2.8 For the hook support of Problem 2.1, knowing that the magnitude of P is 75 N, determine by trigonometry (a) the required magnitude of the force Q if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. PROBLEM 2.1 Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P = 75 N and Q = 125 N, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

Using the triangle rule and law of sines: (a) (b)

Q 75 N = sin 20° sin 35°

Q = 44.7 N

α + 20° + 35° = 180° α = 180° − 20° − 35° = 125° R 75 N = sin125° sin 35°

R = 107.1 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 10

!

PROBLEM 2.9 A trolley that moves along a horizontal beam is acted upon by two forces as shown. (a) Knowing that α = 25°, determine by trigonometry the magnitude of the force P so that the resultant force exerted on the trolley is vertical. (b) What is the corresponding magnitude of the resultant?

SOLUTION

Using the triangle rule and the law of sines: (a) (b)

1600 N P = sin 25° sin 75°

P = 3660 N

25° + β + 75° = 180°

β = 180° − 25° − 75° = 80° 1600 N R = sin 25° sin 80°

R = 3730 N

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 11

PROBLEM 2.10 A trolley that moves along a horizontal beam is acted upon by two forces as shown. Determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 2500 N.

SOLUTION

Using the law of cosines:

Using the law of sines:

P 2 = (1600 N)2 + (2500 N)2 − 2(1600 N)(2500 N) cos 75° P = 2596 N

sin α sin 75° = 1600 N 2596 N α = 36.5°

P is directed 90° − 36.5° or 53.5° below the horizontal.

P = 2600 N

53.5°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 12

PROBLEM 2.11 A steel tank is to be positioned in an excavation. Knowing that α = 20°, determine by trigonometry (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the triangle rule and the law of sines: (a)

β + 50° + 60° = 180° β = 180° − 50° − 60° = 70°

(b)

425 lb P = sin 70° sin 60°

P = 392 lb

425 lb R = sin 70° sin 50°

R = 346 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 13

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PROBLEM 2.12 A steel tank is to be positioned in an excavation. Knowing that the magnitude of P is 500 lb, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the triangle rule and the law of sines: (a)

(α + 30°) + 60° + β = 180°

β = 180° − (α + 30°) − 60° β = 90° − α sin (90° − α ) sin 60° 425 lb

(b)

=

500 lb

90° − α = 47.40°

α = 42.6°

R 500 lb = sin (42.6° + 30°) sin 60°

R = 551 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 14

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PROBLEM 2.13 For the hook support of Problem 2.7, determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied to the support is horizontal, (b) the corresponding magnitude of R.

SOLUTION

The smallest force P will be perpendicular to R. (a)

P = (50 N)sin 25°

(b)

R = (50 N) cos 25°

P = 21.1 N R = 45.3 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 15

PROBLEM 2.14 For the steel tank of Problem 2.11, determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied at A is vertical, (b) the corresponding magnitude of R. PROBLEM 2.11 A steel tank is to be positioned in an excavation. Knowing that α = 20°, determine by trigonometry (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

The smallest force P will be perpendicular to R. (a)

P = (425 lb) cos 30°

(b)

R = (425 lb)sin 30°

P = 368 lb R = 213 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 16

PROBLEM 2.15 Solve Problem 2.2 by trigonometry. PROBLEM 2.2 Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P = 60 lb and Q = 25 lb, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION Using the triangle rule and the law of cosines: 20° + 35° + α = 180° α = 125° R 2 = P 2 + Q 2 − 2 PQ cos α R 2 = (60 lb)2 + (25 lb) 2 − 2(60 lb)(25 lb) cos125° R 2 = 3600 + 625 + 3000(0.5736) R = 77.108 lb

Using the law of sines:

sin β sin125° = 25 lb 77.108 lb β = 15.402° 70° + β = 85.402°

R = 77.1 lb

85.4°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 17

PROBLEM 2.16 Solve Problem 2.3 by trigonometry. PROBLEM 2.3 The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

8 10 α = 38.66° 6 tan β = 10 β = 30.96° tan α =

Using the triangle rule:

Using the law of cosines:

Using the law of sines:

α + β + ψ = 180° 38.66° + 30.96° + ψ = 180° ψ = 110.38° R 2 = (120 lb)2 + (40 lb) 2 − 2(120 lb)(40 lb) cos110.38° R = 139.08 lb sin γ sin110.38° = 40 lb 139.08 lb

γ = 15.64° φ = (90° − α ) + γ φ = (90° − 38.66°) + 15.64° φ = 66.98°

R = 139.1 lb

67.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 18

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PROBLEM 2.17 Solve Problem 2.4 by trigonometry. PROBLEM 2.4 Two forces are applied at Point B of beam AB. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION Using the law of cosines:

R 2 = (2 kN)2 + (3 kN)2 − 2(2 kN)(3 kN) cos80° R = 3.304 kN

Using the law of sines:

sin γ sin 80° = 2 kN 3.304 kN

γ = 36.59° β + γ + 80° = 180° γ = 180° − 80° − 36.59° γ = 63.41° φ = 180° − β + 50° φ = 66.59°

R = 3.30 kN

66.6°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 19

PROBLEM 2.18 Two structural members A and B are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 15 kN in member A and 10 kN in member B, determine by trigonometry the magnitude and direction of the resultant of the forces applied to the bracket by members A and B.

SOLUTION Using the force triangle and the laws of cosines and sines: We have

γ = 180° − (40° + 20°) = 120°

Then

R 2 = (15 kN) 2 + (10 kN)2 − 2(15 kN)(10 kN) cos120° = 475 kN 2 R = 21.794 kN

and

Hence:

10 kN 21.794 kN = sin α sin120° 10 kN ! sin α = " # sin120° $ 21.794 kN % = 0.39737 α = 23.414

φ = α + 50° = 73.414

R = 21.8 kN

73.4°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 20

PROBLEM 2.19 Two structural members A and B are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 10 kN in member A and 15 kN in member B, determine by trigonometry the magnitude and direction of the resultant of the forces applied to the bracket by members A and B.

SOLUTION Using the force triangle and the laws of cosines and sines We have

γ = 180° − (40° + 20°) = 120°

Then

R 2 = (10 kN) 2 + (15 kN)2 − 2(10 kN)(15 kN) cos120° = 475 kN 2 R = 21.794 kN

and

Hence:

15 kN 21.794 kN = sin α sin120° 15 kN ! sin α = " # sin120° $ 21.794 kN % = 0.59605 α = 36.588°

φ = α + 50° = 86.588°

R = 21.8 kN

86.6°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 21

PROBLEM 2.20 For the hook support of Problem 2.7, knowing that P = 75 N and α = 50°, determine by trigonometry the magnitude and direction of the resultant of the two forces applied to the support. PROBLEM 2.7 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R.

SOLUTION Using the force triangle and the laws of cosines and sines: We have

β = 180° − (50° + 25°) = 105°

Then

R 2 = (75 N) 2 + (50 N) 2 − 2(75 N)(50 N) cos 105° 2

R = 10066.1 N 2 R = 100.330 N

and

Hence:

sin γ sin105° = 75 N 100.330 N sin γ = 0.72206 γ = 46.225°

γ − 25° = 46.225° − 25° = 21.225°

R = 100.3 N

21.2°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 22

PROBLEM 2.21 Determine the x and y components of each of the forces shown.

SOLUTION Compute the following distances: OA = (600) 2 + (800) 2

= 1000 mm OB = (560)2 + (900) 2

= 1060 mm OC = (480) 2 + (900)2

= 1020 mm

800-N Force:

424-N Force:

408-N Force:

Fx = + (800 N)

800 1000

Fx = +640 N

!

Fy = +(800 N)

600 1000

Fy = +480 N

!

Fx = −(424 N)

560 1060

Fx = −224 N

Fy = −(424 N)

900 1060

Fy = −360 N

Fx = + (408 N)

480 1020

Fx = +192.0 N

Fy = −(408 N)

900 1020

Fy = −360 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 23

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PROBLEM 2.22 Determine the x and y components of each of the forces shown.

SOLUTION Compute the following distances: OA = (84) 2 + (80) 2

= 116 in. OB = (28)2 + (96)2

= 100 in. OC = (48)2 + (90)2

= 102 in.

29-lb Force:

50-lb Force:

51-lb Force:

Fx = + (29 lb)

84 116

Fx = +21.0 lb

Fy = +(29 lb)

80 116

Fy = +20.0 lb

Fx = −(50 lb)

28 100

Fx = −14.00 lb

Fy = +(50 lb)

96 100

Fy = + 48.0 lb

Fx = + (51 lb)

48 102

Fx = +24.0 lb

Fy = −(51 lb)

90 102

Fy = −45.0 lb

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 24

PROBLEM 2.23 Determine the x and y components of each of the forces shown.

SOLUTION 40-lb Force:

50-lb Force:

60-lb Force:

Fx = + (40 lb) cos 60°

Fx = 20.0 lb

Fy = −(40 lb)sin 60°

Fy = −34.6 lb

Fx = −(50 lb)sin 50°

Fx = −38.3 lb

Fy = −(50 lb) cos 50°

Fy = −32.1 lb

Fx = + (60 lb) cos 25°

Fx = 54.4 lb

Fy = +(60 lb)sin 25°

Fy = 25.4 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 25

PROBLEM 2.24 Determine the x and y components of each of the forces shown.

SOLUTION 80-N Force:

120-N Force:

150-N Force:

Fx = + (80 N) cos 40°

Fx = 61.3 N

Fy = + (80 N) sin 40°

Fy = 51.4 N

Fx = + (120 N) cos 70°

Fx = 41.0 N

Fy = +(120 N) sin 70°

Fy = 112.8 N

Fx = −(150 N) cos 35°

Fx = −122. 9 N

Fy = +(150 N) sin 35°

Fy = 86.0 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 26

PROBLEM 2.25 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 300-lb horizontal component, determine (a) the magnitude of the force P, (b) its vertical component.

SOLUTION

P sin 35° = 300 lb

(a)

P=

(b)

Vertical component

300 lb sin 35°

P = 523 lb

Pv = P cos 35°

= (523 lb) cos 35°

Pv = 428 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 27

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PROBLEM 2.26 The hydraulic cylinder BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 750-N component perpendicular to member ABC, determine (a) the magnitude of the force P, (b) its component parallel to ABC.

SOLUTION

(a)

750 N = P sin 20° P = 2193 N

(b)

P = 2190 N

PABC = P cos 20°

= (2193 N) cos 20°

PABC = 2060 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 28

PROBLEM 2.27 The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P must have a 120-N component perpendicular to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC.

SOLUTION

(a)

Px sin 38° 120 N = sin 38°

P=

= 194.91 N

(b)

or

P = 194.9 N

or

Py = 153.6 N

Px tan 38° 120 N = tan 38°

Py =

= 153.59 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 29

PROBLEM 2.28 The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P has a 180-N component along line AC, determine (a) the magnitude of the force P, (b) its component in a direction perpendicular to AC.

SOLUTION

(a)

P=

Py

cos 38° 180 N = cos 38° = 228.4 N

(b)

P = 228 N

Px = Py tan 38° = (180 N) tan 38° = 140.63 N

Px = 140.6 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 30

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PROBLEM 2.29 Member CB of the vise shown exerts on block B a force P directed along line CB. Knowing that P must have a 1200-N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component.

SOLUTION We note: CB exerts force P on B along CB, and the horizontal component of P is Px = 1200 N: Then (a)

Px = P sin 55°

Px sin 55° 1200 N = sin 55° = 1464.9 N

P=

(b)

P = 1465 N

Px = Py tan 55° Px tan 55° 1200 N = tan 55° = 840.2 N

Py =

Py = 840 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 31

PROBLEM 2.30 Cable AC exerts on beam AB a force P directed along line AC. Knowing that P must have a 350-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component.

SOLUTION

(a)

P=

Py cos 55°

350 lb cos 55° = 610.2 lb =

(b)

P = 610 lb

Px = P sin 55° = (610.2 lb) sin 55° = 499.8 lb

Px = 500 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 32

PROBLEM 2.31 Determine the resultant of the three forces of Problem 2.22. PROBLEM 2.22 Determine the x and y components of each of the forces shown.

SOLUTION Components of the forces were determined in Problem 2.22: Force

x Comp. (lb)

y Comp. (lb)

29 lb

+21.0

+20.0

50 lb

–14.00

+48.0

51 lb

+24.0

–45.0

Rx = +31.0

Ry = + 23.0

R = Rx i + R y j

= (31.0 lb) i + (23.0 lb) j Ry tan α = Rx 23.0 31.0 α = 36.573° 23.0 lb R= sin (36.573°) =

= 38.601 lb

R = 38.6 lb

36.6°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 33

PROBLEM 2.32 Determine the resultant of the three forces of Problem 2.24. PROBLEM 2.24 Determine the x and y components of each of the forces shown.

SOLUTION Components of the forces were determined in Problem 2.24: Force

x Comp. (N)

y Comp. (N)

80 N

+61.3

+51.4

120 N

+41.0

+112.8

150 N

–122.9

+86.0

Rx = −20.6

Ry = + 250.2

R = Rx i + Ry j

= ( −20.6 N)i + (250.2 N) j Ry tan α = Rx 250.2 N 20.6 N tan α = 12.1456 α = 85.293° tan α =

R=

250.2 N sin 85.293°

R = 251 N

85.3°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 34

PROBLEM 2.33 Determine the resultant of the three forces of Problem 2.23. PROBLEM 2.23 Determine the x and y components of each of the forces shown.

SOLUTION Force

x Comp. (lb)

y Comp. (lb)

40 lb

+20.00

–34.64

50 lb

–38.30

–32.14

60 lb

+54.38

+25.36

Rx = +36.08

Ry = −41.42

R = Rx i + Ry j

= ( +36.08 lb)i + (−41.42 lb) j Ry tan α = Rx 41.42 lb 36.08 lb tan α = 1.14800 α = 48.942° tan α =

R=

41.42 lb sin 48.942°

R = 54.9 lb

48.9°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 35

PROBLEM 2.34 Determine the resultant of the three forces of Problem 2.21. PROBLEM 2.21 Determine the x and y components of each of the forces shown.

SOLUTION Components of the forces were determined in Problem 2.21: Force

x Comp. (N)

y Comp. (N)

800 lb

+640

+480

424 lb

–224

–360

408 lb

+192

–360

Rx = +608

Ry = −240

R = Rx i + Ry j

= (608 lb)i + (−240 lb) j tan α =

Ry Rx

240 608 α = 21.541° =

240 N sin(21.541°) = 653.65 N

R=

R = 654 N

21.5°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 36

PROBLEM 2.35 Knowing that α = 35°, determine the resultant of the three forces shown.

SOLUTION Fx = +(100 N) cos 35° = +81.915 N

100-N Force:

Fy = −(100 N)sin 35° = −57.358 N Fx = +(150 N) cos 65° = +63.393 N

150-N Force:

Fy = −(150 N) sin 65° = −135.946 N Fx = −(200 N) cos 35° = −163.830 N

200-N Force:

Fy = −(200 N)sin 35° = −114.715 N

Force

x Comp. (N)

y Comp. (N)

100 N

+81.915

−57.358

150 N

+63.393

−135.946

200 N

−163.830

−114.715

Rx = −18.522

Ry = −308.02

R = Rx i + Ry j = (−18.522 N)i + (−308.02 N) j tan α =

Ry Rx

308.02 18.522 α = 86.559° =

R=

308.02 N sin 86.559

R = 309 N

86.6°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 37

!

PROBLEM 2.36 Knowing that the tension in cable BC is 725 N, determine the resultant of the three forces exerted at Point B of beam AB.

SOLUTION Cable BC Force:

500-N Force:

780-N Force:

and

840 = −525 N 1160 840 Fy = (725 N) = 500 N 1160 Fx = −(725 N)

3 Fx = −(500 N) = −300 N 5 4 Fy = −(500 N) = −400 N 5 12 = 720 N 13 5 Fy = −(780 N) = −300 N 13 Fx = (780 N)

Rx = ΣFx = −105 N R y = ΣFy = −200 N R = (−105 N)2 + (−200 N) 2 = 225.89 N

Further:

tan α =

200 105

α = tan −1

200 105

= 62.3°

R = 226 N

Thus:

62.3°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 38

PROBLEM 2.37 Knowing that α = 40°, determine the resultant of the three forces shown.

SOLUTION 60-lb Force:

Fx = (60 lb) cos 20° = 56.38 lb Fy = (60 lb)sin 20° = 20.52 lb

80-lb Force:

Fx = (80 lb) cos 60° = 40.00 lb Fy = (80 lb)sin 60° = 69.28 lb

120-lb Force:

Fx = (120 lb) cos 30° = 103.92 lb Fy = −(120 lb)sin 30° = −60.00 lb

and

Rx = ΣFx = 200.30 lb R y = ΣFy = 29.80 lb R = (200.30 lb) 2 + (29.80 lb) 2

= 202.50 lb

Further:

tan α =

29.80 200.30

α = tan −1

29.80 200.30

R = 203 lb

= 8.46°

8.46°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 39

!

PROBLEM 2.38 Knowing that α = 75°, determine the resultant of the three forces shown.

SOLUTION 60-lb Force:

Fx = (60 lb) cos 20° = 56.38 lb Fy = (60 lb) sin 20° = 20.52 lb

80-lb Force:

Fx = (80 lb) cos 95 ° = −6.97 lb Fy = (80 lb)sin 95° = 79.70 lb

120-lb Force:

Fx = (120 lb) cos 5 ° = 119.54 lb Fy = (120 lb) sin 5° = 10.46 lb

Then

Rx = ΣFx = 168.95 lb R y = ΣFy = 110.68 lb

and

R = (168.95 lb) 2 + (110.68 lb) 2

= 201.98 lb 110.68 168.95 tan α = 0.655 α = 33.23° tan α =

R = 202 lb

33.2°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 40

PROBLEM 2.39 For the collar of Problem 2.35, determine (a) the required value of α if the resultant of the three forces shown is to be vertical, (b) the corresponding magnitude of the resultant.

SOLUTION Rx = ΣFx

= (100 N) cos α + (150 N) cos (α + 30°) − (200 N) cos α Rx = −(100 N) cos α + (150 N) cos (α + 30°)

(1)

Ry = ΣFy

= −(100 N) sin α − (150 N)sin (α + 30°) − (200 N)sin α Ry = −(300 N) sin α − (150 N)sin (α + 30°)

(a)

(2)

For R to be vertical, we must have Rx = 0. We make Rx = 0 in Eq. (1): −100 cos α + 150cos (α + 30°) = 0 −100cos α + 150 (cos α cos 30° − sin α sin 30°) = 0 29.904cos α = 75sin α 29.904 75 = 0.3988 α = 21.74°

tan α =

(b)

α = 21.7°

Substituting for α in Eq. (2): Ry = −300sin 21.74° − 150sin 51.74°

= −228.9 N R = | Ry | = 228.9 N

R = 229 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 41

PROBLEM 2.40 For the beam of Problem 2.36, determine (a) the required tension in cable BC if the resultant of the three forces exerted at Point B is to be vertical, (b) the corresponding magnitude of the resultant.

SOLUTION Rx = ΣFx = − Rx = −

21 TBC + 420 N 29

Ry = ΣFy = Ry =

(a)

(1)

800 5 4 TBC − (780 N) − (500 N) 1160 13 5

20 TBC − 700 N 29

(2)

For R to be vertical, we must have Rx = 0 Set Rx = 0 in Eq. (1)

(b)

840 12 3 TBC + (780 N) − (500 N) 1160 13 5

−

21 TBC + 420 N = 0 29

TBC = 580 N

Substituting for TBC in Eq. (2): 20 (580 N) − 700 N 29 Ry = −300 N Ry =

R = | Ry | = 300 N

R = 300 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 42

PROBLEM 2.41 Determine (a) the required tension in cable AC, knowing that the resultant of the three forces exerted at Point C of boom BC must be directed along BC, (b) the corresponding magnitude of the resultant.

SOLUTION

Using the x and y axes shown: Rx = ΣFx = TAC sin10° + (50 lb) cos 35° + (75 lb) cos 60°

= TAC sin10° + 78.46 lb

(1)

Ry = ΣFy = (50 lb)sin 35° + (75 lb)sin 60° − TAC cos10° Ry = 93.63 lb − TAC cos10°

(a)

(2)

Set Ry = 0 in Eq. (2): 93.63 lb − TAC cos10° = 0 TAC = 95.07 lb

(b)

TAC = 95.1 lb

Substituting for TAC in Eq. (1): Rx = (95.07 lb) sin10° + 78.46 lb = 94.97 lb R = Rx

R = 95.0 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 43

!

PROBLEM 2.42 For the block of Problems 2.37 and 2.38, determine (a) the required value of α if the resultant of the three forces shown is to be parallel to the incline, (b) the corresponding magnitude of the resultant.

SOLUTION

Select the x axis to be along a a′. Then Rx = ΣFx = (60 lb) + (80 lb) cos α + (120 lb)sin α

(1)

Ry = ΣFy = (80 lb)sin α − (120 lb) cos α

(2)

and

(a)

Set Ry = 0 in Eq. (2). (80 lb) sin α − (120 lb) cos α = 0

Dividing each term by cos α gives: (80 lb) tan α = 120 lb 120 lb 80 lb α = 56.310°

tanα =

(b)

α = 56.3°

Substituting for α in Eq. (1) gives: Rx = 60 lb + (80 lb) cos 56.31° + (120 lb)sin 56.31° = 204.22 lb

Rx = 204 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 44

PROBLEM 2.43 Two cables are tied together at C and are loaded as shown. Knowing that α = 20°, determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION Free-Body Diagram

Law of sines:

Force Triangle

TAC T 1962 N = BC = sin 70° sin 50° sin 60°

(a)

TAC =

1962 N sin 70° = 2128.9 N sin 60°

TAC = 2.13 kN

!

(b)

TBC =

1962 N sin 50° = 1735.49 N sin 60°

TBC = 1.735 kN

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 45

PROBLEM 2.44 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION Free-Body Diagram

Law of sines:

Force Triangle

TAC T 500 N = BC = sin 60° sin 40° sin 80°

(a)

TAC =

500 N sin 60° = 439.69 N sin 80°

TAC = 440 N

(b)

TBC =

500 N sin 40° = 326.35 N sin 80°

TBC = 326 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 46

PROBLEM 2.45 Two cables are tied together at C and are loaded as shown. Knowing that P = 500 N and α = 60°, determine the tension in (a) in cable AC, (b) in cable BC.

SOLUTION Free-Body Diagram

Law of sines:

Force Triangle

TAC T 500 N = BC = sin 35° sin 75° sin 70°

(a)

TAC =

500 N sin 35° sin 70°

TAC = 305 N

(b)

TBC =

500 N sin 75° sin 70°

TBC = 514 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 47

PROBLEM 2.46 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION Free-Body Diagram

Force Triangle

W = mg = (200 kg)(9.81 m/s 2 ) = 1962 N

Law of sines:

TAC TBC 1962 N = = sin 15° sin 105° sin 60°

(a)

TAC =

(1962 N) sin 15° sin 60°

TAC = 586 N

(b)

TBC =

(1962 N) sin 105° sin 60°

TBC = 2190 N

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 48

PROBLEM 2.47 Knowing that α = 20°, determine the tension (a) in cable AC, (b) in rope BC.

SOLUTION Free-Body Diagram

Law of sines:

Force Triangle

TAC T 1200 lb = BC = sin 110° sin 5° sin 65°

(a)

TAC =

1200 lb sin 110° sin 65°

TAC = 1244 lb

(b)

TBC =

1200 lb sin 5° sin 65°

TBC = 115.4 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 49

PROBLEM 2.48 Knowing that α = 55° and that boom AC exerts on pin C a force directed along line AC, determine (a) the magnitude of that force, (b) the tension in cable BC.

SOLUTION Free-Body Diagram

Law of sines:

Force Triangle

FAC T 300 lb = BC = sin 35° sin 50° sin 95°

(a)

FAC =

300 lb sin 35° sin 95°

FAC = 172.7 lb

(b)

TBC =

300 lb sin 50° sin 95°

TBC = 231 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 50

PROBLEM 2.49 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that P = 500 lb and Q = 650 lb, determine the magnitudes of the forces exerted on the rods A and B.

SOLUTION Free-Body Diagram Resolving the forces into x- and y-directions: R = P + Q + FA + FB = 0

Substituting components:

R = −(500 lb) j + [(650 lb) cos 50°]i

− [(650 lb) sin 50°] j + FB i − ( FA cos 50°)i + ( FA sin 50°) j = 0

In the y-direction (one unknown force) −500 lb − (650 lb)sin 50° + FA sin 50° = 0

Thus,

FA =

500 lb + (650 lb) sin 50° sin 50°

= 1302.70 lb

In the x-direction: Thus,

FA = 1303 lb

(650 lb) cos 50° + FB − FA cos 50° = 0

FB = FA cos 50° − (650 lb) cos50° = (1302.70 lb) cos 50° − (650 lb) cos 50° = 419.55 lb

FB = 420 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 51

PROBLEM 2.50 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that the magnitudes of the forces exerted on rods A and B are FA = 750 lb and FB = 400 lb, determine the magnitudes of P and Q.

SOLUTION Free-Body Diagram Resolving the forces into x- and y-directions: R = P + Q + FA + FB = 0

Substituting components:

R = − Pj + Q cos 50°i − Q sin 50° j − [(750 lb) cos 50°]i

+ [(750 lb)sin 50°] j + (400 lb)i

In the x-direction (one unknown force) Q cos 50° − [(750 lb) cos 50°] + 400 lb = 0 Q=

(750 lb) cos 50° − 400 lb cos 50°

= 127.710 lb

In the y-direction:

− P − Q sin 50° + (750 lb) sin 50° = 0 P = −Q sin 50° + (750 lb) sin 50°

= −(127.710 lb)sin 50° + (750 lb) sin 50° = 476.70 lb

P = 477 lb; Q = 127.7 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 52

!

PROBLEM 2.51 A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA = 8 kN and FB = 16 kN, determine the magnitudes of the other two forces.

SOLUTION Free-Body Diagram of Connection

ΣFx = 0:

With

3 3 FB − FC − FA = 0 5 5

FA = 8 kN FB = 16 kN FC =

4 4 (16 kN) − (8 kN) 5 5

Σ Fy = 0: − FD +

With FA and FB as above:

FC = 6.40 kN

3 3 FB − FA = 0 ! 5 5

3 3 FD = (16 kN) − (8 kN) ! 5 5

FD = 4.80 kN

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 53

PROBLEM 2.52 A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA = 5 kN and FD = 6 kN, determine the magnitudes of the other two forces.

SOLUTION Free-Body Diagram of Connection

3 3 ΣFy = 0: − FD − FA + FB = 0 5 5 3 FA 5

or

FB = FD +

With

FA = 5 kN, FD = 8 kN FB =

5' 3 ( 6 kN + (5 kN) * 3 )+ 5 ,

ΣFx = 0: − FC +

FB = 15.00 kN

4 4 FB − FA = 0 5 5

4 ( FB − FA ) 5 4 = (15 kN − 5 kN) 5

FC =

FC = 8.00 kN

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 54

PROBLEM 2.53 Two cables tied together at C are loaded as shown. Knowing that Q = 60 lb, determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION

ΣFy = 0: TCA − Q cos 30° = 0

With

Q = 60 lb

(a)

TCA = (60 lb)(0.866)

(b)

ΣFx = 0: P − TCB − Q sin 30° = 0

With

TCA = 52.0 lb

P = 75 lb TCB = 75 lb − (60 lb)(0.50)

or TCB = 45.0 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 55

PROBLEM 2.54 Two cables tied together at C are loaded as shown. Determine the range of values of Q for which the tension will not exceed 60 lb in either cable.

SOLUTION Free-Body Diagram

ΣFx = 0: −TBC − Q cos 60° + 75 lb = 0 TBC = 75 lb − Q cos 60°

(1)

ΣFy = 0: TAC − Q sin 60° = 0 TAC = Q sin 60°

(2)

TAC # 60 lb:

Requirement

Q sin 60° # 60 lb

From Eq. (2):

Q # 69.3 lb TBC # 60 lb:

Requirement From Eq. (1):

75 lb − Q sin 60° # 60 lb

Q $ 30.0 lb

30.0 lb # Q # 69.3 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 56

PROBLEM 2.55 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that α = 30° and β = 10° and that the combined weight of the boatswain’s chair and the sailor is 900 N, determine the tension (a) in the support cable ACB, (b) in the traction cable CD.

SOLUTION Free-Body Diagram

ΣFx = 0: TACB cos 10° − TACB cos 30° − TCD cos 30° = 0 TCD = 0.137158TACB

(1)

ΣFy = 0: TACB sin 10° + TACB sin 30° + TCD sin 30° − 900 = 0 0.67365TACB + 0.5TCD = 900

(a)

Substitute (1) into (2):

0.67365 TACB + 0.5(0.137158 TACB ) = 900

TACB = 1212.56 N

(b)

From (1):

(2)

TCD = 0.137158(1212.56 N)

TACB = 1213 N TCD = 166.3 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 57

PROBLEM 2.56 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that α = 25° and β = 15° and that the tension in cable CD is 80 N, determine (a) the combined weight of the boatswain’s chair and the sailor, (b) in tension in the support cable ACB.

SOLUTION Free-Body Diagram

ΣFx = 0: TACB cos 15° − TACB cos 25° − (80 N) cos 25° = 0 TACB = 1216.15 N

ΣFy = 0: (1216.15 N) sin 15° + (1216.15 N) sin 25° + (80 N) sin 25° − W = 0 W = 862.54 N

(a)

W = 863 N

(b) TACB = 1216 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 58

PROBLEM 2.57 For the cables of Problem 2.45, it is known that the maximum allowable tension is 600 N in cable AC and 750 N in cable BC. Determine (a) the maximum force P that can be applied at C, (b) the corresponding value of α.

SOLUTION Free-Body Diagram

(a)

Law of cosines

Force Triangle

P 2 = (600) 2 + (750)2 − 2(600)(750) cos (25° + 45°) P = 784.02 N

(b)

Law of sines

P = 784 N

sin β sin (25° + 45°) = 600 N 784.02 N

β = 46.0°

α = 46.0° + 25° = 71.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 59

PROBLEM 2.58 For the situation described in Figure P2.47, determine (a) the value of α for which the tension in rope BC is as small as possible, (b) the corresponding value of the tension. PROBLEM 2.47 Knowing that α = 20°, determine the tension (a) in cable AC, (b) in rope BC.

SOLUTION Free-Body Diagram

Force Triangle

To be smallest, TBC must be perpendicular to the direction of TAC . (a) (b)

Thus,

α = 5°

= 5.00°

TBC = (1200 lb) sin 5°

TBC = 104.6 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 60

PROBLEM 2.59 For the structure and loading of Problem 2.48, determine (a) the value of α for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension.

SOLUTION TBC must be perpendicular to FAC to be as small as possible.

Free-Body Diagram: C

Force Triangle is a right triangle

To be a minimum, TBC must be perpendicular to FAC . (a)

We observe:

α = 90° − 30°

α = 60.0°

TBC = (300 lb)sin 50°

(b) or

TBC = 229.81 lb

TBC = 230 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 61

PROBLEM 2.60 Knowing that portions AC and BC of cable ACB must be equal, determine the shortest length of cable that can be used to support the load shown if the tension in the cable is not to exceed 870 N.

SOLUTION Free-Body Diagram: C (For T = 725 N) ΣFy = 0: 2Ty − 1200 N = 0

Ty = 600 N Tx2 + Ty2 = T 2 Tx2 + (600 N)2 = (870 N) 2 Tx = 630 N

By similar triangles:

BC 2.1 m = 870 N 630 N BC = 2.90 m L = 2( BC ) = 5.80 m

L = 5.80 m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 62

PROBLEM 2.61 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension in each cable is 800 N, determine (a) the magnitude of the largest force P that can be applied at C, (b) the corresponding value of α.

SOLUTION Free-Body Diagram: C

Force Triangle

Force triangle is isosceles with 2 β = 180° − 85° β = 47.5° P = 2(800 N)cos 47.5° = 1081 N

(a)

P = 1081 N

Since P . 0, the solution is correct. (b)

α = 180° − 50° − 47.5° = 82.5°

α = 82.5°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 63

!

PROBLEM 2.62 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension is 1200 N in cable AC and 600 N in cable BC, determine (a) the magnitude of the largest force P that can be applied at C, (b) the corresponding value of α.

SOLUTION Free-Body Diagram

(a)

Law of cosines:

Force Triangle

P 2 = (1200 N) 2 + (600 N) 2 − 2(1200 N)(600 N) cos 85° P = 1294 N

Since P . 1200 N, the solution is correct. P = 1294 N

(b)

Law of sines: sin β sin 85° = 1200 N 1294 N β = 67.5° α = 180° − 50° − 67.5°

α = 62.5°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 64

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PROBLEM 2.63 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the magnitude of the force P required to maintain the equilibrium of the collar when (a) x = 4.5 in., (b) x = 15 in.

SOLUTION (a)

Free Body: Collar A

Force Triangle P 50 lb = 4.5 20.5

(b)

Free Body: Collar A

P = 10.98 lb

Force Triangle P 50 lb = 15 25

P = 30.0 lb

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 65

PROBLEM 2.64 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the distance x for which the collar is in equilibrium when P = 48 lb.

SOLUTION Free Body: Collar A

Force Triangle

N 2 = (50) 2 − (48) 2 = 196 N = 14.00 lb

Similar Triangles x 48 lb = 20 in. 14 lb x = 68.6 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 66

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PROBLEM 2.65 A 160-kg load is supported by the rope-and-pulley arrangement shown. Knowing that β = 20°, determine the magnitude and direction of the force P that must be exerted on the free end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Chapter 4.)

SOLUTION Free-Body Diagram: Pulley A

ΣFx = 0: 2P sin 20° − P cos α = 0

and

cos α = 0.8452 or α = ± 46.840°

α = + 46.840 For

ΣFy = 0: 2P cos 20° + P sin 46.840° − 1569.60 N = 0

or

P = 602 N

46.8°

α = −46.840

For

ΣFy = 0: 2P cos 20° + P sin( −46.840°) − 1569.60 N = 0

or

P = 1365 N

46.8°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 67

PROBLEM 2.66 A 160-kg load is supported by the rope-and-pulley arrangement shown. Knowing that α = 40°, determine (a) the angle β, (b) the magnitude of the force P that must be exerted on the free end of the rope to maintain equilibrium. (See the hint for Problem 2.65.)

SOLUTION Free-Body Diagram: Pulley A (a)

ΣFx = 0: 2 P sin sin β − P cos 40° = 0

1 cos 40° 2 β = 22.52°

sin β =

β = 22.5° (b)

ΣFy = 0: P sin 40° + 2 P cos 22.52° − 1569.60 N = 0 P = 630 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 68

PROBLEM 2.67 A 600-lb crate is supported by several rope-andpulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.65.)

SOLUTION Free-Body Diagram of Pulley (a)

ΣFy = 0: 2T − (600 lb) = 0 T=

1 (600 lb) 2 T = 300 lb

(b)

ΣFy = 0: 2T − (600 lb) = 0 T=

1 (600 lb) 2 T = 300 lb

(c)

ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb

(d)

ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb

(e)

ΣFy = 0: 4T − (600 lb) = 0 T=

1 (600 lb) 4 T = 150.0 lb

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 69

!

PROBLEM 2.68 Solve Parts b and d of Problem 2.67, assuming that the free end of the rope is attached to the crate. PROBLEM 2.67 A 600-lb crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.65.)

SOLUTION Free-Body Diagram of Pulley and Crate

(b)

ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb

(d)

ΣFy = 0: 4T − (600 lb) = 0 T=

1 (600 lb) 4 T = 150.0 lb

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 70

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PROBLEM 2.69 A load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Knowing that P = 750 N, determine (a) the tension in cable ACB, (b) the magnitude of load Q.

SOLUTION Free-Body Diagram: Pulley C ΣFx = 0: TACB (cos 25° − cos 55°) − (750 N) cos 55° = 0

(a)

TACB = 1292.88 N

Hence:

TACB = 1293 N

ΣFy = 0: TACB (sin 25° + sin 55°) + (750 N) sin 55° − Q = 0

(b) !

(1292.88 N)(sin 25° + sin 55°) + (750 N) sin 55° − Q = 0 Q = 2219.8 N

or

Q = 2220 N

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 71

PROBLEM 2.70 An 1800-N load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Determine (a) the tension in cable ACB, (b) the magnitude of load P.

SOLUTION Free-Body Diagram: Pulley C ΣFx = 0: TACB (cos 25° − cos 55°) − P cos 55° = 0 P = 0.58010TACB

or

(1)

ΣFy = 0: TACB (sin 25° + sin 55°) + P sin 55° − 1800 N = 0 1.24177TACB + 0.81915 P = 1800 N

or (a)

(2)

Substitute Equation (1) into Equation (2): 1.24177TACB + 0.81915(0.58010TACB ) = 1800 N TACB = 1048.37 N

Hence:

TACB = 1048 N

(b)

P = 0.58010(1048.37 N) = 608.16 N

Using (1),

P = 608 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 72

PROBLEM 2.71 Determine (a) the x, y, and z components of the 750-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes.

SOLUTION Fh = F sin 35° = (750 N)sin 35° Fh = 430.2 N

(a) Fx = Fh cos 25° = (430.2 N) cos 25° Fx = +390 N,

(b)

Fy = F cos 35° = (750 N) cos 35° Fy = +614 N,

cos θ x = cos θ y = cos θ z =

Fx +390 N = 750 N F Fy F

=

+614 N 750 N

Fz +181.8 N = 750 N F

Fz = Fh sin 25° = (430.2 N) sin 25° Fz = +181.8 N

θ x = 58.7° θ y = 35.0° θ z = 76.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 73

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PROBLEM 2.72 Determine (a) the x, y, and z components of the 900-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes.

SOLUTION Fh = F cos 65° = (900 N) cos 65° Fh = 380.4 N

(a) Fx = Fh sin 20° = (380.4 N)sin 20° Fx = −130.1 N,

(b)

Fy = F sin 65° = (900 N)sin 65° Fy = +816 N, cos θ x = cos θ y = cos θ z =

Fx −130.1 N = 900 N F Fy

Fz = Fh cos 20° = (380.4 N) cos 20° Fz = +357 N

θ x = 98.3°

+816 N 900 N

θ y = 25.0°

Fz +357 N = 900 N F

θ z = 66.6°

F

=

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 74

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PROBLEM 2.73 A horizontal circular plate is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Knowing that the x component of the force exerted by wire AD on the plate is 110.3 N, determine (a) the tension in wire AD, (b) the angles θ x, θ y, and θ z that the force exerted at A forms with the coordinate axes.

SOLUTION (a)

Fx = F sin 30° sin 50° = 110.3 N (Given) F=

(b)

cos θ x =

110.3 N = 287.97 N sin 30° sin 50°

F = 288 N

Fx 110.3 N = = 0.38303 F 287.97 N

θ x = 67.5°

Fy = F cos 30° = 249.39 cos θ y =

Fy F

=

249.39 N = 0.86603 287.97 N

θ y = 30.0°

Fz = − F sin 30° cos 50° = −(287.97 N)sin 30°cos 50° = −92.552 N cos θ z =

Fz −92.552 N = = −0.32139 F 287.97 N

θ z = 108.7°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 75

PROBLEM 2.74 A horizontal circular plate is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Knowing that the z component of the force exerted by wire BD on the plate is –32.14 N, determine (a) the tension in wire BD, (b) the angles θx, θy, and θz that the force exerted at B forms with the coordinate axes.

SOLUTION (a)

Fz = − F sin 30° sin 40° = 32.14 N (Given) F=

(b)

32.14 = 100.0 N sin 30° sin 40°

F = 100.0 N

Fx = − F sin 30° cos 40° = −(100.0 N)sin 30°cos 40° = −38.302 N cos θ x =

Fx 38.302 N = = −0.38302 100.0 N F

θ x = 112.5°

Fy = F cos 30° = 86.603 N cos θ y =

Fy F

=

86.603 N = 0.86603 100 N

θ y = 30.0°

Fz = −32.14 N cos θ z =

Fz −32.14 N = = −0.32140 F 100 N

θ z = 108.7°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 76

PROBLEM 2.75 A horizontal circular plate is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Knowing that the tension in wire CD is 60 lb, determine (a) the components of the force exerted by this wire on the plate, (b) the angles θ x , θ y , and θ z that the force forms with the coordinate axes.

SOLUTION (a)

(b)

Fx = −(60 lb) sin 30°cos 60° = −15 lb

Fx = −15.00 lb

Fy = (60 lb) cos 30° = 51.96 lb

Fy = +52.0 lb

Fz = (60 lb) sin 30° sin 60° = 25.98 lb

Fz = +26.0 lb

cos θ x = cos θ y = cos θ z =

Fx −15.0 lb = = −0.25 60 lb F Fy

θ x = 104.5°

51.96 lb = 0.866 60 lb

θ y = 30.0°

Fz 25.98 lb = = 0.433 F 60 lb

θ z = 64.3°

F

=

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 77

PROBLEM 2.76 A horizontal circular plate is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Knowing that the x component of the force exerted by wire CD on the plate is –20.0 lb, determine (a) the tension in wire CD, (b) the angles θx, θy, and θz that the force exerted at C forms with the coordinate axes.

SOLUTION (a)

Fx = − F sin 30° cos 60° = −20 lb (Given) F=

(b)

cos θ x =

20 lb = 80 lb sin 30°cos 60°

F = 80.0 lb

Fx −20 lb = = − 0.25 80 lb F

θ x = 104.5°

Fy = (80 lb) cos 30° = 69.282 lb cos θ y =

Fy F

=

69.282 lb = 0.86615 80 lb

θ y = 30.0°

Fz = (80 lb)sin 30° sin 60° = 34.641 lb cos θ z =

Fz 34.641 = = 0.43301 80 F

θ z = 64.3°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 78

PROBLEM 2.77 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AC is 120 lb, determine (a) the components of the force exerted by this wire on the pole, (b) the angles θx, θy, and θz that the force forms with the coordinate axes.

SOLUTION (a)

Fx = (120 lb) cos 60° cos 20° Fx = 56.382 lb

Fx = +56.4 lb

Fy = −(120 lb)sin 60° Fy = −103.923 lb

Fy = −103.9 lb

Fz = −(120 lb) cos 60° sin 20° Fz = −20.521 lb

(b)

cos θ x = cos θ y = cos θ z =

Fz = −20.5 lb

Fx 56.382 lb = F 120 lb Fy F

=

−103.923 lb 120 lb

Fz −20.52 lb = F 120 lb

θ x = 62.0° θ y = 150.0° θ z = 99.8°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 79

PROBLEM 2.78 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AD is 85 lb, determine (a) the components of the force exerted by this wire on the pole, (b) the angles θx, θy, and θz that the force forms with the coordinate axes.

SOLUTION (a)

Fx = (85 lb)sin 36° sin 48° = 37.129 lb

Fx = 37.1 lb

Fy = −(85 lb) cos 36° = −68.766 lb

Fy = −68.8 lb

Fz = (85 lb)sin 36° cos 48° Fz = 33.4 lb

= 33.431 lb

(b)

cos θ x = cos θ y = cos θ z =

Fx 37.129 lb = F 85 lb Fy F

=

−68.766 lb 85 lb

Fz 33.431 lb = F 85 lb

θ x = 64.1° θ y = 144.0° θ z = 66.8°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 80

PROBLEM 2.79 Determine the magnitude and direction of the force F = (320 N)i + (400 N)j − (250 N)k.

SOLUTION F = Fx2 + Fy2 + Fz2 F = (320 N)2 + (400 N) 2 + (−250 N) 2 cos θ x = cos θ y = cos θ y =

Fx 320 N = F 570 N Fy F

=

F = 570 N

θ x = 55.8°

400 N 570 N

θ y = 45.4°

Fz −250 N = F 570 N

θ z = 116.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 81

PROBLEM 2.80 Determine the magnitude and direction of the force F = (240 N)i − (270 N)j + (680 N)k.

SOLUTION F = Fx2 + Fy2 + Fz2 F = (240 N) 2 + (−270 N) 2 + (680 N) cos θ x = cos θ y = cos θ z =

Fx 240 N = F 770 N Fy F

=

F = 770 N

θ x = 71.8°

−270 N 770 N

Fz 680 N = F 770 N

θ y = 110.5° θ z = 28.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 82

PROBLEM 2.81 A force acts at the origin of a coordinate system in a direction defined by the angles θx = 70.9° and θy = 144.9°. Knowing that the z component of the force is –52.0 lb, determine (a) the angle θz, (b) the other components and the magnitude of the force.

SOLUTION (a)

We have (cos θ x ) 2 + (cos θ y )2 + (cos θ z ) 2 = 1

(cos θ y ) 2 = 1 − (cos θ y ) 2 − (cos θ z ) 2

Since Fz , 0 we must have cos θ z , 0 Thus, taking the negative square root, from above, we have: cos θ z = − 1 − (cos 70.9°) 2 − (cos144.9°) 2 = 0.47282

(b)

θ z = 118.2°

Then: F=

and

Fz 52.0 lb = = 109.978 lb cos θ z 0.47282

Fx = F cos θ x = (109.978 lb) cos 70.9°

Fx = 36.0 lb

Fy = F cos θ y = (109.978 lb) cos144.9°

Fy = −90.0 lb F = 110.0 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 83

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PROBLEM 2.82 A force acts at the origin of a coordinate system in a direction defined by the angles θy = 55° and θz = 45°. Knowing that the x component of the force is − 500 lb, determine (a) the angle θx, (b) the other components and the magnitude of the force.

SOLUTION (a)

We have (cos θ x ) 2 + (cos θ y )2 + (cos θ z ) 2 = 1

(cos θ y ) 2 = 1 − (cos θ y ) 2 − (cos θ z ) 2

Since Fx , 0 we must have cos θ x , 0 Thus, taking the negative square root, from above, we have: cos θ x = − 1 − (cos 55) 2 − (cos 45) 2 = 0.41353

(b)

θ x = 114.4°

Then: Fx 500 lb = = 1209.10 lb cos θ x 0.41353

F = 1209 lb

Fy = F cos θ y = (1209.10 lb) cos 55°

Fy = 694 lb

Fz = F cos θ z = (1209.10 lb) cos 45°

Fz = 855 lb

F=

and

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 84

PROBLEM 2.83 A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N, θz = 151.2°, and Fy < 0, determine (a) the components Fy and Fz, (b) the angles θx and θy.

SOLUTION Fz = F cos θ z = (210 N) cos151.2°

(a)

= −184.024 N

Then: So: Hence:

F 2 = Fx2 + Fy2 + Fz2 (210 N) 2 = (80 N) 2 + ( Fy ) 2 + (184.024 N)2 Fy = − (210 N) 2 − (80 N) 2 − (184.024 N) 2 = −61.929 N

(b)

Fz = −184.0 N

cos θ x = cos θ y =

Fy = −62.0 lb

Fx 80 N = = 0.38095 F 210 N Fy F

=

61.929 N = −0.29490 210 N

θ x = 67.6° θ y = 107.2

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 85

PROBLEM 2.84 A force F of magnitude 230 N acts at the origin of a coordinate system. Knowing that θx = 32.5°, Fy = − 60 N, and Fz > 0, determine (a) the components Fx and Fz, (b) the angles θy and θz.

SOLUTION (a)

We have Fx = F cos θ x = (230 N) cos 32.5°

Then:

Fx = −194.0 N

Fx = 193.980 N F 2 = Fx2 + Fy2 + Fz2

So: Hence: (b)

(230 N) 2 = (193.980 N) 2 + (−60 N) 2 + Fz2 Fz = + (230 N) 2 − (193.980 N) 2 − (−60 N) 2

Fz = 108.0 N

Fz = 108.036 N Fy

−60 N = − 0.26087 F 230 N F 108.036 N cos θ z = z = = 0.46972 F 230 N cos θ y =

=

θ y = 105.1° θ z = 62.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 86

PROBLEM 2.85 A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AB is 525 lb, determine the components of the force exerted by the wire on the bolt at B.

SOLUTION ! BA = (20 ft)i + (100 ft) j − (25 ft)k BA = (20 ft) 2 + (100 ft) 2 + (−25 ft)2 = 105 ft

F=F

BA

! BA =F BA 525 lb [(20 ft)i + (100 ft) j − (25 ft)k ] = 105 ft F = (100.0 lb)i + (500 lb) j − (125.0 lb)k Fx = +100.0 lb, Fy = +500 lb, Fz = −125.0 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 87

PROBLEM 2.86 A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AD is 315 lb, determine the components of the force exerted by the wire on the bolt at D.

SOLUTION ! DA = (20 ft)i + (100 ft) j + (70 ft)k DA = (20 ft) 2 + (100 ft) 2 + ( +70 ft) 2 = 126 ft

F=F

DA

! DA =F DA 315 lb [(20 ft)i + (100 ft) j + (74 ft)k ] = 126 ft F = (50 lb)i + (250 lb) j + (185 lb)k Fx = +50 lb, Fy = +250 lb, Fz = +185.0 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 88

PROBLEM 2.87 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.

SOLUTION ! DB = (480 mm)i − (510 mm) j + (320 mm)k DB = (480 mm) 2 + (510 mm 2 ) + (320 mm) 2 = 770 mm

F=F

DB

! DB =F DB 385 N = [(480 mm)i − (510 mm)j + (320 mm)k ] 770 mm = (240 N)i − (255 N) j + (160 N)k Fx = +240 N, Fy = −255 N, Fz = +160.0 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 89

PROBLEM 2.88 For the frame and cable of Problem 2.87, determine the components of the force exerted by the cable on the support at E. PROBLEM 2.87 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.

SOLUTION ! EB = (270 mm)i − (400 mm) j + (600 mm)k EB = (270 mm) 2 + (400 mm) 2 + (600 mm) 2 = 770 mm

F=F

EB

! EB =F EB 385 N = [(270 mm)i − (400 mm)j + (600 mm)k ] 770 mm F = (135 N)i − (200 N) j + (300 N)k Fx = +135.0 N, Fy = −200 N, Fz = +300 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 90

PROBLEM 2.89 Knowing that the tension in cable AB is 1425 N, determine the components of the force exerted on the plate at B.

SOLUTION ! BA = −(900 mm)i + (600 mm) j + (360 mm)k BA = (900 mm) 2 + (600 mm) 2 + (360 mm) 2 = 1140 mm

TBA = TBA

BA

! BA BA 1425 N = [ −(900 mm)i + (600 mm) j + (360 mm)k ] 1140 mm = −(1125 N)i + (750 N) j + (450 N)k = TBA

TBA

(TBA ) x = −1125 N, (TBA ) y = 750 N, (TBA ) z = 450 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 91

PROBLEM 2.90 Knowing that the tension in cable AC is 2130 N, determine the components of the force exerted on the plate at C.

SOLUTION ! CA = −(900 mm)i + (600 mm) j − (920 mm)k CA = (900 mm)2 + (600 mm)2 + (920 mm) 2 = 1420 mm TCA = TCA λCA ! CA = TCA CA 2130 N TCA = [−(900 mm)i + (600 mm) j − (920 mm)k ] 1420 mm = −(1350 N)i + (900 N) j − (1380 N)k (TCA ) x = −1350 N, (TCA ) y = 900 N, (TCA ) z = −1380 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 92

PROBLEM 2.91 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 300 N and Q = 400 N.

SOLUTION P = (300 N)[− cos 30° sin15°i + sin 30° j + cos 30° cos15°k ] = − (67.243 N)i + (150 N) j + (250.95 N)k Q = (400 N)[cos 50° cos 20°i + sin 50° j − cos 50° sin 20°k ] = (400 N)[0.60402i + 0.76604 j − 0.21985] = (241.61 N)i + (306.42 N) j − (87.939 N)k R = P+Q = (174.367 N)i + (456.42 N) j + (163.011 N)k R = (174.367 N)2 + (456.42 N)2 + (163.011 N) 2 = 515.07 N cos θ x = cos θ y = cos θ z =

R = 515 N

Rx 174.367 N = = 0.33853 515.07 N R Ry

θ x = 70.2°

456.42 N = 0.88613 515.07 N

θ y = 27.6°

Rz 163.011 N = = 0.31648 R 515.07 N

θ z = 71.5°

R

=

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 93

!

PROBLEM 2.92 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 400 N and Q = 300 N.

SOLUTION P = (400 N)[ − cos 30° sin15°i + sin 30° j + cos 30° cos15°k ] = − (89.678 N)i + (200 N) j + (334.61 N)k Q = (300 N)[cos 50° cos 20°i + sin 50° j − cos 50° sin 20°k ] = (181.21 N)i + (229.81 N)j − (65.954 N)k R = P+Q = (91.532 N)i + (429.81 N) j + (268.66 N)k R = (91.532 N)2 + (429.81 N) 2 + (268.66 N) 2 = 515.07 N cos θ x = cos θ y = cos θ z =

R = 515 N

Rx 91.532 N = = 0.177708 R 515.07 N Ry

θ x = 79.8°

429.81 N = 0.83447 515.07 N

θ y = 33.4°

Rz 268.66 N = = 0.52160 R 515.07 N

θ z = 58.6°

R

=

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 94

!

PROBLEM 2.93 Knowing that the tension is 425 lb in cable AB and 510 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

SOLUTION ! AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.) 2 + (45 in.) 2 + (60 in.) 2 = 85 in. ! AC = (100 in.)i − (45 in.) j + (60 in.)k AC = (100 in.) 2 + (45 in.) 2 + (60 in.) 2 = 125 in. ! (40 in.)i − (45 in.) j + (60 in.)k ! AB = (425 lb) " TAB = TAB AB = TAB # 85 in. AB $ % TAB = (200 lb)i − (225 lb) j + (300 lb)k ! AC (100 in.)i − (45 in.) j + (60 in.)k ! = (510 lb) " TAC = TAC AC = TAC # AC 125 in. $ % TAC = (408 lb)i − (183.6 lb) j + (244.8 lb)k R = TAB + TAC = (608)i − (408.6 lb) j + (544.8 lb)k

Then: and

R = 912.92 lb

R = 913 lb

cos θ x =

608 lb = 0.66599 912.92 lb

cos θ y =

408.6 lb = −0.44757 912.92 lb

θ y = 116.6°

cos θ z =

544.8 lb = 0.59677 912.92 lb

θ z = 53.4°

θ x = 48.2°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 95

!

PROBLEM 2.94 Knowing that the tension is 510 lb in cable AB and 425 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

SOLUTION ! AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.) 2 + (45 in.) 2 + (60 in.) 2 = 85 in. ! AC = (100 in.)i − (45 in.) j + (60 in.)k AC = (100 in.) 2 + (45 in.) 2 + (60 in.) 2 = 125 in. ! (40 in.)i − (45 in.) j + (60 in.)k ! AB = (510 lb) " TAB = TAB AB = TAB # 85 in. AB $ % TAB = (240 lb)i − (270 lb) j + (360 lb)k ! AC (100 in.)i − (45 in.) j + (60 in.)k ! = (425 lb) " TAC = TAC AC = TAC # AC 125 in. $ % TAC = (340 lb)i − (153 lb) j + (204 lb)k R = TAB + TAC = (580 lb)i − (423 lb) j + (564 lb)k

Then: and

R = 912.92 lb

R = 913 lb

cos θ x =

580 lb = 0.63532 912.92 lb

cos θ y =

−423 lb = −0.46335 912.92 lb

cos θ z =

564 lb = 0.61780 912.92 lb

θ x = 50.6° θ y = 117.6° θ z = 51.8°

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 96

PROBLEM 2.95 For the frame of Problem 2.87, determine the magnitude and direction of the resultant of the forces exerted by the cable at B knowing that the tension in the cable is 385 N. PROBLEM 2.87 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.

SOLUTION ! BD = −(480 mm)i + (510 mm) j − (320 mm)k BD = (480 mm) 2 + (510 mm) 2 + (320 mm) 2 = 770 mm ! BD FBD = TBD BD = TBD BD (385 N) = [−(480 mm)i + (510 mm) j − (320 mm)k ] (770 mm) = −(240 N)i + (255 N) j − (160 N)k ! BE = −(270 mm)i + (400 mm) j − (600 mm)k

BE = (270 mm) 2 + (400 mm) 2 + (600 mm) 2 = 770 mm ! BE FBE = TBE BE = TBE BE (385 N) = [−(270 mm)i + (400 mm) j − (600 mm)k ] (770 mm) = −(135 N)i + (200 N) j − (300 N)k R = FBD + FBE = −(375 N)i + (455 N) j − (460 N)k

R = (375 N)2 + (455 N) 2 + (460 N) 2 = 747.83 N

R = 748 N

cos θ x =

−375 N 747.83 N

θ x = 120.1°

cos θ y =

455 N 747.83 N

θ y = 52.5°

cos θ z =

−460 N 747.83 N

θ z = 128.0°

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 97

PROBLEM 2.96 For the cables of Problem 2.89, knowing that the tension is 1425 N in cable AB and 2130 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

SOLUTION TAB = −TBA

(use results of Problem 2.89)

(TAB ) x = +1125 N (TAB ) y = −750 N (TAB ) z = − 450 N TAC = −TCA

(use results of Problem 2.90)

(TAC ) x = +1350 N (TAC ) y = −900 N (TAC ) z = +1380 N

Resultant:

Rx = ΣFx = +1125 + 1350 = +2475 N Ry = ΣFy = −750 − 900 = −1650 N Rz = ΣFz = −450 + 1380 = + 930 N R = Rx2 + Ry2 + Rz2 = (+2475) 2 + (−1650) 2 + ( +930)2 = 3116.6 N cos θ x = cos θ y = cos θ z =

R = 3120 N

Rx +2475 = R 3116.6 Ry

θ x = 37.4°

−1650 3116.6

θ y = 122.0°

Rz + 930 = R 3116.6

θ z = 72.6°

R

=

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 98

!

PROBLEM 2.97 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in AC is 150 lb and that the resultant of the forces exerted at A by wires AC and AD must be contained in the xy plane, determine (a) the tension in AD, (b) the magnitude and direction of the resultant of the two forces.

SOLUTION R = TAC + TAD = (150 lb)(cos 60° cos 20°i − sin 60° j − cos 60° sin 20°k ) + TAD (sin 36° sin 48°i − cos 36° j + sin 36° cos 48°k )

(a)

(1)

Since Rz = 0, The coefficient of k must be zero. (150 lb)( − cos 60° sin 20°) + TAD (sin 36° cos 48°) = 0 TAD = 65.220 lb

(b)

TAD = 65.2 lb

Substituting for TAD into Eq. (1) gives: R = [(150 lb) cos 60° cos 20° + (65.220 lb) sin 36° sin 48°)]i − [(150 lb) sin 60° + (65.220 lb) cos 36°]j + 0 R = (98.966 lb)i − (182.668 lb) j R = (98.966 lb)2 + (182.668 lb) 2 = 207.76 lb

R = 208 lb

cos θ x =

98.966 lb 207.76 lb

θ x = 61.6°

cos θ y =

182.668 lb 207.76 lb

θ y = 151.6°

cos θ z = 0

θ z = 90.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 99

!

PROBLEM 2.98 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in AD is 125 lb and that the resultant of the forces exerted at A by wires AC and AD must be contained in the xy plane, determine (a) the tension in AC, (b) the magnitude and direction of the resultant of the two forces.

SOLUTION R = TAC + TAD = TAC (cos 60° cos 20°i − sin 60° j − cos 60° sin 20°k ) + (125 lb)(sin 36° sin 48°i − cos 36° j + sin 36° cos 48°k )

(a)

(1)

Since Rz = 0, The coefficient of k must be zero. TAC (− cos 60° sin 20°) + (125 lb)(sin 36° cos 48°) = 0 TAC = 287.49 lb

(b)

TAC = 287 lb

Substituting for TAC into Eq. (1) gives: R = [(287.49 lb) cos 60° cos 20° + (125 lb) sin 36° sin 48°]i − [(287.49 lb) sin 60° + (125 lb) cos 36°]j + 0 R = (189.677 lb)i − (350.10 lb) j R = (189.677 lb)2 + (350.10 lb) 2 R = 398 lb

= 398.18 lb cos θ x =

189.677 lb 398.18 lb

θ x = 61.6°

cos θ y =

350.10 lb 398.18 lb

θ y = 151.6°

cos θ z = 0

θ z = 90.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 100

!

PROBLEM 2.99 Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AB is 259 N.

SOLUTION

The forces applied at A are:

TAB , TAC , TAD , and P

where P = Pj. To express the other forces in terms of the unit vectors i, j, k, we write ! AB = − (4.20 m)i − (5.60 m) j AB = 7.00 m ! AC = (2.40 m)i − (5.60 m) j + (4.20 m)k AC = 7.40 m ! AD = − (5.60 m) j − (3.30 m)k AD = 6.50 m ! AB and TAB = TAB AB = TAB = (− 0.6i − 0.8 j)TAB AB! AC TAC = TAC AC = TAC = (0.32432 − 0.75676 j + 0.56757k )TAC AC! AD TAD = TAD AD = TAD = ( − 0.86154 j − 0.50769k )TAD AD

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 101

PROBLEM 2.99 (Continued)

Equilibrium condition

ΣF = 0: TAB + TAC + TAD + Pj = 0

Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k: (− 0.6TAB + 0.32432TAC )i + (−0.8TAB − 0.75676TAC − 0.86154TAD + P ) j + (0.56757TAC − 0.50769TAD )k = 0

Equating to zero the coefficients of i, j, k: − 0.6TAB + 0.32432TAC = 0

(1)

− 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0

(2)

0.56757TAC − 0.50769TAD = 0

(3)

Setting TAB = 259 N in (1) and (2), and solving the resulting set of equations gives TAC = 479.15 N TAD = 535.66 N

P = 1031 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 102

PROBLEM 2.100 Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AC is 444 N.

SOLUTION See Problem 2.99 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 0.6TAB + 0.32432TAC = 0

(1)

− 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0

(2)

0.56757TAC − 0.50769TAD = 0

(3)

Substituting TAC = 444 N in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives TAB = 240 N TAD = 496.36 N

P = 956 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 103

PROBLEM 2.101 Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AD is 481 N.

SOLUTION See Problem 2.99 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3). − 0.6TAB + 0.32432TAC = 0

(1)

− 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0

(2)

0.56757TAC − 0.50769TAD = 0

(3)

Substituting TAD = 481 N in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives TAC = 430.26 N TAB = 232.57 N

P = 926 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 104

PROBLEM 2.102 Three cables are used to tether a balloon as shown. Knowing that the balloon exerts an 800-N vertical force at A, determine the tension in each cable.

SOLUTION See Problem 2.99 for the figure and analysis leading to the linear algebraic Equations (1), (2), and (3). − 0.6TAB + 0.32432TAC = 0

(1)

− 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0

(2)

0.56757TAC − 0.50769TAD = 0

(3)

From Eq. (1)

TAB = 0.54053TAC

From Eq. (3)

TAD = 1.11795TAC

Substituting for TAB and TAD in terms of TAC into Eq. (2) gives: − 0.8(0.54053TAC ) − 0.75676TAC − 0.86154(1.11795TAC ) + P = 0 2.1523TAC = P ; P = 800 N 800 N 2.1523 = 371.69 N

TAC =

Substituting into expressions for TAB and TAD gives: TAB = 0.54053(371.69 N) TAD = 1.11795(371.69 N) TAB = 201 N, TAC = 372 N, TAD = 416 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 105

PROBLEM 2.103 A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AB is 750 lb.

SOLUTION The forces applied at A are: TAB , TAC , TAD and W

where P = Pj. To express the other forces in terms of the unit vectors i, j, k, we write ! AB = − (36 in.)i + (60 in.) j − (27 in.)k AB = 75 in. ! AC = (60 in.) j + (32 in.)k AC = 68 in. ! AD = (40 in.)i + (60 in.) j − (27 in.)k AD = 77 in. ! AB TAB = TAB AB = TAB and AB = (− 0.48i + 0.8 j − 0.36k )TAB ! AC TAC = TAC AC = TAC AC = (0.88235 j + 0.47059k )TAC ! AD TAD = TAD AD = TAD AD = (0.51948i + 0.77922 j − 0.35065k )TAD Equilibrium Condition with

W = − Wj ΣF = 0: TAB + TAC + TAD − Wj = 0

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PROBLEM 2.103 (Continued)

Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k: (− 0.48TAB + 0.51948TAD )i + (0.8TAB + 0.88235TAC + 0.77922TAD − W ) j + (− 0.36TAB + 0.47059TAC − 0.35065TAD )k = 0

Equating to zero the coefficients of i, j, k: −0.48TAB + 0.51948TAD = 0 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 −0.36TAB + 0.47059TAC − 0.35065TAD = 0

Substituting TAB = 750 lb in Equations (1), (2), and (3) and solving the resulting set of equations, using conventional algorithms for solving linear algebraic equations, gives: TAC = 1090.1 lb TAD = 693 lb

W = 2100 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 107

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PROBLEM 2.104 A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AD is 616 lb.

SOLUTION See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 0.48TAB + 0.51948TAD = 0 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 −0.36TAB + 0.47059TAC − 0.35065TAD = 0

Substituting TAD = 616 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: TAB = 667.67 lb TAC = 969.00 lb

W = 1868 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 108

PROBLEM 2.105 A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AC is 544 lb.

SOLUTION See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 0.48TAB + 0.51948TAD = 0 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 − 0.36TAB + 0.47059TAC − 0.35065TAD = 0

Substituting TAC = 544 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: TAB = 374.27 lb TAD = 345.82 lb

W = 1049 lb

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PROBLEM 2.106 A 1600-lb crate is supported by three cables as shown. Determine the tension in each cable.

SOLUTION See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: −0.48TAB + 0.51948TAD = 0 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 −0.36TAB + 0.47059TAC − 0.35065TAD = 0

Substituting W = 1600 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: TAB = 571 lb TAC = 830 lb TAD = 528 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 110

!

PROBLEM 2.107 Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that Q = 0, find the value of P for which the tension in cable AD is 305 N.

SOLUTION

!

ΣFA = 0: TAB + TAC + TAD + P = 0 where P = Pi ! AB = −(960 mm)i − (240 mm)j + (380 mm)k AB = 1060 mm ! AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm ! ! AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm ! 48 12 19 ! AB = TAB " − i − j + k # TAB = TAB AB = TAB 53 53 53 % AB $ ! 12 3 4 ! AC TAC = TAC AC = TAC = TAC " − i − j − k # 13 13 13 AC $ % 305 N [( −960 mm)i + (720 mm) j − (220 mm)k ] TAD = TAD AD = 1220 mm = −(240 N)i + (180 N) j − (55 N)k

Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives: i: P =

48 12 TAB + TAC + 240 N 53 13

(1)

j:

12 3 TAB + TAC = 180 N 53 13

(2)

k:

19 4 TAB − TAC = 55 N 53 13

(3)

Solving the system of linear equations using conventional algorithms gives: TAB = 446.71 N TAC = 341.71 N

P = 960 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 111

!

PROBLEM 2.108 Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that P = 1200 N, determine the values of Q for which cable AD is taut.

SOLUTION We assume that TAD = 0 and write ΣFA = 0: TAB + TAC + Qj + (1200 N)i = 0 ! AB = −(960 mm)i − (240 mm)j + (380 mm)k AB = 1060 mm ! AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm ! AB 48 12 19 ! TAB = TAB AB = TAB = " − i − j + k # TAB AB $ 53 53 53 % ! AC 12 3 4 ! TAC = TAC AC = TAC = " − i − j − k # TAC AC $ 13 13 13 % Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives: i: −

48 12 TAB − TAC + 1200 N = 0 53 13

(1)

j: −

12 3 TAB − TAC + Q = 0 53 13

(2)

k:

19 4 TAB − TAC = 0 53 13

(3)

Solving the resulting system of linear equations using conventional algorithms gives: TAB = 605.71 N TAC = 705.71 N Q = 300.00 N

0 # Q , 300 N

Note: This solution assumes that Q is directed upward as shown (Q $ 0), if negative values of Q are considered, cable AD remains taut, but AC becomes slack for Q = −460 N. !

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 112

PROBLEM 2.109 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AB is 630 lb, determine the vertical force P exerted by the tower on the pin at A.

SOLUTION Free Body A:

We write

ΣF = 0: TAB + TAC + TAD + Pj = 0 ! AB = −45i − 90 j + 30k AB = 105 ft ! AC = 30i − 90 j + 65k AC = 115 ft ! AD = 20i − 90 j − 60k AD = 110 ft ! AB TAB = TAB AB = TAB AB 3 6 2 ! = " − i − j + k # TAB 7 7 % $ 7 ! AC TAC = TAC AC = TAC AC 6 18 13 ! j + k # TAC =" i− 23 23 % $ 23 ! AD TAD = TAD AD = TAD AD 2 9 6 ! = " i − j − k # TAD 11 11 11 $ %

Substituting into the Eq. ΣF = 0 and factoring i, j, k : 3 6 2 ! " − TAB + TAC + TAD # i 7 23 11 $ % 6 18 9 ! + " − TAB − TAC − TAD + P # j 23 11 $ 7 % 2 13 6 ! + " TAB + TAC − TAD # k = 0 23 11 $7 % PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 113

PROBLEM 2.109 (Continued)

Setting the coefficients of i, j, k , equal to zero: i:

3 6 2 − TAB + TAC + TAD = 0 7 23 11

(1)

j:

6 18 9 − TAB − TAC − TAD + P = 0 7 23 11

(2)

k:

2 13 6 TAB + TAC − TAD = 0 7 23 11

(3)

Set TAB = 630 lb in Eqs. (1) – (3): 6 2 TAC + TAD = 0 23 11

(1′)

18 9 TAC − TAD + P = 0 23 11

(2′)

13 6 TAC − TAD = 0 23 11

(3′)

−270 lb + −540 lb −

180 lb +

Solving,

TAC = 467.42 lb TAD = 814.35 lb P = 1572.10 lb

P = 1572 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 114

!

PROBLEM 2.110 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 920 lb, determine the vertical force P exerted by the tower on the pin at A.

SOLUTION See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 3 6 2 − TAB + TAC + TAD = 0 7 23 11

(1)

6 18 9 − TAB − TAC − TAD + P = 0 7 23 11

(2)

2 13 6 TAB + TAC − TAD = 0 7 23 11

(3)

Substituting for TAC = 920 lb in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives:

Solving,

3 2 − TAB + 240 lb + TAD = 0 7 11

(1′)

6 9 − TAB − 720 lb − TAD + P = 0 7 11

(2′)

2 6 TAB + 520 lb − TAD = 0 7 11

(3′)

TAB = 1240.00 lb TAD = 1602.86 lb P = 3094.3 lb

P = 3090 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 115

PROBLEM 2.111 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AC is 60 N, determine the weight of the plate.

SOLUTION We note that the weight of the plate is equal in magnitude to the force P exerted by the support on Point A. Free Body A: ΣF = 0: TAB + TAC + TAD + Pj = 0

We have: ! AB = −(320 mm)i − (480 mm)j + (360 mm)k AB = 680 mm ! AC = (450 mm)i − (480 mm) j + (360 mm)k AC = 750 mm ! AD = (250 mm)i − (480 mm) j − ( 360 mm ) k AD = 650 mm

Thus: TAB = TAB

AB

TAC = TAC

AC

TAD = TAD

AD

! AB 8 12 9 ! = − i − j + k TAB AB "$ 17 17 17 #% ! AC = TAC = ( 0.6i − 0.64 j + 0.48k ) TAC AC ! AD 5 9.6 7.2 ! = TAD =" i− j− k TAD AD $ 13 13 13 #%

= TAB

Dimensions in mm

Substituting into the Eq. ΣF = 0 and factoring i, j, k : 8 5 ! " − TAB + 0.6TAC + TAD # i 13 $ 17 % 12 9.6 ! TAD + P # j + " − TAB − 0.64TAC − 17 13 $ % 9 7.2 ! TAD # k = 0 + " TAB + 0.48TAC − 13 $ 17 %

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PROBLEM 2.111 (Continued) Setting the coefficient of i, j, k equal to zero: i:

−

8 5 TAB + 0.6TAC + TAD = 0 17 13

(1)

j:

−

12 9.6 TAB − 0.64TAC − TAD + P = 0 7 13

(2)

9 7.2 TAB + 0.48TAC − TAD = 0 17 13

(3)

8 5 TAB + 36 N + TAD = 0 17 13

(1′)

9 7.2 TAB + 28.8 N − TAD = 0 17 13

(3′)

k:

Making TAC = 60 N in (1) and (3): −

Multiply (1′) by 9, (3′) by 8, and add: 554.4 N −

12.6 TAD = 0 TAD = 572.0 N 13

Substitute into (1′) and solve for TAB : TAB =

17 5 ! 36 + × 572 # " 8 $ 13 %

TAB = 544.0 N

Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (544 N) + 0.64(60 N) + (572 N) 17 13 = 844.8 N

P=

Weight of plate = P = 845 N

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PROBLEM 2.112 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 520 N, determine the weight of the plate.

SOLUTION See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 8 5 TAB + 0.6TAC + TAD = 0 17 13

(1)

12 9.6 TAB + 0.64 TAC − TAD + P = 0 17 13

(2)

9 7.2 TAB + 0.48TAC − TAD = 0 17 13

(3)

− −

Making TAD = 520 N in Eqs. (1) and (3): 8 TAB + 0.6TAC + 200 N = 0 17

(1′)

9 TAB + 0.48TAC − 288 N = 0 17

(3′)

−

Multiply (1′) by 9, (3′) by 8, and add: 9.24TAC − 504 N = 0 TAC = 54.5455 N

Substitute into (1′) and solve for TAB : TAB =

17 (0.6 × 54.5455 + 200) TAB = 494.545 N 8

Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (494.545 N) + 0.64(54.5455 N) + (520 N) 17 13 Weight of plate = P = 768 N = 768.00 N

P=

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PROBLEM 2.113 For the transmission tower of Problems 2.109 and 2.110, determine the tension in each guy wire knowing that the tower exerts on the pin at A an upward vertical force of 2100 lb.

SOLUTION See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 3 6 2 − TAB + TAC + TAD = 0 7 23 11

(1)

6 18 9 − TAB − TAC − TAD + P = 0 7 23 11

(2)

2 13 6 TAB + TAC − TAD = 0 7 23 11

(3)

Substituting for P = 2100 lb in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives: 3 6 2 − TAB + TAC + TAD = 0 7 23 11

(1′)

6 18 9 − TAB − TAC − TAD + 2100 lb = 0 7 23 11

(2′)

2 13 6 TAB + TAC − TAD = 0 7 23 11

(3′)

TAB = 841.55 lb TAC = 624.38 lb TAD = 1087.81 lb

TAB = 842 lb TAC = 624 lb TAD = 1088 lb

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!

PROBLEM 2.114 A horizontal circular plate weighing 60 lb is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Determine the tension in each wire.

SOLUTION ΣFx = 0: −TAD (sin 30°)(sin 50°) + TBD (sin 30°)(cos 40°) + TCD (sin 30°)(cos 60°) = 0

Dividing through by sin 30° and evaluating: −0.76604TAD + 0.76604TBD + 0.5TCD = 0

(1)

ΣFy = 0: −TAD (cos 30°) − TBD (cos 30°) − TCD (cos 30°) + 60 lb = 0 TAD + TBD + TCD = 69.282 lb

or

(2)

ΣFz = 0: TAD sin 30° cos 50° + TBD sin 30° sin 40° − TCD sin 30° sin 60° = 0

!

or

0.64279TAD + 0.64279TBD − 0.86603TCD = 0

(3)

Solving Equations (1), (2), and (3) simultaneously: TAD = 29.5 lb TBD = 10.25 lb

!

!

!

TCD = 29.5 lb

!

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PROBLEM 2.115 For the rectangular plate of Problems 2.111 and 2.112, determine the tension in each of the three cables knowing that the weight of the plate is 792 N.

SOLUTION See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below. Setting P = 792 N gives: 8 5 TAB + 0.6TAC + TAD = 0 17 13

(1)

12 9.6 TAB − 0.64TAC − TAD + 792 N = 0 17 13

(2)

9 7.2 TAB + 0.48TAC − TAD = 0 17 13

(3)

− −

Solving Equations (1), (2), and (3) by conventional algorithms gives TAB = 510.00 N

TAB = 510 N

TAC = 56.250 N

TAC = 56.2 N

TAD = 536.25 N

TAD = 536 N

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PROBLEM 2.116 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = 0.

SOLUTION ΣFA = 0: TAB + TAC + TAD + P + Q = 0

Where

P = Pi and Q = Qj ! AB = −(960 mm)i − (240 mm) j + (380 mm)k AB = 1060 mm ! AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm ! AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm ! 48 12 19 ! AB TAB = TAB AB = TAB = TAB " − i − j + k # 53 53 53 % AB $ ! 12 3 4 ! AC TAC = TAC AC = TAC = TAC " − i − j − k # 13 13 13 AC $ % ! 48 36 11 ! AD = TAD " − i + TAD = TAD AD = TAD j− k# 61 61 61 % AD $

Substituting into ΣFA = 0, setting P = (2880 N)i and Q = 0, and setting the coefficients of i, j, k equal to 0, we obtain the following three equilibrium equations: i: −

48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61

(1)

j: −

12 3 36 TAB − TAC + TAD = 0 53 13 61

(2)

k:

19 4 11 TAB − TAC − TAD = 0 53 13 61

(3)

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 122

PROBLEM 2.116 (Continued)

Solving the system of linear equations using conventional algorithms gives: TAB = 1340.14 N TAC = 1025.12 N TAD = 915.03 N

TAB = 1340 N TAC = 1025 N TAD = 915 N

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PROBLEM 2.117 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = 576 N.

SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below: −

48 12 48 TAB − TAC − TAD + P = 0 53 13 61

(1)

−

12 3 36 TAB − TAC + TAD + Q = 0 53 13 61

(2)

19 4 11 TAB − TAC − TAD = 0 53 13 61

(3)

Setting P = 2880 N and Q = 576 N gives: −

48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61

(1′)

12 3 36 TAB − TAC + TAD + 576 N = 0 53 13 61

(2′)

19 4 11 TAB − TAC − TAD = 0 53 13 61

(3′)

−

Solving the resulting set of equations using conventional algorithms gives: TAB = 1431.00 N TAC = 1560.00 N TAD = 183.010 N

TAB = 1431 N

!

TAC = 1560 N TAD = 183.0 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 124

PROBLEM 2.118 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = −576 N. (Q is directed downward).

SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:! !

−

48 12 48 TAB − TAC − TAD + P = 0 53 13 61

(1)

−

12 3 36 TAB − TAC + TAD + Q = 0 53 13 61

(2)

19 4 11 TAB − TAC − TAD = 0 53 13 61

(3)

Setting P = 2880 N and Q = −576 N gives: −

48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61

(1′)

12 3 36 TAB − TAC + TAD − 576 N = 0 53 13 61

(2′)

19 4 11 TAB − TAC − TAD = 0 53 13 61

(3′)

−

Solving the resulting set of equations using conventional algorithms gives: TAB = 1249.29 N TAC = 490.31 N TAD = 1646.97 N

TAB = 1249 N TAC = 490 N TAD = 1647 N

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PROBLEM 2.119 Using two ropes and a roller chute, two workers are unloading a 200-lb cast-iron counterweight from a truck. Knowing that at the instant shown the counterweight is kept from moving and that the positions of Points A, B, and C are, respectively, A(0, –20 in., 40 in.), B(–40 in., 50 in., 0), and C(45 in., 40 in., 0), and assuming that no friction exists between the counterweight and the chute, determine the tension in each rope. (Hint: Since there is no friction, the force exerted by the chute on the counterweight must be perpendicular to the chute.)

SOLUTION From the geometry of the chute: N=

N

(2 j + k ) 5 = N (0.8944 j + 0.4472k )

The force in each rope can be written as the product of the magnitude of the force and the unit vector along the cable. Thus, with ! AB = (40 in.)i + (70 in.)j − (40 in.)k

and

AB = (40 in.) 2 + (70 in.) 2 + (40 in.)2 = 90 in. ! AB TAB = T AB = TAB AB TAB [(−40 in.)i + (70 in.)j − (40 in.)k ] = 90 in. 4 7 4 ! TAB = TAB " − i + j − k # 9 9 % $ 9 ! AC = (45 in.)i + (60 in.)j − (40 in.)k AC = (45 in.) 2 + (60 in.)2 + (40 in.)2 = 85 in. ! AC TAC = T AC = TAC AC TAC [(45 in.)i + (60 in.)j − (40 in.)k ] = 85 in. 9 12 8 ! TAC = TAC " i + j − k # $ 17 17 17 %

Then:

ΣF = 0: N + TAB + TAC + W = 0

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PROBLEM 2.119 (Continued)

With W = 200 lb, and equating the factors of i, j, and k to zero, we obtain the linear algebraic equations: i: j:

k:

4 9 − TAB + TAC = 0 9 17

(1)

7 12 2 TAB + TAC + − 200 lb = 0 9 17 5

(2)

4 8 1 − TAB − TAC + N =0 9 17 5

(3)

Using conventional methods for solving linear algebraic equations we obtain: TAB = 65.6 lb TAC = 55.1 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 127

PROBLEM 2.120 Solve Problem 2.119 assuming that a third worker is exerting a force P = −(40 lb)i on the counterweight. PROBLEM 2.119 Using two ropes and a roller chute, two workers are unloading a 200-lb cast-iron counterweight from a truck. Knowing that at the instant shown the counterweight is kept from moving and that the positions of Points A, B, and C are, respectively, A(0, –20 in., 40 in.), B(–40 in., 50 in., 0), and C(45 in., 40 in., 0), and assuming that no friction exists between the counterweight and the chute, determine the tension in each rope. (Hint: Since there is no friction, the force exerted by the chute on the counterweight must be perpendicular to the chute.)

SOLUTION See Problem 2.119 for the analysis leading to the vectors describing the tension in each rope. 4 7 4 ! TAB = TAB " − i + j − k # 9 9 % $ 9 9 12 8 ! TAC = TAC " i + j − k # 17 17 17 $ %

Then: Where and

ΣFA = 0: N + TAB + TAC + P + W = 0 P = −(40 lb)i W = (200 lb)j

Equating the factors of i, j, and k to zero, we obtain the linear equations: 4 9 i : − TAB + TAC − 40 lb = 0 9 17 j:

k:

2

7 12 N + TAB + TAC − 200 lb = 0 9 17 5

1 5

4 8 N − TAB − TAC = 0 9 17

Using conventional methods for solving linear algebraic equations we obtain TAB = 24.8 lb TAC = 96.4 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 128

PROBLEM 2.121 A container of weight W is suspended from ring A. Cable BAC passes through the ring and is attached to fixed supports at B and C. Two forces P = Pi and Q = Qk are applied to the ring to maintain the container in the position shown. Knowing that W = 376 N, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.)

SOLUTION Free Body A:

TAB = T

AB

" AB =T AB (−130 mm)i + (400 mm) j + (160 mm)k =T 450 mm 13 40 16 ! =T "− i + j+ k# 45 45 % $ 45 TAC = T

AC

" AC =T AC ( −150 mm)i + (400 mm) j + (−240 mm)k =T 490 mm 15 40 24 ! = T "− i + j− k# 49 49 49 % $ ΣF = 0: TAB + TAC + Q + P + W = 0

Setting coefficients of i, j, k equal to zero: i: −

13 15 T − T +P=0 45 49

0.59501T = P

(1)

j: +

40 40 T + T −W = 0 45 49

1.70521T = W

(2)

k: +

16 24 T − T +Q =0 45 49

0.134240 T = Q

(3)

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PROBLEM 2.121 (Continued)

Data:

W = 376 N 1.70521T = 376 N T = 220.50 N 0.59501(220.50 N) = P

P = 131.2 N

0.134240(220.50 N) = Q

Q = 29.6 N

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PROBLEM 2.122 For the system of Problem 2.121, determine W and Q knowing that P = 164 N. PROBLEM 2.121 A container of weight W is suspended from ring A. Cable BAC passes through the ring and is attached to fixed supports at B and C. Two forces P = Pi and Q = Qk are applied to the ring to maintain the container in the position shown. Knowing that W = 376 N, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.)

SOLUTION Refer to Problem 2.121 for the figure and analysis resulting in Equations (1), (2), and (3) for P, W, and Q in terms of T below. Setting P = 164 N we have: Eq. (1):

0.59501T = 164 N

T = 275.63 N

Eq. (2):

1.70521(275.63 N) = W

W = 470 N

Eq. (3):

0.134240(275.63 N) = Q

Q = 37.0 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 131

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PROBLEM 2.123 A container of weight W is suspended from ring A, to which cables AC and AE are attached. A force P is applied to the end F of a third cable that passes over a pulley at B and through ring A and that is attached to a support at D. Knowing that W = 1000 N, determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.)

SOLUTION The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector along the cable. That is, with ! AB = −(0.78 m)i + (1.6 m) j + (0 m)k AB = (−0.78 m) 2 + (1.6 m) 2 + (0)2 = 1.78 m TAB = T

AB

= TAB

! AB AB

TAB [−(0.78 m)i + (1.6 m) j + (0 m)k ] 1.78 m = TAB (−0.4382i + 0.8989 j + 0k ) =

and

TAB ! AC = (0)i + (1.6 m) j + (1.2 m)k

and

AC = (0 m) 2 + (1.6 m) 2 + (1.2 m) 2 = 2 m ! AC TAC TAC = T AC = TAC = [(0)i + (1.6 m) j + (1.2 m)k ] AC 2 m TAC = TAC (0.8 j + 0.6k ) ! AD = (1.3 m)i + (1.6 m) j + (0.4 m)k

AD = (1.3 m)2 + (1.6 m)2 + (0.4 m) 2 = 2.1 m ! T AD TAD = T AD = TAD = AD [(1.3 m)i + (1.6 m) j + (0.4 m)k ] AD 2.1 m TAD = TAD (0.6190i + 0.7619 j + 0.1905k )

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PROBLEM 2.123 (Continued)

! AE = −(0.4 m)i + (1.6 m) j − (0.86 m)k

Finally,

AE = (−0.4 m) 2 + (1.6 m) 2 + (−0.86 m) 2 = 1.86 m ! AE TAE = T AE = TAE AE T = AE [−(0.4 m)i + (1.6 m) j − (0.86 m)k ] 1.86 m TAE = TAE (−0.2151i + 0.8602 j − 0.4624k )

With the weight of the container

W = −W j, at A we have: ΣF = 0: TAB + TAC + TAD − Wj = 0

Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations: −0.4382TAB + 0.6190TAD − 0.2151TAE = 0

(1)

0.8989TAB + 0.8TAC + 0.7619TAD + 0.8602TAE − W = 0

(2)

0.6TAC + 0.1905TAD − 0.4624TAE = 0

(3)

Knowing that W = 1000 N and that because of the pulley system at B TAB = TAD = P, where P is the externally applied (unknown) force, we can solve the system of linear Equations (1), (2) and (3) uniquely for P. P = 378 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 133

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PROBLEM 2.124 Knowing that the tension in cable AC of the system described in Problem 2.123 is 150 N, determine (a) the magnitude of the force P, (b) the weight W of the container. PROBLEM 2.123 A container of weight W is suspended from ring A, to which cables AC and AE are attached. A force P is applied to the end F of a third cable that passes over a pulley at B and through ring A and that is attached to a support at D. Knowing that W = 1000 N, determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.)

SOLUTION Here, as in Problem 2.123, the support of the container consists of the four cables AE, AC, AD, and AB, with the condition that the force in cables AB and AD is equal to the externally applied force P. Thus, with the condition TAB = TAD = P

and using the linear algebraic equations of Problem 2.131 with TAC = 150 N, we obtain (a)

P = 454 N

(b) W = 1202 N

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PROBLEM 2.125 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. If a 60-lb force Q is applied to collar B as shown, determine (a) the tension in the wire when x = 9 in., (b) the corresponding magnitude of the force P required to maintain the equilibrium of the system.

SOLUTION Free Body Diagrams of Collars: A:

B:

AB

ΣF = 0: Pi + N y j + N z k + TAB λ AB = 0

Collar A: Substitute for

! AB − xi − (20 in.) j + zk = = AB 25 in.

AB

and set coefficient of i equal to zero: P−

AB

and set coefficient of k equal to zero: 60 lb − x = 9 in.

(a)

From Eq. (2): (b)

(1)

ΣF = 0: (60 lb)k + N x′ i + N y′ j − TAB λ AB = 0

Collar B: Substitute for

TAB x =0 25 in.

From Eq. (1):

TAB z =0 25 in.

(2)

(9 in.)2 + (20 in.) 2 + z 2 = (25 in.) 2 z = 12 in. 60 lb − TAB (12 in.) 25 in. P=

(125.0 lb)(9 in.) ! 25 in.

TAB = 125.0 lb P = 45.0 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 135

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PROBLEM 2.126 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. Determine the distances x and z for which the equilibrium of the system is maintained when P = 120 lb and Q = 60 lb.

SOLUTION See Problem 2.125 for the diagrams and analysis leading to Equations (1) and (2) below: P=

TAB x =0 25 in.

(1)

60 lb −

TAB z =0 25 in.

(2)

For P = 120 lb, Eq. (1) yields

TAB x = (25 in.)(20 lb)

(1′)

From Eq. (2)

TAB z = (25 in.) (60 lb)

(2′)

x =2 z

Dividing Eq. (1′) by (2′): Now write

(3)

x 2 + z 2 + (20 in.) 2 = (25 in.) 2

(4)

Solving (3) and (4) simultaneously 4 z 2 + z 2 + 400 = 625 z 2 = 45 z = 6.708 in.

From Eq. (3)

x = 2 z = 2(6.708 in.) = 13.416 in. x = 13.42 in., z = 6.71 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 136

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PROBLEM 2.127 The direction of the 75-lb forces may vary, but the angle between the forces is always 50°. Determine the value of for which the resultant of the forces acting at A is directed horizontally to the left.

SOLUTION We must first replace the two 75-lb forces by their resultant R1 using the triangle rule.

R1 = 2(75 lb) cos 25° = 135.946 lb R1 = 135.946 lb

α + 25°

Next we consider the resultant R 2 of R1 and the 240-lb force where R 2 must be horizontal and directed to the left. Using the triangle rule and law of sines, sin (α + 25°) sin (30°) = 240 lb 135.946 sin (α + 25°) = 0.88270

α + 25° = 61.970° α = 36.970°

α = 37.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 137

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PROBLEM 2.128 A stake is being pulled out of the ground by means of two ropes as shown. Knowing the magnitude and direction of the force exerted on one rope, determine the magnitude and direction of the force P that should be exerted on the other rope if the resultant of these two forces is to be a 40-lb vertical force.

SOLUTION Triangle rule: Law of cosines:

Law of sines:

P 2 = (30) 2 + (40) 2 − 2(30)(40) cos 25° P = 18.0239 lb

sin α sin 25° = 30 lb 18.0239 lb α = 44.703° 90° − α = 45.297°

P = 18.02 lb

45.3°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 138

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PROBLEM 2.129 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 240-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component.

SOLUTION

(a)

P=

(b)

Px =

Py

sin 35° Py

tan 40°

=

240 lb sin 40°

or P = 373 lb

=

240 lb tan 40°

or Px = 286 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 139

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PROBLEM 2.130 Two cables are tied together at C and loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION Free Body Diagram at C: ΣFx = 0: −

12 ft 7.5 ft TAC + TBC = 0 12.5 ft 8.5 ft TBC = 1.08800TAC

ΣFy = 0:

(a)

3.5 ft 4 ft TAC + TBC − 396 lb = 0 12 ft 8.5 ft 3.5 ft 4 ft (1.08800TAC ) − 396 lb = 0 TAC + 12.5 ft 8.5 ft (0.28000 + 0.51200)TAC = 396 lb TAC = 500.0 lb

(b)

TBC = (1.08800)(500.0 lb)

TAC = 500 lb TBC = 544 lb

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PROBLEM 2.131 Two cables are tied together at C and loaded as shown. Knowing that P = 360 N, determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION Free Body: C

(a)

ΣFx = 0: −

(b)

ΣFy = 0:

12 4 TAC + (360 N) = 0 13 5

TAC = 312 N

5 3 (312 N) + TBC + (360 N) − 480 N = 0 13 5 TBC = 480 N − 120 N − 216 N

TBC = 144 N

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PROBLEM 2.132 Two cables are tied together at C and loaded as shown. Determine the range of values of P for which both cables remain taut.

SOLUTION Free Body: C ΣFx = 0: −

12 4 TAC + P = 0 13 5 TAC =

ΣFy = 0:

Substitute for TAC from (1):

13 P 15

(1)

5 3 TAC + TBC + P − 480 N = 0 13 5

3 ! 5 "! 13 " # 13 $# 15 $ P + TBC + 5 P − 480 N = 0 % &% & TBC = 480 N −

14 P 15

(2)

From (1), TAC . 0 requires P . 0. From (2), TBC . 0 requires

14 P , 480 N, P , 514.29 N 15 0 , P , 514 N

Allowable range:

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PROBLEM 2.133 A force acts at the origin of a coordinate system in a direction defined by the angles θ x = 69.3° and θ z = 57.9°. Knowing that the y component of the force is –174.0 lb, determine (a) the angle θ y , (b) the other components and the magnitude of the force.

SOLUTION (a)

To determine θ y , use the relation cos 2 θ y + cos 2 θ y + cos 2 θ z = 1

or

cos 2 θ y = 1 − cos 2 θ x − cos 2 θ y

Since Fy , 0, we must have cos θ y , 0 cos θ y = − 1 − cos 2 69.3° − cos 2 57.9°

θ y = 140.3°

= − 0.76985

(b)

Fy

−174.0 lb = 226.02 lb −0.76985

F = 226 lb

Fx = F cos θ x = (226.02 lb) cos 69.3°

Fx = 79.9 lb

Fz = F cos θ z = (226.02 lb) cos 57.9°

Fz = 120.1 lb

F=

cos θ y

=

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PROBLEM 2.134 Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles θ x , θ y , and θ z defining the direction of that force.

SOLUTION 56 ft 65 ft = 0.86154

cos θ y =

From triangle AOB:

θ y = 30.51° Fx = − F sin θ y cos 20°

(a)

= −(3900 lb)sin 30.51° cos 20° Fx = −1861 lb Fy = + F cos θ y = (3900 lb)(0.86154)

!

Fz = + (3900 lb)sin 30.51° sin 20°

(b)

cos θ x =

From above:

Fx 1861 lb =− = − 0.4771 3900 lb F

θ y = 30.51° cos θ z =

Fz 677 lb =+ = + 0.1736 3900 lb F

Fy = +3360 lb Fz = +677 lb

θ x = 118.5° θ y = 30.5° θ z = 80.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 144

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PROBLEM 2.135 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension is 10 kN in cable AB and 7.5 kN in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

SOLUTION !!!" AB = −15.588i + 15 j + 12k AB = 24.739 m !!!" AC = −15.588i + 18.60 j − 15k AC = 28.530 m !!!" AB TAB = TAB !AB = TAB AB −15.588i + 15 j + 12k TAB = (10 kN) 24.739 TAB = (6.301 kN)i + (6.063 kN) j + (4.851 kN)k !!!" −15.588i + 18.60 j − 15k AC TAC = TAC !AC = TAC (7.5 kN) 28.530 AC TAC = −(4.098 kN)i + (4.890 kN) j − (3.943 kN)k R = TAB + TAC = −(10.399 kN)i + (10.953 kN) j + (0.908 kN)k R = (10.399)2 + (10.953)2 + (0.908) 2 R = 15.13 kN

= 15.130 kN cos θ x = cos θ y = cos θ z =

Rx −10.399 kN = = − 0.6873 15.130 kN R Ry

θ z = 133.4°

10.953 kN = 0.7239 15.130 kN

θ y = 43.6°

Rz 0.908 kN = = 0.0600 R 15.130 kN

θ z = 86.6°

R

=

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PROBLEM 2.136 A container of weight W = 1165 N is supported by three cables as shown. Determine the tension in each cable.

SOLUTION Free Body: A

We have:

ΣF = 0: TAB + TAC + TAD + W = 0 !!!" AB = (450 mm)i + (600 mm) j AB = 750 mm !!!" AC = (600 mm) j − (320 mm)k AC = 680 mm !!!" AD = (500 mm)i + (600 mm) j + (360 mm)k AD = 860 mm !!!" AB ! 450 600 " =# TAB = TAB λ AB = TAB i+ j TAB AB % 750 750 $& = (0.6i + 0.8 j)TAB !!!" AC ! 600 320 " 8 " ! 15 TAC = TAC λ AC = TAC =# j− k $ TAC = # j − k $ TAC 680 & AC % 680 % 17 17 & !!!" 600 360 " AD ! 500 = #− TAD = TAD λ AD = TAD i+ j+ k TAD AD % 860 860 860 $& 30 18 " ! 25 TAD = # − i + j + k $ TAD 43 43 & % 43

Substitute into ΣF = 0, factor i, j, k, and set their coefficient equal to zero: 0.6TAB − 0.8TAB + −

25 TAD = 0 43

TAB = 0.96899TAD

15 30 TAC + TAD − 1165 N = 0 17 43

8 18 TAC + TAD = 0 17 43

TAC = 0.88953TAD

(1) (2) (3)

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PROBLEM 2.136 (Continued)

Substitute for TAB and TAC from (1) and (3) into (2): 15 30 " ! # 0.8 × 0.96899 + 17 × 0.88953 + 53 $ TAD − 1165 N = 0 % & 2.2578TAD − 1165 N = 0

TAD = 516 N

From (1):

TAB = 0.96899(516 N)

TAB = 500 N

From (3):

TAC = 0.88953(516 N)

TAC = 459 N

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PROBLEM 2.137 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force P = (341 N)j is applied to collar A, determine (a) the tension in the wire when y = 155 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the system.

SOLUTION For both Problems 2.137 and 2.138:

Free Body Diagrams of Collars: ( AB) 2 = x 2 + y 2 + z 2

Here

(0.525 m) 2 = (.20 m) 2 + y 2 + z 2 y 2 + z 2 = 0.23563 m 2

or Thus, why y given, z is determined, Now

!AB

!!!" AB = AB 1 (0.20i − yj + zk )m 0.525 m = 0.38095i − 1.90476 yj + 1.90476 zk =

Where y and z are in units of meters, m. From the F.B. Diagram of collar A:

ΣF = 0: N x i + N z k + Pj + TAB λ AB = 0

Setting the j coefficient to zero gives:

P − (1.90476 y )TAB = 0 P = 341 N

With

TAB =

341 N 1.90476 y

Now, from the free body diagram of collar B:

ΣF = 0: N x i + N y j + Qk − TAB !AB = 0

Setting the k coefficient to zero gives:

Q − TAB (1.90476 z ) = 0

And using the above result for TAB we have

Q = TAB z =

341 N (341 N)( z ) (1.90476 z ) = y (1.90476) y

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PROBLEM 2.137 (Continued)

Then, from the specifications of the problem, y = 155 mm = 0.155 m z 2 = 0.23563 m 2 − (0.155 m) 2 z = 0.46 m

and 341 N 0.155(1.90476) = 1155.00 N

TAB =

(a)

TAB = 1.155 kN

or and 341 N(0.46 m)(0.866) (0.155 m) = (1012.00 N)

Q=

(b)

Q = 1.012 kN

or

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PROBLEM 2.138 Solve Problem 2.137 assuming that y = 275 mm. PROBLEM 2.137 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force P = (341 N)j is applied to collar A, determine (a) the tension in the wire when y = 155 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the system.

SOLUTION From the analysis of Problem 2.137, particularly the results: y 2 + z 2 = 0.23563 m 2 341 N TAB = 1.90476 y 341 N Q= z y

With y = 275 mm = 0.275 m, we obtain: z 2 = 0.23563 m 2 − (0.275 m) 2 z = 0.40 m

and TAB =

(a)

341 N = 651.00 (1.90476)(0.275 m) TAB = 651 N

or and Q=

(b)

341 N(0.40 m) (0.275 m) Q = 496 N

or

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CHAPTER 3

PROBLEM 3.1 A foot valve for a pneumatic system is hinged at B. Knowing that a = 28°, determine the moment of the 16-N force about Point B by resolving the force into horizontal and vertical components.

SOLUTION Note that and

θ = α − 20° = 28° − 20° = 8° Fx = (16 N) cos8° = 15.8443 N Fy = (16 N) sin 8° = 2.2268 N

Also

x = (0.17 m) cos 20° = 0.159748 m y = (0.17 m) sin 20° = 0.058143 m

Noting that the direction of the moment of each force component about B is counterclockwise, MB = xFy + yFx = (0.159748 m)(2.2268 N) + (0.058143 m)(15.8443 N) = 1.277 N ⋅ m

or M B = 1.277 N ⋅ m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 153

PROBLEM 3.2 A foot valve for a pneumatic system is hinged at B. Knowing that a = 28°, determine the moment of the 16-N force about Point B by resolving the force into components along ABC and in a direction perpendicular to ABC.

SOLUTION First resolve the 4-lb force into components P and Q, where Q = (16 N) sin 28° = 7.5115 N

Then

M B = rA/B Q

= (0.17 m)(7.5115 N)

or M B = 1.277 N ⋅ m

= 1.277 N ⋅ m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 154

PROBLEM 3.3 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the smallest force applied at B that creates the same moment about D.

SOLUTION (a)

Fx = (300 N) cos 25°

= 271.89 N Fy = (300 N) sin 25° = 126.785 N F = (271.89 N)i + (126.785 N) j ! r = DA = −(0.1 m)i − (0.2 m) j MD = r × F M D = [−(0.1 m)i − (0.2 m) j] × [(271.89 N)i + (126.785 N) j] = −(12.6785 N ⋅ m)k + (54.378 N ⋅ m)k = (41.700 N ⋅ m)k M D = 41.7 N ⋅ m

(b)

The smallest force Q at B must be perpendicular to ! DB at 45° ! M D = Q ( DB ) 41.700 N ⋅ m = Q (0.28284 m)

Q = 147.4 N

45°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 155

PROBLEM 3.4 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the magnitude and sense of the horizontal force applied at C that creates the same moment about D, (c) the smallest force applied at C that creates the same moment about D.

SOLUTION (a)

See Problem 3.3 for the figure and analysis leading to the determination of MD M D = 41.7 N ⋅ m

(b)

Since C is horizontal C = C i ! r = DC = (0.2 m)i − (0.125 m) j M D = r × C i = C (0.125 m)k

41.7 N ⋅ m = (0.125 m)(C ) C = 333.60 N

(c)

C = 334 N

The smallest force C must be perpendicular to DC; thus, it forms α with the vertical

0.125 m 0.2 m α = 32.0°

tan α =

M D = C ( DC ); DC = (0.2 m) 2 + (0.125 m) 2

= 0.23585 m 41.70 N ⋅ m = C (0.23585 m)

C = 176.8 N

58.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 156

PROBLEM 3.5 An 8-lb force P is applied to a shift lever. Determine the moment of P about B when a is equal to 25°.

SOLUTION First note

Px = (8 lb) cos 25°

= 7.2505 lb Py = (8 lb)sin 25°

= 3.3809 lb

Noting that the direction of the moment of each force component about B is clockwise, have M B = − xPy − yPx

= −(8 in.)(3.3809 lb) − (22 in.)(7.2505 lb) = −186.6 lb ⋅ in.

or M B = 186.6 lb ⋅ in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 157

PROBLEM 3.6 For the shift lever shown, determine the magnitude and the direction of the smallest force P that has a 210-lb ⋅ in. clockwise moment about B.

SOLUTION For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus,

α =θ 8 in. 22 in. = 19.98° = tan −1

and Where

M B = dPmin d = rA/B

= (8 in.) 2 + (22 in.)2 = 23.409 in.

Then

210 lb ⋅ in. 23.409 in. = 8.97 lb

Pmin =

Pmin = 8.97 lb

19.98°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 158

PROBLEM 3.7 An 11-lb force P is applied to a shift lever. The moment of P about B is clockwise and has a magnitude of 250 lb ⋅ in. Determine the value of a.

SOLUTION By definition

M B = rA/B P sin θ

where

θ = α + (90° − φ )

and

φ = tan −1

also

8 in. = 19.9831° 22 in.

rA/B = (8 in.) 2 + (22 in.) 2 = 23.409 in.

Then

or

250 lb ⋅ in = (23.409 in.)(11 lb) x sin(α + 90° − 19.9831°) sin (α + 70.0169°) = 0.97088

or

α + 70.0169° = 76.1391°

and

α + 70.0169° = 103.861°

α = 6.12° 33.8°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 159

PROBLEM 3.8 It is known that a vertical force of 200 lb is required to remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P that creates the same moment about B if a = 10°, (c) the smallest force P that creates the same moment about B.

SOLUTION (a)

M B = rC/B FN

We have

= (4 in.)(200 lb) = 800 lb ⋅ in.

or MB = 800 lb ⋅ in. (b)

By definition

M B = rA/B P sin θ

θ = 10° + (180° − 70°) = 120°

Then

800 lb ⋅ in. = (18 in.) × P sin120°

or P = 51.3 lb (c)

For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus, P must be directed as shown. Thus

M B = dPmin d = rA/B

or or

800 lb ⋅ in. = (18 in.)Pmin Pmin = 44.4 lb

Pmin = 44.4 lb

20°

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PROBLEM 3.9 A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and length d is 1.90 m, determine the moment about D of the force exerted by the cable at C by resolving that force into horizontal and vertical components applied (a) at Point C, (b) at Point E.

SOLUTION (a)

Slope of line

EC =

12 (TAB ) 13 12 = (1040 N) 13 = 960 N

Then

TABx =

and

TABy =

Then

0.875 m 5 = 1.90 m + 0.2 m 12

5 (1040 N) 13 = 400 N

M D = TABx (0.875 m) − TABy (0.2 m) = (960 N)(0.875 m) − (400 N)(0.2 m) = 760 N ⋅ m

(b)

We have

or M D = 760 N ⋅ m

M D = TABx ( y ) + TABx ( x) = (960 N)(0) + (400 N)(1.90 m) = 760 N ⋅ m

or M D = 760 N ⋅ m

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PROBLEM 3.10 It is known that a force with a moment of 960 N ⋅ m about D is required to straighten the fence post CD. If d = 2.80 m, determine the tension that must be developed in the cable of winch puller AB to create the required moment about Point D.

SOLUTION

Slope of line

EC =

0.875 m 7 = 2.80 m + 0.2 m 24

Then

TABx =

24 TAB 25

and

TABy =

7 TAB 25

We have

M D = TABx ( y ) + TABy ( x) 24 7 TAB (0) + TAB (2.80 m) 25 25 = 1224 N

960 N ⋅ m = TAB

or TAB = 1224 N

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PROBLEM 3.11 It is known that a force with a moment of 960 N ⋅ m about D is required to straighten the fence post CD. If the capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified moment about Point D.

SOLUTION

The minimum value of d can be found based on the equation relating the moment of the force TAB about D: M D = (TAB max ) y (d )

where

M D = 960 N ⋅ m (TAB max ) y = TAB max sin θ = (2400 N) sin θ

Now

sin θ =

0.875m (d + 0.20) 2 + (0.875) 2 m

0.875 960 N ⋅ m = 2400 N " " (d + 0.20)2 + (0.875) 2 $

or

(d + 0.20) 2 + (0.875) 2 = 2.1875d

or

(d + 0.20) 2 + (0.875) 2 = 4.7852d 2

or

! # (d ) # %

3.7852d 2 − 0.40d − .8056 = 0

Using the quadratic equation, the minimum values of d are 0.51719 m and − .41151 m. Since only the positive value applies here, d = 0.51719 m or d = 517 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 163

PROBLEM 3.12 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-lb force directed along its centerline on the ball and socket at B, determine the moment of the force about A.

SOLUTION First note

dCB = (12.0 in.) 2 + (2.33 in.) 2 = 12.2241 in.

Then

and

12.0 in. 12.2241 in. 2.33 in. sin θ = 12.2241 in.

cos θ =

FCB = FCB cos θ i − FCB sin θ j =

125 lb [(12.0 in.) i − (2.33 in.) j] 12.2241 in.

Now

M A = rB/A × FCB

where

rB/A = (15.3 in.) i − (12.0 in. + 2.33 in.) j = (15.3 in.) i − (14.33 in.) j

Then

M A = [(15.3 in.)i − (14.33 in.) j] ×

125 lb (12.0i − 2.33j) 12.2241 in.

= (1393.87 lb ⋅ in.)k = (116.156 lb ⋅ ft)k

or M A = 116.2 lb ⋅ ft

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 164

PROBLEM 3.13 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-lb force directed along its centerline on the ball and socket at B, determine the moment of the force about A.

SOLUTION First note

dCB = (17.2 in.) 2 + (7.62 in.) 2 = 18.8123 in.

Then

and

17.2 in. 18.8123 in. 7.62 in. sin θ = 18.8123 in.

cos θ =

FCB = ( FCB cos θ )i − ( FCB sin θ ) j

=

125 lb [(17.2 in.)i + (7.62 in.) j] 18.8123 in.

Now

M A = rB/A × FCB

where

rB/A = (20.5 in.)i − (4.38 in.) j

Then

MA = [(20.5 in.)i − (4.38 in.) j] ×

= (1538.53 lb ⋅ in.)k = (128.2 lb ⋅ ft)k

125 lb (17.2i − 7.62 j) 18.8123 in.

or M A = 128.2 lb ⋅ ft

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 165

PROBLEM 3.14 A mechanic uses a piece of pipe AB as a lever when tightening an alternator belt. When he pushes down at A, a force of 485 N is exerted on the alternator at B. Determine the moment of that force about bolt C if its line of action passes through O.

SOLUTION We have

M C = rB/C × FB

Noting the direction of the moment of each force component about C is clockwise. M C = xFBy + yFBx

Where

and

x = 120 mm − 65 mm = 55 mm y = 72 mm + 90 mm = 162 mm FBx = FBy =

65 2

(65) + (72) 2 72 2

(65) + (72) 2

(485 N) = 325 N (485 N) = 360 N

M C = (55 mm)(360 N) + (162)(325 N) = 72450 N ⋅ mm = 72.450 N ⋅ m

or M C = 72.5 N ⋅ m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 166

PROBLEM 3.15 Form the vector products B × C and B′ × C, where B = B′, and use the results obtained to prove the identity sin a cos β =

1 1 sin ( a + β ) + sin ( a − β ). 2 2

SOLUTION Note:

B = B(cos β i + sin β j) B′ = B(cos β i − sin β j) C = C (cos α i + sin α j)

By definition:

Now

| B × C | = BC sin (α − β )

(1)

| B′ × C | = BC sin (α + β )

(2)

B × C = B(cos β i + sin β j) × C (cos α i + sin α j)

= BC (cos β sin α − sin β cos α )k

and

(3)

B′ × C = B(cos β i − sin β j) × C (cos α i + sin α j)

= BC (cos β sin α + sin β cos α ) k

(4)

Equating the magnitudes of B × C from Equations (1) and (3) yields: BC sin(α − β ) = BC (cos β sin α − sin β cos α )

(5)

Similarly, equating the magnitudes of B′ × C from Equations (2) and (4) yields: BC sin(α + β ) = BC (cos β sin α + sin β cos α )

(6)

Adding Equations (5) and (6) gives: sin(α − β ) + sin(α + β ) = 2cos β sin α

or sin α cos β =

1 1 sin(α + β ) + sin(α − β ) 2 2

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PROBLEM 3.16 A line passes through the Points (20 m, 16 m) and (−1 m, −4 m). Determine the perpendicular distance d from the line to the origin O of the system of coordinates.

SOLUTION d AB = [20 m − ( −1 m)]2 + [16 m − ( −4 m)]2 = 29 m

Assume that a force F, or magnitude F(N), acts at Point A and is directed from A to B. Then, Where

F = F λ AB

λ AB = =

By definition Where Then

rB − rA d AB 1 (21i + 20 j) 29

M O = | rA × F | = dF rA = −(1 m)i − (4 m) j M O = [ −(−1 m)i − (4 m) j] ×

F [(21 m)i + (20 m) j] 29 m

F [−(20)k + (84)k ] 29 & 64 ' = ( F )k N ⋅ m * 29 + =

Finally

& 64 ' ( 29 F ) = d ( F ) * + 64 d= m 29

d = 2.21 m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 168

PROBLEM 3.17 The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when (a) P = −7i + 3j − 3k and Q = 2i + 2j + 5k, (b) P = 6i − 5j − 2k and Q = −2i + 5j − k.

SOLUTION (a)

We have

A = |P × Q|

where

P = −7i + 3j − 3k Q = 2i + 2 j + 5k

Then

i j k P × Q = −7 3 −3 2 2 5 = [(15 + 6)i + ( −6 + 35) j + ( −14 − 6)k ] = (21)i + (29) j(−20)k A = (20) 2 + (29) 2 + (−20) 2

(b)

We have

A = |P × Q|

where

P = 6i − 5 j − 2k

or A = 41.0

Q = −2i + 5 in. j − 1k

Then

i j k P × Q = 6 −5 −2 −2 5 −1 = [(5 + 10)i + (4 + 6) j + (30 − 10)k ] = (15)i + (10) j + (20)k A = (15) 2 + (10) 2 + (20) 2

or A = 26.9

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PROBLEM 3.18 A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) i + 2j − 5k and 4i − 7j − 5k, (b) 3i − 3j + 2k and −2i + 6j − 4k.

SOLUTION (a)

We have where

=

A×B |A × B|

A = 1i + 2 j − 5k B = 4i − 7 j − 5k

Then

i j k A × B = 1 +2 −5 4 −7 −5 = (−10 − 35)i + (20 + 5) j + (−7 − 8)k = 15(3i − 1j − 1k )

and

|A × B | = 15 (−3)2 + (−1) 2 + (−1)2 = 15 11 =

(b)

We have where

=

15(−3i − 1j − 1k ) 15 11

=

or

1 11

(−3i − j − k )

A×B |A × B|

A = 3i − 3 j + 2k B = −2i + 6 j − 4k

Then

i j k A × B = 3 −3 2 −2 6 −4 = (12 − 12)i + (−4 + 12) j + (18 − 6)k = (8 j + 12k )

and

|A × B| = 4 (2) 2 + (3) 2 = 4 13 =

4(2 j + 3k ) 4 13

or

=

1 13

(2 j + 3k )

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 170

PROBLEM 3.19 Determine the moment about the origin O of the force F = 4i + 5j − 3k that acts at a Point A. Assume that the position vector of A is (a) r = 2i − 3j + 4k, (b) r = 2i + 2.5j − 1.5k, (c) r = 2i + 5j + 6k.

SOLUTION (a)

i j k M O = 2 −3 4 4 5 −3 = (9 − 20)i + (16 + 6) j + (10 + 12)k

(b)

i j k M O = 2 2.5 −1.5 4 5 −3 = (−7.5 + 7.5)i + (−6 + 6) j + (10 − 10)k

(c)

M O = −11i + 22 j + 22k

MO = 0

i j k MO = 2 5 6 4 5 −3 = (−15 − 30)i + (24 + 6) j + (10 − 20)k

M O = −45i + 30 j − 10k

Note: The answer to Part b could have been anticipated since the elements of the last two rows of the determinant are proportional.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 171

PROBLEM 3.20 Determine the moment about the origin O of the force F = −2i + 3j + 5k that acts at a Point A. Assume that the position vector of A is (a) r = i + j + k, (b) r = 2i + 3j − 5k, (c) r = −4i + 6j + 10k.

SOLUTION

(a)

i j k MO = 1 1 1 −2 3 5 = (5 − 3)i + (−2 − 5) j + (3 + 2)k

(b)

i j k MO = 2 3 − 5 −2 3 5 = (15 + 15)i + (10 − 10) j + (6 + 6)k

(c)

M O = 2i − 7 j + 5k

M O = 30i + 12k

i j k M O = −4 6 10 −2 3 5 = (30 − 30)i + ( −20 + 20) j + (−12 + 12)k

MO = 0

Note: The answer to Part c could have been anticipated since the elements of the last two rows of the determinant are proportional.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 172

PROBLEM 3.21 A 200-N force is applied as shown to the bracket ABC. Determine the moment of the force about A.

SOLUTION We have

M A = rC/A × FC

where

rC/A = (0.06 m)i + (0.075 m) j FC = −(200 N) cos 30° j + (200 N)sin 30°k

Then

i j k M A = 200 0.06 0.075 0 − cos 30° sin 30° 0 = 200[(0.075sin 30°)i − (0.06sin 30°) j − (0.06 cos 30°)k ]

or M A = (7.50 N ⋅ m)i − (6.00 N ⋅ m) j − (10.39 N ⋅ m)k

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 173

PROBLEM 3.22 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B.

SOLUTION We have

M O = rB/O × FB

where

rB/O = (7 m) j FB = TAB + TBC TAB = = TBC = =

BATAB

−(0.75 m)i − (7 m) j + (6 m)k (.75) 2 + (7) 2 + (6) 2 m

(555 N)

BC TBC

(4.25 m)i − (7 m) j + (1 m)k (4.25) 2 + (7) 2 + (1) 2 m

(660 N)

FB = [−(45.00 N)i − (420.0 N) j + (360.0 N)k ] + [(340.0 N)i − (560.0 N) j + (80.00 N)k ] = (295.0 N)i − (980.0 N) j + (440.0 N)k

and

i j k MO = 0 7 0 N⋅m 295 980 440 = (3080 N ⋅ m)i − (2070 N ⋅ m)k

or M O = (3080 N ⋅ m)i − (2070 N ⋅ m)k

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 174

PROBLEM 3.23 The 6-m boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a Point C located on the vertical wall. If the tension in the cable is 2.5 kN, determine the moment about A of the force exerted by the cable at B.

SOLUTION First note

d BC = (−6)2 + (2.4) 2 + (−4) 2 = 7.6 m 2.5 kN (−6i + 2.4 j − 4k ) 7.6

Then

TBC =

We have

M A = rB/A × TBC

where

rB/A = (6 m)i

Then

M A = (6 m)i ×

2.5 kN (−6i + 2.4 j − 4k ) 7.6

or M A = (7.89 kN ⋅ m) j + (4.74 kN ⋅ m)k

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 175

PROBLEM 3.24 A wooden board AB, which is used as a temporary prop to support a small roof, exerts at Point A of the roof a 57-lb force directed along BA. Determine the moment about C of that force.

SOLUTION We have

M C = rA/C × FBA

where

rA/C = (48 in.)i − (6 in.) j + (36 in.)k

and

FBA =

BA FBA

−(5 in.)i + (90 in.) j − (30 in.)k ! # (57 lb) =" 2 2 2 " # (5) (90) (30) in. + + $ % = −(3 lb)i + (54 lb) j − (18 lb)k i j k M C = 48 6 36 lb ⋅ in. 3 54 18

= −(1836 lb ⋅ in.)i + (756 lb ⋅ in.) j + (2574 lb ⋅ in.)

or M C = −(153.0 lb ⋅ ft)i + (63.0 lb ⋅ ft) j + (215 lb ⋅ ft)k

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 176

PROBLEM 3.25 The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 810 N. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C.

SOLUTION (a)

We have

M A = rE/A × TDE

where

rE/A = (2.3 m) j TDE = =

DE TDE

(0.6 m)i + (3.3 m) j − (3 m)k (0.6) 2 + (3.3)2 + (3)2 m

(810 N)

= (108 N)i + (594 N) j − (540 N)k i j k MA = 0 2.3 0 N⋅m 108 594 −540 = −(1242 N ⋅ m)i − (248.4 N ⋅ m)k

or M A = −(1242 N ⋅ m)i − (248 N ⋅ m)k (b)

We have

M A = rG/A × TCG

where

rG/A = (2.7 m)i + (2.3 m) j TCG = =

CG TCG

−(.6 m)i + (3.3 m) j − (3 m)k (.6) 2 + (3.3) 2 + (3) 2 m

(810 N)

= −(108 N)i + (594 N) j − (540 N)k i j k M A = 2.7 2.3 0 N⋅m −108 594 −540 = −(1242 N ⋅ m)i + (1458 N ⋅ m) j + (1852 N ⋅ m)k

or M A = −(1242 N ⋅ m)i + (1458 N ⋅ m) j + (1852 N ⋅ m)k

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 177

PROBLEM 3.26 A small boat hangs from two davits, one of which is shown in the figure. The tension in line ABAD is 82 lb. Determine the moment about C of the resultant force RA exerted on the davit at A.

SOLUTION We have

R A = 2FAB + FAD

where and

FAB = −(82 lb) j ! AD 6i − 7.75 j − 3k = (82 lb) FAD = FAD AD 10.25 FAD = (48 lb)i − (62 lb) j − (24 lb)k

Thus

R A = 2FAB + FAD = (48 lb)i − (226 lb) j − (24 lb)k

Also

rA/C = (7.75 ft) j + (3 ft)k

Using Eq. (3.21):

i j k M C = 0 7.75 3 48 − 226 −24 = (492 lb ⋅ ft)i + (144 lb ⋅ ft) j − (372 lb ⋅ ft)k M C = (492 lb ⋅ ft)i + (144.0 lb ⋅ ft) j − (372 lb ⋅ ft)k

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 178

PROBLEM 3.27 In Problem 3.22, determine the perpendicular distance from Point O to cable AB. PROBLEM 3.22 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B.

SOLUTION We have where

| M O | = TBA d d = perpendicular distance from O to line AB.

Now

M O = rB/O × TBA

and

rB/O = (7 m) j TBA = =

BATAB

−(0.75 m)i − (7 m) j + (6 m)k (0.75) 2 + (7) 2 + (6) 2 m

(555 N)

= −(45.0 N)i − (420 N) j + (360 N)k i j k MO = 0 7 0 N⋅m −45 −420 360 = (2520.0 N ⋅ m)i + (315.00 N ⋅ m)k

and

| M O | = (2520.0) 2 + (315.00) 2 = 2539.6 N ⋅ m 2539.6 N ⋅ m = (555 N)d

or

d = 4.5759 m

or d = 4.58 m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 179

PROBLEM 3.28 In Problem 3.22, determine the perpendicular distance from Point O to cable BC. PROBLEM 3.22 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B.

SOLUTION We have where

| M O | = TBC d d = perpendicular distance from O to line BC. M O = rB/O × TBC rB/O = 7 mj TBC = =

BC TBC

(4.25 m)i − (7 m) j + (1 m)k (4.25)2 + (7) 2 + (1) 2 m

(660 N)

= (340 N)i − (560 N) j + (80 N)k i j k MO = 0 7 0 340 −560 80 = (560 N ⋅ m)i − (2380 N ⋅ m)k

and

| M O | = (560)2 + (2380) 2 = 2445.0 N ⋅ m 2445.0 N ⋅ m = (660 N)d d = 3.7045 m

or d = 3.70 m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 180

PROBLEM 3.29 In Problem 3.24, determine the perpendicular distance from Point D to a line drawn through Points A and B. PROBLEM 3.24 A wooden board AB, which is used as a temporary prop to support a small roof, exerts at Point A of the roof a 57-lb force directed along BA. Determine the moment about C of that force.

SOLUTION

We have where

| M D | = FBA d d = perpendicular distance from D to line AB.

M D = rA/D × FBA rA/D = −(6 in.) j + (36 in.)k FBA = =

BA FBA

(−(5 in.)i + (90 in.) j − (30 in.)k ) (5)2 + (90) 2 + (30) 2 in.

(57 lb)

= −(3 lb)i + (54 lb) j − (18 lb)k i j k M D = 0 −6 36 lb ⋅ in. −3 54 −18 = −(1836.00 lb ⋅ in.)i − (108.000 lb ⋅ in.) j − (18.0000 lb ⋅ in.)k

and

| M D | = (1836.00) 2 + (108.000)2 + (18.0000)2 = 1839.26 lb ⋅ in. 1839.26 lb ⋅ in = (57 lb)d d = 32.268 in.

or d = 32.3 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 181

PROBLEM 3.30 In Problem 3.24, determine the perpendicular distance from Point C to a line drawn through Points A and B. PROBLEM 3.24 A wooden board AB, which is used as a temporary prop to support a small roof, exerts at Point A of the roof a 57-lb force directed along BA. Determine the moment about C of that force.

SOLUTION We have where

| M C | = FBA d d = perpendicular distance from C to line AB. M C = rA/C × FBA rA/C = (48 in.)i − (6 in.) j + (36 in.)k FBA = =

BA FBA

( −(5 in.)i + (90 in.) j − (30 in.)k ) (5) 2 + (90) 2 + (30) 2 in.

(57 lb)

= −(3 lb)i + (54 lb) j − (18 lb)k i j k M C = 48 −6 36 lb ⋅ in. −3 54 −18 = −(1836.00lb ⋅ in.)i − (756.00 lb ⋅ in.) j + (2574.0 lb ⋅ in.)k

and

| M C | = (1836.00) 2 + (756.00) 2 + (2574.0)2 = 3250.8 lb ⋅ in. 3250.8 lb ⋅ in. = 57 lb d = 57.032 in.

or d = 57.0 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 182

PROBLEM 3.31 In Problem 3.25, determine the perpendicular distance from Point A to portion DE of cable DEF. PROBLEM 3.25 The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 810 N. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C.

SOLUTION We have where

| M A | = TDE d d = perpendicular distance from A to line DE. M A = rE/A × TDE rE/A = (2.3 m) j TDE = =

DE TDE

(0.6 m)i + (3.3 m) j − (3 m)k (0.6) 2 + (3.3) 2 + (3) 2 m

(810 N)

= (108 N)i + (594 N) j − (540 N)k i j k MA = 0 2.3 0 N⋅m 108 594 540 = − (1242.00 N ⋅ m)i − (248.00 N ⋅ m)k

and

|M A | = (1242.00)2 + (248.00) 2 = 1266.52 N ⋅ m 1266.52 N ⋅ m = (810 N)d d = 1.56360 m

or d = 1.564 m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 183

PROBLEM 3.32 In Problem 3.25, determine the perpendicular distance from Point A to a line drawn through Points C and G. PROBLEM 3.25 The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 810 N. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C.

SOLUTION We have where

|M A | = TCG d

d = perpendicular distance from A to line CG. M A = rG/A × TCG rG/A = (2.7 m)i + (2.3 m) j TCG = =

CG TCG

−(0.6 m) i + (3.3 m) j − (3 m) k (0.6) 2 + (3.3) 2 + (3) 2 m

(810 N)

= −(108 N) i + (594 N) j − (540 N) k i j k 0 N⋅m M A = 2.7 2.3 −108 594 −540 = −(1242.00 N ⋅ m)i + (1458.00 N ⋅ m) j + (1852.00 N ⋅ m)k

and

|M A | = (1242.00) 2 + (1458.00) 2 + (1852.00)2 = 2664.3 N ⋅ m 2664.3 N ⋅ m = (810 N)d d = 3.2893 m

or d = 3.29 m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 184

PROBLEM 3.33 In Problem 3.26, determine the perpendicular distance from Point C to portion AD of the line ABAD. PROBLEM 3.26 A small boat hangs from two davits, one of which is shown in the figure. The tension in line ABAD is 82 lb. Determine the moment about C of the resultant force RA exerted on the davit at A.

SOLUTION First compute the moment about C of the force FDA exerted by the line on D: From Problem 3.26:

FDA = − FAD = −(48 lb) i + (62 lb) j + (24 lb)k M C = rD/C × FDA = + (6 ft)i × [−(48 lb)i + (62 lb) j + (24 lb)k ] = −(144 lb ⋅ ft) j + (372 lb ⋅ ft)k M C = (144) 2 + (372)2 = 398.90 lb ⋅ ft

Then

M C = FDA d

Since

FDA = 82 lb d= =

MC FDA 398.90 lb ⋅ ft 82 lb

d = 4.86 ft

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 185

PROBLEM 3.34 Determine the value of a that minimizes the perpendicular distance from Point C to a section of pipeline that passes through Points A and B.

SOLUTION Assuming a force F acts along AB, |M C | = |rA / C × F| = F ( d )

d = perpendicular distance from C to line AB

Where

F= =

AB F

(24 ft) i + (24 ft) j − (28) k (24) 2 + (24)2 + (18) 2 ft

F

F (6) i + (6) j − (7) k 11 = (3 ft)i − (10 ft) j − (a − 10 ft)k =

rA/C

i j k F M C = 3 −10 10a 11 6 6 −7 = [(10 + 6a )i + (81 − 6a) j + 78 k ] |M C | = |rA/C × F 2 |

Since

or

F 11 |rA/C × F 2 | = ( dF ) 2

1 (10 + 6a) 2 + (81 − 6a) 2 + (78)2 = d 2 121

Setting

d da

(d 2 ) = 0 to find a to minimize d 1 [2(6)(10 + 6a) + 2(−6)(81 − 6a)] = 0 121

Solving

a = 5.92 ft

or a = 5.92 ft

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 186

PROBLEM 3.35 Given the vectors P = 3i − j + 2k , Q = 4i + 5 j − 3k , and S = −2i + 3j − k , compute the scalar products P ⋅ Q, P ⋅ S, and Q ⋅ S.

SOLUTION P ⋅ Q = (3i − 1j + 2k ) ⋅ (4i − 5 j − 3k ) = (3)(4) + (−1)(−5) + (2)(−3) =1

or

P ⋅Q =1

or

P ⋅ S = −11

or

Q ⋅ S = 10

!

P ⋅ S = (3i − 1j + 2k ) ⋅ (−2i + 3j − 1k )

= (3)(−2) + (−1)(3) + (2)( −1) = −11

Q ⋅ S = (4i − 5 j − 3k ) ⋅ ( −2i + 3j − 1k ) = (4)(−2) + (5)(3) + ( −3)(−1) = 10

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 187

PROBLEM 3.36 Form the scalar products B ⋅ C and B′ ⋅ C, where B = B ′, and use the results obtained to prove the identity cos a cos β =

1 1 cos (a + β ) + cos (a − β ). 2 2

SOLUTION By definition

B ⋅ C = BC cos(α − β ) B = B [(cos β )i + (sin β ) j] C = C [(cos α )i + (sin α ) j]

where

(B cos β )( C cos α ) + ( B sin β )(C sin α ) = BC cos (α − β ) cos β cos α + sin β sin α = cos(α − β )

or

(1)

B′ ⋅ C = BC cos (α + β )

By definition

B′ = [(cos β )i − (sin β ) j]

where

(B cos β ) (C cos α ) + (− B sin β )(C sin α ) = BC cos (α + β )

or

cos β cos α − sin β sin α = cos (α + β )

(2)

Adding Equations (1) and (2), 2 cos β cos α = cos (α − β ) + cos (α + β )

or cos α cos β =

1 1 cos (α + β ) + cos (α − β ) 2 2

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 188

PROBLEM 3.37 Section AB of a pipeline lies in the yz plane and forms an angle of 37° with the z axis. Branch lines CD and EF join AB as shown. Determine the angle formed by pipes AB and CD.

SOLUTION First note

! AB = AB (sin 37° j − cos 37°k ) CD = CD(− cos 40° cos 55° j + sin 40° j − cos 40° sin 55°k )

Now

! ! AB ⋅ CD = ( AB)(CD ) cos θ

or

AB (sin 37° j − cos 37°k ) ⋅ CD (− cos 40° cos 55°i + sin 40° j − cos 40° sin 55°k )

= (AB)(CD) cos θ

or

cos θ = (sin 37°)(sin 40°) + (− cos 37°)(− cos 40° sin 55°) = 0.88799

or θ = 27.4°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 189

PROBLEM 3.38 Section AB of a pipeline lies in the yz plane and forms an angle of 37° with the z axis. Branch lines CD and EF join AB as shown. Determine the angle formed by pipes AB and EF.

SOLUTION First note

Now or

! AB = AB (sin 37° j − cos 37°k ) ! EF = EF (cos 32° cos 45°i + sin 32° j − cos 32° sin 45°k )

! ! AB ⋅ EF = ( AB )( EF ) cos θ AB (sin 37° j − cos 37°k ) ⋅ EF (cos 32° cos 45° j + sin 32° j − cos 32° sin 45°k )

= ( AB )( EF ) cos θ

or

cos θ = (sin 37°)(sin 32°) + ( − cos 37°)( − cos 32° sin 45°) = 0.79782

or θ = 37.1°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 190

PROBLEM 3.39 Consider the volleyball net shown. Determine the angle formed by guy wires AB and AC.

SOLUTION First note

AB = (−6.5)2 + (−8)2 + (2) 2 = 10.5 ft AC = (0) 2 + (−8) 2 + (6)2 = 10 ft

and

By definition or

or

! AB = −(6.5 ft)i − (8 ft) j + (2 ft)k ! AC = −(8 ft) j + (6 ft)k ! ! AB ⋅ AC = ( AB )( AC ) cos θ (−6.5i − 8 j + 2k ) ⋅ (−8 j + 6k ) = (10.5)(10) cos θ (−6.5)(0) + ( −8)( −8) + (2)(6) = 105cos θ cos θ = 0.72381

or θ = 43.6°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 191

PROBLEM 3.40 Consider the volleyball net shown. Determine the angle formed by guy wires AC and AD.

SOLUTION First note

AC = (0)2 + (−8) 2 + (6) 2 = 10 ft AD = (4) 2 + ( −8) 2 + (1) 2

and

By definition or

= 9 ft ! AC = −(8 ft)j + (6 ft)k ! AD = (4 ft) j − (8 ft) j + (1 ft)k ! ! AC ⋅ AD = ( AC )( AD ) cos θ (−8 j + 6k ) ⋅ (4i − 8 j + k ) = (10)(9) cos θ (0)(4) + ( −8)( −8) + (6)(1) = 90 cos θ

or

cos θ = 0.777 78

or θ = 38.9°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 192

PROBLEM 3.41 Knowing that the tension in cable AC is 1260 N, determine (a) the angle between cable AC and the boom AB, (b) the projection on AB of the force exerted by cable AC at Point A.

SOLUTION (a)

First note

AC = (−2.4) 2 + (0.8)2 + (1.2) 2 = 2.8 m AB = (−2.4) 2 + (−1.8)2 + (0) 2

and

By definition or

(b)

= 3.0 m ! AC = −(2.4 m)i + (0.8 m) j + (1.2 m)k ! AB = −(2.4 m)i − (1.8 m) j ! ! AC ⋅ AB = ( AC )( AB) cos θ (−2.4i + 0.8 j + 1.2k ) ⋅ (−2.4i − 1.8 j) = (2.8)(30) × cos θ

or

(−2.4)(−2.4) + (0.8)(−1.8) + (1.2)(0) = 8.4cos θ

or

cos θ = 0.514 29

We have

or θ = 59.0°

(TAC ) AB = TAC ⋅ λ AB = TAC cos θ = (1260 N)(0.51429)

or (TAC ) AB = 648 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 193

PROBLEM 3.42 Knowing that the tension in cable AD is 405 N, determine (a) the angle between cable AD and the boom AB, (b) the projection on AB of the force exerted by cable AD at Point A.

SOLUTION (a)

First note

AD = (−2.4) 2 + (1.2) 2 + (−2.4) 2 = 3.6 m AB = (−2.4) 2 + (−1.8) 2 + (0) 2 = 3.0 m

and

# = −(2.4 m)i + (1.2 m) j − (2.4 m)k AD AB = −(2.4 m)i − (1.8 m) j

# ⋅ AB = ( AD )( AB) cos θ AD

By definition,

(−2.4i + 1.2 j − 2.4k ) ⋅ (−2.4i − 1.8 j) = (3.6)(3.0) cos θ (−2.4)(−2.4) + (1.2)(−1.8) + (−2.4)(0) = 10.8cos θ cos θ =

(b)

1 3

θ = 70.5°

(TAD ) AB = TAD ⋅ λ AB = TAD cos θ &1' = (405 N) ( ) * 3+

(TAD ) AB = 135.0 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 194

PROBLEM 3.43 Slider P can move along rod OA. An elastic cord PC is attached to the slider and to the vertical member BC. Knowing that the distance from O to P is 6 in. and that the tension in the cord is 3 lb, determine (a) the angle between the elastic cord and the rod OA, (b) the projection on OA of the force exerted by cord PC at Point P.

SOLUTION First note

OA = (12)2 + (12)2 + (−6) 2 = 18 in. OA 1 = (12i + 12 j − 6k ) OA 18 1 = (2i + 2 j − k ) 3

Then

λ OA =

Now

1 OP = 6 in. - OP = (OA) 3

The coordinates of Point P are (4 in., 4 in., −2 in.) ! PC = (5 in.)i + (11 in.) j + (14 in.)k so that PC = (5) 2 + (11) 2 + (14)2 = 342 in.

and (a)

We have or or

! PC ⋅ λ OA = ( PC ) cos θ 1 (5i + 11j + 14k ) ⋅ (2i + 2 j − k ) = 342 cos θ 3 1 [(5)(2) + (11)(2) + (14)( −1)] 3 342 = 0.324 44

cos θ =

or θ = 71.1° (b)

We have

(TPC ) OA = TPC ⋅ λ OA = (TPC λ PC ) ⋅ λ OA PC ⋅ λ OA PC = TPC cos θ = TPC

= (3 lb)(0.324 44)

or (TPC )OA = 0.973 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 195

PROBLEM 3.44 Slider P can move along rod OA. An elastic cord PC is attached to the slider and to the vertical member BC. Determine the distance from O to P for which cord PC and rod OA are perpendicular.

SOLUTION First note

OA = (12)2 + (12)2 + (−6)2 = 18 in.

Then

λ OA =

OA 1 = (12i + 12 j − 6k ) OA 18 1 = (2i + 2 j − k ) 3

Let the coordinates of Point P be (x in., y in., z in.). Then ! PC = [(9 − x)in.]i + (15 − y )in.] j + [(12 − z )in.]k Also, and

! d OP = dOP λ OA = OP (2i + 2 j − k ) 3 ! OP = ( x in.)i + ( y in.) j + ( z in.)k 2 2 1 x = dOP y = dOP z = dOP 3 3 3

The requirement that OA and PC be perpendicular implies that ! λ OA ⋅ PC = 0 or

1 (2 j + 2 j − k ) ⋅ [(9 − x)i + (15 − y ) j + (12 − z )k ] = 0 3

or

2 2 & ' & ' & 1 '! (2) ( 9 − dOP ) + (2) (15 − dOP ) + (−1) "12 − ( − dOP ) # = 0 3 3 * + * + * 3 +% $

or dOP = 12.00 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 196

PROBLEM 3.45 Determine the volume of the parallelepiped of Fig. 3.25 when (a) P = 4i − 3j + 2k, Q = −2i − 5j + k, and S = 7i + j − k, (b) P = 5i − j + 6k, Q = 2i + 3j + k, and S = −3i − 2j + 4k.

SOLUTION Volume of a parallelepiped is found using the mixed triple product. (a)

Vol = P ⋅ (Q × S) 4 −3 2 = −2 −5 1 in.3 7 1 −1 = (20 − 21 − 4 + 70 + 6 − 4) = 67

or Volume = 67.0 (b)

Vol = P ⋅ (Q × S) 5 −1 6 = 2 3 1 in.3 −3 −2 4 = (60 + 3 − 24 + 54 + 8 + 10) = 111

or Volume = 111.0

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 197

PROBLEM 3.46 Given the vectors P = 4i − 2 j + 3k , Q = 2i + 4 j − 5k , and S = S x i − j + 2k , determine the value of S x for which the three vectors are coplanar.

SOLUTION If P, Q, and S are coplanar, then P must be perpendicular to (Q × S). P ⋅ (Q × S) = 0

(or, the volume of a parallelepiped defined by P, Q, and S is zero). −2 3 4 −5 = 0 −1 2

Then

4 2 Sx

or

32 + 10S x − 6 − 20 + 8 − 12S x = 0

Sx = 7

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 198

PROBLEM 3.47 The 0.61 × 1.00-m lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 66 N, determine the moment about each of the coordinate axes of the force exerted by the cord at D.

SOLUTION First note

z = (0.61) 2 − (0.11)2

= 0.60 m

Then

d DE = (0.3) 2 + (0.6) 2 + (−0.6) 2

= 0.9 m 66 N (0.3i + 0.6 j − 0.6k ) 0.9 = 22[(1 N)i + (2 N) j − (2 N)k ]

and

TDE =

Now

M A = rD/A × TDE

where

rD/A = (0.11 m) j + (0.60 m)k

Then

i j k M A = 22 0 0.11 0.60 −2 1 2 = 22[(−0.22 − 1.20)i + 0.60 j − 0.11k ] = − (31.24 N ⋅ m)i + (13.20 N ⋅ m) j − (2.42 N ⋅ m)k M x = −31.2 N ⋅ m, M y = 13.20 N ⋅ m, M z = −2.42 N ⋅ m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 199

PROBLEM 3.48 The 0.61 × 1.00-m lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 66 N, determine the moment about each of the coordinate axes of the force exerted by the cord at C.

SOLUTION First note

z = (0.61) 2 − (0.11)2

= 0.60 m

Then

dCE = (−0.7) 2 + (0.6) 2 + (−0.6) 2

= 1.1 m 66 N (−0.7i + 0.6 j − 0.6k ) 1.1 = 6[−(7 N)i + (6 N) j − (6 N)k ]

and

TCE =

Now

M A = rE/A × TCE

where

rE/A = (0.3 m)i + (0.71 m) j

Then

i j k M A = 6 0.3 0.71 0 −7 −6 6 = 6[ −4.26i + 1.8 j + (1.8 + 4.97)k ] = − (25.56 N ⋅ m)i + (10.80 N ⋅ m) j + (40.62 N ⋅ m)k M x = −25.6 N ⋅ m, M y = 10.80 N ⋅ m, M z = 40.6 N ⋅ m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 200

PROBLEM 3.49 To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the moments about the y and the z axes of the force exerted at B by portion AB of the rope are, respectively, 120 N ⋅ m and −460 N ⋅ m, determine the distance a.

SOLUTION First note

! BA = (2.2 m)i − (3.2 m) j − ( a m)k

Now

M D = rA/D × TBA

where

rA/D = (2.2 m)i + (1.6 m) j

Then

TBA =

TBA (2.2i − 3.2 j − ak ) (N) d BA

MD =

i j k TBA 2.2 1.6 0 d BA 2.2 −3.2 − a

=

Thus

TBA {−1.6a i + 2.2a j + [(2.2)(−3.2) − (1.6)(2.2)]k} d BA

M y = 2.2

TBA a d BA

M z = −10.56

Then forming the ratio

TBA d BA

(N ⋅ m) (N ⋅ m)

My Mz T

2.2 dBA (N ⋅ m) 120 N ⋅ m BA = −460 N ⋅ m −10.56 TdBA (N ⋅ m)

or a = 1.252 m

BA

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 201

PROBLEM 3.50 To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the man applies a 195-N force to end A of the rope and that the moment of that force about the y axis is 132 N ⋅ m, determine the distance a.

SOLUTION d BA = (2.2) 2 + (−3.2) 2 + (−a ) 2

First note

= 15.08 + a 2 m 195 N (2.2i − 3.2 j − a k ) d BA

and

TBA =

Now

M y = j ⋅ (rA/D × TBA )

where

rA/0 = (2.2 m)i + (1.6 m) j

Then

0 1 0 195 My = 2.2 1.6 0 d BA 2.2 −3.2 − a =

195 (2.2a ) (N ⋅ m) d BA

Substituting for My and dBA 132 N ⋅ m =

or

195 15.08 + a 2

(2.2a )

0.30769 15.08 + a 2 = a

Squaring both sides of the equation 0.094675(15.08 + a 2 ) = a 2

or a = 1.256 m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 202

PROBLEM 3.51 A small boat hangs from two davits, one of which is shown in the figure. It is known that the moment about the z axis of the resultant force RA exerted on the davit at A must not exceed 279 lb ⋅ ft in absolute value. Determine the largest allowable tension in line ABAD when x = 6 ft.

SOLUTION First note

R A = 2TAB + TAD

Also note that only TAD will contribute to the moment about the z axis. Now

AD = (6) 2 + (−7.75) 2 + (−3) 2 = 10.25 ft ! AD =T AD T (6i − 7.75 j − 3k ) = 10.25

Then,

TAD

Now

M z = k ⋅ (rA/C × TAD )

where

rA/C = (7.75 ft) j + (3 ft)k

Then for Tmax ,

0 0 1 Tmax 279 = 0 7.75 3 10.25 6 −7.75 −3 =

Tmax | − (1)(7.75)(6)| 10.25

or Tmax = 61.5 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 203

PROBLEM 3.52 For the davit of Problem 3.51, determine the largest allowable distance x when the tension in line ABAD is 60 lb.

SOLUTION From the solution of Problem 3.51, TAD is now TAD = T =

AD AD 60 lb x 2 + ( −7.75) 2 + (−3)2

( xi − 7.75 j − 3k )

Then M z = k ⋅ (rA / C × TAD ) becomes 279 =

279 =

60 x 2 + (−7.75) 2 + ( −3) 2 60

0 0 1 0 7.75 3 x −7.75 −3

| − (1)(7.75)( x) |

2

x + 69.0625 279 x 2 + 69.0625 = 465 x 0.6 x 2 + 69.0625 = x

Squaring both sides:

0.36 x 2 + 24.8625 = x 2 x 2 = 38.848

x = 6.23 ft

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 204

PROBLEM 3.53 To loosen a frozen valve, a force F of magnitude 70 lb is applied to the handle of the valve. Knowing that θ = 25°, Mx = −61 lb ⋅ ft, and M z = −43 lb ⋅ ft, determine φ and d.

SOLUTION We have

ΣM O : rA/O × F = M O

where

rA/O = −(4 in.)i + (11 in.) j − (d )k F = F (cos θ cos φ i − sin θ j + cos θ sin φ k )

For

F = 70 lb, θ = 25°

F = (70 lb)[(0.90631cos φ )i − 0.42262 j + (0.90631sin φ )k ] i M O = (70 lb) −4 −0.90631cos φ

j k 11 −d in. −0.42262 0.90631sin φ

= (70 lb)[(9.9694sin φ − 0.42262d ) i + (−0.90631d cos φ + 3.6252sin φ ) j + (1.69048 − 9.9694cos φ )k ] in.

and

M x = (70 lb)(9.9694sin φ − 0.42262d )in. = −(61 lb ⋅ ft)(12 in./ft)

(1)

M y = (70 lb)(−0.90631d cos φ + 3.6252sin φ ) in.

(2)

M z = (70 lb)(1.69048 − 9.9694cos φ ) in. = −43 lb ⋅ ft(12 in./ft)

(3)

From Equation (3)

& 634.33 ' φ = cos −1 ( ) = 24.636° * 697.86 +

or

From Equation (1)

& 1022.90 ' d =( ) = 34.577 in. * 29.583 +

or d = 34.6 in.

φ = 24.6°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 205

PROBLEM 3.54 When a force F is applied to the handle of the valve shown, its moments about the x and z axes are, respectively, M x = −77 lb ⋅ ft and M z = −81 lb ⋅ ft. For d = 27 in., determine the moment My of F about the y axis.

SOLUTION We have

ΣM O : rA/O × F = M O

Where

rA/O = −(4 in.)i + (11 in.) j − (27 in.)k F = F (cos θ cos φ i − sin θ j + cos θ sin φ k )

MO = F

i −4 cos θ cos φ

j 11 − sin θ

k −27 lb ⋅ in. cos θ sin φ

= F [(11cos θ sin φ − 27 sin θ )i + (−27 cos θ cos φ + 4 cos θ sin φ ) j + (4sin θ − 11cos θ cos φ )k ](lb ⋅ in.)

and

M x = F (11cos θ sin φ − 27sin θ )(lb ⋅ in.)

(1)

M y = F (−27 cos θ cos φ + 4cos θ sin φ ) (lb ⋅ in.)

(2)

M z = F (4sin θ − 11cos θ cos φ ) (lb ⋅ in.)

(3)

Now, Equation (1)

cos θ sin φ =

1 & Mx ' + 27sin θ ) 11 (* F +

(4)

and Equation (3)

cos θ cos φ =

M ' 1& 4sin θ − z ) ( 11 * F +

(5)

Substituting Equations (4) and (5) into Equation (2), .0 M '! 1& 1 &M ' ! /0 M y = F 1−27 " ( 4sin θ − z ) # + 4 " ( x + 27 sin θ ) # 2 F +% + % 04 03 $11 * $11 * F

or

My =

1 (27 M z + 4 M x ) 11

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 206

PROBLEM 3.54 (Continued)

Noting that the ratios 27 and 114 are the ratios of lengths, have 11 27 4 (−81 lb ⋅ ft) + (−77 lb ⋅ ft) 11 11 = 226.82 lb ⋅ ft

My =

or M y = −227 lb ⋅ ft

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 207

PROBLEM 3.55 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable.

SOLUTION M AD = λ AD ⋅ (rB/A × TBH )

Where

and

1 λ AD = (4i − 3k ) 5 rB/A = (0.5 m)i d BH = (0.375)2 + (0.75) 2 + (−0.75) 2 = 1.125 m

Then

Finally

450 N (0.375i + 0.75 j − 0.75k ) 1.125 = (150 N)i + (300 N) j − (300 N)k

TBH =

MAD

4 0 −3 1 = 0.5 0 0 5 150 300 −300 1 = [(−3)(0.5)(300)] 5

or M AD = − 90.0 N ⋅ m

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PROBLEM 3.56 In Problem 3.55, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable.

SOLUTION M AD = λ AD ⋅ (rB/A × TBG )

Where

and

1 λ AD = (4i − 3k ) 5 rB/A = (0.5 m) j BG = (−0.5) 2 + (0.925)2 + (−0.4)2 = 1.125 m

Then

Finally

$ 450 N ( −0.5i + 0.925 j − 0.4k ) T BG = 1.125 = −(200 N)i + (370 N) j − (160 N)k

MAD

4 0 −3 1 0.5 0 0 = 5 −200 370 −160 1 = [(−3)(0.5)(370)] 5

M AD = −111.0 N ⋅ m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 209

PROBLEM 3.57 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B.

SOLUTION First note

d AE = (0.9) 2 + (−0.6) 2 + (0.2) 2 = 1.1 m

Then

TAE =

Also

Then

Now where

Then

55 N (0.9i − 0.6 j + 0.2k ) 1.1 = 5[(9 N)i − (6 N) j + (2 N)k ]

DB = (1.2) 2 + ( −0.35) 2 + (0)2

λ DB

= 1.25 m ! DB = DB 1 (1.2i − 0.35 j) = 1.25 1 = (24i − 7 j) 25

M DB = λ DB ⋅ (rA/D × TAE )

TDA = −(0.1 m) j + (0.2 m)k

M DB

24 −7 0 1 = (5) 0 −0.1 0.2 25 −6 9 2 1 = (−4.8 − 12.6 + 28.8) 5

or M DB = 2.28 N ⋅ m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 210

PROBLEM 3.58 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B.

SOLUTION First note

dCF = (0.6) 2 + (−0.9) 2 + (−0.2) 2 = 1.1 m

Then

TCF =

Also

DB = (1.2) 2 + ( −0.35) 2 + (0)2

33 N (0.6i − 0.9 j + 0.2k ) 1.1 = 3[(6 N)i − (9 N) j − (2 N)k ]

= 1.25 m ! DB = DB 1 (1.2i − 0.35 j) = 1.25 1 = (24i − 7 j) 25

Then

λ DB

Now

M DB = λ DB ⋅ (rC/D × TCF )

where

rC/D = (0.2 m) j − (0.4 m)k

Then

M DB

24 −7 0 1 = (3) 0 0.2 −0.4 25 6 −9 −2 =

3 (−9.6 + 16.8 − 86.4) 25

or M DB = −9.50 N ⋅ m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 211

PROBLEM 3.59 A regular tetrahedron has six edges of length a. A force P is directed as shown along edge BC. Determine the moment of P about edge OA.

SOLUTION We have

M OA = λOA ⋅ (rC/O × P)

where From triangle OBC

(OA) x =

a 2

(OA) z = (OA) x tan 30° =

Since

a& 1 ' a ( )= 2* 3 + 2 3

(OA) 2 = (OA) 2x + (OA) 2y + (OAz )2 2

or

& a ' &a' a 2 = ( ) + (OA) 2y + ( ) *2+ *2 3+

(OA) y = a 2 −

Then

and

rA/O =

OA

=

2

a2 a2 2 − =a 4 12 3

2 a a i+a j+ k 2 3 2 3 1 2 1 i+ j+ k 2 3 2 3

P = λBC P =

(a sin 30°)i − (a cos 30°)k P ( P) = (i − 3k ) 2 a

rC/O = ai

M OA

1 2 = 1

2 3 0

2 3 &P' (a) ( ) 0 *2+

1

0

− 3

=

1

aP & 2 ' aP (( − )) (1)(− 3) = 2 * 3+ 2

M OA =

aP 2

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 212

PROBLEM 3.60 A regular tetrahedron has six edges of length a. (a) Show that two opposite edges, such as OA and BC, are perpendicular to each other. (b) Use this property and the result obtained in Problem 3.59 to determine the perpendicular distance between edges OA and BC.

SOLUTION (a)

For edge OA to be perpendicular to edge BC, ! ! OA ⋅ BC = 0 where From triangle OBC

(OA) x =

a 2

a& 1 ' a ( )= 2* 3 + 2 3 ! &a' & a ' OA = ( ) i + (OA) y j + ( )k *2+ *2 3+ ! BC = ( a sin 30°) i − (a cos 30°) k

(OA) z = (OA) x tan 30° =

and

=

Then

or

so that (b)

Have M OA

a a 3 a i− k = (i − 3 k ) 2 2 2

& a ' ! a a " i + (OA) y j + ( ) k # ⋅ (i − 3k ) = 0 2 *2 3+ % $2

a2 a2 + (OA) y (0) − =0 4 4 ! ! OA ⋅ BC = 0 ! ! OA is perpendicular to BC . ! ! = Pd , with P acting along BC and d the perpendicular distance from OA to BC .

From the results of Problem 3.57 M OA = Pa 2

Pa 2

or d =

= Pd

a 2

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!

PROBLEM 3.61 A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the moment of that force about the line joining Points A and D.

SOLUTION First note that BC = (48) 2 + (36) 2 = 60 in. and that

BE BC

=

45 60

= 34 . The coordinates of Point E are

then ( 34 × 48, 96, 34 × 36 ) or (36 in., 96 in., 27 in.). Then d EF = (−15) 2 + (−110) 2 + (30)2 = 115 in.

46 lb ( −15i − 110 j + 30k ) 115 = 2[−(3 lb)i − (22 lb) j + (6 lb)k ]

Then

TEF =

Also

AD = (48)2 + ( −12)2 + (36)2

Then

Now

λ AD

= 12 26 in. ! AD = AD 1 (48i − 12 j + 36k ) = 12 26 1 = (4i − j + 3k ) 26

M AD = λ AD ⋅ (rE/A × TEF )

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PROBLEM 3.61 (Continued)

where

Then

rE/A = (36 in.)i + (96 in.) j + (27 in.)k

M AD =

=

−1 3 4 (2) 36 96 27 26 −3 −22 6

1

2 26

(2304 + 81 − 2376 + 864 + 216 + 2376)

or M AD = 1359 lb ⋅ in.

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PROBLEM 3.62 A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EG at E is 54 lb, determine the moment of that force about the line joining Points A and D.

SOLUTION First note that BC = (48) 2 + (36) 2 = 60 in. and that

BE BC

=

45 60

= 43 . The coordinates of Point E are

then ( 34 × 48, 96, 34 × 36 ) or (36 in., 96 in., 27 in.). Then d EG = (11) 2 + (−88) 2 + (−44) 2 = 99 in.

Then

Also

54 lb (11i − 88 j − 44k ) 99 = 6[(1 lb)i − (8 lb) j − (4 lb)k ]

TEG =

AD = (48)2 + ( −12)2 + (36)2 = 12 26 in. ! AD = AD 1 (48i − 12 j + 36k ) = 12 26 1 (4i − j + 3k ) = 26

Then

λ AD

Now

M AD = λ AD ⋅ (rE/A × TEG )

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PROBLEM 3.62 (Continued)

where

Then

rE/A = (36 in.)i + (96 in.) j + (27 in.)k

M AD =

=

4 −1 3 (6) 36 96 27 26 1 − 8 −4

1

6 26

(−1536 − 27 − 864 − 288 − 144 + 864)

or M AD = −2350 lb ⋅ in.

!

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PROBLEM 3.63 Two forces F1 and F2 in space have the same magnitude F. Prove that the moment of F1 about the line of action of F2 is equal to the moment of F2 about the line of action of F1 .

SOLUTION

First note that

F1 = F1λ1

and F2 = F2 λ 2

Let M1 = moment of F2 about the line of action of M1 and M 2 = moment of F1 about the line of action of M 2 Now, by definition

M1 = λ1 ⋅ (rB/A × F2 ) = λ1 ⋅ (rB/A × λ 2 ) F2 M 2 = λ2 ⋅ (rA/B × F1 ) = λ2 ⋅ (rA/B × λ1 ) F1

Since

F1 = F2 = F

and rA/B = −rB/A

M1 = λ1 ⋅ (rB/A × λ 2 ) F M 2 = λ 2 ⋅ (−rB/A × λ1 ) F

Using Equation (3.39) so that

λ1 ⋅ (rB/A × λ 2 ) = λ 2 ⋅ ( −rB/A × λ1 ) M 2 = λ 1⋅ (rB/A × λ 2 ) F !

M12 = M 21

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PROBLEM 3.64 In Problem 3.55, determine the perpendicular distance between portion BH of the cable and the diagonal AD. PROBLEM 3.55 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable.

SOLUTION From the solution to Problem 3.55:

TBH = 450 N TBH = (150 N)i + (300 N) j − (300 N)k | M AD | = 90.0 N ⋅ m 1 λ AD = (4i − 3k ) 5

Based on the discussion of Section 3.11, it! follows that only the perpendicular component of TBH will contribute to the moment of TBH about line AD. Now

(TBH )parallel = TBH ⋅ λ AD 1 = (150i + 300 j − 300k ) ⋅ (4i − 3k ) 5 1 = [(150)(4) + (−300)(−3)] 5 = 300 N

Also so that

TBH = (TBH ) parallel + (TBH )perpendicular (TBH )perpendicular = (450) 2 − (300) 2 = 335.41 N

Since λ AD and (TBH )perpendicular are perpendicular, it follows that M AD = d (TBH ) perpendicular

or

90.0 N ⋅ m = d (335.41 N) d = 0.26833 m

d = 0.268 m

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PROBLEM 3.65 In Problem 3.56, determine the perpendicular distance between portion BG of the cable and the diagonal AD. PROBLEM 3.56 In Problem 3.55, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable.

SOLUTION From the solution to Problem 3.56:

ΤBG = 450 N

TBG = −(200 N)i + (370 N) j − (160 N)k | M AD | = 111 N ⋅ m 1 λ AD = (4i − 3k ) 5

Based on the discussion of Section 3.11, it" follows that only the perpendicular component of TBG will contribute to the moment of TBG about line AD. Now

(TBG ) parallel = TBG ⋅ λ AD 1 = ( −200i + 370 j − 160k ) ⋅ (4i − 3k ) 5 1 = [(−200)(4) + (−160)(−3)] 5 = −64 N

Also so that

TBG = (TBG ) parallel + (TBG ) perpendicular (TBG ) perpendicular = (450) 2 − ( −64) 2 = 445.43 N

Since λ AD and (TBG ) perpendicular are perpendicular, it follows that M AD = d (TBG ) perpendicular

or

111 N ⋅ m = d (445.43 N) d = 0.24920 m

d = 0.249 m

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PROBLEM 3.66 In Problem 3.57, determine the perpendicular distance between cable AE and the line joining Points D and B. PROBLEM 3.57 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B.

SOLUTION From the solution to Problem 3.57

ΤAE = 55 N

TAE = 5[(9 N)i − (6 N) j + (2 N)k ] | M DB | = 2.28 N ⋅ m λ DB =

1 (24i − 7 j) 25

Based on the discussion of Section 3.11, it! follows that only the perpendicular component of TAE will contribute to the moment of TAE about line DB. Now

(TAE )parallel = TAE ⋅ λ DB = 5(9i − 6 j + 2k ) ⋅

1 (24i − 7 j) 25

1 = [(9)(24) + (−6)(−7)] 5 = 51.6 N

Also so that

TAE = (TAE ) parallel + (TAE ) perpendicular (TAE )perpendicular = (55) 2 + (51.6)2 = 19.0379 N

Since λ DB and (TAE )perpendicular are perpendicular, it follows that M DB = d (TAE ) perpendicular

or

2.28 N ⋅ m = d (19.0379 N) d = 0.119761

d = 0.1198 m

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PROBLEM 3.67 In Problem 3.58, determine the perpendicular distance between cable CF and the line joining Points D and B. PROBLEM 3.58 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B.

SOLUTION From the solution to Problem 3.58

ΤCF = 33 N

TCF = 3[(6 N)i − (9 N) j − (2 N)k ] | M DB | = 9.50 N ⋅ m λ DB =

1 (24i − 7 j) 25

Based on the discussion of Section 3.11, it! follows that only the perpendicular component of TCF will contribute to the moment of TCF about line DB. Now

(TCF )parallel = TCF ⋅ λ DB = 3(6i − 9 j − 2k ) ⋅

1 (24i − 7 j) 25

3 [(6)(24) + (−9)( −7)] 25 = 24.84 N =

Also so that

TCF = (TCF ) parallel + (TCF ) perpendicular

(TCF )perpendicular = (33) 2 − (24.84) 2 = 21.725 N

Since λ DB and (TCF )perpendicular are perpendicular, it follows that | M DB | = d (TCF ) perpendicular

or

9.50 N ⋅ m = d × 21.725 N

or d = 0.437 m

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PROBLEM 3.68 In Problem 3.61, determine the perpendicular distance between cable EF and the line joining Points A and D. PROBLEM 3.61 A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the moment of that force about the line joining Points A and D.

SOLUTION From the solution to Problem 3.61

TEF = 46 lb

TEF = 2[−(3 lb)i − (22 lb) j + (6 lb)k ] | M AD | = 1359 lb ⋅ in. λ AD =

1 26

(4i − j + 3k )

Based on the discussion of Section 3.11, it! follows that only the perpendicular component of TEF will contribute to the moment of TEF about line AD. Now

(TEF ) parallel = TEF ⋅ λ AD = 2(−3i − 22 j + 6k ) ⋅ =

1 26

(4i − j + 3k )

2

[(−3)(4) + (−22)(−1) + (6)(3)] 26 = 10.9825 lb

Also so that

TEF = (TEF ) parallel + (TEF )perpendicular (TEF ) perpendicular = (46) 2 − (10.9825) 2 = 44.670 lb

Since λ AD and (TEF ) perpendicular are perpendicular, it follows that M AD = d (TEF ) perpendicular

or

1359 lb ⋅ in. = d × 44.670 lb

or d = 30.4 in.

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!

PROBLEM 3.69 In Problem 3.62, determine the perpendicular distance between cable EG and the line joining Points A and D. PROBLEM 3.62 A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EG at E is 54 lb, determine the moment of that force about the line joining Points A and D.

SOLUTION From the solution to Problem 3.62

TEG = 54 lb TEG = 6[(1 lb)i − (8 lb) j − (4 lb)k ] | M AD | = 2350 lb ⋅ in. 1

λ AD =

26

(4i − j + 3k )

Based on the discussion of Section 3.11, it! follows that only the perpendicular component of TEG will contribute to the moment of TEG about line AD. Now

(TEG ) parallel = TEG ⋅ λ AD = 6(i − 8 j − 4k ) ⋅ =

Thus,

6 26

1 26

(4i − j + 3k )

[(1)(4) + ( −8)( −1) + (−4)(3)] = 0

(TEG ) perpendicular = TEG = 54 lb

Since λ AD and (TEG ) perpendicular are perpendicular, it follows that | M AD | = d (TEG ) perpendicular

or

2350 lb ⋅ in. = d × 54 lb

or d = 43.5 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 224

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PROBLEM 3.70 Two parallel 60-N forces are applied to a lever as shown. Determine the moment of the couple formed by the two forces (a) by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples, (b) by using the perpendicular distance between the two forces, (c) by summing the moments of the two forces about Point A.

SOLUTION (a)

We have where

ΣM B : − d1C x + d 2 C y = M d1 = (0.360 m) sin 55° = 0.29489 m d 2 = (0.360 m) sin 55° = 0.20649 m C x = (60 N) cos 20° = 56.382 N C y = (60 N)sin 20° = 20.521 N M = −(0.29489 m)(56.382 N)k + (0.20649 m)(20.521 N)k = −(12.3893 N ⋅ m)k

(b)

We have

M = Fd (−k ) = 60 N[(0.360 m) sin(55° − 20°)]( −k ) = −(12.3893 N ⋅ m)k

(c)

We have

or M = 12.39 N ⋅ m

or M = 12.39 N ⋅ m

ΣM A : Σ(rA × F) = rB/A × FB + rC/A × FC = M

i j k sin 55° 0 M = (0.520 m)(60 N) cos 55° − cos 20° − sin 20° 0 i j k + (0.800 m)(60 N) cos 55° sin 55° 0 cos 20° sin 20° 0 = (17.8956 N ⋅ m − 30.285 N ⋅ m)k = −(12.3892 N ⋅ m)k

or M = 12.39 N ⋅ m

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PROBLEM 3.71 A plate in the shape of a parallelogram is acted upon by two couples. Determine (a) the moment of the couple formed by the two 21-lb forces, (b) the perpendicular distance between the 12-lb forces if the resultant of the two couples is zero, (c) the value of α if the resultant couple is 72 lb ⋅ in. clockwise and d is 42 in.

SOLUTION (a)

M1 = d1 F1

We have

d1 = 16 in.

where

F1 = 21 lb M1 = (16 in.)(21 lb) = 336 lb ⋅ in.

(b)

(c)

M1 + M 2 = 0

We have or

336 lb ⋅ in. − d 2 (12 lb) = 0

d 2 = 28.0 in.

M total = M1 + M 2

We have or

or M1 = 336 lb ⋅ in.

−72 lb ⋅ in. = 336 lb ⋅ in. − (42 in.)(sin α )(12 lb) sin α = 0.80952

α = 54.049°

and

or α = 54.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 226

PROBLEM 3.72 A couple M of magnitude 18 N ⋅ m is applied to the handle of a screwdriver to tighten a screw into a block of wood. Determine the magnitudes of the two smallest horizontal forces that are equivalent to M if they are applied (a) at corners A and D, (b) at corners B and C, (c) anywhere on the block.

SOLUTION (a)

M = Pd

We have or

18 N ⋅ m = P(.24 m) P = 75.0 N

or Pmin = 75.0 N

d BC = ( BE ) 2 + ( EC ) 2

(b)

= (.24 m) 2 + (.08 m) 2 = 0.25298 m M = Pd

We have

18 N ⋅ m = P(0.25298 m) P = 71.152 N

or

P = 71.2 N

d AC = ( AD) 2 + ( DC ) 2

(c)

= (0.24 m) 2 + (0.32 m) 2 = 0.4 m M = Pd AC

We have

18 N ⋅ m = P(0.4 m) P = 45.0 N

or

P = 45.0 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 227

PROBLEM 3.73 Four 1-in.-diameter pegs are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. (a) Determine the resultant couple acting on the board. (b) If only one string is used, around which pegs should it pass and in what directions should it be pulled to create the same couple with the minimum tension in the string? (c) What is the value of that minimum tension?

SOLUTION M = (35 lb)(7 in.) + (25 lb)(9 in.) = 245 lb ⋅ in. + 225 lb ⋅ in.

(a)

M = 470 lb ⋅ in.

(b)

With only one string, pegs A and D, or B and C should be used. We have 6 8

tan θ =

θ = 36.9°

90° − θ = 53.1°

Direction of forces:

(c)

With pegs A and D:

θ = 53.1°

With pegs B and C:

θ = 53.1°

The distance between the centers of the two pegs is 82 + 62 = 10 in.

Therefore, the perpendicular distance d between the forces is 1 ! d = 10 in. + 2 " in. # $2 % = 11 in. M = Fd

We must have

470 lb ⋅ in. = F (11 in.)

F = 42.7 lb

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PROBLEM 3.74 Four pegs of the same diameter are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. Determine the diameter of the pegs knowing that the resultant couple applied to the board is 485 lb·in. counterclockwise.

SOLUTION M = d AD FAD + d BC FBC

485 lb ⋅ in. = [(6 + d )in.](35 lb) + [(8 + d )in.](25 lb)

d = 1.250 in.

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PROBLEM 3.75 The shafts of an angle drive are acted upon by the two couples shown. Replace the two couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION Based on where

M = M1 + M 2 M1 = −(8 lb ⋅ ft)j M 2 = −(6 lb ⋅ ft)k M = −(8 lb ⋅ ft)j − (6 lb ⋅ ft)k

and

|M| = (8) 2 + (6) 2 = 10 lb ⋅ ft

or M = 10.00 lb ⋅ ft

M |M| −(8 lb ⋅ ft)j − (6 lb ⋅ ft)k = 10 lb ⋅ ft = −0.8 j − 0.6k =

or

M = |M| = (10 lb ⋅ ft)(−0.8 j − 0.6k ) cos θ x = 0 cos θ y = −0.8

θ x = 90° θ y = 143.130°

cos θ z = −0.6

θ z = 126.870° or θ x = 90.0° θ y = 143.1° θ z = 126.9°

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PROBLEM 3.76 If P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION We have where

M = M1 + M 2 M1 = rG/C × F1 rG/C = −(0.3 m)i F1 = (18 N)k M1 = −(0.3 m)i × (18 N)k = (5.4 N ⋅ m) j

Also,

M 2 = rD/F × F2 rD/F = −(.15 m)i + (.08 m) j F2 = λED F2 =

(.15 m)i + (.08 m) j + (.17 m)k (.15)2 + (.08) 2 + (.17) 2 m

(34 N)

= 141.421 N ⋅ m(.15i + .08j + .17k ) i j k M 2 = 141.421 N ⋅ m −.15 .08 0 −.15 .08 .17 = 141.421(.0136i + 0.0255 j)N ⋅ m

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PROBLEM 3.76 (Continued) and

M = [(5.4 N ⋅ m)j] + [141.421(.0136i + .0255 j) N ⋅ m] = (1.92333 N ⋅ m)i + (9.0062 N ⋅ m)j

| M | = (M x )2 + (M y )2 = (1.92333)2 + (9.0062) 2 = 9.2093 N ⋅ m

λ=

or M = 9.21 N ⋅ m

M (1.92333 N ⋅ m)i + (9.0062 N ⋅ m) j = |M| 9.2093 N ⋅ m

= 0.20885 + 0.97795 cos θ x = 0.20885

θ x = 77.945°

or

θ x = 77.9°

or

θ y = 12.05°

or

θ z = 90.0°

cos θ y = 0.97795

θ y = 12.054° cos θ z = 0.0

θ z = 90°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 232

PROBLEM 3.77 If P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION M = M1 + M 2 ; F1 = 16 lb, F2 = 40 lb M1 = rC × F1 = (30 in.)i × [−(16 lb) j] = −(480 lb ⋅ in.)k M 2 = rE/B × F2 ; rE/B = (15 in.)i − (5 in.) j d DE = (0) 2 + (5) 2 + (10) 2 = 5 5 in. F2 =

40 lb 5 5

(5 j − 10k )

= 8 5[(1 lb) j − (2 lb)k ] i j k M 2 = 8 5 15 −5 0 0 1 −2 = 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ] M = −(480 lb ⋅ in.)k + 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ] = (178.885 lb ⋅ in.)i + (536.66 lb ⋅ in.) j − (211.67 lb ⋅ in.)k M = (178.885) 2 + (536.66) 2 + (−211.67) 2 = 603.99 lb ⋅ in

M = 604 lb ⋅ in.

M = 0.29617i + 0.88852 j − 0.35045k M cos θ x = 0.29617 λ axis =

cos θ y = 0.88852

θ x = 72.8° θ y = 27.3° θ z = 110.5°

cos θ z = −0.35045

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PROBLEM 3.78 If P = 20 lb, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION From the solution to Problem. 3.77 16 lb force:

M1 = −(480 lb ⋅ in.)k

40 lb force:

M 2 = 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ]

P = 20 lb

M 3 = rC × P

= (30 in.)i × (20 lb)k = (600 lb ⋅ in.) j M = M1 + M 2 + M 3 = −(480)k + 8 5 (10i + 30 j + 15k ) + 600 j = (178.885 lb ⋅ in)i + (1136.66 lb ⋅ in.) j − (211.67 lb ⋅ in.)k M = (178.885) 2 + (113.66) 2 + (211.67) 2

= 1169.96 lb ⋅ in

M = 1170 lb ⋅ in.

M = 0.152898i + 0.97154 j − 0.180921k M cos θ x = 0.152898 λ axis =

cos θ y = 0.97154

θ x = 81.2° θ y = 13.70° θ z = 100.4°

cos θ z = −0.180921

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PROBLEM 3.79 If P = 20 N, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION We have

M = M1 + M 2 + M 3

where i j k M1 = rG/C × F1 = −0.3 0 0 N ⋅ m = (5.4 N ⋅ m)j 0 0 18

M 2 = rD/F

i j k × F2 = −.15 .08 0 141.421 N ⋅ m −.15 .08 .17

= 141.421(.0136i + .0255 j)N ⋅ m

(See Solution to Problem 3.76.) i j k M 3 = rC/A × F3 = 0.3 0 0.17 N ⋅ m 0 20 0 = −(3.4 N ⋅ m)i + (6 N ⋅ m)k M = [(1.92333 − 3.4)i + (5.4 + 3.6062) j + (6)k ] N ⋅ m = −(1.47667 N ⋅ m)i + (9.0062 N ⋅ m) j + (6 N ⋅ m)

| M | = M x2 + M y2 + M z2 = (1.47667) + (9.0062) + (6)2

or M = 10.92 N ⋅ m

= 10.9221 N ⋅ m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 235

PROBLEM 3.79 (Continued)

M −1.47667 + 9.0062 + 6 = |M| 10.9221 = −0.135200i + 0.82459 j + 0.54934k =

cos θ x = −0.135200 θ x = 97.770

or θ x = 97.8°

cos θ y = 0.82459

θ y = 34.453

or θ y = 34.5°

cos θ z = 0.54934

θ z = 56.678

or θ z = 56.7°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 236

PROBLEM 3.80 Shafts A and B connect the gear box to the wheel assemblies of a tractor, and shaft C connects it to the engine. Shafts A and B lie in the vertical yz plane, while shaft C is directed along the x axis. Replace the couples applied to the shafts with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION Represent the given couples by the following couple vectors: M A = −1600sin 20° j + 1600cos 20°k

= −(547.232 N ⋅ m) j + (1503.51 N ⋅ m)k M B = 1200sin 20° j + 1200 cos 20°k = (410.424 N ⋅ m) j + (1127.63 N ⋅ m)k M C = −(1120 N ⋅ m)i

The single equivalent couple is M = M A + M B + MC

= −(1120 N ⋅ m)i − (136.808 N ⋅ m) j + (2631.1 N ⋅ m)k M = (1120) 2 + (136.808)2 + (2631.1)2 = 2862.8 N ⋅ m −1120 cos θ x = 2862.8 −136.808 cos θ y = 2862.8 2631.1 cos θ z = 2862.8 M = 2860 N ⋅ m θ x = 113.0° θ y = 92.7° θ z = 23.2°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 237

PROBLEM 3.81 The tension in the cable attached to the end C of an adjustable boom ABC is 560 lb. Replace the force exerted by the cable at C with an equivalent forcecouple system (a) at A, (b) at B.

SOLUTION (a)

Based on

ΣF : FA = T = 560 lb FA = 560 lb

or

20°

ΣM A : M A = (T sin 50°)(d A ) = (560 lb)sin 50°(18 ft) = 7721.7 lb ⋅ ft M A = 7720 lb ⋅ ft

or (b)

Based on

ΣF : FB = T = 560 lb FB = 560 lb

or

20°

ΣM B : M B = (T sin 50°)(d B ) = (560 lb) sin 50°(10 ft) = 4289.8 lb ⋅ ft M B = 4290 lb ⋅ ft

or

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PROBLEM 3.82 A 160-lb force P is applied at Point A of a structural member. Replace P with (a) an equivalent force-couple system at C, (b) an equivalent system consisting of a vertical force at B and a second force at D.

SOLUTION (a)

Based on

ΣF : PC = P = 160 lb

or PC = 160 lb

60°

ΣM C : M C = − Px d cy + Py dCx

where

Px = (160 lb) cos 60° = 80 lb Py = (160 lb)sin 60° dCx

= 138.564 lb = 4 ft

dCy = 2.75 ft M C = (80 lb)(2.75 ft) + (138.564 lb)(4 ft) = 220 lb ⋅ ft + 554.26 lb ⋅ ft = 334.26 lb ⋅ ft

(b)

Based on

or M C = 334 lb ⋅ ft

ΣFx : PDx = P cos 60° = (160 lb) cos 60° = 80 lb

ΣM D : ( P cos 60°)( d DA ) = PB (d DB ) [(160 lb) cos 60°](1.5 ft) = PB (6 ft) PB = 20.0 lb

or PB = 20.0 lb

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PROBLEM 3.82 (Continued) ΣFy : P sin 60° = PB + PDy (160 lb)sin 60° = 20.0 lb + PDy PDy = 118.564 lb PD = ( PDx ) 2 + ( PDy ) 2 = (80) 2 + (118.564) 2 = 143.029 lb PDy ! # $ PDx % 118.564 ! = tan −1 " # $ 80 % = 55.991°

θ = tan −1 "

or PD = 143.0 lb

56.0°

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PROBLEM 3.83 The 80-N horizontal force P acts on a bell crank as shown. (a) Replace P with an equivalent force-couple system at B. (b) Find the two vertical forces at C and D that are equivalent to the couple found in Part a.

SOLUTION (a)

ΣF : FB = F = 80 N

Based on

or FB = 80.0 N

ΣM : M B = Fd B = 80 N (.05 m) = 4.0000 N ⋅ m M B = 4.00 N ⋅ m

or (b)

If the two vertical forces are to be equivalent to MB, they must be a couple. Further, the sense of the moment of this couple must be counterclockwise. Then, with FC and FD acting as shown, ΣM : M D = FC d 4.0000 N ⋅ m = FC (.04 m) FC = 100.000 N

or FC = 100.0 N

ΣFy : 0 = FD − FC FD = 100.000 N

or FD = 100.0 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 241

PROBLEM 3.84 A dirigible is tethered by a cable attached to its cabin at B. If the tension in the cable is 1040 N, replace the force exerted by the cable at B with an equivalent system formed by two parallel forces applied at A and C.

SOLUTION Require the equivalent forces acting at A and C be parallel and at an angle of α with the vertical. Then for equivalence, ΣFx : (1040 N)sin 30° = FA sin α + FB sin α

(1)

ΣFy : −(1040 N) cos 30° = − FA cos α − FB cos α

(2)

Dividing Equation (1) by Equation (2), ( FA + FB ) sin α (1040 N) sin 30° = −(1040 N) cos 30° −( FA + FB ) cos α

Simplifying yields α = 30° Based on ΣM C : [(1040 N) cos 30°](4 m) = ( FA cos 30°)(10.7 m) FA = 388.79 N FA = 389 N

or

60°

Based on ΣM A : − [(1040 N) cos 30°](6.7 m) = ( FC cos 30°)(10.7 m) FC = 651.21 N FC = 651 N

or

60°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 242

PROBLEM 3.85 The force P has a magnitude of 250 N and is applied at the end C of a 500-mm rod AC attached to a bracket at A and B. Assuming α = 30° and β = 60°, replace P with (a) an equivalent force-couple system at B, (b) an equivalent system formed by two parallel forces applied at A and B.

SOLUTION (a)

ΣF : F = P or F = 250 N

Equivalence requires

60°

ΣM B : M = −(0.3 m)(250 N) = −75 N ⋅ m

The equivalent force-couple system at B is F = 250 N

(b)

M = 75.0 N ⋅ m

60°

Require

Equivalence then requires ΣFx : 0 = FA cos φ + FB cos φ FA = − FB

or cos φ = 0

ΣFy : − 250 = − FA sin φ − FB sin φ

Now if

FA = − FB & −250 = 0 reject cos φ = 0

or and Also

φ = 90° FA + FB = 250 ΣM B : − (0.3 m)(250 N) = (0.2m) FA

or

FA = −375 N

and

FB = 625 N FA = 375 N

60°

FB = 625 N

60°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 243

PROBLEM 3.86 Solve Problem 3.85, assuming α = β = 25°.

SOLUTION

(a)

Equivalence requires ΣF : FB = P or FB = 250 N

25.0°

ΣM B : M B = −(0.3 m)[(250 N)sin 50°] = −57.453 N ⋅ m

The equivalent force-couple system at B is FB = 250 N

(b)

25.0°

M B = 57.5 N ⋅ m

Require

Equivalence requires M B = d AE Q (0.3 m)[(250 N) sin 50°] = [(0.2 m) sin 50°]Q Q = 375 N

Adding the forces at B:

FA = 375 N

25.0°

FB = 625 N

25.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 244

PROBLEM 3.87 A force and a couple are applied as shown to the end of a cantilever beam. (a) Replace this system with a single force F applied at Point C, and determine the distance d from C to a line drawn through Points D and E. (b) Solve Part a if the directions of the two 360-N forces are reversed.

SOLUTION (a)

We have

ΣF : F = (360 N) j − (360 N) j − (600 N)k

or F = −(600 N)k ΣM D : (360 N)(0.15 m) = (600 N)(d )

and

d = 0.09 m

or d = 90.0 mm below ED (b)

We have from Part a

F = −(600 N)k

ΣM D : −(360 N)(0.15 m) = −(600 N)( d )

and

d = 0.09 m

or d = 90.0 mm above ED

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 245

PROBLEM 3.88 The shearing forces exerted on the cross section of a steel channel can be represented by a 900-N vertical force and two 250-N horizontal forces as shown. Replace this force and couple with a single force F applied at Point C, and determine the distance x from C to line BD. (Point C is defined as the shear center of the section.)

SOLUTION Replace the 250-N forces with a couple and move the 900-N force to Point C such that its moment about H is equal to the moment of the couple

M H = (0.18)(250 N) = 45 N ⋅ m

Then or

M H = x(900 N) 45 N ⋅ m = x(900 N) x = 0.05 m F = 900 N

x = 50.0 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 246

PROBLEM 3.89 While tapping a hole, a machinist applies the horizontal forces shown to the handle of the tap wrench. Show that these forces are equivalent to a single force, and specify, if possible, the point of application of the single force on the handle.

SOLUTION Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-couple. Have FB = 2.9 lb − 2.65 lb = 0.25 lb, where the 2.65 lb force be part of the couple. Combining the two parallel forces, M couple = (2.65 lb)[(3.2 in. + 2.8 in.) cos 25°] = 14.4103 lb ⋅ in.

and

M couple = 14.4103 lb ⋅ in.

A single equivalent force will be located in the negative z-direction Based on

ΣM B : −14.4103 lb ⋅ in. = [(.25 lb) cos 25°](a ) a = 63.600 in.

F′ = (.25 lb)(cos 25°i + sin 25°k ) F′ = (0.227 lb)i + (0.1057 lb)k and is applied on an extension of handle BD at a distance of 63.6 in. to the right of B

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 247

PROBLEM 3.90 Three control rods attached to a lever ABC exert on it the forces shown. (a) Replace the three forces with an equivalent force-couple system at B. (b) Determine the single force that is equivalent to the force-couple system obtained in Part a, and specify its point of application on the lever.

SOLUTION (a)

First note that the two 20-lb forces form A couple. Then F = 48 lb

θ

where

θ = 180° − (60° + 55°) = 65°

and

M = ΣM B = (30 in.)(48 lb) cos55° − (70 in.)(20 lb) cos 20° = −489.62 lb ⋅ in

The equivalent force-couple system at B is F = 48.0 lb

(b)

65°

M = 490 lb ⋅ in.

The single equivalent force F ′ is equal to F. Further, since the sense of M is clockwise, F ′ must be applied between A and B. For equivalence. ΣM B : M = − aF ′ cos 55°

where a is the distance from B to the point of application of F′. Then −489.62 lb ⋅ in. = −a (48.0 lb) cos 55°

or

a = 17.78 in.

F ′ = 48.0 lb

65.0°

and is applied to the lever 17.78 in. To the left of pin B

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 248

PROBLEM 3.91 A hexagonal plate is acted upon by the force P and the couple shown. Determine the magnitude and the direction of the smallest force P for which this system can be replaced with a single force at E.

SOLUTION From the statement of the problem, it follows that ΣM E = 0 for the given force-couple system. Further, for Pmin, must require that P be perpendicular to rB/E . Then ΣM E : (0.2 sin 30° + 0.2)m × 300 N + (0.2 m)sin 30° × 300 N − (0.4 m) Pmin = 0

or

Pmin = 300 N Pmin = 300 N

30.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 249

PROBLEM 3.92 A rectangular plate is acted upon by the force and couple shown. This system is to be replaced with a single equivalent force. (a) For α = 40°, specify the magnitude and the line of action of the equivalent force. (b) Specify the value of α if the line of action of the equivalent force is to intersect line CD 300 mm to the right of D.

SOLUTION (a)

The given force-couple system (F, M) at B is F = 48 N

and

M = ΣM B = (0.4 m)(15 N) cos 40° + (0.24 m)(15 N)sin 40°

or

M = 6.9103 N ⋅ m

The single equivalent force F′ is equal to F. Further for equivalence ΣM B : M = dF ′ 6.9103 N ⋅ m = d × 48 N

or

d = 0.14396 m

or

F ′ = 48 N

and the line of action of F′ intersects line AB 144 mm to the right of A. (b)

Following the solution to Part a but with d = 0.1 m and α unknown, have ΣM B : (0.4 m)(15 N) cos α + (0.24 m)(15 N) sin α = (0.1 m)(48 N)

or

5cos α + 3sin α = 4

Rearranging and squaring

25 cos 2 α = (4 − 3 sin α )2

Using cos 2 α = 1 − sin 2 α and expanding 25(1 − sin 2 α ) = 16 − 24 sin α + 9 sin 2 α

or Then

34 sin 2 α − 24 sin α − 9 = 0 24 ± (−24) 2 − 4(34)(−9) 2(34) sin α = 0.97686 or sin α = −0.27098 sin α =

α = 77.7° or α = −15.72°

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PROBLEM 3.93 An eccentric, compressive 1220-N force P is applied to the end of a cantilever beam. Replace P with an equivalent force-couple system at G.

SOLUTION We have ΣF : − (1220 N)i = F F = − (1220 N)i

Also, we have ΣM G : rA/G × P = M i j k 1220 0 −.1 −.06 N ⋅ m = M −1 0 0 M = (1220 N ⋅ m)[(−0.06)(−1) j − ( −0.1)( −1)k ]

or M = (73.2 N ⋅ m) j − (122 N ⋅ m)k

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 251

PROBLEM 3.94 To keep a door closed, a wooden stick is wedged between the floor and the doorknob. The stick exerts at B a 175-N force directed along line AB. Replace that force with an equivalent force-couple system at C.

SOLUTION We have ΣF : PAB = FC

where PAB = =

AB PAB

(33 mm)i + (990 mm) j − (594 mm)k (175 N) 1155.00 mm

or FC = (5.00 N)i + (150 N) j − (90.0 N)k We have

ΣM C : rB/C × PAB = M C i j k M C = 5 0.683 −0.860 0 N ⋅ m 1 30 −18 = (5){(− 0.860)(−18)i − (0.683)(−18) j + [(0.683)(30) − (0.860)(1)]k}

or M C = (77.4 N ⋅ m)i + (61.5 N ⋅ m) j + (106.8 N ⋅ m)k

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 252

PROBLEM 3.95 An antenna is guyed by three cables as shown. Knowing that the tension in cable AB is 288 lb, replace the force exerted at A by cable AB with an equivalent force-couple system at the center O of the base of the antenna.

SOLUTION We have

d AB = (−64)2 + (−128) 2 + (16) 2 = 144 ft

Then

TAB =

Now

288 lb ( −64i − 128 j + 16k ) 144 = (32 lb)( −4i − 8 j + k )

M = M O = rA / O × TAB = 128 j × 32(−4i − 8 j + k ) = (4096 lb ⋅ ft)i + (16,384 lb ⋅ ft)k

The equivalent force-couple system at O is F = −(128.0 lb)i − (256 lb) j + (32.0 lb)k M = (4.10 kip ⋅ ft)i + (16.38 kip ⋅ ft)k

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 253

PROBLEM 3.96 An antenna is guyed by three cables as shown. Knowing that the tension in cable AD is 270 lb, replace the force exerted at A by cable AD with an equivalent force-couple system at the center O of the base of the antenna.

SOLUTION We have

d AD = ( −64) 2 + (−128)2 + (−128) 2 = 192 ft

Then

Now

270 lb (−64i − 128 j + 128k ) 192 = (90 lb)(−i − 2 j − 2k )

TAD =

M = M O = rA/O × TAD = 128 j × 90(−i − 2 j − 2k ) = −(23, 040 lb ⋅ ft)i + (11,520 lb ⋅ ft)k

The equivalent force-couple system at O is F = −(90.0 lb)i − (180.0 lb) j − (180.0 lb)k M = −(23.0 kip ⋅ ft)i + (11.52 kip ⋅ ft)k

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 254

PROBLEM 3.97 Replace the 150-N force with an equivalent force-couple system at A.

SOLUTION Equivalence requires

ΣF : F = (150 N)(− cos 35° j − sin 35°k ) = −(122.873 N) j − (86.036 N)k ΣMA : M = rD/A × F

where

Then

rD/A = (0.18 m)i − (0.12 m) j + (0.1 m)k i j k 0.1 N⋅m −0.12 M = 0.18 0 −122.873 −86.036 = [( −0.12)(−86.036) − (0.1)(−122.873)]i + [−(0.18)(−86.036)]j + [(0.18)(−122.873)]k = (22.6 N ⋅ m)i + (15.49 N ⋅ m) j − (22.1 N ⋅ m)k

The equivalent force-couple system at A is F = −(122.9 N) j − (86.0 N)k M = (22.6 N ⋅ m)i + (15.49 N ⋅ m) j − (22.1 N ⋅ m)k

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 255

!

PROBLEM 3.98 A 77-N force F1 and a 31-N ⋅ m couple M1 are applied to corner E of the bent plate shown. If F1 and M1 are to be replaced with an equivalent force-couple system (F2, M2) at corner B and if (M2)z = 0, determine (a) the distance d, (b) F2 and M2.

SOLUTION (a)

ΣM Bz : M 2 z = 0

We have

k ⋅ (rH /B × F1 ) + M 1z = 0

where

(1)

rH /B = (0.31 m)i − (0.0233) j F1 =

EH F1

(0.06 m)i + (0.06 m) j − (0.07 m)k (77 N) 0.11 m = (42 N)i + (42 N) j − (49 N)k =

M1z = k ⋅ M1 M1 = =

EJ M 1

− di + (0.03 m) j − (0.07 m)k

d 2 + 0.0058 m

(31 N ⋅ m)

Then from Equation (1), 0 0 1 ( −0.07 m)(31 N ⋅ m) 0.31 −0.0233 0 + =0 2 0.0058 + d 42 42 −49

Solving for d, Equation (1) reduces to (13.0200 + 0.9786) −

From which

2.17 N ⋅ m d 2 + 0.0058

d = 0.1350 m

=0

or d = 135.0 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 256

PROBLEM 3.98 (Continued)

(b)

F2 = F1 = (42i + 42 j − 49k )N

or F2 = (42 N)i + (42 N) j − (49 N)k

M 2 = rH /B × F1 + M1 i j k (0.1350)i + 0.03j − 0.07k (31 N ⋅ m) = 0.31 −0.0233 0 + 0.155000 −49 42 42 = (1.14170i + 15.1900 j + 13.9986k ) N ⋅ m + (−27.000i + 6.0000 j − 14.0000k ) N ⋅ m M 2 = − (25.858 N ⋅ m)i + (21.190 N ⋅ m) j

or M 2 = − (25.9 N ⋅ m)i + (21.2 N ⋅ m) j

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PROBLEM 3.99 A 46-lb force F and a 2120-lb ⋅ in. couple M are applied to corner A of the block shown. Replace the given force-couple system with an equivalent force-couple system at corner H.

SOLUTION We have Then

Also

d AJ = (18) 2 + (−14) 2 + (−3) 2 = 23 in.

46 lb (18i − 14 j − 3k ) 23 = (36 lb)i − (28 lb) j − (6 lb)k

F=

d AC = (−45) 2 + (0)2 + (−28)2 = 53 in.

2120 lb ⋅ in. ( −45i − 28k ) 53 = −(1800 lb ⋅ in.)i − (1120 lb ⋅ in.)k

Then

M=

Now

M ′ = M + rA/H × F

where

Then

rA/H = (45 in.)i + (14 in.) j i j k 0 M ′ = (−1800i − 1120k ) + 45 14 36 −28 −6

= (−1800i − 1120k ) + {[(14)(−6)]i + [−(45)( −6)]j + [(45)(−28) − (14)(36)]k} = (−1800 − 84)i + (270) j + (−1120 − 1764)k = −(1884 lb ⋅ in.)i + (270 lb ⋅ in.)j − (2884 lb ⋅ in.)k = −(157 lb ⋅ ft)i + (22.5 lb ⋅ ft) j − (240 lb ⋅ ft)k F′ = (36.0 lb)i − (28.0 lb) j − (6.00 lb)k

The equivalent force-couple system at H is

M ′ = −(157 lb ⋅ ft)i + (22.5 lb ⋅ ft) j − (240 lb ⋅ ft)k

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PROBLEM 3.100 The handpiece for a miniature industrial grinder weighs 0.6 lb, and its center of gravity is located on the y axis. The head of the handpiece is offset in the xz plane in such a way that line BC forms an angle of 25° with the x direction. Show that the weight of the handpiece and the two couples M1 and M2 can be replaced with a single equivalent force. Further, assuming that M1 = 0.68 lb ⋅ in. and M2 = 0.65 lb ⋅ in., determine (a) the magnitude and the direction of the equivalent force, (b) the point where its line of action intersects the xz plane.

SOLUTION First assume that the given force W and couples M1 and M2 act at the origin. Now

W = − Wj

and

M = M1 + M 2 = − ( M 2 cos 25°)i + ( M 1 − M 2 sin 25°)k

Note that since W and M are perpendicular, it follows that they can be replaced with a single equivalent force. F = W or F = − Wj = − (0.6 lb) j

or F = − (0.600 lb)j

(a)

We have

(b)

Assume that the line of action of F passes through Point P(x, 0, z). Then for equivalence M = rP/0 × F

where

rP/0 = xi + zk − ( M 2 cos 25°)i + ( M1 − M 2 sin 25°)k i j = x 0 0 −W

Equating the i and k coefficients, (b)

For

z=

k z = (Wz )i − (Wx)k 0 − M z cos 25° W

and

x = −" $

M 1 − M 2 sin 25° ! # W %

W = 0.6 lb M1 = 0.68 lb ⋅ in. M 2 = 0.65 lb ⋅ in. x=

0.68 − 0.65sin 25° = 0.67550 in. − 0.6

or

z=

− 0.65cos 25° = − 0.98183 in. 0.6

or z = − 0.982 in.

x = 0.675 in.

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PROBLEM 3.101 A 4-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent forcecouple system at end A of the beam. (b) Which of the loadings are equivalent?

SOLUTION

(a)

(a)

We have

ΣFy : −400 N − 200 N = Ra

or and

R a = 600 N

ΣM A : 1800 N ⋅ m − (200 N)(4 m) = M a

or M a = 1000 N ⋅ m (b)

We have

ΣFy : − 600 N = Rb

or and

ΣM A : − 900 N ⋅ m = M b

or (c)

We have

M b = 900 N ⋅ m

ΣFy : 300 N − 900 N = Rc

or and

R b = 600 N

R c = 600 N

ΣM A : 4500 N ⋅ m − (900 N)(4 m) = M c

or

M c = 900 N ⋅ m

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PROBLEM 3.101 (Continued)

(d)

We have

and

(e)

We have

ΣFy : − 400 N + 800 N = Rd

or

R d = 400 N

or

M d = 900 N ⋅ m

ΣM A : (800 N)(4 m) − 2300 N ⋅ m = M d

ΣFy : − 400 N − 200 N = Re

or and

ΣM A : 200 N ⋅ m + 400 N ⋅ m − (200 N)(4 m) = M e

or ( f ) We have

M e = 200 N ⋅ m

ΣFy : − 800 N + 200 N = R f

or and

R e = 600 N

R f = 600 N

ΣM A : − 300 N ⋅ m + 300 N ⋅ m + (200 N)(4 m) = M f

or M f = 800 N ⋅ m (g)

We have

ΣFy : − 200 N − 800 N = Rg

or and

R g = 1000 N

ΣM A : 200 N ⋅ m + 4000 N ⋅ m − (800 N)(4 m) = M g

or M g = 1000 N ⋅ m (h)

We have

ΣFy : − 300 N − 300 N = Rh

or and

ΣM A : 2400 N ⋅ m − 300 N ⋅ m − (300 N)(4 m) = M h

or (b)

R h = 600 N

M h = 900 N ⋅ m

!

Therefore, loadings (c) and (h) are equivalent.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 261

PROBLEM 3.102 A 4-m-long beam is loaded as shown. Determine the loading of Problem 3.101 which is equivalent to this loading.

SOLUTION We have and

ΣFy : − 200 N − 400 N = R

or R = 600 N

ΣM A : −400 N ⋅ m + 2800 N ⋅ m − (400 N)(4 m) = M M = 800 N ⋅ m

or

Equivalent to case ( f ) of Problem 3.101 Problem 3.101 Equivalent force-couples at A

Case

R

(a)

600 N

1000 N ⋅ m

(b)

600 N

900 N ⋅ m

(c)

600 N

900 N ⋅ m

(d)

400 N

900 N ⋅ m

(e)

600 N

200 N ⋅ m

(f )

600 N

800 N ⋅ m

(g)

1000 N

1000 N ⋅ m

(h)

600 N

900 N ⋅ m

M

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PROBLEM 3.103 Determine the single equivalent force and the distance from Point A to its line of action for the beam and loading of (a) Problem 3.101b, (b) Problem 3.101d, (c) Problem 3.101e. PROBLEM 3.101 A 4-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent force-couple system at end A of the beam. (b) Which of the loadings are equivalent?

SOLUTION (a)

For equivalent single force at distance d from A ΣFy : −600 N = R

We have

or R = 600 N and

ΣM C : (600 N)(d ) − 900 N ⋅ m = 0

or d = 1.500 m (b)

ΣFy : − 400 N + 800 N = R

We have

or R = 400 N and

ΣM C : (400 N)( d ) + (800 N)(4 − d ) − 2300 N ⋅ m = 0

(c)

d = 2.25 m

or

R = 600 N

ΣFy : − 400 N − 200 N = R

We have

and

or

ΣM C : 200 N ⋅ m + (400 N)(d ) − (200 N)(4 − d ) + 400 N ⋅ m = 0

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PROBLEM 3.104 Five separate force-couple systems act at the corners of a piece of sheet metal, which has been bent into the shape shown. Determine which of these systems is equivalent to a force F = (10 lb)i and a couple of moment M = (15 lb ⋅ ft)j + (15 lb ⋅ ft)k located at the origin.

SOLUTION First note that the force-couple system at F cannot be equivalent because of the direction of the force [The force of the other four systems is (10 lb)i]. Next move each of the systems to the origin O; the forces remain unchanged. A: M A = ΣM O = (5 lb ⋅ ft) j + (15 lb ⋅ ft)k + (2 ft)k × (10 lb)i

= (25 lb ⋅ ft) j + (15 lb ⋅ ft)k D : M D = ΣM O = −(5 lb ⋅ ft) j + (25 lb ⋅ ft)k

+ [(4.5 ft) j + (1 ft) j + (2 ft)k ] × 10 lb)i = (15 lb ⋅ ft)i + (15 lb ⋅ ft)k G : M G = ΣM O = (15 lb ⋅ ft)i + (15 lb ⋅ ft) j I : M I = ΣM I = (15 lb ⋅ ft) j − (5 lb ⋅ ft)k

+ [(4.5 ft)i + (1 ft) j] × (10 lb) j = (15 lb ⋅ ft) j − (15 lb ⋅ ft)k

The equivalent force-couple system is the system at corner D.

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!

PROBLEM 3.105 The weights of two children sitting at ends A and B of a seesaw are 84 lb and 64 lb, respectively. Where should a third child sit so that the resultant of the weights of the three children will pass through C if she weighs (a) 60 lb, (b) 52 lb.

SOLUTION

(a)

For the resultant weight to act at C, Then

ΣM C = 0 WC = 60 lb

(84 lb)(6 ft) − 60 lb(d ) − 64 lb(6 ft) = 0 d = 2.00 ft to the right of C

(b)

For the resultant weight to act at C, Then

ΣM C = 0 WC = 52 lb

(84 lb)(6 ft) − 52 lb(d ) − 64 lb(6 ft) = 0 d = 2.31 ft to the right of C

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 265

!

PROBLEM 3.106 Three stage lights are mounted on a pipe as shown. The lights at A and B each weigh 4.1 lb, while the one at C weighs 3.5 lb. (a) If d = 25 in., determine the distance from D to the line of action of the resultant of the weights of the three lights. (b) Determine the value of d so that the resultant of the weights passes through the midpoint of the pipe.

SOLUTION

For equivalence ΣFy : − 4.1 − 4.1 − 3.5 = − R or R = 11.7 lb ΣFD : − (10 in.)(4.1 lb) − (44 in.)(4.1 lb) −[(4.4 + d )in.](3.5 lb) = −( L in.)(11.7 lb) 375.4 + 3.5d = 11.7 L (d , L in in.)

or

d = 25 in.

(a) We have

375.4 + 3.5(25) = 11.7 L or

L = 39.6 in.

The resultant passes through a Point 39.6 in. to the right of D. L = 42 in.

(b) We have

375.4 + 3.5d = 11.7(42)

or d = 33.1 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 266

PROBLEM 3.107 A beam supports three loads of given magnitude and a fourth load whose magnitude is a function of position. If b = 1.5 m and the loads are to be replaced with a single equivalent force, determine (a) the value of a so that the distance from support A to the line of action of the equivalent force is maximum, (b) the magnitude of the equivalent force and its point of application on the beam.

SOLUTION

For equivalence ΣFy : −1300 + 400

a − 400 − 600 = − R b

a! R = " 2300 − 400 # N b $ %

or

ΣM A :

a a! 400 # − a (400) − (a + b)(600) = − LR " b% 2$

1000a + 600b − 200 L=

or

(1)

2300 − 400

a2 b

a b

10a + 9 −

Then with

b = 1.5 m L =

4 2 a 3

(2)

8 23 − a 3

Where a, L are in m (a)

Find value of a to maximize L dL "$ = da

8 ! 8 ! 4 ! 8! 10 − a #" 23 − a # − "10a + 9 − a 2 #" − # 3 %$ 3 % $ 3 %$ 3 % 8 ! " 23 − 3 a # $ %

2

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PROBLEM 3.107 (Continued)

230 −

or

16a 2 − 276a + 1143 = 0 276 ± (−276) 2 − 4(16)(1143) 2(16)

Then

a=

or

a = 10.3435 m and a = 6.9065 m

Since (b)

184 80 64 2 80 32 a− a+ a + a + 24 − a 2 = 0 3 3 9 3 9

or

AB = 9 m, a must be less than 9 m

a = 6.91 m

6.9065 1.5

Using Eq. (1)

R = 2300 − 400

and using Eq. (2)

4 10(6.9065) + 9 − (6.9065)2 3 = 3.16 m L= 8 23 − (6.9065) 3

or R = 458 N

R is applied 3.16 m to the right of A.

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!

!

PROBLEM 3.108 Gear C is rigidly attached to arm AB. If the forces and couple shown can be reduced to a single equivalent force at A, determine the equivalent force and the magnitude of the couple M.

SOLUTION We have

For equivalence ΣFx : −18sin 30° + 25cos 40° = Rx

or

Rx = 10.1511 lb ΣFy : −18cos 30° − 40 − 25sin 40° = Ry

or Then

Ry = −71.658 lb R = (10.1511) 2 + (71.658) 2 = 72.416 tanθ =

or Also

71.658 10.1511

θ = 81.9°

R = 72.4 lb

81.9°

ΣM A : M − (22 in.)(18 lb)sin 35° − (32 in.)(40 lb) cos 25° − (48 in.)(25 lb) sin 65° = 0 M = 2474.8 lb ⋅ in.

or M = 206 lb ⋅ ft

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PROBLEM 3.109 A couple of magnitude M = 54 lb ⋅ in. and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resultant intersects line AB and line BC.

SOLUTION (a)

We have

ΣF : R = (−10 j) + (30 cos 60°)i + 30 sin 60° j + (−45i ) = −(30 lb)i + (15.9808 lb) j

or R = 34.0 lb (b)

28.0°

First reduce the given forces and couple to an equivalent force-couple system (R , M B ) at B. We have

ΣM B : M B = (54 lb ⋅ in) + (12 in.)(10 lb) − (8 in.)(45 lb) = −186 lb ⋅ in.

Then with R at D or and with R at E or

ΣM B : −186 lb ⋅ in = a(15.9808 lb)

a = 11.64 in. ΣM B : −186 lb ⋅ in = C (30 lb)

C = 6.2 in.

The line of action of R intersects line AB 11.64 in. to the left of B and intersects line BC 6.20 in. below B.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 270

PROBLEM 3.110 A couple M and the three forces shown are applied to an angle bracket. Find the moment of the couple if the line of action of the resultant of the force system is to pass through (a) Point A, (b) Point B, (c) Point C.

SOLUTION In each case, must have M iR = 0 (a)

M AB = ΣM A = M + (12 in.)[(30 lb) sin 60°] − (8 in.)(45 lb) = 0 M = +48.231 lb ⋅ in.

(b)

M = 48.2 lb ⋅ in.

M BR = ΣM B = M + (12 in.)(10 lb) − (8 in.)(45 lb) = 0 M = +240 lb ⋅ in.

(c)

M = 240 lb ⋅ in.

M CR = ΣM C = M + (12 in.)(10 lb) − (8 in.)[(30 lb) cos 60°] = 0 M =0

M=0

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 271

PROBLEM 3.111 Four forces act on a 700 × 375-mm plate as shown. (a) Find the resultant of these forces. (b) Locate the two points where the line of action of the resultant intersects the edge of the plate.

SOLUTION (a)

R = ΣF = (−400 N + 160 N − 760 N)i + (600 N + 300 N + 300 N) j = −(1000 N)i + (1200 N) j

R = (1000 N) 2 + (1200 N)2 = 1562.09 N 1200 N ! tan θ = " − # $ 1000 N % = −1.20000 θ = −50.194°

(b)

R = 1562 N

50.2°

M CR = Σr × F

= (0.5 m)i × (300 N + 300 N) j = (300 N ⋅ m)k (300 N ⋅ m)k = xi × (1200 N) j x = 0.25000 m x = 250 mm (300 N ⋅ m) = yj × ( −1000 N)i

y = 0.30000 m y = 300 mm

Intersection 250 mm to right of C and 300 mm above C

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PROBLEM 3.112 Solve Problem 3.111, assuming that the 760-N force is directed to the right. PROBLEM 3.111 Four forces act on a 700 × 375-mm plate as shown. (a) Find the resultant of these forces. (b) Locate the two points where the line of action of the resultant intersects the edge of the plate.

SOLUTION R = ΣF

(a)

= ( −400N + 160 N + 760 N)i + (600 N + 300 N + 300 N) j = (520 N)i + (1200 N) j

R = (520 N) 2 + (1200 N) 2 = 1307.82 N 1200 N ! tan θ = " # = 2.3077 $ 520 N % θ = 66.5714°

R = 1308 N

66.6°

M CR = Σr × F

(b)

= (0.5 m)i × (300 N + 300 N) j = (300 N ⋅ m)k (300 N ⋅ m)k = xi × (1200 N) j x = 0.25000 m

or

x = 0.250 mm (300 N ⋅ m)k = [ x′i + (0.375 m) j] × [(520 N)i + (1200 N) j] = (1200 x′ − 195)k

x′ = 0.41250 m

or

x′ = 412.5 mm

Intersection 412 mm to the right of A and 250 mm to the right of C

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 273

PROBLEM 3.113 A truss supports the loading shown. Determine the equivalent force acting on the truss and the point of intersection of its line of action with a line drawn through Points A and G.

SOLUTION We have

R = ΣF R = (240 lb)(cos 70°i − sin 70° j) − (160 lb) j + (300 lb)(− cos 40°i − sin 40° j) − (180 lb) j

R = −(147.728 lb)i − (758.36 lb) j R = Rx2 + Ry2 = (147.728) 2 + (758.36) 2 = 772.62 lb Ry ! # $ Rx % −758.36 ! = tan −1 " # $ −147.728 % = 78.977°

θ = tan −1 "

or R = 773 lb

We have

ΣM A = dRy

where

ΣM A = −[240 lb cos 70°](6 ft) − [240 lbsin 70°](4 ft)

79.0°

− (160 lb)(12 ft) + [300 lb cos 40°](6 ft) − [300 lb sin 40°](20 ft) − (180 lb)(8 ft) = −7232.5 lb ⋅ ft −7232.5 lb ⋅ ft −758.36 lb = 9.5370 ft

d=

or d = 9.54 ft to the right of A

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 274

PROBLEM 3.114 Pulleys A and B are mounted on bracket CDEF. The tension on each side of the two belts is as shown. Replace the four forces with a single equivalent force, and determine where its line of action intersects the bottom edge of the bracket.

SOLUTION Equivalent force-couple at A due to belts on pulley A We have

ΣF : − 120 lb − 160 lb = RA R A = 280 lb

We have

ΣM A :

−40 lb(2 in.) = M A M A = 80 lb ⋅ in.

Equivalent force-couple at B due to belts on pulley B We have

ΣF: (210 lb + 150 lb)

25° = R B R B = 360 lb

We have

25°

ΣM B : − 60 lb(1.5 in.) = M B M B = 90 lb ⋅ in.

Equivalent force-couple at F We have

ΣF: R F = ( − 280 lb) j + (360 lb)(cos 25°i + sin 25° j)

= (326.27 lb)i − (127.857 lb) j R = RF 2 2 = RFx + RFy

= (326.27)2 + (127.857)2 = 350.43 lb RFy ! # $ RFx % −127.857 ! = tan −1 " # $ 326.27 % = −21.399°

θ = tan −1 "

or R F = R = 350 lb

21.4°

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PROBLEM 3.114 (Continued)

We have

ΣM F : M F = − (280 lb)(6 in.) − 80 lb ⋅ in. − [(360 lb) cos 25°](1.0 in.) + [(360 lb) sin 25°](12 in.) − 90 lb ⋅ in. M F = − (350.56 lb ⋅ in.)k

To determine where a single resultant force will intersect line FE, M F = dRy d=

MF Ry

−350.56 lb ⋅ in. −127 ⋅ 857 lb = 2.7418 in. =

or d = 2.74 in.

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 276

PROBLEM 3.115 A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. For P = 0, determine the location of the rivet hole if it is to be located (a) on line FG, (b) on line GH.

SOLUTION We have

First replace the applied forces and couples with an equivalent force-couple system at G. ΣFx : 200cos 15° − 120 cos 70° + P = Rx

Thus

Rx = (152.142 + P) N

or

ΣFy : − 200sin 15° − 120sin 70° − 80 = Ry Ry = −244.53 N

or

ΣM G : − (0.47 m)(200 N) cos15° + (0.05 m)(200 N)sin15° + (0.47 m)(120 N) cos 70° − (0.19 m)(120 N)sin 70° − (0.13 m)( P N) − (0.59 m)(80 N) + 42 N ⋅ m + 40 N ⋅ m = M G M G = −(55.544 + 0.13P) N ⋅ m

or

(1)

Setting P = 0 in Eq. (1): Now with R at I

ΣM G : − 55.544 N ⋅ m = − a(244.53 N) a = 0.227 m

or and with R at J

ΣM G : − 55.544 N ⋅ m = −b(152.142 N) b = 0.365 m

or (a)

The rivet hole is 0.365 m above G.

(b)

The rivet hole is 0.227 m to the right of G.

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PROBLEM 3.116 Solve Problem 3.115, assuming that P = 60 N. PROBLEM 3.115 A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. For P = 0, determine the location of the rivet hole if it is to be located (a) on line FG, (b) on line GH.

SOLUTION See the solution to Problem 3.115 leading to the development of Equation (1) M G = −(55.544 + 0.13P) N ⋅ m Rx = (152.142 + P) N

and

P = 60 N

For We have

Rx = (152.142 + 60) = 212.14 N M G = −[55.544 + 0.13(60)] = −63.344 N ⋅ m

Then with R at I

ΣM G : −63.344 N ⋅ m = −a(244.53 N) a = 0.259 m

or and with R at J

ΣM G : −63.344 N ⋅ m = −b(212.14 N) b = 0.299 m

or (a)

The rivet hole is 0.299 m above G.

(b)

The rivet hole is 0.259 m to the right of G.

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PROBLEM 3.117 A 32-lb motor is mounted on the floor. Find the resultant of the weight and the forces exerted on the belt, and determine where the line of action of the resultant intersects the floor.

SOLUTION We have ΣF : (60 lb)i − (32 lb) j + (140 lb)(cos 30°i + sin 30° j) = R R = (181.244 lb)i + (38.0 lb) j

or R = 185.2 lb We have

11.84°

ΣM O : ΣM O = xRy

− [(140 lb) cos 30°][(4 + 2 cos 30°)in.] − [(140 lb) sin 30°][(2 in.)sin 30°] − (60 lb)(2 in.) = x(38.0 lb) x=

and

1 (− 694.97 − 70.0 − 120) in. 38.0

x = −23.289 in.

Or, resultant intersects the base (x axis) 23.3 in. to the left of the vertical centerline (y axis) of the motor.

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PROBLEM 3.118 As follower AB rolls along the surface of member C, it exerts a constant force F perpendicular to the surface. (a) Replace F with an equivalent force-couple system at the Point D obtained by drawing the perpendicular from the point of contact to the x axis. (b) For a = 1 m and b = 2 m, determine the value of x for which the moment of the equivalent forcecouple system at D is maximum.

SOLUTION (a)

The slope of any tangent to the surface of member C is dy d ( x2 = *b ""1 − 2 dx dx *, $ a

! ) −2b ## + = 2 x % -+ a

Since the force F is perpendicular to the surface, dy ! tan α = − " # $ dx %

−1

=

a2 1 ! 2b "$ x #%

For equivalence ΣF : F = R ΣM D : ( F cos α )( y A ) = M D

where cos α =

2bx (a 2 ) 2 + (2bx)2

x2 ! y A = b ""1 − 2 ## $ a % x3 ! 2 Fb 2 " x − 2 # a % $ MD = 4 2 2 a + 4b x

Therefore, the equivalent force-couple system at D is R=F

a2 ! tan −1 "" ## $ 2bx %

x3 ! 2 Fb2 " x − 2 # a % $ M= a 4 + 4b 2 x 2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 280

PROBLEM 3.118 (Continued)

(b)

To maximize M, the value of x must satisfy

dM =0 dx

a = 1 m, b = 2 m

where, for M=

8F ( x − x3 ) 1 + 16 x 2

(1 ) 1 + 16 x 2 (1 − 3x 2 ) − ( x − x3 ) * (32 x)(1 + 16 x 2 ) −1/ 2 + dM 2 , - =0 = 8F 2 dx (1 + 16 x ) (1 + 16 x 2 )(1 − 3x 2 ) − 16 x( x − x3 ) = 0 32 x 4 + 3x 2 − 1 = 0

or x2 =

−3 ± 9 − 4(32)(−1) = 0.136011 m 2 2(32)

Using the positive value of x2

x = 0.36880 m

and − 0.22976 m 2

or x = 369 mm

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!

PROBLEM 3.119 Four forces are applied to the machine component ABDE as shown. Replace these forces by an equivalent force-couple system at A.

SOLUTION R = −(50 N) j − (300 N)i − (120 N)i − (250 N)k R = −(420 N)i − (50 N)j − (250 N)k rB = (0.2 m)i rD = (0.2 m)i + (0.16 m)k rE = (0.2 m)i − (0.1 m) j + (0.16 m)k M RA = rB × [−(300 N)i − (50 N) j] + rD × (−250 N)k + r × ( − 120 N)i

i j k i j k = 0.2 m 0 0 + 0.2 m 0 0.16 m −300 N −50 N 0 0 0 −250 N i j k + 0.2 m −0.1 m 0.16 m −120 N 0 0 = −(10 N ⋅ m)k + (50 N ⋅ m) j − (19.2 N ⋅ m) j − (12 N ⋅ m)k

Force-couple system at A is R = −(420 N)i − (50 N) j − (250 N)k M RA = (30.8 N ⋅ m) j − (220 N ⋅ m)k

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PROBLEM 3.120 Two 150-mm-diameter pulleys are mounted on line shaft AD. The belts at B and C lie in vertical planes parallel to the yz plane. Replace the belt forces shown with an equivalent forcecouple system at A.

SOLUTION Equivalent force-couple at each pulley Pulley B

R B = (145 N)(− cos 20° j + sin 20°k ) − 215 Nj = − (351.26 N) j + (49.593 N)k M B = − (215 N − 145 N)(0.075 m)i = − (5.25 N ⋅ m)i

Pulley C

R C = (155 N + 240 N)(− sin10° j − cos10°k ) = − (68.591 N) j − (389.00 N)k M C = (240 N − 155 N)(0.075 m)i = (6.3750 N ⋅ m)i

Then

R = R B + R C = − (419.85 N) j − (339.41)k

or R = (420 N) j − (339 N)k

M A = M B + M C + rB/ A × R B + rC/ A × R C i j k 0 0 N⋅m = − (5.25 N ⋅ m)i + (6.3750 N ⋅ m)i + 0.225 0 −351.26 49.593 i j k + 0.45 0 0 N⋅m 0 −68.591 −389.00 = (1.12500 N ⋅ m)i + (163.892 N ⋅ m) j − (109.899 N ⋅ m)k

or M A = (1.125 N ⋅ m)i + (163.9 N ⋅ m) j − (109.9 N ⋅ m)k

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!

PROBLEM 3.121 While using a pencil sharpener, a student applies the forces and couple shown. (a) Determine the forces exerted at B and C knowing that these forces and the couple are equivalent to a force-couple system at A consisting of the force R = (2.6 lb)i + Ry j − (0.7 lb)k and the couple M RA = M x i + (1.0 lb · ft)j − (0.72 lb · ft)k. (b) Find the corresponding values of Ry and M x .

SOLUTION (a)

From the statement of the problem, equivalence requires ΣF : B + C = R

or

ΣFx : Bx + C x = 2.6 lb

(1)

ΣFy : − C y = R y

(2)

ΣFz : − C z = −0.7 lb or C z = 0.7 lb

and

ΣM A : (rB/A × B + M B ) + rC/A × C = M AR

or

1.75 ! ΣM x : (1 lb ⋅ ft) + " ft # (C y ) = M x $ 12 %

(3)

3.75 ! 1.75 ! 3.5 ! ΣM y : " ft # ( Bx ) + " ft # (C x ) + " ft # (0.7 lb) = 1 lb ⋅ ft $ 12 % $ 12 % $ 12 %

or Using Eq. (1)

3.75Bx + 1.75C x = 9.55 3.75Bx + 1.75(2.6 Bx ) = 9.55

or

Bx = 2.5 lb

and

C x = 0.1 lb 3.5 ! ΣM z : − " ft # (C y ) = −0.72 lb ⋅ ft $ 12 % C y = 2.4686 lb

or

B = (2.5 lb)i C = (0.1000 lb)i − (2.47 lb) j − (0.700 lb)k

(b)

Eq. (2) & Using Eq. (3)

Ry = −2.47 lb 1.75 ! 1+ " # (2.4686) = M x $ 12 %

or M x = 1.360 lb ⋅ ft PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 284

PROBLEM 3.122 A mechanic uses a crowfoot wrench to loosen a bolt at C. The mechanic holds the socket wrench handle at Points A and B and applies forces at these points. Knowing that these forces are equivalent to a force-couple system at C consisting of the force C = (8 lb)i + (4 lb)k and the couple M C = (360 lb · in.)i, determine the forces applied at A and at B when Az = 2 lb.

SOLUTION We have

ΣF :

or

Fx : Ax + Bx = 8 lb

A+B=C

Bx = −( Ax + 8 lb)

(1)

ΣFy : Ay + By = 0

Ay = − By

or

(2)

ΣFz : 2 lb + Bz = 4 lb

Bz = 2 lb

or

(3)

ΣM C : rB/C × B + rA/C × A = M C

We have

i

j

8 Bx

0 By

k

i

j

2 + 8 2 Ax

0 Ay

k

8 lb ⋅ in. = (360 lb ⋅ in.)i 2

(2 By − 8 Ay )i + (2 Bx − 16 + 8 Ax − 16) j

or

+ (8By + 8 Ay )k = (360 lb ⋅ in.)i

From

i-coefficient j-coefficient k-coefficient

2 By − 8 Ay = 360 lb ⋅ in. −2 Bx + 8 Ax = 32 lb ⋅ in.

8 By + 8 Ay = 0

(4) (5) (6)

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PROBLEM 3.122 (Continued)

From Equations (2) and (4):

2 By − 8(− By ) = 360 By = 36 lb

From Equations (1) and (5):

Ay = 36 lb

2(− Ax − 8) + 8 Ax = 32 Ax = 1.6 lb

From Equation (1):

Bx = −(1.6 + 8) = −9.6 lb A = (1.600 lb)i − (36.0 lb) j + (2.00 lb)k B = −(9.60 lb)i + (36.0 lb) j + (2.00 lb)k

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PROBLEM 3.123 As an adjustable brace BC is used to bring a wall into plumb, the force-couple system shown is exerted on the wall. Replace this force-couple system with an equivalent force-couple system at A if R = 21.2 lb and M = 13.25 lb · ft.

SOLUTION

We have where

ΣF : R = R A = RλBC BC

=

RA =

or We have where

(42 in.)i − (96 in.) j − (16 in.)k 106 in.

21.2 lb (42i − 96 j − 16k ) 106

R A = (8.40 lb)i − (19.20 lb) j − (3.20 lb)k ΣM A : rC/A × R + M = M A

rC/A = (42 in.)i + (48 in.)k =

1 (42i + 48k )ft 12

= (3.5 ft)i + (4.0 ft)k R = (8.40 lb)i − (19.50 lb) j − (3.20 lb)k M = −λBC M −42i + 96 j + 16k (13.25 lb ⋅ ft) 106 = −(5.25 lb ⋅ ft)i + (12 lb ⋅ ft) j + (2lb ⋅ ft)k =

Then

i j k 3.5 0 4.0 lb ⋅ ft + (−5.25i + 12 j + 2k )lb ⋅ ft = M A 8.40 −19.20 −3.20 M A = (71.55 lb ⋅ ft)i + (56.80 lb ⋅ ft)j − (65.20 lb ⋅ ft)k

or M A = (71.6 lb ⋅ ft)i + (56.8 lb ⋅ ft)j − (65.2 lb ⋅ ft)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 287

PROBLEM 3.124 A mechanic replaces a car’s exhaust system by firmly clamping the catalytic converter FG to its mounting brackets H and I and then loosely assembling the mufflers and the exhaust pipes. To position the tailpipe AB, he pushes in and up at A while pulling down at B. (a) Replace the given force system with an equivalent force-couple system at D. (b) Determine whether pipe CD tends to rotate clockwise or counterclockwise relative to muffler DE, as viewed by the mechanic.

SOLUTION (a)

Equivalence requires ΣF : R = A + B = (100 N)(cos 30° j − sin 30° k ) − (115 N) j = −(28.4 N) j − (50 N)k

and where

ΣM D : M D = rA/D × FA + rB/D × FB rA/D = −(0.48 m)i − (0.225 m) j + (1.12 m)k rB/D = −(0.38 m)i + (0.82 m)k

Then i j k i j k M D = 100 −0.48 −0.225 1.12 + 115 −0.38 0 0.82 0 cos 30° − sin 30° 0 −1 0 = 100[(0.225sin 30° − 1.12cos 30°)i + (−0.48sin 30°) j + (−0.48cos 30°)k ] + 115[(0.82)i + (0.38)k ] = 8.56i − 24.0 j + 2.13k

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PROBLEM 3.124 (Continued) The equivalent force-couple system at D is R = −(28.4 N) j − (50.0 N)k M D = (8.56 N ⋅ m)i − (24.0 N ⋅ m) j + (2.13 N ⋅ m)k

(b)

Since ( M D ) z is positive, pipe CD will tend to rotate counterclockwise relative to muffler DE.

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PROBLEM 3.125 For the exhaust system of Problem 3.124, (a) replace the given force system with an equivalent force-couple system at F, where the exhaust pipe is connected to the catalytic converter, (b) determine whether pipe EF tends to rotate clockwise or counterclockwise, as viewed by the mechanic.

SOLUTION (a)

Equivalence requires ΣF : R = A + B = (100 N)(cos 30° j − sin 30° k ) − (115 N) j = −(28.4 N) j − (50 N)k

and where

M F : M F = rA/F × A + rB/F × B rA/F = −(0.48 m)i − (0.345 m) j + (2.10 m)k rB/F = −(0.38 m)i − (0.12 m) j + (1.80 m)k

Then i j k i j k M F = 100 −0.48 −0.345 2.10 + 115 −0.38 0.12 1.80 0 cos 30° − sin 30° 0 −1 0 M F = 100[(0.345 sin 30° − 2.10 cos 30°)i + (−0.48 sin 30°) j + (−0.48 cos 30°)k ] + 115[(1.80)i + (0.38)k ] = 42.4i − 24.0 j + 2.13k

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PROBLEM 3.125 (Continued) The equivalent force-couple system at F is R = −(28.4 N) j − (50 N)k M F = (42.4 N ⋅ m)i − (24.0 N ⋅ m) j + (2.13 N ⋅ m)k

(b)

Since ( M F ) z is positive, pipe EF will tend to rotate counterclockwise relative to the mechanic.

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PROBLEM 3.126 The head-and-motor assembly of a radial drill press was originally positioned with arm AB parallel to the z axis and the axis of the chuck and bit parallel to the y axis. The assembly was then rotated 25° about the y axis and 20° about the centerline of the horizontal arm AB, bringing it into the position shown. The drilling process was started by switching on the motor and rotating the handle to bring the bit into contact with the workpiece. Replace the force and couple exerted by the drill press with an equivalent forcecouple system at the center O of the base of the vertical column.

SOLUTION We have

R = F = (11 lb)[( sin 20° cos 25°)]i − (cos 20°) j − (sin 20° sin 25°)k ] = (3.4097 lb)i − (10.3366 lb)j − (1.58998 lb)k

or

R = (3.41 lb)i − (10.34 lb) j − (1.590 lb)k

We have

M O = rB/O × F × M C

where rB/O = [(14 in.) sin 25°]i + (15 in.) j + [(14 in.) cos 25°]k = (5.9167 in.)i + (15 in.) j + (12.6883 in.)k M C = (90 lb ⋅ in.)[(sin 20° cos 25°)i − (cos 20°) j − (sin 20° sin 25°)k ] = (27.898 lb ⋅ in.)i − (84.572 lb ⋅ in.) j − (13.0090 lb ⋅ in.)k i j k 15 12.6883 lb ⋅ in. M O = 5.9167 3.4097 −10.3366 1.58998 + (27.898 − 84.572 − 13.0090) lb ⋅ in. = (135.202 lb ⋅ in.)i − (31.901 lb ⋅ in.) j − (125.313 lb ⋅ in.)k

or M O = (135.2 lb ⋅ in.)i − (31.9 lb ⋅ in.) j − (125.3 lb ⋅ in.)k

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PROBLEM 3.127 Three children are standing on a 5 × 5-m raft. If the weights of the children at Points A, B, and C are 375 N, 260 N, and 400 N, respectively, determine the magnitude and the point of application of the resultant of the three weights.

SOLUTION

We have

ΣF : FA + FB + FC = R −(375 N) j − (260 N) j − (400 N) j = R −(1035 N) j = R or

We have

R = 1035 N

ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) = R ( z D ) (375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m) = (1035 N)(z D ) z D = 3.0483 m

We have

or

z D = 3.05 m

ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) = R ( xD ) 375 N(1 m) + (260 N)(1.5 m) + (400 N)(4.75 m) = (1035 N)( xD ) xD = 2.5749 m

or

xD = 2.57 m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 293

PROBLEM 3.128 Three children are standing on a 5 × 5-m raft. The weights of the children at Points A, B, and C are 375 N, 260 N, and 400 N, respectively. If a fourth child of weight 425 N climbs onto the raft, determine where she should stand if the other children remain in the positions shown and the line of action of the resultant of the four weights is to pass through the center of the raft.

SOLUTION

We have

ΣF : FA + FB + FC = R −(375 N) j − (260 N) j − (400 N) j − (425 N) j = R R = −(1460 N) j

We have

ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R( z H ) (375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m) + (425 N)(z D ) = (1460 N)(2.5 m) z D = 1.16471 m

We have

or

z D = 1.165 m

ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xH ) (375 N)(1 m) + (260 N)(1.5 m) + (400 N)(4.75 m) + (425 N)(xD ) = (1460 N)(2.5 m) xD = 2.3235 m

or

xD = 2.32 m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 294

PROBLEM 3.129 Four signs are mounted on a frame spanning a highway, and the magnitudes of the horizontal wind forces acting on the signs are as shown. Determine the magnitude and the point of application of the resultant of the four wind forces when a = 1 ft and b = 12 ft.

SOLUTION We have

Assume that the resultant R is applied at Point P whose coordinates are (x, y, 0). Equivalence then requires ΣFz : − 105 − 90 − 160 − 50 = − R

or R = 405 lb ΣM x : (5 ft)(105 lb) − (1 ft)(90 lb) + (3 ft)(160 lb) + (5.5 ft)(50 lb) = − y (405 lb)

or

y = −2.94 ft

ΣM y : (5.5 ft)(105 lb) + (12 ft)(90 lb) + (14.5 ft)(160 lb) + (22.5 ft)(50 lb) = − x(405 lb) x = 12.60 ft

or

R acts 12.60 ft to the right of member AB and 2.94 ft below member BC.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 295

PROBLEM 3.130 Four signs are mounted on a frame spanning a highway, and the magnitudes of the horizontal wind forces acting on the signs are as shown. Determine a and b so that the point of application of the resultant of the four forces is at G.

SOLUTION Since R acts at G, equivalence then requires that ΣM G of the applied system of forces also be zero. Then at

G : ΣM x : − (a + 3) ft × (90 lb) + (2 ft)(105 lb)

+ (2.5 ft)(50 lb) = 0

or a = 0.722 ft !

ΣM y : − (9 ft)(105 ft) − (14.5 − b) ft × (90 lb)

!

+ (8 ft)(50 lb) = 0

or b = 20.6 ft

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PROBLEM 3.131* A group of students loads a 2 × 3.3-m flatbed trailer with two 0.66 × 0.66 × 0.66-m boxes and one 0.66 × 0.66 × 1.2-m box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second 0.66 × 0.66 × 1.2-m box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.)

SOLUTION

For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the intersection with the center line of the trailer, the added 0.66 × 0.66 × 1.2-m box should be placed adjacent to one of the edges of the trailer with the 0.66 × 0.66-m side on the bottom. The edges to be considered are based on the location of the resultant for the three given weights. We have

ΣF : − (224 N) j − (392 N) j − (176 N) j = R R = −(792 N) j

We have

ΣM z : − (224 N)(0.33 m) − (392 N)(1.67 m) − (176 N)(1.67 m) = ( −792 N)( x) xR = 1.29101 m

We have

ΣM x : (224 N)(0.33 m) + (392 N)(0.6 m) + (176 N)(2.0 m) = (792 N)( z ) z R = 0.83475 m

From the statement of the problem, it is known that the resultant of R from the original loading and the lightest load W passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0. Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G as possible without the box overhanging the trailer. These two requirements imply (0.33 m # x # 1 m)(1.5 m # z # 2.97 m)

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PROBLEM 3.131* (Continued) xL = 0.33 m

With at

G : ΣM Z : (1 − 0.33) m × WL − (1.29101 − 1) m × (792 N) = 0 WL = 344.00 N

or

Now must check if this is physically possible, at

G : ΣM x : ( Z L − 1.5)m × 344 N) − (1.5 − 0.83475)m × (792 N) = 0 Z L = 3.032 m

or which is not acceptable.

Z L = 2.97 m:

With at

G : ΣM x : (2.97 − 1.5)m × WL − (1.5 − 0.83475)m × (792 N) = 0 WL = 358.42 N

or Now check if this is physically possible at

G : ΣM z : (1 − X L )m × (358.42 N) − (1.29101 − 1)m × (792 N) = 0

or

X L = 0.357 m ok! WL = 358 N

The minimum weight of the fourth box is

And it is placed on end (A 0.66 × 0.66-m side down) along side AB with the center of the box 0.357 m from side AD.

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PROBLEM 3.132* Solve Problem 3.131 if the students want to place as much weight as possible in the fourth box and at least one side of the box must coincide with a side of the trailer. PROBLEM 3.131* A group of students loads a 2 × 3.3-m flatbed trailer with two 0.66 × 0.66 × 0.66-m boxes and one 0.66 × 0.66 × 1.2-m box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second 0.66 × 0.66 × 1.2-m box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.)

SOLUTION First replace the three known loads with a single equivalent force R applied at coordinate ( X R , 0, Z R ) Equivalence requires ΣFy : − 224 − 392 − 176 = − R

or

R = 792 N ΣM x : (0.33 m)(224 N) + (0.6 m)(392 N) + (2 m)(176 N) = z R (792 N)

or

z R = 0.83475 m

ΣM z : − (0.33 m)(224 N) − (1.67 m)(392 N) − (1.67 m)(176 N) = xR (792 N)

or

xR = 1.29101 m

From the statement of the problem, it is known that the resultant of R and the heaviest loads WH passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0

Further, since WH is to be as large as possible, the fourth box should be placed as close to G as possible while keeping one of the sides of the box coincident with a side of the trailer. Thus, the two limiting cases are xH = 0.6 m or

z H = 2.7 m

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PROBLEM 3.132* (Continued) Now consider these two possibilities With xH = 0.6 m: at

G : ΣM z : (1 − 0.6)m × WH − (1.29101 − 1)m × (792 N) = 0 WH = 576.20 N

or Checking if this is physically possible at or

G : ΣM x : ( z H − 1.5)m × (576.20 N) − (1.5 − 0.83475)m × (792 N) = 0 z H = 2.414 m

which is acceptable. With z H = 2.7 m at or

G : ΣM x : (2.7 − 1.5) WH − (1.5 − 0.83475)m × (792 N) = 0 WH = 439 N

Since this is less than the first case, the maximum weight of the fourth box is WH = 576 N

and it is placed with a 0.66 × 1.2-m side down, a 0.66-m edge along side AD, and the center 2.41 m from side DC.

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PROBLEM 3.133 Three forces of the same magnitude P act on a cube of side a as shown. Replace the three forces by an equivalent wrench and determine (a) the magnitude and direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench.

SOLUTION Force-couple system at O: R = Pi + Pj + Pk = P (i + j + k ) M OR = aj × Pi + ak × Pj + ai × Pk = − Pak − Pai − Paj M OR = − Pa (i + j + k )

Since R and M OR have the same direction, they form a wrench with M1 = M OR . Thus, the axis of the wrench is the diagonal OA. We note that cos θ x = cos θ y = cos θ z =

a a 3

=

1 3

R = P 3 θ x = θ y = θ z = 54.7° M1 = M OR = − Pa 3 Pitch = p =

(a) (b) (c)

M 1 − Pa 3 = = −a R P 3

R = P 3 θ x = θ y = θ z = 54.7°

–a Axis of the wrench is diagonal OA

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PROBLEM 3.134* A piece of sheet metal is bent into the shape shown and is acted upon by three forces. If the forces have the same magnitude P, replace them with an equivalent wrench and determine (a) the magnitude and the direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench.

SOLUTION

(

)

First reduce the given forces to an equivalent force-couple system R, M OR at the origin. We have ΣF : − Pj + Pj + Pk = R R = Pk

or

"5 # ! ΣM O : − (aP) j + $ −( aP)i + & aP ' k % = M OR (2 ) + * 5 # " M OR = aP & −i − j + k ' 2 ) (

or (a)

Then for the wrench R=P

and

axis

=

R =k R

cos θ x = 0 cos θ y = 0 cos θ z = 1

or (b)

θ x = 90° θ y = 90° θ z = 0°

Now M1 = λ axis ⋅ M OR 5 # " = k ⋅ aP & −i − j + k ' 2 ) ( 5 = aP 2

Then

P=

M1 25 aP = R P

or P =

5 a 2

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PROBLEM 3.134* (Continued) (c)

The components of the wrench are (R , M1 ), where M1 = M1 λ axis , and the axis of the wrench is assumed to intersect the xy plane at Point Q whose coordinates are (x, y, 0). Thus require M z = rQ × R R

Where

M z = M O × M1

Then 5 # 5 " aP & −i − j + k ' − aPk = ( xi + yj) + Pk 2 ) 2 (

Equating coefficients i : − aP = yP

or

j: − aP = − xP or

y = −a x=a

The axis of the wrench is parallel to the z axis and intersects the xy plane at

x = a, y = −a.

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PROBLEM 3.135* The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLUTION First, reduce the given force system to a force-couple system. We have

ΣF : − (20 N)i − (15 N) j = R

We have

ΣM O : Σ(rO × F ) + ΣM C = M OR

R = 25 N

M OR = −20 N(0.1 m)j − (4 N ⋅ m)i − (1 N ⋅ m)j

= −(4 N ⋅ m)i − (3 N ⋅ m) j R = −(20.0 N)i − (15.0 N)j

(a) (b)

We have

R R = (−0.8i − 0.6 j) ⋅ [−(4 N ⋅ m)]i − (3 N ⋅ m)j]

M1 = λR ⋅ M OR

λ=

= 5 N⋅m

Pitch

p=

M1 5 N ⋅ m = = 0.200 m R 25 N

or p = 0.200 m (c)

From above note that M1 = M OR

Therefore, the axis of the wrench goes through the origin. The line of action of the wrench lies in the xy plane with a slope of y=

3 x 4

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PROBLEM 3.136* The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLUTION

First, reduce the given force system to a force-couple at the origin. We have

ΣF : − (10 lb) j − (11 lb) j = R R = − (21 lb) j

We have

ΣM O : Σ(rO × F ) + ΣM C = M OR M OR

i j k i j k = 0 0 20 lb ⋅ in. + 0 0 −15 lb ⋅ in. − (12 lb ⋅ in) j 0 −10 0 0 −11 0 = (35 lb ⋅ in.)i − (12 lb ⋅ in.) j

R = − (21 lb) j

(a) (b)

We have

or R = − (21.0 lb) j

R R = (− j) ⋅ [(35 lb ⋅ in.)i − (12 lb ⋅ in.) j]

M1 =

R

⋅ M OR

R

=

= 12 lb ⋅ in. and M1 = −(12 lb ⋅ in.) j

and pitch

p=

M 1 12 lb ⋅ in. = = 0.57143 in. R 21 lb

or

p = 0.571 in.

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PROBLEM 3.136* (Continued)

(c)

We have

M OR = M1 + M 2 M 2 = M OR − M1 = (35 lb ⋅ in.)i

Require

M 2 = rQ/O × R (35 lb ⋅ in.)i = ( xi + zk ) × [ −(21 lb) j] 35i = −(21x)k + (21z )i

From i:

From k:

35 = 21z z = 1.66667 in. 0 = − 21x z=0

The axis of the wrench is parallel to the y axis and intersects the xz plane at x = 0, z = 1.667 in.

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PROBLEM 3.137* Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLUTION

First, reduce the given force system to a force-couple at the origin. We have

ΣF : − (84 N) j − (80 N)k = R

R = 116 N

and

ΣM O : Σ(rO × F) + ΣM C = M OR i j k i j k 0.6 0 .1 + 0.4 0.3 0 + ( −30 j − 32k )N ⋅ m = M OR 0 84 0 0 0 80 M OR = − (15.6 N ⋅ m)i + (2 N ⋅ m) j − (82.4 N ⋅ m)k R = − (84.0 N) j − (80.0 N)k

(a) (b)

We have

M1 =

R

⋅ M OR

R

=

R R

−84 j − 80k ⋅ [− (15.6 N ⋅ m)i + (2 N ⋅ m) j − (82.4 N ⋅ m)k ] 116 = 55.379 N ⋅ m =−

and Then pitch

M1 = M1λR = − (40.102 N ⋅ m) j − (38.192 N ⋅ m)k p=

M 1 55.379 N ⋅ m = = 0.47741 m R 116 N

or p = 0.477 m

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PROBLEM 3.137* (Continued)

(c)

We have

M OR = M1 + M 2 M 2 = M OR − M1 = [(−15.6i + 2 j − 82.4k ) − (40.102 j − 38.192k )] N ⋅ m = − (15.6 N ⋅ m)i + (42.102 N ⋅ m) j − (44.208 N ⋅ m)k

Require

M 2 = rQ/O × R (−15.6i + 42.102 j − 44.208k ) = ( xi + zk ) × (84 j − 80k ) = (84 z )i + (80 x) j − (84 x)k

From i:

or From k:

or

−15.6 = 84 z z = − 0.185714 m z = − 0.1857 m

−44.208 = −84 x x = 0.52629 m x = 0.526 m

The axis of the wrench intersects the xz plane at x = 0.526 m

y = 0 z = − 0.1857 m

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PROBLEM 3.138* Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLUTION

First, reduce the given force system to a force-couple at the origin at B. (a)

We have

15 # " 8 ΣF : − (26.4 lb)k − (17 lb) & i + j ' = R 17 17 ( ) R = − (8.00 lb)i − (15.00 lb) j − (26.4 lb)k

and We have

R = 31.4 lb

ΣM B : rA/B × FA + M A + M B = M RB

M RB

i j k 15 # " 8 0 − 220k − 238 & i + j ' = 264i − 220k − 14(8i + 15 j) = 0 −10 ( 17 17 ) 0 0 − 26.4

M RB = (152 lb ⋅ in.)i − (210 lb ⋅ in.)j − (220 lb ⋅ in.)k

(b)

We have

R R −8.00i − 15.00 j − 26.4k = ⋅ [(152 lb ⋅ in.)i − (210 lb ⋅ in.) j − (220 lb ⋅ in.)k ] 31.4 = 246.56 lb ⋅ in.

M1 =

R

⋅ M OR

R

=

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PROBLEM 3.138* (Continued)

and Then pitch (c)

We have

M1 = M1λR = − (62.818 lb ⋅ in.)i − (117.783 lb ⋅ in.) j − (207.30 lb ⋅ in.)k p=

M 1 246.56 lb ⋅ in. = = 7.8522 in. 31.4 lb R

or p = 7.85 in.

M RB = M1 + M 2 M 2 = M RB − M1 = (152i − 210 j − 220k ) − ( − 62.818i − 117.783j − 207.30k ) = (214.82 lb ⋅ in.)i − (92.217 lb ⋅ in.) j − (12.7000 lb ⋅ in.)k

Require

M 2 = rQ/B × R i j k 214.82i − 92.217 j − 12.7000k = x 0 z −8 −15 −26.4 = (15 z )i − (8 z ) j + (26.4 x) j − (15 x)k

From i:

214.82 = 15 z

From k:

−12.7000 = −15 x

The axis of the wrench intersects the xz plane at

z = 14.3213 in. x = 0.84667 in. x = 0.847 in. y = 0 z = 14.32 in.

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PROBLEM 3.139* Two ropes attached at A and B are used to move the trunk of a fallen tree. Replace the forces exerted by the ropes with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the yz plane.

SOLUTION (a)

(

)

First replace the given forces with an equivalent force-couple system R, M OR at the origin. We have d AC = (6) 2 + (2) 2 + (9) 2 = 11 m d BD = (14) 2 + (2)2 + (5) 2 = 15 m

Then 1650 N = (6i + 2 j + 9k ) 11 = (900 N)i + (300 N) j + (1350 N)k

TAC =

and 1500 N = (14i + 2 j + 5k ) 15 = (1400 N)i + (200 N) j + (500 N)k

TBD =

Equivalence then requires ΣF : R = TAC + TBD = (900i + 300 j + 1350k ) +(1400i + 200 j + 500k ) = (2300 N)i + (500 N) j + (1850 N)k ΣM O : M OR = rA × TAC + rB × TBD = (12 m)k × [(900 N)i + (300 N)j + (1350 N)k ] + (9 m)i × [(1400 N)i + (200 N)j + (500 N)k ] = −(3600)i + (10800 − 4500) j + (1800)k = −(3600 N ⋅ m)i + (6300 N ⋅ m)j + (1800 N ⋅ m)k

The components of the wrench are (R , M1 ), where R = (2300 N)i + (500 N) j + (1850 N)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 311

PROBLEM 3.139* (Continued) ! (b)

We have R = 100 (23)2 + (5) 2 + (18.5) 2 = 2993.7 N

Let λ axis =

Then

1 R (23i + 5 j + 18.5k ) = R 29.937

M1 = λ axis ⋅ M OR 1 (23i + 5 j + 18.5k ) ⋅ (−3600i + 6300 j + 1800k ) 29.937 1 [(23)( −36) + (5)(63) + (18.5)(18)] = 0.29937 = −601.26 N ⋅ m =

Finally

P=

M1 −601.26 N ⋅ m = R 2993.7 N

or P = − 0.201 m (c)

We have

M1 = M 1

axis

= (−601.26 N ⋅ m) ×

1 (23i + 5 j + 18.5k ) 29.937

or

M1 = −(461.93 N ⋅ m)i − (100.421 N ⋅ m) j − (371.56 N ⋅ m)k

Now

M 2 = M OR − M1 = (−3600i + 6300 j + 1800k ) − ( −461.93i − 100.421j − 371.56k ) = − (3138.1 N ⋅ m)i + (6400.4 N ⋅ m)j + (2171.6 N ⋅ m)k

For equivalence

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PROBLEM 3.139* (Continued)

Thus require

M 2 = rP × R

r = ( yj + zk )

Substituting i j k −3138.1i + 6400.4 j + 2171.6k = 0 y z 2300 500 1850

Equating coefficients j : 6400.4 = 2300 z

or

k : 2171.6 = −2300 y or

The axis of the wrench intersects the yz plane at

z = 2.78 m y = − 0.944 m

y = −0.944 m

z = 2.78 m

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PROBLEM 3.140* A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P, replace the forces exerted on the pole with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLUTION

(a)

First reduce the given force system to a force-couple at the origin. We have

ΣF : P

BA

+P

DC

+P

DE

=R

3 # "3 4 # " −9 4 12 # ! "4 R = P $& j − k ' + & i − j ' + & i − j + k ' % 5 ) (5 5 ) ( 25 5 25 ) + *( 5 R=

R=

We have

3P (2i − 20 j − k ) 25

3P 27 5 (2) 2 + (20) 2 + (1)2 = P 25 25 ΣM : Σ(rO × P) = M OR

3P # 4P # 4P 12 P # " −4 P " 3P " −9 P j− k ' + (20a) j × & i− j ' + (20a) j × & i− j+ k ' = M OR (24a) j × & 5 5 5 5 25 5 25 ( ) ( ) ( ) M OR =

24 Pa ( −i − k ) 5

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PROBLEM 3.140* (Continued)

(b)

We have

M1 = λR ⋅ M OR

where

λR =

Then

M1 =

and pitch

R 3P 25 1 = = (2i − 20 j − k ) (2i − 20 j − k ) R 25 27 5 P 9 5

p=

1 9 5

(2i − 20 j − k ) ⋅

M 1 −8Pa " 25 # −8a = & '= R 15 5 ( 27 5 P ) 81

M1 = M1λR =

(c)

Then

M 2 = M OR − M1 =

24 Pa −8 Pa (−i − k ) = 5 15 5

or p = − 0.0988a

−8 Pa " 1 # 8Pa (−2i + 20 j + k ) & ' (2i − 20 j − k ) = 675 15 5 ( 9 5 )

24 Pa 8Pa 8 Pa ( −i − k ) − (−2i + 20 j + k ) = (−430i − 20 j − 406k ) 5 675 675 M 2 = rQ/O × R

Require

" 8Pa # " 3P # & 675 ' (−403i − 20 j − 406k ) = ( xi + zk ) × & 25 ' (2i − 20 j − k ) ( ) ( ) " 3P # =& ' [20 zi + ( x + 2 z ) j − 20 xk ] ( 25 )

From i:

8(− 403)

Pa " 3P # = 20 z & ' z = −1.99012a 675 ( 25 )

From k:

8(−406)

Pa " 3P # = −20 x & ' x = 2.0049a 675 ( 25 )

The axis of the wrench intersects the xz plane at x = 2.00a, z = −1.990a

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PROBLEM 3.141* Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz plane.

SOLUTION First, reduce the given force system to a force-couple at the origin. We have

ΣF : FA + FG = R (40 mm)i + (60 mm) j − (120 mm)k ! R = (50 N)k + 70 N $ % 140 mm * + = (20 N)i + (30 N) j − (10 N)k

and We have

R = 37.417 N ΣM O : Σ(rO × F) + ΣM C = M OR M OR = [(0.12 m) j × (50 N)k ] + {(0.16 m)i × [(20 N)i + (30 N) j − (60 N)k ]} + (10 N ⋅ m) $ *

(160 mm)i − (120 mm) j ! % 200 mm +

(40 mm)i − (120 mm) j + (60 mm)k ! + (14 N ⋅ m) $ % 140 mm * + R M 0 = (18 N ⋅ m)i − (8.4 N ⋅ m) j + (10.8 N ⋅ m)k

To be able to reduce the original forces and couples to a single equivalent force, R and M must be perpendicular. Thus, R ⋅ M = 0. Substituting ?

(20i + 30 j − 10k ) ⋅ (18i − 8.4 j + 10.8k ) = 0 ?

or

(20)(18) + (30)(−8.4) + (−10)(10.8) = 0

or

0=0

R and M are perpendicular so that the given system can be reduced to the single equivalent force R = (20.0 N)i + (30.0 N) j − (10.00 N)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 316

PROBLEM 3.141* (Continued) ! Then for equivalence

Thus require

M OR = rp × R rp = yj + zk

Substituting i j k 18i − 8.4 j + 10.8k = 0 y z 20 30 −10

Equating coefficients j: − 8.4 = 20 z k:

or

z = −0.42 m

10.8 = −20 y or

y = −0.54 m

The line of action of R intersects the yz plane at

x=0

y = −0.540 m z = −0.420 m

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PROBLEM 3.142* Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz plane.

SOLUTION First determine the resultant of the forces at D. We have d DA = (−12) 2 + (9) 2 + (8) 2 = 17 in. d ED = (−6) 2 + (0)2 + (−8)2 = 10 in.

Then 34 lb = (−12i + 9 j + 8k ) 17 = −(24 lb)i + (18 lb) j + (16 lb)k

FDA =

and 30 lb = (−6i − 8k ) 10 = −(18 lb)i − (24 lb)k

FED =

Then ΣF : R = FDA + FED = (−24i + 18 j + 16k + ( −18i − 24k ) = −(42 lb)i + (18 lb)j − (8 lb)k

For the applied couple d AK = ( −6) 2 + (−6) 2 + (18) 2 = 6 11 in.

Then M=

160 lb ⋅ in.

( −6i − 6 j + 18k ) 6 11 160 = [−(1 lb ⋅ in.)i − (1 lb ⋅ in.)j + (3 lb ⋅ in.)k ] 11

To be able to reduce the original forces and couple to a single equivalent force, R and M must be perpendicular. Thus ?

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PROBLEM 3.142* (Continued)

Substituting (−42i + 18 j − 8k ) ⋅ 160

or

11

160 11

?

(−i − j + 3k ) = 0 ?

[(−42)(−1) + (18)(−1) + (−8)(3)] = 0

0 =0

or

R and M are perpendicular so that the given system can be reduced to the single equivalent force R = −(42.0 lb)i + (18.00 lb) j − (8.00 lb)k

Then for equivalence

M = rP/D × R

Thus require where

rP/D = −(12 in.)i + [( y − 3)in.] j + ( z in.)k

Substituting i j k ( −i − j + 3k ) = −12 ( y − 3) z 11 −42 −8 18 = [( y − 3)( −8) − ( z )(18)]i

160

+ [( z )(−42) − (−12)(−8)]j + [( −12)(18) − ( y − 3)(−42)]k

Equating coefficients j: − k:

160

= − 42 z − 96 or 11 480 = −216 + 42( y − 3) or 11

The line of action of R intersects the yz plane at

x=0

z = −1.137 in. y = 11.59 in.

y = 11.59 in. z = −1.137 in.

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PROBLEM 3.143* Replace the wrench shown with an equivalent system consisting of two forces perpendicular to the y axis and applied respectively at A and B.

SOLUTION Express the forces at A and B as A = Ax i + Az k B = Bx i + Bz k

Then, for equivalence to the given force system ΣFx : Ax + Bx = 0

(1)

ΣFz : Az + Bz = R

(2)

ΣM x : Az ( a) + Bz ( a + b) = 0

(3)

ΣM z : − Ax (a) − Bx (a + b) = M

(4)

Bx = − Ax

From Equation (1), Substitute into Equation (4)

− Ax ( a) + Ax ( a + b) = M M M and Bx = − Ax = b b

From Equation (2),

Bz = R − Az

and Equation (3),

Az a + ( R − Az )(a + b) = 0

" a# Az = R &1 + ' ( b)

and

a# " Bz = R − R &1 + ' ( b) a Bz = − R b

"M A=& ( b

Then

a# # " ' i + R &1 + b ' k ) ( )

"M B = −& ( b

# "a # 'i − & b R'k ) ( )

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PROBLEM 3.144* Show that, in general, a wrench can be replaced with two forces chosen in such a way that one force passes through a given point while the other force lies in a given plane.

SOLUTION

First, choose a coordinate system so that the xy plane coincides with the given plane. Also, position the coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a. Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the scalar components of R and M are known relative to the shown coordinate system. A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the given Point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B. The known components of the wrench can be expressed as R = Rx i + Ry j + Rz k

and M = M x i + M y j + M z k

while the unknown forces A and B can be expressed as A = Ax i + Ay j + Az k and B = Bx i + Bz k

Since the position vector of Point P is given, it follows that the scalar components (x, y, z) of the position vector rP are also known. Then, for equivalence of the two systems ΣFx : Rx = Ax + Bx

(1)

ΣFy : Ry = Ay

(2)

ΣFz : Rz = Az + Bz

(3)

ΣM x : M x = yAz − zAy

(4)

ΣM y : M y = zAx − xAz − bBz

(5)

ΣM z : M z = xAy − yAx

(6)

Based on the above six independent equations for the six unknowns ( Ax , Ay , Az , Bx , By , Bz , b), there exists a unique solution for A and B. Ay = Ry

From Equation (2)

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PROBLEM 3.144* (Continued)

Equation (6)

"1# Ax = & ' ( xRy − M z ) ( y)

Equation (1)

"1# Bx = Rx − & ' ( xRy − M z ) ( y)

Equation (4)

"1# Az = & ' ( M x + zRy ) ( y)

Equation (3)

"1# Bz = Rz − & ' ( M x + zRy ) ( y) b=

Equation (5)

( xM x + yM y + zM z ) ( M x − yRz + zRy )

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!

!

PROBLEM 3.145* Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given point.

SOLUTION

First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular coordinate system with the axis of the wrench while one of the other axes passes through the given point. See Figures a and b. We have

R = Rj and M = Mj

and are known.

The unknown forces A and B can be expressed as A = Ax i + Ay j + Az k

and B = Bx i + By j + Bz k

The distance a is known. It is assumed that force B intersects the xz plane at (x, 0, z). The for equivalence ΣFx :

0 = Ax + Bx

(1)

ΣFy :

R = Ay + By

(2)

ΣFz :

0 = Az + Bz

(3)

0 = − zBy

(4)

ΣM x :

ΣM y : M = − aAz − xBz + zBx ΣM z :

(5)

0 = aAy + xBy

(6)

Since A and B are made perpendicular, A ⋅ B = 0 or

There are eight unknowns:

Ax Bx + Ay B y + Az Bz = 0

(7)

Ax , Ay , Az , Bx , By , Bz , x, z

But only seven independent equations. Therefore, there exists an infinite number of solutions. 0 = − zBy

Next consider Equation (4): If By = 0, Equation (7) becomes

Ax Bx + Az Bz = 0 Ax2 + Az2 = 0

Using Equations (1) and (3) this equation becomes

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PROBLEM 3.145* (Continued) Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that By ≠ 0, so that from Equation (4), z = 0. To obtain one possible solution, arbitrarily let Ax = 0. (Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.) The defining equations then become 0 = Bx

(1)′

R = Ay + By

(2)

0 = Az + Bz

(3)

M = − aAz − xBz

(5)′

0 = aAy + xBy

(6)

Ay By + Az Bz = 0

(7)′

Then Equation (2) can be written

Ay = R − By

Equation (3) can be written

Bz = − Az

Equation (6) can be written

x=−

aAy By

Substituting into Equation (5)′, " R − By M = − aAz − & − a & By ( M Az = − By aR

or

# ' ( − Az ) ' )

(8)

Substituting into Equation (7)′, " M #" M # ( R − By ) By + & − By '& By ' = 0 ( aR )( aR )

or

By =

a 2 R3 a2 R2 + M 2

Then from Equations (2), (8), and (3) a2 R2 RM 2 = a2 R2 + M 2 a2 R2 + M 2 # M " a 2 R3 aR 2 M Az = − = − && 2 2 ' aR ( a R + M 2 ') a2 R2 + M 2

Ay = R −

Bz =

aR 2 M a2 R2 + M 2

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PROBLEM 3.145* (Continued)

In summary A=

RM ( Mj − aRk ) a R2 + M 2

B=

aR 2 (aRj + Mk ) a2 R2 + M 2

2

Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at a given point. Lastly, if R . 0 and M . 0, it follows from the equations found for A and B that Ay . 0 and By . 0. From Equation (6), x , 0 (assuming a . 0). Then, as a consequence of letting Ax = 0, force A lies in a plane parallel to the yz plane and to the right of the origin, while force B lies in a plane parallel to the yz plane but to the left to the origin, as shown in the figure below.

!

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PROBLEM 3.146* Show that a wrench can be replaced with two forces, one of which has a prescribed line of action.

SOLUTION

First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and another axis intersects the prescribed line of action (AA′). Note that it has been assumed that the line of action of force B intersects the xz plane at Point P(x, 0, z). Denoting the known direction of line AA′ by

λ A = λx i + λ y j + λz k it follows that force A can be expressed as A = Aλ A = A(λx i + λ y j + λz k )

Force B can be expressed as B = Bx i + By j + Bz k

Next, observe that since the axis of the wrench and the prescribed line of action AA′ are known, it follows that the distance a can be determined. In the following solution, it is assumed that a is known. Then, for equivalence ΣFx : 0 = Aλx + Bx

(1)

ΣFy : R = Aλ y + By

(2)

ΣFz : 0 = Aλz + Bz

(3)

ΣM x : 0 = − zBy

(4)

ΣM y : M = − aAλz + zBx − xBz

(5)

ΣM x : 0 = − aAλ y + xBy

(6)

Since there are six unknowns (A, Bx, By, Bz, x, z) and six independent equations, it will be possible to obtain a solution.

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PROBLEM 3.146* (Continued) Case 1: Let z = 0 to satisfy Equation (4) Now Equation (2) Equation (3) Equation (6)

Aλ y = R − By Bz = − Aλz x=−

aAλ y By

" a = −& & By (

# ' ( R − By ) ' )

Substitution into Equation (5) " a M = − aAλz − $ − & $* &( By A=−

! # ' ( R − By )(− Aλz ) % ' %+ )

1 "M #

B λz &( aR ') y

Substitution into Equation (2) R=− By =

Then

1 "M # B λ + By λz &( aR ') y y

λz aR 2 λz aR − λ y M

MR R = λz aR − λ y M λ − aR λ y z M λx MR Bx = − Aλx = λz aR − λ y M A=−

Bz = − Aλz =

λz MR λz aR − λ y M A=

In summary

B=

and

R

λz aR − λ y M

P λA aR λy − λz M

(λx Mi + λz aRj + λz M k )

" R # x = a &1 − ' & By ') ( " λz aR − λ y M = a $1 − R & & λ aR 2 $* z (

#! '' % ) %+

or x =

λy M λz R

Note that for this case, the lines of action of both A and B intersect the x axis. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 327

PROBLEM 3.146* (Continued)

Case 2: Let By = 0 to satisfy Equation (4) Now Equation (2)

A=

R

λy

Equation (1)

"λ Bx = − R & x & λy (

# ' ' )

Equation (3)

"λ Bz = − R & z & λy (

# ' ' )

Equation (6)

aAλ y = 0

which requires a = 0

Substitution into Equation (5) "λ M = z $− R & x $* &( λ y

#! "λ '% − x $− R & z '% $* &( λ y )+

#! "M # ' % or λz x − λx z = & ' λ y '% ( R) )+

This last expression is the equation for the line of action of force B. In summary " R A=& & λy (

# ' λA ' )

" R B=& & λy (

# ' ( −λ x i − λx k ) ' )

Assuming that λx , λ y , λz . 0, the equivalent force system is as shown below.

Note that the component of A in the xz plane is parallel to B.!

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PROBLEM 3.147 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at B that creates a moment of equal magnitude and opposite sense about E.

SOLUTION

(a)

By definition We have

W = mg = 80 kg(9.81 m/s 2 ) = 784.8 N

ΣM E : M E = (784.8 N)(0.25 m) M E = 196.2 N ⋅ m

(b)

For the force at B to be the smallest, resulting in a moment (ME) about E, the line of action of force FB must be perpendicular to the line connecting E to B. The sense of FB must be such that the force produces a counterclockwise moment about E. Note: We have

d = (0.85 m) 2 + (0.5 m) 2 = 0.98615 m

ΣM E : 196.2 N ⋅ m = FB (0.98615 m) FB = 198.954 N

and

" 0.85 m # ' = 59.534° ( 0.5 m )

θ = tan −1 &

FB = 199.0 N

or

59.5°

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PROBLEM 3.148 It is known that the connecting rod AB exerts on the crank BC a 1.5-kN force directed down and to the left along the centerline of AB. Determine the moment of the force about C.

SOLUTION Using (a)

M C = y1 ( FAB ) x + x1 ( FAB ) y " 7 # " 24 # = (0.028 m) & × 1500 N ' + (0.021 m) & × 1500 N ' ( 25 ) ( 25 ) = 42 N ⋅ m

or M C = 42.0 N ⋅ m Using (b) M C = y2 ( FAB ) x

" 7 # = (0.1 m) & × 1500 N ' = 42 N ⋅ m ( 25 )

or M C = 42.0 N ⋅ m

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PROBLEM 3.149 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B.

SOLUTION We have Then

Txz = (6 lb) cos 8° = 5.9416 lb Tx = Txz sin 30° = 2.9708 lb Ty = TBC sin 8° = − 0.83504 lb Tz = Txz cos 30° = −5.1456 lb

Now

M A = rB/A × TBC

where

rB/A = (6sin 45°) j − (6cos 45°)k =

6 ft 2

(j − k)

i j k −1 0 1 2 2.9708 −0.83504 −5.1456 6 6 6 = (−5.1456 − 0.83504)i − (2.9708) j − (2.9708)k 2 2 2 6

Then

MA =

or

M A = −(25.4 lb ⋅ ft)i − (12.60 lb ⋅ ft) j − (12.60 lb ⋅ ft)k

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PROBLEM 3.150 Ropes AB and BC are two of the ropes used to support a tent. The two ropes are attached to a stake at B. If the tension in rope AB is 540 N, determine (a) the angle between rope AB and the stake, (b) the projection on the stake of the force exerted by rope AB at Point B.

SOLUTION First note

BA = (−3) 2 + (3)2 + (−1.5) 2 = 4.5 m BD = (−0.08) 2 + (0.38)2 + (0.16) 2 = 0.42 m

λ BD

(a)

TBA (−3i + 3j − 1.5k ) 4.5 T = BA (−2i + 2 j − k ) 3 BD 1 = = ( −0.08i + 0.38 j + 0.16k ) BD 0.42 1 = (−4i + 19 j + 8k ) 21

TBA =

Then

We have or

or

TBA ⋅ λ BD = TBA cos θ TBA 1 ( −2i + 2 j − k ) ⋅ (−4i + 19 j + 8k ) = TBA cos θ 3 21 1 [(−2)( −4) + (2)(19) + (−1)(8)] 63 = 0.60317

cos θ =

or θ = 52.9° (b)

We have

(TBA ) BD = TBA ⋅ λ BD = TBA cos θ = (540 N)(0.60317)

or (TBA ) BD = 326 N

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PROBLEM 3.151 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x axis of the forces exerted by the cables on the barn at Points A and D is equal to 4728 lb ⋅ ft, determine the magnitude of TDE when TAB = 255 lb.

SOLUTION The moment about the x axis due to the two cable forces can be found using the z components of each force acting at their intersection with the xy-plane (A and D). The x components of the forces are parallel to the x axis, and the y components of the forces intersect the x axis. Therefore, neither the x or y components produce a moment about the x axis. We have

ΣM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x

where

(TAB ) z = k ⋅ TAB = k ⋅ (TAB λ AB )

" − i − 12 j + 12k # ! = k ⋅ $ 255 lb & '% 17 ( )+ * = 180 lb (TDE ) z = k ⋅ TDE = k ⋅ (TDE λDE ) " 1.5i − 14 j + 12k # ! = k ⋅ $TDE & '% 18.5 ( )+ * = 0.64865TDE y A = 12 ft yD = 14 ft M x = 4728 lb ⋅ ft (180 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft

and

TDE = 282.79 lb

or TDE = 283 lb

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PROBLEM 3.152 Solve Problem 3.151 when the tension in cable AB is 306 lb. PROBLEM 3.151 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x axis of the forces exerted by the cables on the barn at Points A and D is equal to 4728 lb ⋅ ft, determine the magnitude of TDE when TAB = 255 lb.

SOLUTION The moment about the x axis due to the two cable forces can be found using the z components of each force acting at the intersection with the xy plane (A and D). The x components of the forces are parallel to the x axis, and the y components of the forces intersect the x axis. Therefore, neither the x or y components produce a moment about the x axis. We have

ΣM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x

Where

(TAB ) z = k ⋅ TAB = k ⋅ (TAB

AB )

" − i − 12 j + 12k # ! = k ⋅ $306 lb & '% 17 ( )+ * = 216 lb (TDE ) z = k ⋅ TDE = k ⋅ (TDE

DE )

" 1.5i − 14 j + 12k # ! = k ⋅ $TDE & '% 18.5 ( )+ * = 0.64865TDE y A = 12 ft yD = 14 ft M x = 4728 lb ⋅ ft (216 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft

and

TDE = 235.21 lb

or TDE = 235 lb

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PROBLEM 3.153 A wiring harness is made by routing either two or three wires around 2-in.-diameter pegs mounted on a sheet of plywood. If the force in each wire is 3 lb, determine the resultant couple acting on the plywood when a = 18 in. and (a) only wires AB and CD are in place, (b) all three wires are in place.

SOLUTION In general, M = ΣdF, where d is the perpendicular distance between the lines of action of the two forces acting on a given wire. (a)

We have

M = d AB FAB + dCD FCD 4 " # = (2 + 24) in. × 3 lb + & 2 + × 28 ' in. × 3 lb 5 ( )

or M = 151.2 lb ⋅ in.

(b)

We have

M = [d AB FAB + dCD FCD ] + d EF FEF = 151.2 lb ⋅ in. − 28 in. × 3 lb

or M = 67.2 lb ⋅ in.

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PROBLEM 3.154 A worker tries to move a rock by applying a 360-N force to a steel bar as shown. (a) Replace that force with an equivalent force-couple system at D. (b) Two workers attempt to move the same rock by applying a vertical force at A and another force at D. Determine these two forces if they are to be equivalent to the single force of Part a.

SOLUTION (a)

(a)

We have

(b)

ΣF : 360 N(− sin 40°i − cos 40° j) = −(231.40 N)i − (275.78 N) j = F or F = 360 N

We have where

50°

ΣM D : rB/D × R = M

rB/D = −[(0.65 m) cos 30°]i + [(0.65 m)sin 30°]j = −(0.56292 m)i + (0.32500 m) j

i

j

k

M = −0.56292 0.32500 0 N ⋅ m −231.40 −275.78 0 = [155.240 + 75.206)N ⋅ m]k = (230.45 N ⋅ m)k

(b)

We have where

or M = 230 N ⋅ m

ΣM D : M = rA/D × FA

rB/D = −[(1.05 m) cos 30°]i + [(1.05 m)sin 30°]j = −(0.90933 m)i + (0.52500 m) j

i

j

k

FA = −0.90933 0.52500 0 N ⋅ m

0 −1 = [230.45 N ⋅ m]k

0

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PROBLEM 3.154 (Continued) (0.90933FA )k = 230.45k

or

FA = 253.42 N

We have

or FA = 253 N

ΣF : F = FA + FD

−(231.40 N)i − (275.78 N) j = −(253.42 N) j + FD (− cos θ i − sin θ j)

From

i : 231.40 N = FD cos θ

(1)

j: 22.36 N = FD sin θ

(2)

Equation (2) divided by Equation (1) tan θ = 0.096629 θ = 5.5193° or θ = 5.52°

Substitution into Equation (1) FD =

231.40 = 232.48 N cos 5.5193°

or FD = 232 N

5.52°

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PROBLEM 3.155 A 110-N force acting in a vertical plane parallel to the yz plane is applied to the 220-mm-long horizontal handle AB of a socket wrench. Replace the force with an equivalent forcecouple system at the origin O of the coordinate system.

SOLUTION We have

ΣF : PB = F

where

PB = 110 N[− (sin15°) j + (cos15°)k ] = −(28.470 N) j + (106.252 N)k

or F = −(28.5 N) j + (106.3 N)k

We have where

ΣM O : rB/O × PB = M O

rB/O = [(0.22 cos 35°)i + (0.15) j − (0.22sin 35°)k ]m = (0.180213 m)i + (0.15 m) j − (0.126187 m)k

i j k 0.180213 0.15 0.126187 N ⋅ m = M O 0 106.3 −28.5 M O = [(12.3487)i − (19.1566) j − (5.1361)k ]N ⋅ m or M O = (12.35 N ⋅ m)i − (19.16 N ⋅ m)j − (5.13 N ⋅ m)k

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PROBLEM 3.156 Four ropes are attached to a crate and exert the forces shown. If the forces are to be replaced with a single equivalent force applied at a point on line AB, determine (a) the equivalent force and the distance from A to the point of application of the force when α = 30°, (b) the value of α so that the single equivalent force is applied at Point B.

SOLUTION We have

(a)

For equivalence

ΣFx : −100 cos 30° + 400 cos 65° + 90 cos 65° = Rx

or

Rx = 120.480 lb ΣFy : 100 sin α + 160 + 400 sin 65° + 90 sin 65° = R y

or

R y = (604.09 + 100sin α ) lb

With α = 30°

R y = 654.09 lb

Then

R = (120.480) 2 + (654.09) 2

(1)

= 665 lb

Also

654.09 120.480 or θ = 79.6°

tan θ =

ΣM A : (46 in.)(160 lb) + (66 in.)(400 lb) sin 65° + (26 in.)(400 lb) cos 65° + (66 in.)(90 lb) sin 65° + (36 in.)(90 lb) cos 65° = d (654.09 lb)

or

ΣM A = 42, 435 lb ⋅ in. and d = 64.9 in.

R = 665 lb

79.6°

and R is applied 64.9 in. To the right of A. (b)

We have d = 66 in. Then or Using Eq. (1)

ΣM A : 42, 435 lb ⋅ in = (66 in.) Ry Ry = 642.95 lb 642.95 = 604.09 + 100sin α

or α = 22.9°

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PROBLEM 3.157 A blade held in a brace is used to tighten a screw at A. (a) Determine the forces exerted at B and C, knowing that these forces are equivalent to a force-couple system at A consisting of R = −(30 N)i + Ry j + Rz k and M RA = − (12 N · m)i. (b) Find the corresponding values of Ry and Rz . (c) What is the orientation of the slot in the head of the screw for which the blade is least likely to slip when the brace is in the position shown?

SOLUTION (a)

Equivalence requires or Equating the i coefficients Also or Equating coefficients

ΣF : R = B + C −(30 N)i + Ry j + Rz k = − Bk + (−Cx i + C y j + C z k ) i : − 30 N = −Cx

or Cx = 30 N

ΣM A : M RA = rB/A × B + rC/A × C −(12 N ⋅ m)i = [(0.2 m)i + (0.15 m)j] × (− B)k +(0.4 m)i × [−(30 N)i + C y j + Cz k ] i : − 12 N ⋅ m = −(0.15 m) B k : 0 = (0.4 m)C y

or B = 80 N or C y = 0

j: 0 = (0.2 m)(80 N) − (0.4 m)C z or Cz = 40 N B = −(80.0 N)k C = −(30.0 N)i + (40.0 N)k

(b)

Now we have for the equivalence of forces −(30 N)i + Ry j + Rz k = −(80 N)k + [(−30 N)i + (40 N)k ]

Equating coefficients

j: R y = 0

Ry = 0

k : Rz = −80 + 40

(c)

or

!

Rz = −40.0 N

First note that R = −(30 N)i − (40 N)k. Thus, the screw is best able to resist the lateral force Rz when the slot in the head of the screw is vertical.

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PROBLEM 3.158 A concrete foundation mat in the shape of a regular hexagon of side 12 ft supports four column loads as shown. Determine the magnitudes of the additional loads that must be applied at B and F if the resultant of all six loads is to pass through the center of the mat.

SOLUTION From the statement of the problem it can be concluded that the six applied loads are equivalent to the resultant R at O. It then follows that ΣM O = 0 or ΣM x = 0 ΣM z = 0

For the applied loads.

Then

ΣM x = 0: (6 3 ft) FB + (6 3 ft)(10 kips) − (6 3 ft)(20 kips) − (6 3 ft) FF = 0 FB − FF = 10

or

(1)

ΣM z = 0: (12 ft)(15 kips) + (6 ft) FB − (6 ft)(10 kips) − (12 ft)(30 kips) − (6 ft)(20 kips) + (6 ft) FF = 0 FB + FF = 60

or

(2)

Then (1) + (2) -

FB = 35.0 kips

and

FF = 25.0 kips

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CHAPTER 4

PROBLEM 4.1 A 2100-lb tractor is used to lift 900 lb of gravel. Determine the reaction at each of the two (a) rear wheels A, (b) front wheels B.

SOLUTION

(a)

Rear wheels

ΣM B = 0: + (2100 lb)(40 in.) − (900 lb)(50 in.) + 2 A(60 in.) = 0

A = +325 lb

(b)

Front wheels

ΣM A : − (2100 lb)(20 in.) − (900 lb)(110 in.) − 2 B(60 in.) = 0 B = +1175 lb

Check:

A = 325 lb

B = 1175 lb

+ΣFy = 0: 2 A + 2 B − 2100 lb − 900 lb = 0 2(325 lb) + 2(1175 lb) − 2100 lb − 900 = 0 0 = 0 (Checks) !

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PROBLEM 4.2 A gardener uses a 60-N wheelbarrow to transport a 250-N bag of fertilizer. What force must she exert on each handle?

SOLUTION Free-Body Diagram: !

ΣM A = 0: (2 F )(1 m) − (60 N)(0.15 m) − (250 N)(0.3 m) = 0 F = 42.0 N

!

! !

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PROBLEM 4.3 The gardener of Problem 4.2 wishes to transport a second 250-N bag of fertilizer at the same time as the first one. Determine the maximum allowable horizontal distance from the axle A of the wheelbarrow to the center of gravity of the second bag if she can hold only 75 N with each arm. PROBLEM 4.2 A gardener uses a 60-N wheelbarrow to transport a 250-N bag of fertilizer. What force must she exert on each handle?

SOLUTION Free-Body Diagram: !

ΣMA = 0: 2(75 N)(1 m) − (60 N)(0.15 m) − (250 N)(0.3 m) − (250 N) x = 0 x = 0.264 m

!

! !

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PROBLEM 4.4 For the beam and loading shown, determine (a) the reaction at A, (b) the tension in cable BC.

SOLUTION Free-Body Diagram:

(a)

Reaction at A:

ΣFx = 0: Ax = 0

ΣMB = 0: (15 lb)(28 in.) + (20 lb)(22 in.) + (35 lb)(14 in.) + (20 lb)(6 in.) − Ay (6 in.) = 0 Ay = +245 lb

(b)

Tension in BC

A = 245 lb

ΣM A = 0: (15 lb)(22 in.) + (20 lb)(16 in.) + (35 lb)(8 in.) − (15 lb)(6 in.) − FBC (6 in.) = 0 FBC = +140.0 lb

Check:

FBC = 140.0 lb

ΣFy = 0: − 15 lb − 20 lb = 35 lb − 20 lb + A − FBC = 0 −105 lb + 245 lb − 140.0 = 0 0 = 0 (Checks) !

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PROBLEM 4.5 Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B.

SOLUTION Free-Body Diagram:

W = (350 kg)(9.81 m/s2 ) = 3.434 kN Wt = (1400 kg)(9.81 m/s 2 ) = 13.734 kN

(a)

Rear wheels

ΣM B = 0: W (1.7 m + 2.05 m) + W (2.05 m) + Wt (1.2 m) − 2 A(3 m) = 0 (3.434 kN)(3.75 m) + (3.434 kN)(2.05 m) + (13.734 kN)(1.2 m) − 2 A(3 m) = 0 A = +6.0663 kN

(b)

Front wheels

A = 6.07 kN

ΣFy = 0: − W − W − Wt + 2 A + 2 B = 0 −3.434 kN − 3.434 kN − 13.734 kN + 2(6.0663 kN) + 2B = 0 B = + 4.2347 kN

B = 4.23 kN

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PROBLEM 4.6 Solve Problem 4.5, assuming that crate D is removed and that the position of crate C is unchanged. PROBLEM 4.5 Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B.

SOLUTION Free-Body Diagram:

W = (350 kg)(9.81 m/s2 ) = 3.434 kN Wt = (1400 kg)(9.81 m/s 2 ) = 13.734 kN

(a)

Rear wheels

ΣM B = 0: W (1.7 m + 2.05 m) + Wt (1.2 m) − 2 A(3 m) = 0 (3.434 kN)(3.75 m) + (13.734 kN)(1.2 m) − 2 A(3 m) = 0 A = + 4.893 kN

(b)

Front wheels

A = 4.89 kN

ΣM y = 0: − W − Wt + 2 A + 2 B = 0 −3.434 kN − 13.734 kN + 2(4.893 kN) + 2B = 0 B = +3.691 kN

B = 3.69 kN

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PROBLEM 4.7 A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if a = 10 in., (b) if a = 7 in.

SOLUTION Free-Body Diagram:

ΣFx = 0: Bx = 0 ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0

(40a − 160) 12

A=

(1)

ΣM A = 0: − (40 lb)(6 in.) − (50 lb)(12 in.) − (30 lb)(a + 12 in.) − (10 lb)(a + 20 in.) + (12 in.) B y = 0 By =

Bx = 0 B =

Since (a)

(b)

(1400 + 40a) 12 (1400 + 40a) 12

(2)

For a = 10 in.

Eq. (1):

A=

(40 × 10 − 160) = +20.0 lb 12

Eq. (2):

B=

(1400 + 40 × 10) = +150.0 lb 12

B = 150.0 lb

Eq. (1):

A=

(40 × 7 − 160) = +10.00 lb 12

A = 10.00 lb

Eq. (2):

B=

(1400 + 40 × 7) = +140.0 lb 12

B = 140.0 lb

A = 20.0 lb

For a = 7 in.

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PROBLEM 4.8 For the bracket and loading of Problem 4.7, determine the smallest distance a if the bracket is not to move. PROBLEM 4.7 A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if a = 10 in., (b) if a = 7 in.

SOLUTION Free-Body Diagram:

For no motion, reaction at A must be downward or zero; smallest distance a for no motion corresponds to A = 0. ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0 A=

(40a − 160) 12

A = 0: (40a − 160) = 0

a = 4.00 in.

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!

PROBLEM 4.9 The maximum allowable value of each of the reactions is 180 N. Neglecting the weight of the beam, determine the range of the distance d for which the beam is safe.

SOLUTION

ΣFx = 0: Bx = 0 B = By ΣM A = 0: (50 N) d − (100 N)(0.45 m − d ) − (150 N)(0.9 m − d ) + B (0.9 m − d ) = 0 50d − 45 + 100d − 135 + 150d + 0.9 B − Bd d=

180 N ⋅ m − (0.9 m) B 300 A − B

(1)

ΣM B = 0: (50 N)(0.9 m) − A(0.9 m − d ) + (100 N)(0.45 m) = 0 45 − 0.9 A + Ad + 45 = 0 (0.9 m) A − 90 N ⋅ m A

(2)

d$

180 − (0.9)180 18 = = 0.15 m 300 − 180 120

d $ 150.0 mm "

d#

(0.9)180 − 90 72 = = 0.40 m 180 180

d=

Since B # 180 N, Eq. (1) yields.

Since A # 180 N, Eq. (2) yields.

Range:

d # 400 mm "

!

150.0 mm # d # 400 mm

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PROBLEM 4.10 Solve Problem 4.9 if the 50-N load is replaced by an 80-N load. PROBLEM 4.9 The maximum allowable value of each of the reactions is 180 N. Neglecting the weight of the beam, determine the range of the distance d for which the beam is safe.

SOLUTION

ΣFx = 0: Bx = 0 B = By

ΣM A = 0: (80 N)d − (100 N)(0.45 m − d ) − (150 N)(0.9 m − d ) + B(0.9 m − d ) = 0 80d − 45 + 100d − 135 + 150d + 0.9 B − Bd = 0 d=

180 N ⋅ m − 0.9 B 330N − B

(1)

ΣM B = 0: (80 N)(0.9 m) − A(0.9 m − d ) + (100 N)(0.45 m) = 0 d=

0.9 A − 117 A

(2)

Since B # 180 N, Eq. (1) yields. d $ (180 − 0.9 × 180)/(330 − 180) =

18 = 0.12 m 150

d = 120.0 mm "

Since A # 180 N, Eq. (2) yields. d # (0.9 × 180 − 112)/180 =

Range:

45 = 0.25 m 180

d = 250 mm "

120.0 mm # d # 250 mm

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PROBLEM 4.11 For the beam of Sample Problem 4.2, determine the range of values of P for which the beam will be safe, knowing that the maximum allowable value of each of the reactions is 30 kips and that the reaction at A must be directed upward.

SOLUTION ΣFx = 0: Bx = 0 B = By

ΣM A = 0: − P(3 ft) + B (9 ft) − (6 kips)(11 ft) − (6 kips)(13 ft) = 0 P = 3B − 48 kips

(1)

ΣM B = 0: − A(9 ft) + P (6 ft) − (6 kips)(2 ft) − (6 kips)(4 ft) = 0 P = 1.5 A + 6 kips

(2)

Since B # 30 kips, Eq. (1) yields. P # (3)(30 kips) − 48 kips

P # 42.0 kips "

Since 0 # A # 30 kips. Eq. (2) yields. 0 + 6 kips # P # (1.5)(30 kips)1.6 kips 6.00 kips # P # 51.0 kips

"

Range of values of P for which beam will be safe: 6.00 kips # P # 42.0 kips

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PROBLEM 4.12 The 10-m beam AB rests upon, but is not attached to, supports at C and D. Neglecting the weight of the beam, determine the range of values of P for which the beam will remain in equilibrium.

SOLUTION Free-Body Diagram: ΣM C = 0: P(2 m) − (4 kN)(3 m) − (20 kN)(8 m) + D(6 m) = 0 P = 86 kN − 3D

(1)

ΣM D = 0: P(8 m) + (4 kN)(3 m) − (20 kN)(2 m) − C (6 m) = 0 P = 3.5 kN + 0.75C

(2)

For no motion C $ 0 and D $ 0 For C $ 0 from (2) P # 3.50 kN For D $ 0 from (1) P # 86.0 kN Range of P for no motion: 3.50 kN # P # 86.0 kN

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PROBLEM 4.13 The maximum allowable value of each of the reactions is 50 kN, and each reaction must be directed upward. Neglecting the weight of the beam, determine the range of values of P for which the beam is safe.

SOLUTION Free-Body Diagram: ΣM C = 0: P(2 m) − (4 kN)(3 m) − (20 kN)(8 m) + D(6 m) = 0 P = 86 kN − 3D

(1)

ΣM D = 0: P(8 m) + (4 kN)(3 m) − (20 kN)(2 m) − C (6 m) = 0 P = 3.5 kN + 0.75C

(2)

For C $ 0, from (2):

P $ 3.50 kN

"

For D $ 0, from (1):

P # 86.0 kN

"

P # 3.5 kN + 0.75(50 kN) P # 41.0 kN

"

P $ 86 kN − 3(50 kN) P $ −64.0 kN

"

For C # 50 kN, from (2):

For D # 50 kN, from (1):

Comparing the four criteria, we find 3.50 kN # P # 41.0 kN

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 357

PROBLEM 4.14 For the beam and loading shown, determine the range of the distance a for which the reaction at B does not exceed 100 lb downward or 200 lb upward.

SOLUTION Assume B is positive when directed

Sketch showing distance from D to forces.

ΣM D = 0: (300 lb)(8 in. − a ) − (300 lb)(a − 2 in.) − (50 lb)(4 in.) + 16 B = 0 −600a + 2800 + 16B = 0 (2800 + 16B) 600

(1)

[2800 + 16(−100)] 1200 = = 2 in. 600 600

a $ 2.00 in. "

a=

For B = 100 lb = −100 lb, Eq. (1) yields: a$

For B = 200 = +200 lb, Eq. (1) yields: a#

Required range:

[2800 + 16(200)] 6000 = = 10 in. 600 600

a # 10.00 in. "

2.00 in. # a # 10.00 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 358

PROBLEM 4.15 Two links AB and DE are connected by a bell crank as shown. Knowing that the tension in link AB is 720 N, determine (a) the tension in link DE, (b) the reaction at C.

SOLUTION Free-Body Diagram:

ΣM C = 0: FAB (100 mm) − FDE (120 mm) = 0 FDE =

(a)

For

(1)

FAB = 720 N FDE =

(b)

5 FAB 6

5 (720 N) 6

FDE = 600 N

3 ΣFx = 0: − (720 N) + C x = 0 5 C x = +432 N 4 ΣFy = 0: − (720 N) + C y − 600 N = 0 5 C y = +1176 N C = 1252.84 N α = 69.829° C = 1253 N

69.8°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 359

PROBLEM 4.16 Two links AB and DE are connected by a bell crank as shown. Determine the maximum force that may be safely exerted by link AB on the bell crank if the maximum allowable value for the reaction at C is 1600 N.

SOLUTION See solution to Problem 4.15 for F. B. D. and derivation of Eq. (1) FDE =

5 FAB 6

(1)

3 3 ΣFx = 0: − FAB + C x = 0 C x = FAB 5 5 ΣFy = 0: −

4 FAB + C y − FDE = 0 5

4 5 − FAB + C y − FAB = 0 5 6 49 Cy = FAB 30

C = C x2 + C y2 1 (49) 2 + (18) 2 FAB 30 C = 1.74005FAB =

For

C = 1600 N, 1600 N = 1.74005FAB

FAB = 920 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 360

PROBLEM 4.17 The required tension in cable AB is 200 lb. Determine (a) the vertical force P that must be applied to the pedal, (b) the corresponding reaction at C.

SOLUTION Free-Body Diagram:

BC = 7 in.

(a)

ΣM C = 0: P(15 in.) − (200 lb)(6.062 in.) = 0 P = 80.83 lb

(b)

P = 80.8 lb

ΣFy = 0: C x − 200 lb = 0

C x = 200 lb

ΣFy = 0: C y − P = 0 C y − 80.83 lb = 0

C y = 80.83 lb

α = 22.0° C = 215.7 lb C = 216 lb

22.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 361

PROBLEM 4.18 Determine the maximum tension that can be developed in cable AB if the maximum allowable value of the reaction at C is 250 lb.

SOLUTION Free-Body Diagram:

BC = 7 in. ΣM C = 0: P (15 in.) − T (6.062 in.) = 0

P = 0.40415T

ΣFy = 0: − P + C y = 0 − 0.40415 P + C y = 0 C y = 0.40415T ΣFx = 0: −T + C x = 0

Cx = T

C = C x2 + C y2 = T 2 + (0.40415T ) 2 C = 1.0786T

For

C = 250 lb 250 lb = 1.0786T

T = 232 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 362

PROBLEM 4.19 The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION

At B:

Ty Tx

=

0.18 m 0.24 m

3 Ty = Tx 4

(a)

(1)

ΣM C = 0: Tx (0.18 m) − (240 N)(0.4 m) − (240 N)(0.8 m) = 0 Tx = +1600 N

Eq. (1)

Ty =

3 (1600 N) = 1200 N 4

T = Tx2 + Ty2 = 16002 + 12002 = 2000 N

(b)

T = 2.00 kN

ΣFx = 0: Cx − Tx = 0 Cx − 1600 N = 0 C x = +1600 N

C x = 1600 N

ΣFy = 0: C y − Ty − 240 N − 240 N = 0 C y − 1200 N − 480 N = 0 C y = +1680 N

C y = 1680 N

α = 46.4° C = 2320 N

C = 2.32 kN

46.4°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 363

PROBLEM 4.20 Solve Problem 4.19, assuming that a = 0.32 m. PROBLEM 4.19 The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION

At B:

Ty Tx

=

0.32 m 0.24 m

4 Ty = Tx 3 ΣM C = 0: Tx (0.32 m) − (240 N)(0.4 m) − (240 N)(0.8 m) = 0 Tx = 900 N

Eq. (1)

Ty =

4 (900 N) = 1200 N 3

T = Tx2 + Ty2 = 9002 + 12002 = 1500 N

T = 1.500 kN

ΣFx = 0: C x − Tx = 0 C x − 900 N = 0 C x = +900 N

C x = 900 N

ΣFy = 0: C y − Ty − 240 N − 240 N = 0 C y − 1200 N − 480 N = 0 C y = +1680 N

C y = 1680 N

α = 61.8° C = 1906 N

C = 1.906 kN

61.8°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 364

PROBLEM 4.21 Determine the reactions at A and B when (a) h = 0, (b) h = 200 mm.

SOLUTION Free-Body Diagram:

ΣM A = 0: ( B cos 60°)(0.5 m) − ( B sin 60°)h − (150 N)(0.25 m) = 0 37.5 B= 0.25 − 0.866h

(a)

(1)

When h = 0: B=

Eq. (1):

37.5 = 150 N 0.25

B = 150.0 N

30.0°

ΣFy = 0: Ax − B sin 60° = 0 Ax = (150)sin 60° = 129.9 N

A x = 129.9 N

ΣFy = 0: Ay − 150 + B cos 60° = 0 Ay = 150 − (150) cos 60° = 75 N

A y = 75 N

α = 30° A = 150.0 N

(b)

A = 150.0 N

30.0°

When h = 200 mm = 0.2 m Eq. (1):

B=

37.5 = 488.3 N 0.25 − 0.866(0.2)

B = 488 N

30.0°

ΣFx = 0: Ax − B sin 60° = 0 Ax = (488.3) sin 60° = 422.88 N

A x = 422.88 N

ΣFy = 0: Ay − 150 + B cos 60° = 0 Ay = 150 − (488.3) cos 60° = −94.15 N

A y = 94.15 N

α = 12.55° A = 433.2 N

A = 433 N

12.6°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 365

PROBLEM 4.22 For the frame and loading shown, determine the reactions at A and E when (a) α = 30°, (b) α = 45°.

SOLUTION Free-Body Diagram:

ΣM A = 0: ( E sin α )(8 in.) + ( E cos α )(5 in.) − (20 lb)(10 in.) − (20 lb)(3 in.) = 0

(a)

When α = 30°:

E=

260 8sin α + 5cos α

E=

260 = 31.212 lb 8sin 30° + 5cos 30°

E = 31.2 lb

60.0°

ΣFx = 0: Ax − 20 lb + (31.212 lb) sin 30° = 0 Ax = +4.394 lb

A x = 4.394 lb

ΣFy = 0: Ay − 20° + (31.212 lb) cos 30° = 0 Ay = −7.03 lb

A y = 7.03 lb A = 8.29 lb

58.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 366

PROBLEM 4.22 (Continued)

(b)

When α = 45° : E=

260 = 28.28 lb 8sin 45° + 5cos α

E = 28.3 lb

45.0°

ΣFx = 0: Ax − 20 lb + (28.28 lb)sin 45° = 0 Ax = 0

Ax = 0

ΣFy = 0: Ay − 20 lb + (28.28 lb) cos 45° = 0 Ay = 0

Ay = 0

A=0

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 367

PROBLEM 4.23 For each of the plates and loadings shown, determine the reactions at A and B.

SOLUTION (a)

Free-Body Diagram:

ΣM A = 0: B(20 in.) − (50 lb)(4 in.) − (40 lb)(10 in.) = 0 B = +30 lb

B = 30.0 lb

ΣFx = 0: Ax + 40 lb = 0 Ax = −40 lb

A x = 40.0 lb

ΣFy = 0: Ay + B − 50 lb = 0 Ay + 30 lb − 50 lb = 0 Ay = +20 lb

A y = 20.0 lb

α = 26.56° A = 44.72 lb

A = 44.7 lb

26.6°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 368

PROBLEM 4.23 (Continued)

(b)

Free-Body Diagram:

ΣM A = 0: ( B cos 30°)(20 in.) − (40 lb)(10 in.) − (50 lb)(4 in.) = 0 B = 34.64 lb

B = 34.6 lb

60.0°

ΣFx = 0: Ax − B sin 30° + 40 lb Ax − (34.64 lb) sin 30° + 40 lb = 0 Ax = −22.68 lb

A x = 22.68 lb

ΣFy = 0: Ay + B cos 30° − 50 lb = 0 Ay + (34.64 lb) cos 30° − 50 lb = 0 Ay = +20 lb

A y = 20.0 lb

α = 41.4° A = 30.24 lb

A = 30.2 lb

41.4°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 369

PROBLEM 4.24 For each of the plates and loadings shown, determine the reactions at A and B.

SOLUTION (a)

Free-Body Diagram:

ΣM B = 0: A(20 in.) + (50 lb)(16 in.) − (40 lb)(10 in.) = 0 A = +20 lb

A = 20.0 lb

ΣFx = 0: 40 lb + Bx = 0 Bx = −40 lb

B x = 40 lb

ΣFy = 0: A + By − 50 lb = 0 20 lb + By − 50 lb = 0 By = +30 lb

α = 36.87°

B = 50 lb

B y = 30 lb B = 50.0 lb

36.9°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 370

PROBLEM 4.24 (Continued)

(b)

ΣM A = 0: − ( A cos 30°)(20 in.) − (40 lb)(10 in.) + (50 lb)(16 in.) = 0 A = 23.09 lb

A = 23.1 lb

60.0°

ΣFx = 0: A sin 30° + 40 lb + Bx = 0 (23.09 lb) sin 30° + 40 lb + 8 x = 0 Bx = −51.55 lb

B x = 51.55 lb

ΣFy = 0: A cos 30° + By − 50 lb = 0 (23.09 lb) cos 30° + By − 50 lb = 0

By = +30 lb

B y = 30 lb

α = 30.2° B = 59.64 lb

B = 59.6 lb

30.2°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 371

PROBLEM 4.25 Determine the reactions at A and B when (a) α = 0, (b) α = 90°, (c) α = 30°.

SOLUTION (a)

α =0 ΣM A = 0: B(20 in.) − 750 lb ⋅ in. = 0 B = 37.5 lb ΣFx = 0: Ax = 0 ΣFy = 0: Ay + 37.5 lb = 0 Ay = −37.5 lb B = 37.5 lb

A = 37.5 lb

(b)

α = 90° ΣM A = 0: B (12 in.) − 750 lb ⋅ in. = 0 B = 62.5 lb ΣFA = 0: Ax − 62.5 lb = 0,

Ax = 62.5 lb

ΣFy = 0: Ay = 0 A = 62.5 lb

(c)

B = 62.5 lb

α = 30° ΣM A = 0: ( B cos 30°)(20 in.) + ( B sin 30°)(12 in.) − 750 lb ⋅ in. = 0 B = 32.16 lb ΣFx = 0: Ax − (32.16 lb) sin 30° = 0 Ax = 16.08 lb ΣFy = 0: Ay + (32.16 lb) cos 30° = 0

Ay = −27.85 lb

A = 32.16 lb α = 60.0° A = 32.2 lb

60.0°

B = 32.2 lb

60.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 372

PROBLEM 4.26 A rod AB hinged at A and attached at B to cable BD supports the loads shown. Knowing that d = 200 mm, determine (a) the tension in cable BD, (b) the reaction at A.

SOLUTION Free-Body Diagram:

(a)

Move T along BD until it acts at Point D. ΣM A = 0: (T sin 45°)(0.2 m) + (90 N)(0.1 m) + (90 N)(0.2 m) = 0 T = 190.92 N

(b)

T = 190.9 N

ΣFx = 0: Ax − (190.92 N) cos 45° = 0 Ax = +135 N

A x = 135.0 N

ΣFy = 0: Ay − 90 N − 90 N + (190.92 N) sin 45° = 0 Ay = +45 N

A y = 45.0 N A = 142.3 N

18.43°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 373

PROBLEM 4.27 A rod AB hinged at A and attached at B to cable BD supports the loads shown. Knowing that d = 150 mm, determine (a) the tension in cable BD, (b) the reaction at A.

SOLUTION Free-Body Diagram:

tan α =

(a)

10 ; α = 33.69° 15

Move T along BD until it acts at Point D.

ΣM A = 0: (T sin 33.69°)(0.15 m) − (90 N)(0.1 m) − (90 N)(0.2 m) = 0 T = 324.5 N

(b)

T = 324 N

ΣFx = 0: Ax − (324.99 N) cos 33.69° = 0 Ax = +270 N

A x = 270 N

ΣFy = 0: Ay − 90 N − 90 N + (324.5 N)sin 33.69° = 0 Ay = 0

Ay = 0

A = 270 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 374

PROBLEM 4.28 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 500-N horizontal force at B, determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION

Triangle ACD is isosceles with

C = 90° + 30° = 120°

A=

D=

1 (180° − 120°) = 30° 2

Thus DA forms angle of 60° with horizontal. (a)

We resolve FAD into components along AB and perpendicular to AB.

ΣM C = 0: ( FAD sin 30°)(250 mm) − (500 N)(100 mm) = 0

(b)

FAD = 400 N

ΣFx = 0: − (400 N) cos 60° + C x − 500 N = 0

C x = +300 N

ΣFy = 0: − (400 N) sin 60° + C y = 0

C y = +346.4 N C = 458 N

49.1°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 375

PROBLEM 4.29 A force P of magnitude 280 lb is applied to member ABCD, which is supported by a frictionless pin at A and by the cable CED. Since the cable passes over a small pulley at E, the tension may be assumed to be the same in portions CE and ED of the cable. For the case when a = 3 in., determine (a) the tension in the cable, (b) the reaction at A.

SOLUTION Free-Body Diagram: (a)

ΣM A = 0: − (280 lb)(8 in.) 7 T (12 in.) 25 24 − T (8 in.) = 0 25

T (12 in.) −

(12 − 11.04)T = 840

(b)

ΣFx = 0:

T = 875 lb

7 (875 lb) + 875 lb + Ax = 0 25 Ax = −1120

ΣFy = 0: Ay − 280 lb − Ay = +1120

A x = 1120 lb

24 (875 lb) = 0 25 A y = 1120 lb A = 1584 lb

45.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 376

PROBLEM 4.30 Neglecting friction, determine the tension in cable ABD and the reaction at support C.

SOLUTION Free-Body Diagram:

ΣM C = 0: T (0.25 m) − T (0.1 m) − (120 N)(0.1 m) = 0

T = 80.0 N

ΣFx = 0: C x − 80 N = 0 C x = +80 N

C x = 80.0 N

ΣFy = 0: C y − 120 N + 80 N = 0 C y = +40 N

C y = 40.0 N C = 89.4 N

26.6°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 377

PROBLEM 4.31 Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the θ = 30°, determine the reaction (a) at B, (b) at C.

SOLUTION Free-Body Diagram:

ΣM D = 0: C x ( R) − P( R) = 0 Cx = + P ΣFx = 0: C x − B sin θ = 0 P − B sin θ = 0 B = P/sin θ B=

P sin θ

θ

ΣFy = 0: C y + B cos θ − P = 0 C y + ( P/sin θ ) cos θ − P = 0 1 ! Cy = P 1 − ! tan θ # "

For θ = 30°: (a) (b)

B = P/sin 30° = 2 P

B = 2P

60.0°

Cx = + P Cx = P C y = P(1 − 1/tan 30°) = − 0.732/P

C y = 0.7321P C = 1.239P

36.2°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 378

!

PROBLEM 4.32 Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the θ = 60°, determine the reaction (a) at B, (b) at C.

SOLUTION See the solution to Problem 4.31 for the free-body diagram and analysis leading to the following expressions: Cx = + P 1 % $ Cy = P 1 − tan θ !# " P B= sin θ

For θ = 60°: (a) (b)

B = P/sin 60° = 1.1547 P

B = 1.155P

30.0°

Cx = + P Cx = P C y = P(1 − 1/tan 60°) = + 0.4226 P

C y = 0.4226 P C = 1.086P

22.9°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 379

!

PROBLEM 4.33 Neglecting friction, determine the tension in cable ABD and the reaction at C when θ = 60°.

SOLUTION ΣM C = 0: T (2a + a cos θ ) − Ta + Pa = 0 T=

P 1 + cos θ

(1)

ΣFx = 0: C x − T sin θ = 0 C x = T sin θ =

P sin θ 1 + cos θ

ΣFy = 0: C y + T + T cos θ − P = 0 C y = P − T (1 + cos θ ) = P − P

1 + cos θ 1 + cos θ

Cy = 0 C y = 0, C = C x

Since

C=P

sin θ 1 + cos θ

(2)

For θ = 60°: P P = 1 + cos 60° 1 + 12

Eq. (1):

T=

Eq. (2):

C=P

sin 60° 0.866 =P 1 + cos 60° 1 + 12

T=

2 P 3

C = 0.577P

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 380

!

PROBLEM 4.34 Neglecting friction, determine the tension in cable ABD and the reaction at C when θ = 45°.

SOLUTION Free-Body Diagram:

Equilibrium for bracket: ΣM C = 0: − T (a ) − P(a ) + (T sin 45°)(2a sin 45°) + (T cos 45°)(a + 2a cos 45°) = 0 T = 0.58579

or T = 0.586 P

ΣFx = 0: C x + (0.58579 P) sin 45° = 0 C x = 0.41422 P ΣFy = 0: C y + 0.58579 P − P + (0.58579 P) cos 45° = 0 Cy = 0

or C = 0.414P

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 381

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PROBLEM 4.35 A light rod AD is supported by frictionless pegs at B and C and rests against a frictionless wall at A. A vertical 120-lb force is applied at D. Determine the reactions at A, B, and C.

SOLUTION Free-Body Diagram:

ΣFx = 0: A cos 30° − (120 lb) cos 60° = 0 A = 69.28 lb

A = 69.3 lb

ΣM B = 0: C (8 in.) − (120 lb)(16 in.) cos 30° + (69.28 lb)(8 in.)sin 30° = 0 C = 173.2 lb

C = 173.2 lb

60.0°

B = 34.6 lb

60.0°

ΣM C = 0: B(8 in.) − (120 lb)(8 in.) cos 30° + (69.28 lb)(16 in.) sin 30° = 0 B = 34.6 lb

Check:

ΣFy = 0: 173.2 − 34.6 − (69.28)sin 30° − (120)sin 60° = 0 0 = 0(check) !

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 382

PROBLEM 4.36 A light bar AD is suspended from a cable BE and supports a 50-lb block at C. The ends A and D of the bar are in contact with frictionless vertical walls. Determine the tension in cable BE and the reactions at A and D.

SOLUTION Free-Body Diagram:

ΣFx = 0:

A= D

ΣFy = 0:

TBE = 50.0 lb

We note that the forces shown form two couples. ΣM = 0: A(8 in.) − (50 lb)(3 in.) = 0 A = 18.75 lb A = 18.75 lb

D = 18.75 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 383

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PROBLEM 4.37 Bar AC supports two 400-N loads as shown. Rollers at A and C rest against frictionless surfaces and a cable BD is attached at B. Determine (a) the tension in cable BD, (b) the reaction at A, (c) the reaction at C.

SOLUTION

Similar triangles: ABE and ACD AE BE = ; AD CD

(a)

ΣM A = 0: Tx (0.25 m) −

BE 0.15 m = ; BE = 0.075 m 0.5 m 0.25 m

$ 0.075 % Tx ! (0.5 m) − (400 N)(0.1 m) − (400 N)(0.4 m) = 0 " 0.35 #

Tx = 1400 N

0.075 (1400 N) 0.35 = 300 N

Ty =

T = 1432 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 384

PROBLEM 4.37 (Continued)

(b)

ΣFy = 0: A − 300 N − 400 N − 400 N = 0 A = + 1100 N

(c)

A = 1100 N

ΣFx = 0: −C + 1400 N = 0 C = + 1400 N

C = 1400 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 385

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PROBLEM 4.38 Determine the tension in each cable and the reaction at D.

SOLUTION tan α =

0.08 m 0.2 m

α = 21.80° tan β =

0.08 m 0.1 m

β = 38.66° ΣM B = 0: (600 N)(0.1 m) − (TCF sin 38.66°)(0.1 m) = 0

TCF = 960.47 N

TCF = 96.0 N

ΣM C = 0: (600 N)(0.2 m) − (TBE sin 21.80°)(0.1 m) = 0

TBE = 3231.1 N

TBE = 3230 N

ΣFy = 0: TBE cos α + TCF cos β − D = 0 (3231.1 N) cos 21.80° + (960.47 N) cos 38.66° − D = 0

D = 3750.03 N

D = 3750 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 386

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PROBLEM 4.39 Two slots have been cut in plate DEF, and the plate has been placed so that the slots fit two fixed, frictionless pins A and B. Knowing that P = 15 lb, determine (a) the force each pin exerts on the plate, (b) the reaction at F.

SOLUTION Free-Body Diagram:

ΣFx = 0: 15 lb − B sin 30° = 0

B = 30.0 lb

60.0°

ΣM A = 0: − (30 lb)(4 in.) + B sin 30°(3 in.) + B cos 30°(11 in.) − F (13 in.) = 0 −120 lb ⋅ in. + (30 lb) sin 30°(3 in.) + (30 lb) cos 30°(11 in.) − F (13 in.) = 0

F = + 16.2145 lb

F = 16.21 lb

ΣFy = 0: A − 30 lb + B cos 30° − F = 0

A − 30 lb + (30 lb) cos 30° − 16.2145 lb = 0 A = + 20.23 lb

A = 20.2 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 387

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PROBLEM 4.40 For the plate of Problem 4.39 the reaction at F must be directed downward, and its maximum allowable value is 20 lb. Neglecting friction at the pins, determine the required range of values of P. PROBLEM 4.39 Two slots have been cut in plate DEF, and the plate has been placed so that the slots fit two fixed, frictionless pins A and B. Knowing that P = 15 lb, determine (a) the force each pin exerts on the plate, (b) the reaction at F.

SOLUTION Free-Body Diagram:

ΣFx = 0: P − B sin 30° = 0

B = 2P

60°

ΣM A = 0: − (30 lb)(4 in.) + B sin 30°(3 in.) + B cos 30°(11 in.) − F (13 in.) = 0

−120 lb ⋅ in.+ 2 P sin 30°(3 in.) + 2 P cos 30°(11 in.) − F (13 in.) = 0 −120 + 3P + 19.0525P − 13F = 0 P=

13E + 120 22.0525

For F = 0:

P=

13(0) + 120 = 5.442 lb 22.0525

For P = 20 lb:

P=

13(20) + 120 = 17.232 lb 22.0525

For

0 # F # 20 lb:

(1)

5.44 lb # P # 17.231 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 388

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PROBLEM 4.41 Bar AD is attached at A and C to collars that can move freely on the rods shown. If the cord BE is vertical (α = 0), determine the tension in the cord and the reactions at A and C.

SOLUTION Free-Body Diagram:

ΣFy = 0: − T cos 30° + (80 N) cos 30° = 0 T = 80 N

T = 80.0 N

ΣM C = 0: ( A sin 30°)(0.4 m) − (80 N)(0.2 m) − (80 N)(0.2 m) = 0 A = + 160 N

A = 160.0 N

30.0°

C = 160.0 N

30.0°

ΣM A = 0: (80 N)(0.2 m) − (80 N)(0.6 m) + (C sin 30°)(0.4 m) = 0 C = + 160 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 389

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PROBLEM 4.42 Solve Problem 4.41 if the cord BE is parallel to the rods (α = 30°). PROBLEM 4.41 Bar AD is attached at A and C to collars that can move freely on the rods shown. If the cord BE is vertical (α = 0), determine the tension in the cord and the reactions at A and C.

SOLUTION Free-Body Diagram:

ΣFy = 0: − T + (80 N) cos 30° = 0 T = 69.282 N

T = 69.3 N

ΣM C = 0: − (69.282 N) cos 30°(0.2 m) − (80 N)(0.2 m) + ( A sin 30°)(0.4 m) = 0 A = + 140.000 N

A = 140.0 N

30.0°

C = 180.0 N

30.0°

ΣM A = 0: + (69.282 N) cos 30°(0.2 m) − (80 N)(0.6 m) + (C sin 30°)(0.4 m) = 0 C = + 180.000 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 390

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PROBLEM 4.43 An 8-kg mass can be supported in the three different ways shown. Knowing that the pulleys have a 100-mm radius, determine the reaction at A in each case.

SOLUTION !

W = mg = (8 kg)(9.81m/s 2 ) = 78.48 N

!

(a)

ΣFx = 0: Ax = 0 ΣFy = 0: Ay − W = 0

!

A y = 78.48 N

ΣM A = 0: M A − W (1.6 m) = 0

!

M A = + (78.48 N)(1.6 m)

!

M A = 125.56 N ⋅ m

A = 78.5 N

! (b)

!

M A = 125.6 N ⋅ m

ΣFx = 0: Ax − W = 0

A x = 78.48

ΣFy = 0: Ay − W = 0

A y = 78.48

A = (78.48 N) 2 = 110.99 N

!

45°

ΣM A = 0: M A − W (1.6 m) = 0

!

M A = + (78.48 N)(1.6 m)

! !

A = 111.0 N

!

(c)

45°

M A = 125.56 N ⋅ m M A = 125.6 N ⋅ m

ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 2W = 0

! !

Ay = 2W = 2(78.48 N) = 156.96 N ΣM A = 0: M A − 2W (1.6 m) = 0 M A = + 2(78.48 N)(1.6 m) A = 157.0 N

M A = 125.1 N ⋅ m M A = 125 N ⋅ m

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 391

PROBLEM 4.44 A tension of 5 lb is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 0.4 in., determine the reaction at C.

SOLUTION From f.b.d. of system ΣFx = 0: C x + (5 lb) = 0 Cx = −5 lb ΣFy = 0: C y − (5 lb) = 0 C y = 5 lb

Then

C = ( Cx ) 2 + (C y ) 2 = (5) 2 + (5) 2 = 7.0711 lb

and

θ = tan −1

$ +5 % ! = −45° " −5 #

or C = 7.07 lb

45.0°

ΣM C = 0: M C + (5 lb)(6.4 in.) + (5 lb)(2.2 in.) = 0 M C = − 43.0 lb ⋅ in

or

M C = 43.0 lb ⋅ in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 392

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PROBLEM 4.45 Solve Problem 4.44, assuming that 0.6-in.-radius pulleys are used. PROBLEM 4.44 A tension of 5 lb is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 0.4 in., determine the reaction at C.

SOLUTION From f.b.d. of system ΣFx = 0: C x + (5 lb) = 0 Cx = −5 lb ΣFy = 0: C y − (5 lb) = 0 C y = 5 lb

Then

C = (Cx ) 2 + (C y ) 2 = (5) 2 + (5) 2 = 7.0711 lb

and

θ = tan −1

$ 5 % ! = −45.0° " −5 #

or C = 7.07 lb

45.0°

ΣM C = 0: M C + (5 lb)(6.6 in.) + (5 lb)(2.4 in.) = 0 M C = − 45.0 lb ⋅ in.

or

M C = 45.0 lb ⋅ in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 393

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PROBLEM 4.46 A 6-m telephone pole weighing 1600 N is used to support the ends of two wires. The wires form the angles shown with the horizontal and the tensions in the wires are, respectively, T1 = 600 N and T2 = 375 N. Determine the reaction at the fixed end A.

SOLUTION Free-Body Diagram:

ΣFx = 0: Ax + (375 N) cos 20° − (600 N) cos10° = 0 Ax = +238.50 N ΣFy = 0: Ay − 1600 N − (600 N)sin10° − (375 N) sin 20° = 0 Ay = +1832.45 N A = 238.502 + 1832.452 1832.45 θ = tan −1 238.50

A = 1848 N

82.6°

ΣM A = 0: M A + (600 N) cos10°(6 m) − (375 N) cos 20°(6 m) = 0 M A = −1431.00 N ⋅ m

M A = 1431 N ⋅ m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 394

PROBLEM 4.47 Beam AD carries the two 40-lb loads shown. The beam is held by a fixed support at D and by the cable BE that is attached to the counterweight W. Determine the reaction at D when (a) W = 100 lb, (b) W = 90 lb.

SOLUTION W = 100 lb

(a) From f.b.d. of beam AD

ΣFx = 0: Dx = 0 ΣFy = 0: D y − 40 lb − 40 lb + 100 lb = 0 Dy = −20.0 lb

or

D = 20.0 lb

ΣM D = 0: M D − (100 lb)(5 ft) + (40 lb)(8 ft) + (40 lb)(4 ft) = 0 M D = 20.0 lb ⋅ ft

or M D = 20.0 lb ⋅ ft

W = 90 lb

(b) From f.b.d. of beam AD

ΣFx = 0: Dx = 0 ΣFy = 0: D y + 90 lb − 40 lb − 40 lb = 0 Dy = −10.00 lb

or

D = 10.00 lb

ΣM D = 0: M D − (90 lb)(5 ft) + (40 lb)(8 ft) + (40 lb)(4 ft) = 0 M D = −30.0 lb ⋅ ft

or M D = −30.0 lb ⋅ ft

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 395

PROBLEM 4.48 For the beam and loading shown, determine the range of values of W for which the magnitude of the couple at D does not exceed 40 lb ⋅ ft.

SOLUTION For Wmin , From f.b.d. of beam AD

M D = − 40 lb ⋅ ft ΣM D = 0: (40 lb)(8 ft) − Wmin (5 ft) + (40 lb)(4 ft) − 40 lb ⋅ ft = 0 Wmin = 88.0 lb

For Wmax , From f.b.d. of beam AD

M D = 40 lb ⋅ ft ΣM D = 0: (40 lb)(8 ft) − Wmax (5 ft) + (40 lb)(4 ft) + 40 lb ⋅ ft = 0 Wmax = 104.0 lb

or 88.0 lb # W # 104.0 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 396

PROBLEM 4.49 Knowing that the tension in wire BD is 1300 N, determine the reaction at the fixed support C of the frame shown.

SOLUTION T = 1300 N 5 Tx = T 13 = 500 N 12 Ty = T 13 = 1200 N ΣM x = 0: C x − 450 N + 500 N = 0

C x = −50 N

ΣFy = 0: C y − 750 N − 1200 N = 0 C y = +1950 N

C x = 50 N C y = 1950 N

C = 1951 N

88.5°

ΣM C = 0: M C + (750 N)(0.5 m) + (4.50 N)(0.4 m) − (1200 N)(0.4 m) = 0 M C = −75.0 N ⋅ m

M C = 75.0 N ⋅ m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 397

PROBLEM 4.50 Determine the range of allowable values of the tension in wire BD if the magnitude of the couple at the fixed support C is not to exceed 100 N · m.

SOLUTION

ΣM C = 0: (750 N)(0.5 m) + (450 N)(0.4 m) − −

$ 5 % T ! (0.6 m) " 13 #

$ 12 % T ! (0.15 m) + M C = 0 " 13 #

375 N ⋅ m + 180 N ⋅ m − T=

$ 4.8 % m !T + M C = 0 " 13 #

13 (555 + M C ) 4.8

For

M C = −100 N ⋅ m: T =

13 (555 − 100) = 1232 N 4.8

For

M C = +100 N ⋅ m: T =

13 (555 + 100) = 1774 N 4.8

For

|M C | # 100 N ⋅ m:

1.232 kN # T # 1.774 kN

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 398

PROBLEM 4.51 A vertical load P is applied at end B of rod BC. (a) Neglecting the weight of the rod, express the angle θ corresponding to the equilibrium position in terms of P, l, and the counterweight W. (b) Determine the value of θ corresponding to equilibrium if P = 2W .

SOLUTION (a)

Triangle ABC is isosceles. We have

CD = ( BC ) cos

θ 2

= l cos

θ 2

θ% $ ΣM C = 0: W l cos ! − P(l sin θ ) = 0 2# " Setting

sin θ = 2sin

θ

θ

θ

W − 2 P sin

(b)

For P = 2W :

sin

θ 2

θ 2

or

θ

θ

cos : Wl cos − 2 Pl sin cos = 0 2 2 2 2 2

θ 2

=

θ 2

=0

θ = 2sin −1 $

W % ! 2 " P#

W W = = 0.25 2 P 4W

θ = 29.0°

= 14.5° = 165.5° θ = 331°(discard)

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 399

PROBLEM 4.52 A slender rod AB, of weight W, is attached to blocks A and B, which move freely in the guides shown. The blocks are connected by an elastic cord that passes over a pulley at C. (a) Express the tension in the cord in terms of W and θ. (b) Determine the value of θ for which the tension in the cord is equal to 3W.

SOLUTION (a)

From f.b.d. of rod AB &$ 1 % ' ΣM C = 0: T (l sin θ ) + W ( ! cos θ ) − T (l cos θ ) = 0 2 *" # + T=

W cos θ 2(cos θ − sin θ )

Dividing both numerator and denominator by cos θ, T=

(b)

For T = 3W ,

or

3W =

W$ 1 % 2 " 1 − tan θ !#

or T =

( W2 ) (1 − tan θ )

( W2 )

(1 − tan θ ) 1 1 − tan θ = 6

θ = tan −1

$5% ! = 39.806° "6#

or

θ = 39.8°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 400

PROBLEM 4.53 Rod AB is acted upon by a couple M and two forces, each of magnitude P. (a) Derive an equation in θ, P, M, and l that must be satisfied when the rod is in equilibrium. (b) Determine the value of θ corresponding to equilibrium when M = 150 N · m, P = 200 N, and l = 600 mm.

SOLUTION Free-Body Diagram:

(a)

From free-body diagram of rod AB ΣM C = 0: P(l cos θ ) + P(l sin θ ) − M = 0 or sinθ + cosθ =

(b)

For

M Pl

M = 150 lb ⋅ in., P = 20 lb, and l = 6 in. 150 lb ⋅ in. 5 sin θ + cos θ = = = 1.25 (20 lb)(6 in.) 4 sin 2 θ + cos 2 θ = 1

Using identity

sin θ + (1 − sin 2 θ )1/2 = 1.25 (1 − sin 2 θ )1/2 = 1.25 − sin θ 1 − sin 2 θ = 1.5625 − 2.5sin θ + sin 2 θ 2sin 2 θ − 2.5sin θ + 0.5625 = 0

Using quadratic formula sin θ = =

or

−( −2.5) ± (625) − 4(2)(0.5625) 2(2) 2.5 ± 1.75 4

sin θ = 0.95572 and sin θ = 0.29428 θ = 72.886° and θ = 17.1144° or θ = 17.11° and θ = 72.9°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 401

PROBLEM 4.54 Rod AB is attached to a collar at A and rests against a small roller at C. (a) Neglecting the weight of rod AB, derive an equation in P, Q, a, l, and θ that must be satisfied when the rod is in equilibrium. (b) Determine the value of θ corresponding to equilibrium when P = 16 lb, Q = 12 lb, l = 20 in., and a = 5 in.

SOLUTION Free-Body Diagram:

ΣFy = 0: C cos θ − P − Q = 0 C= ΣM A = 0: C

(a)

P+Q cos θ

a − Pl cos θ = 0 cos θ

P+Q a ⋅ − Pl cos θ = 0 cos θ cos θ

(b)

For

cos3 θ =

a( P + Q) Pl

P = 16 lb, Q = 12 lb, l = 20 in., and a = 5 in. (5 in.)(16 lb + 12 lb) (16 lb)(20 in.) = 0.4375

cos3 θ =

cos θ = 0.75915

θ = 40.6°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 402

PROBLEM 4.55 A collar B of weight W can move freely along the vertical rod shown. The constant of the spring is k, and the spring is unstretched when θ = 0. (a) Derive an equation in θ, W, k, and l that must be satisfied when the collar is in equilibrium. (b) Knowing that W = 300 N, l = 500 mm, and k = 800 N/m, determine the value of θ corresponding to equilibrium.

SOLUTION First note

T = ks

Where

k = spring constant s = elongation of spring l = −l cos θ l (1 − cos θ ) = cos θ kl T= (1 − cos θ ) cos θ

(a)

From f.b.d. of collar B or

(b)

For

ΣFy = 0: T sin θ − W = 0 kl (1 − cos θ )sin θ − W = 0 cos θ

or tan θ − sin θ =

W kl

W = 3 lb l = 6 in. k = 8 lb/ft 6 in. = 0.5 ft l= 12 in./ft tan θ − sin θ =

Solving numerically,

3 lb = 0.75 (8 lb/ft)(0.5 ft)

θ = 57.957°

or

θ = 58.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 403

PROBLEM 4.56 A vertical load P is applied at end B of rod BC. The constant of the spring is k, and the spring is unstretched when θ = 90°. (a) Neglecting the weight of the rod, express the angle θ corresponding to equilibrium in terms of P, k, and l. (b) Determine the value of θ corresponding to equilibrium when P = 14 kl.

SOLUTION First note

T = tension in spring = ks

where

s = elongation of spring = ( AB)θ − ( AB)θ = 90° $θ % $ 90° % ! − 2l sin 2 ! 2 " # " # & $ θ % $ 1 %' = 2l (sin ! − !) * " 2 # " 2 #+ = 2l sin

& $ θ % $ 1 %' T = 2kl (sin !) !− * " 2 # " 2 #+

(a)

From f.b.d. of rod BC

(1)

& $ θ %' ΣM C = 0: T (l cos ! ) − P(l sin θ ) = 0 " 2 #+ *

Substituting T from Equation (1) & $ θ % $ 1 %' & $ θ %' 2kl (sin ! − ! ) (l cos ! ) − P ( l sin θ ) = 0 " 2 #+ * " 2 # " 2 #+ * & $ θ % $ 1 %' & $θ % $θ % $ θ %' 2kl 2 (sin ! − ! ) cos ! − Pl ( 2sin ! cos ! ) = 0 2 2 2 " # " # " 2 #+ * * " # " 2 #+

Factoring out

2l cos

$θ % ! , leaves "2#

& $ θ % $ 1 %' $θ % kl (sin ! ) − P sin ! = 0 !− 2 2 "2# #+ * " # "

or

sin

$ θ % 1 $ kl % != ! 2 " kl − P # "2#

&

' ) * 2(kl − P) +

θ = 2sin −1 (

kl

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 404

PROBLEM 4.56 (Continued)

(b)

P=

kl 4 &

kl ' kl ) kl − 2 (* ( 4 )) + & ' kl $ 4 % = 2sin −1 ( !) * 2 " 3kl # +

θ = 2sin −1 (

= 2sin −1

$ 4 % ! "3 2 #

= 2sin −1 (0.94281) = 141.058°

or θ = 141.1°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 405

PROBLEM 4.57 Solve Sample Problem 4.56, assuming that the spring is unstretched when θ = 90°.

SOLUTION First note

T = tension in spring = ks

where

s = deformation of spring = rβ F = kr β

From f.b.d. of assembly or

ΣM 0 = 0: W (l cos β ) − F (r ) = 0 Wl cos β − kr 2 β = 0 cos β =

For

kr 2 β Wl

k = 250 lb/in. r = 3 in. l = 8 in. W = 400 lb cos β =

or

(250 lb/in.)(3 in.)2 β (400 lb)(8 in.)

cos β = 0.703125β

Solving numerically,

β = 0.89245 rad

or

β = 51.134°

Then

θ = 90° + 51.134° = 141.134°

or θ = 141.1°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 406

PROBLEM 4.58 A slender rod AB, of weight W, is attached to blocks A and B that move freely in the guides shown. The constant of the spring is k, and the spring is unstretched when θ = 0. (a) Neglecting the weight of the blocks, derive an equation in W, k, l, and θ that must be satisfied when the rod is in equilibrium. (b) Determine the value of θ when W = 75 lb, l = 30 in., and k = 3 lb/in.

SOLUTION Free-Body Diagram:

Spring force:

Fs = ks = k (l − l cos θ ) = kl (1 − cos θ ) ΣM D = 0: Fs (l sin θ ) − W

(a)

$l % cos θ ! = 0 "2 #

kl (1 − cos θ )l sin θ − kl (1 − cos θ ) tan θ −

(b)

For given values of

W l cos θ = 0 2

W =0 2

or (1 − cos θ ) tan θ =

W 2kl

W = 75 lb l = 30 in. k = 3 lb/in. (1 − cos θ ) tan θ = tan θ − sin θ 75 lb = 2(3 lb/in.)(30 in.) = 0.41667

Solving numerically:

θ = 49.710°

or

θ = 49.7°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 407

PROBLEM 4.59 Eight identical 500 × 750-mm rectangular plates, each of mass m = 40 kg, are held in a vertical plane as shown. All connections consist of frictionless pins, rollers, or short links. In each case, determine whether (a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate or indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever possible, compute the reactions.

SOLUTION 1.

Three non-concurrent, non-parallel reactions (a)

Plate: completely constrained

(b)

Reactions: determinate

(c)

Equilibrium maintained A = C = 196.2 N

2.

Three non-concurrent, non-parallel reactions (a)

Plate: completely constrained

(b)

Reactions: determinate

(c)

Equilibrium maintained B = 0, C = D = 196.2 N

3.

Four non-concurrent, non-parallel reactions (a)

Plate: completely constrained

(b)

Reactions: indeterminate

(c)

Equilibrium maintained A x = 294 N

,

D x = 294 N

( A y + D y = 392 N )

4.

Three concurrent reactions (through D) (a)

Plate: improperly constrained

(b)

Reactions: indeterminate

(c)

No equilibrium

(ΣM D ≠ 0)

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 408

PROBLEM 4.59 (Continued)

5.

Two reactions (a)

Plate: partial constraint

(b)

Reactions: determinate

(c)

Equilibrium maintained C = R = 196.2 N

6.

Three non-concurrent, non-parallel reactions (a)

Plate: completely constrained

(b)

Reactions: determinate

(c)

Equilibrium maintained B = 294 N

7.

8.

, D = 491 N

53.1°

Two reactions (a)

Plate: improperly constrained

(b)

Reactions determined by dynamics

(c)

No equilibrium

(ΣFy ≠ 0)

For non-concurrent, non-parallel reactions (a)

Plate: completely constrained

(b)

Reactions: indeterminate

(c)

Equilibrium maintained B = D y = 196.2 N

(C + D x = 0)

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 409

PROBLEM 4.60 The bracket ABC can be supported in the eight different ways shown. All connections consist of smooth pins, rollers, or short links. For each case, answer the questions listed in Problem 4.59, and, wherever possible, compute the reactions, assuming that the magnitude of the force P is 100 lb.

SOLUTION 1.

Three non-concurrent, non-parallel reactions (a)

Bracket: complete constraint

(b)

Reactions: determinate

(c)

Equilibrium maintained A = 120.2 lb

2.

3.

56.3°, B = 66.7 lb

Four concurrent, reactions (through A) (a)

Bracket: improper constraint

(b)

Reactions: indeterminate

(c)

No equilibrium

(ΣM A ≠ 0)

Two reactions (a)

Bracket: partial constraint

(b)

Reactions: determinate

(c)

Equilibrium maintained A = 50 lb , C = 50 lb

4.

Three non-concurrent, non-parallel reactions (a)

Bracket: complete constraint

(b)

Reactions: determinate

(c)

Equilibrium maintained A = 50 lb , B = 83.3 lb

36.9°, C = 66.7 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 410

PROBLEM 4.60 (Continued)

5.

6.

Four non-concurrent, non-parallel reactions (a)

Bracket: complete constraint

(b)

Reactions: indeterminate

(c)

Equilibrium maintained

(ΣM C = 0) A y = 50 lb

Four non-concurrent, non-parallel reactions (a)

Bracket: complete constraint

(b)

Reactions: indeterminate

(c)

Equilibrium maintained A x = 66.7 lb , B = 66.7 lb

( A y + B y = 100 lb )

7.

Three non-concurrent, non-parallel reactions (a)

Bracket: complete constraint

(b)

Reactions: determinate

(c)

Equilibrium maintained A = C = 50 lb

8.

Three concurrent, reactions (through A) (a)

Bracket: improper constraint

(b)

Reactions: indeterminate

(c)

No equilibrium

(ΣM A ≠ 0)

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 411

PROBLEM 4.61 Determine the reactions at A and B when a = 180 mm.

SOLUTION Reaction at B must pass through D where A and 300-N load intersect.

∆BCD :

Free-Body Diagram: (Three-force member)

240 180 β = 53.13°

tan β =

Force triangle A = (300 N) tan 53.13° = 400 N

300 N cos 53.13° = 500 N !

A = 400 N

B=

B = 500 N

53.1°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 412

!

PROBLEM 4.62 For the bracket and loading shown, determine the range of values of the distance a for which the magnitude of the reaction at B does not exceed 600 N.

SOLUTION Reaction at B must pass through D where A and 300-N load intersect. Free-Body Diagram: (Three-force member) a=

240 mm tan β

(1)

Force Triangle (with B = 600 N) 300 N = 0.5 600 N β = 60.0°

cos β =

Eq. (1)

240 mm tan 60.0° = 138.56 mm

a=

For B # 600 N a $ 138.6 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 413

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PROBLEM 4.63 Using the method of Section 4.7, solve Problem 4.17. PROBLEM 4.17 The required tension in cable AB is 200 lb. Determine (a) the vertical force P that must be applied to the pedal, (b) the corresponding reaction at C.

SOLUTION Free-Body Diagram: (Three-Force body) Reaction at C must pass through E, where D and 200-lb force intersect. 6.062 in. 15 in. β = 22.005°

tan β =

Force triangle (a)

P = (200 lb)tan 22.005° P = 80.83 lb

(b)

C=

P = 80.8 lb

200 lb = 215.7 lb cos 22.005°

C = 216 lb

22.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 414

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PROBLEM 4.64 Using the method of Section 4.7, solve Problem 4.18. PROBLEM 4.18 Determine the maximum tension that can be developed in cable AB if the maximum allowable value of the reaction at C is 250 lb.

SOLUTION Free-Body Diagram: (Three -Force body)

Reaction at C must pass through E, where D and the force T intersect. 6.062 in. 15 in. β = 22.005°

tan β =

Force triangle T = (250 lb) cos 22.005° T = 231.8 lb !

T = 232 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 415

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PROBLEM 4.65 The spanner shown is used to rotate a shaft. A pin fits in a hole at A, while a flat, frictionless surface rests against the shaft at B. If a 60-lb force P is exerted on the spanner at D, find the reactions at A and B.

SOLUTION Free-Body Diagram: (Three-Force body)

The line of action of A must pass through D, where B and P intersect. 3sin 50° 3cos 50° + 15 = 0.135756 α = 7.7310°

tan α =

Force triangle

60 lb sin 7.7310° = 446.02 lb 60 lb B= tan 7.7310° = 441.97 lb A=

A = 446 lb

7.73°

B = 442 lb

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 416

PROBLEM 4.66 Determine the reactions at B and D when b = 60 mm.

SOLUTION Since CD is a two-force member, the line of action of reaction at D must pass through Points C and D. Free-Body Diagram: (Three-Force body)

Reaction at B must pass through E, where the reaction at D and 80-N force intersect. 220 mm 250 mm β = 41.348°

tan β =

Force triangle

Law of sines 80 N B D = = sin 3.652° sin 45° sin131.348° B = 888.0 N D = 942.8 N ! B = 888 N

41.3°

D = 943 N

45.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 417

PROBLEM 4.67 Determine the reactions at B and D when b = 120 mm.

SOLUTION Since CD is a two-force member, line of action of reaction at D must pass through C and D

.

Free-Body Diagram: (Three-Force body)

Reaction at B must pass through E, where the reaction at D and 80-N force intersect. 280 mm 250 mm β = 48.24°

tan β =

Force triangle

Law of sines

80 N B D = = sin 3.24° sin135° sin 41.76° B = 1000.9 N D = 942.8 N !

B = 1001 N

48.2° D = 943 N

45.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 418

PROBLEM 4.68 Determine the reactions at B and C when a = 1.5 in.

SOLUTION Since CD is a two-force member, the force it exerts on member ABD is directed along DC. Free-Body Diagram of ABD: (Three-Force member)

The reaction at B must pass through E, where D and the 50-lb load intersect. Triangle CFD:

Triangle EAD:

Triangle EGB:

Force triangle

3 = 0.6 5 α = 30.964°

tan α =

AE = 10 tan α = 6 in. GE = AE − AG = 6 − 1.5 = 4.5 in. GB 3 = GE 4.5 β = 33.690°

tan β =

B D 50 lb = = sin120.964° sin 33.690° sin 25.346° B = 100.155 lb D = 64.789 lb

!

B = 100.2 lb

56.3°

C = D = 64.8 lb

31.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 419

!

PROBLEM 4.69 A 50-kg crate is attached to the trolley-beam system shown. Knowing that a = 1.5 m, determine (a) the tension in cable CD, (b) the reaction at B.

SOLUTION

Free-Body Diagram:

Three-Force body: W and TCD intersect at E. 0.7497 m 1.5 m β = 26.56°

tan β =

Force triangle 3 forces intersect at E. W = (50 kg) 9.81 m/s 2 = 490.5 N

Law of sines

TCD 490.5 N B = = sin 61.56° sin 63.44° sin 55° TCD = 498.9 N B = 456.9 N

TCD = 499 N

(a)

B = 457 N

(b)

26.6°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 420

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PROBLEM 4.70 Solve Problem 4.69, assuming that a = 3 m. PROBLEM 4.69 A 50-kg crate is attached to the trolley-beam system shown. Knowing that a = 1.5 m, determine (a) the tension in cable CD, (b) the reaction at B.

SOLUTION W and TCD intersect at E Free-Body Diagram: Three-Force body:

AE 0.301 m = AB 3m β = 5.722°

tan β =

Force Triangle (Three forces intersect at E.) W = (50 kg) 9.81 m/s 2 = 490.5 N

Law of sines

TCD 490.5 N B = = sin 29.278° sin 95.722° sin 55° TCD = 997.99 N B = 821.59 N TCD = 998 N

(a)

B = 822 N

(b)

5.72°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 421

PROBLEM 4.71 One end of rod AB rests in the corner A and the other end is attached to cord BD. If the rod supports a 40-lb load at its midpoint C, find the reaction at A and the tension in the cord.

SOLUTION Free-Body Diagram: (Three-Force body)

The line of action of reaction at A must pass through E, where T and the 40-lb load intersect. EF 23 = AF 12 α = 62.447°

tan α =

5 EH = DH 12 β = 22.620°

tan β =

Force triangle

A T 40 lb = = sin 67.380° sin 27.553° sin 85.067°

A = 37.1 lb

62.4°

T = 18.57 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 422

PROBLEM 4.72 Determine the reactions at A and D when β = 30°.

SOLUTION From f.b.d. of frame ABCD ΣM D = 0: − A(0.18 m) + [(150 N) sin 30°](0.10 m) + [(150 N) cos 30°](0.28 m) = 0 A = 243.74 N

or A = 244 N

!

ΣFx = 0: (243.74 N) + (150 N) sin 30° + Dx = 0 Dx = −318.74 N ΣFy = 0: D y − (150 N) cos 30° = 0 Dy = 129.904 N

Then

D = ( Dx ) 2 + Dx2 = (318.74) 2 + (129.904) 2 = 344.19 N

and

Dy ! # $ Dx % 129.904 ! = tan −1 " # $ −318.74 %

θ = tan −1 "

= −22.174°

or D = 344 N

22.2°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 423

PROBLEM 4.73 Determine the reactions at A and D when β = 60°.

SOLUTION From f.b.d. of frame ABCD ΣM D = 0: − A(0.18 m) + [(150 N) sin 60°](0.10 m) + [(150 N) cos 60°](0.28 m) = 0 A = 188.835 N

or A = 188.8 N

!

ΣFx = 0: (188.835 N) + (150 N) sin 60° + Dx = 0 Dx = −318.74 N ΣFy = 0: D y − (150 N) cos 60° = 0 Dy = 75.0 N

Then

D = ( Dx ) 2 + ( Dy ) 2 = (318.74) 2 + (75.0) 2 = 327.44 N

and

Dy ! # $ Dx %

θ = tan −1 "

75.0 ! = tan −1 " # $ −318.74 % = −13.2409°

or D = 327 N

13.24°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 424

PROBLEM 4.74 A 40-lb roller, of diameter 8 in., which is to be used on a tile floor, is resting directly on the subflooring as shown. Knowing that the thickness of each tile is 0.3 in., determine the force P required to move the roller onto the tiles if the roller is (a) pushed to the left, (b) pulled to the right.

SOLUTION See solution to Problem 4.73 for free-body diagram and analysis leading to the following equations: T=

P 1 + cos θ

C=P

(1)

sin θ 1 + cos θ

(2)

For θ = 45° P P = 1 + cos 45° 1.7071

Eq. (1):

T=

Eq. (2):

C=P

sin 45° 0.7071 =P 1 + cos 45° 1.7071

T = 0.586 P C = 0.444P

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 425

PROBLEM 4.75 Member ABC is supported by a pin and bracket at B and by an inextensible cord attached at A and C and passing over a frictionless pulley at D. The tension may be assumed to be the same in portions AD and CD of the cord. For the loading shown and neglecting the size of the pulley, determine the tension in the cord and the reaction at B.

SOLUTION Free-Body Diagram:

Reaction at B must pass through D. 7 in. 12 in. α = 30.256° 7 in. tan β = 24 in. β = 16.26° tan α =

Force triangle

Law of sines

T T − 72 lb B = = sin 59.744° sin13.996° sin106.26 T (sin13.996°) = (T − 72 lb)(sin 59.744°) T (0.24185) = (T − 72)(0.86378) T = 100.00 lb sin 106.26° sin 59.744 = 111.14 lb

T = 100.0 lb

B = (100 lb)

B = 111.1 lb

30.3°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 426

PROBLEM 4.76 Member ABC is supported by a pin and bracket at B and by an inextensible cord attached at A and C and passing over a frictionless pulley at D. The tension may be assumed to be the same in portions AD and CD of the cord. For the loading shown and neglecting the size of the pulley, determine the tension in the cord and the reaction at B.

SOLUTION Free-Body Diagram: Force triangle

Reaction at B must pass through D. tan α =

120 ; α = 36.9° 160

T T − 75 lb B = = 4 3 5 3T = 4T − 300; T = 300 lb 5 5 B = T = (300 lb) = 375 lb 4 4

B = 375 lb

36.9°!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 427

PROBLEM 4.77 Rod AB is supported by a pin and bracket at A and rests against a frictionless peg at C. Determine the reactions at A and C when a 170-N vertical force is applied at B.

SOLUTION Free-Body Diagram: (Three-Force body) The reaction at A must pass through D where C and 170-N force intersect. 160 mm 300 mm α = 28.07°

tan α =

We note that triangle ABD is isosceles (since AC = BC) and, therefore CAD = α = 28.07°

Also, since CD ⊥ CB, reaction C forms angle α = 28.07° with horizontal.

Force triangle We note that A forms angle 2α with vertical. Thus A and C form angle 180° − (90° − α ) − 2α = 90° − α

Force triangle is isosceles and we have A = 170 N C = 2(170 N)sin α = 160.0 N A = 170.0 N

33.9°

C = 160.0 N

28.1°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 428

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PROBLEM 4.78 Solve Problem 4.77, assuming that the 170-N force applied at B is horizontal and directed to the left. PROBLEM 4.77 Rod AB is supported by a pin and bracket at A and rests against a frictionless peg at C. Determine the reactions at A and C when a 170-N vertical force is applied at B.

SOLUTION Free-Body Diagram: (Three-Force body)

The reaction at A must pass through D, where C and the 170-N force intersect. 160 mm 300 mm α = 28.07°

tan α =

We note that triangle ADB is isosceles (since AC = BC). Therefore Also

A=

B = 90° − α .

ADB = 2α

Force triangle The angle between A and C must be 2α − α = α

α = 28.07°

Thus, force triangle is isosceles and A = 170.0 N C = 2(170 N) cos α = 300 N A = 170.0 N

56.1°

C = 300 N

28.1°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 429

!

PROBLEM 4.79 Using the method of Section 4.7, solve Problem 4.21. PROBLEM 4.21 Determine the reactions at A and B when (a) h = 0, (b) h = 200 mm.

SOLUTION ! !

Free-Body Diagram:

(a)

h=0 Reaction A must pass through C where 150-N weight and B interect. Force triangle is equilateral

(b)

A = 150.0 N

30.0°

B = 150.0 N

30.0°

h = 200 mm 55.662 250 β = 12.552°

!

tan β =

! Force triangle Law of sines

A B 150 N = = sin17.448° sin 60° sin102.552° A = 433.247 N B = 488.31 N

! !

A = 433 N

!

B = 488 N

12.55° 30.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 430

!

PROBLEM 4.80 Using the method of Section 4.7, solve Problem 4.28. PROBLEM 4.28 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 500-N horizontal force at B, determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION Free-Body Diagram: (Three-Force body) Reaction at C must pass through E, where FAD and 500-N force intersect. Since AC = CD = 250 mm, triangle ACD is isosceles. We have

C = 90° + 30° = 120°

and

A=

D=

1 (180° − 120°) = 30° 2

Dimensions in mm

On the other hand, from triangle BCF: CF = ( BC )sin 30° = 200 sin 30° = 100 mm FD = CD − CF = 250 − 100 = 150 mm

From triangle EFD, and since

D = 30° : EF = ( FD ) tan 30° = 150 tan 30° = 86.60 mm

From triangle EFC:

100 mm CF = EF 86.60 mm α = 49.11°

tan α =

Force triangle Law of sines

FAD C 500 N = = sin 49.11° sin 60° sin 70.89° FAD = 400 N, C = 458 N FAD = 400 N

(a)

C = 458 N

(b)

49.1°

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 431

PROBLEM 4.81 Knowing that θ = 30°, determine the reaction (a) at B, (b) at C.

SOLUTION

Reaction at C must pass through D where force P and reaction at B intersect. In ∆ CDE:

Free-Body Diagram: (Three-Force body)

( 3 − 1) R R = 3 −1 β = 36.2°

tan β =

Force triangle

Law of sines

P B C = = sin 23.8° sin126.2° sin 30° B = 2.00 P C = 1.239 P

(a)

B = 2P

60.0°

(b)

C = 1.239P

36.2°

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 432

PROBLEM 4.82 Knowing that θ = 60°, determine the reaction (a) at B, (b) at C.

SOLUTION Reaction at C must pass through D where force P and reaction at B intersect. In ∆CDE:

tan β =

R−

=1−

R 3

Free-Body Diagram: (Three-Force body)

R 1 3

β = 22.9° Force triangle

Law of sines

P B C = = sin 52.9° sin 67.1° sin 60° B = 1.155P C = 1.086 P

(a)

B = 1.155P

30.0°

(b)

C = 1.086P

22.9°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 433

!

PROBLEM 4.83 Rod AB is bent into the shape of an arc of circle and is lodged between two pegs D and E. It supports a load P at end B. Neglecting friction and the weight of the rod, determine the distance c corresponding to equilibrium when a = 20 mm and R = 100 mm.

SOLUTION Free-Body Diagram: yED = xED = a,

Since Slope of ED is slope of HC is

45° 45°

Also

DE = 2 a

and

a 1! DH = HE = " # DE = 2 $2%

For triangles DHC and EHC

sin β =

a 2

R

=

a 2R

Now

c = R sin(45° − β )

For

a = 20 mm and R = 100 mm sin β =

20 mm

2(100 mm) = 0.141421 β = 8.1301°

and

c = (100 mm) sin(45° − 8.1301°) = 60.00 mm

or c = 60.0 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 434

!

PROBLEM 4.84 A slender rod of length L is attached to collars that can slide freely along the guides shown. Knowing that the rod is in equilibrium, derive an expression for the angle θ in terms of the angle β.

SOLUTION As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the force geometry Free-Body Diagram: tan β =

xGB y AB

where y AB = L cos θ

and xGB = tan β =

1 L sin θ 2 1 2

L sin θ

L cos θ 1 = tan θ 2

or tan θ = 2 tan β

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 435

!

PROBLEM 4.85 An 8-kg slender rod of length L is attached to collars that can slide freely along the guides shown. Knowing that the rod is in equilibrium and that β = 30°, determine (a) the angle θ that the rod forms with the vertical, (b) the reactions at A and B.

SOLUTION (a)

As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the geometry of the forces Free-Body Diagram: tan β =

xCB yBC

where xCB =

1 L sin θ 2

and yBC = L cos θ tan β =

1 tan θ 2

or

tan θ = 2 tan β

For

β = 30° tan θ = 2 tan 30° = 1.15470 θ = 49.107°

or

θ = 49.1°

!

W = mg = (8 kg)(9.81 m/s2 ) = 78.480 N

(b) From force triangle

A = W tan β = (78.480 N) tan 30° = 45.310 N

and

B=

or

W 78.480 N = = 90.621 N cos β cos 30°

A = 45.3 N

or B = 90.6 N

!

60.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 436

!

PROBLEM 4.86 A slender uniform rod of length L is held in equilibrium as shown, with one end against a frictionless wall and the other end attached to a cord of length S. Derive an expression for the distance h in terms of L and S. Show that this position of equilibrium does not exist if S . 2L.

SOLUTION !

From the f.b.d. of the three-force member AB, forces must intersect at D. Since the force T intersects Point D, directly above G, yBE = h

For triangle ACE:

S 2 = ( AE ) 2 + (2h) 2

(1)

For triangle ABE:

L2 = ( AE ) 2 + (h) 2

(2)

Subtracting Equation (2) from Equation (1) S 2 − L2 = 3h 2

!

(3)

! h=

S 2 − L2 3

!

or

!

As length S increases relative to length L, angle θ increases until rod AB is vertical. At this vertical position:

!

h+L=S

or h = S − L

Therefore, for all positions of AB h$S − L

or

S 2 − L2 $S −L 3

or

S 2 − L2 $ 3( S − L) 2

(4)

= 3( S 2 − 2 SL + L2 )

!

= 3S 2 − 6 SL + 3L2

or

0 $ 2S 2 − 6SL + 4 L2

and

0 $ S 2 − 3SL + 2 L2 = ( S − L)( S − 2 L)

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 437

PROBLEM 4.86 (Continued)

For

S−L=0 S =L

Minimum value of S is L For

S − 2L = 0 S = 2L

Maximum value of S is 2L Therefore, equilibrium does not exist if S . 2 L

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 438

!

PROBLEM 4.87 A slender uniform rod of length L = 20 in. is held in equilibrium as shown, with one end against a frictionless wall and the other end attached to a cord of length S = 30 in. Knowing that the weight of the rod is 10 lb, determine (a) the distance h, (b) the tension in the cord, (c) the reaction at B.

SOLUTION From the f.b.d. of the three-force member AB, forces must intersect at D. Since the force T intersects Point D, directly above G, yBE = h

For triangle ACE:

S 2 = ( AE ) 2 + (2h) 2

(1)

For triangle ABE:

L2 = ( AE ) 2 + (h) 2

(2)

Subtracting Equation (2) from Equation (1) S 2 − L2 = 3h 2

! !

h=

or

! (a)

S 2 − L2 3

L = 20 in. and S = 30 in.

For

(30) 2 − (20) 2 3 = 12.9099 in.

h=

(b) !

W = 10 lb

We have

2h ! # $ s %

θ = sin −1 "

and

!

or h = 12.91 in.

& 2(12.9099) ' = sin −1 ( ) 30 * + θ = 59.391°

! ! ! ! !

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 439

PROBLEM 4.87 (Continued)

From the force triangle W sin θ 10 lb = sin 59.391° = 11.6190 lb

T=

(c)

or

T = 11.62 lb

W tan θ 10 lb = tan 59.391°

B=

or B = 5.92 lb

= 5.9161 lb

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!

PROBLEM 4.88 A uniform rod AB of length 2R rests inside a hemispherical bowl of radius R as shown. Neglecting friction, determine the angle θ corresponding to equilibrium.

SOLUTION Based on the f.b.d., the uniform rod AB is a three-force body. Point E is the point of intersection of the three forces. Since force A passes through O, the center of the circle, and since force C is perpendicular to the rod, triangle ACE is a right triangle inscribed in the circle. Thus, E is a point on the circle. Note that the angle α of triangle DOA is the central angle corresponding to the inscribed angle θ of triangle DCA.

α = 2θ The horizontal projections of AE , ( x AE ), and AG , ( x AG ), are equal. x AE = x AG = x A

or

( AE ) cos 2θ = ( AG ) cos θ

and

(2 R) cos 2θ = R cos θ cos 2θ = 2 cos 2 θ − 1

Now then or

4 cos 2 θ − 2 = cos θ 4 cos 2 θ − cos θ − 2 = 0

Applying the quadratic equation cos θ = 0.84307 and cos θ = − 0.59307

θ = 32.534° and θ = 126.375°(Discard)

or θ = 32.5°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 441

!

PROBLEM 4.89 A slender rod of length L and weight W is attached to a collar at A and is fitted with a small wheel at B. Knowing that the wheel rolls freely along a cylindrical surface of radius R, and neglecting friction, derive an equation in θ, L, and R that must be satisfied when the rod is in equilibrium.

SOLUTION Free-Body Diagram (Three-Force body) Reaction B must pass through D where B and W intersect. Note that ∆ABC and ∆BGD are similar. AC = AE = L cos θ

In ∆ABC:

(CE ) 2 + ( BE )2 = ( BC )2 (2 L cos θ ) 2 + ( L sin θ )2 = R 2 2

R! 2 2 " # = 4cos θ + sin θ $L% 2

R! 2 2 " # = 4cos θ + 1 − cos θ $L% 2

R! 2 " # = 3cos θ + 1 $L% 2 ' 1& cos 2 θ = (" R !# − 1) 3 *$ L % +

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 442

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PROBLEM 4.90 Knowing that for the rod of Problem 4.89, L = 15 in., R = 20 in., and W = 10 lb, determine (a) the angle θ corresponding to equilibrium, (b) the reactions at A and B.

SOLUTION See the solution to Problem 4.89 for free-body diagram and analysis leading to the following equation 2 ' 1& cos 2 θ = (" R !# − 1) 3 *$ L % +

For L = 15 in., R = 20 in., and W = 10 lb. 2 ' 1 & 20 in. ! cos 2 θ = (" # − 1) ; θ = 59.39° 3 ($ 15 in. % ) * +

(a)

In ∆ABC:

θ = 59.4°

BE L sin θ 1 = = tan θ CE 2 L cos θ 2 1 tan α = tan 59.39° = 0.8452 2 α = 40.2° tan α =

Force triangle

A = W tan α = (10 lb) tan 40.2° = 8.45 lb W (10 lb) = = 13.09 lb B= cos α cos 40.2° A = 8.45 lb

(b)

B = 13.09 lb

(c)

49.8°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 443

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PROBLEM 4.91 A 4 × 8-ft sheet of plywood weighing 34 lb has been temporarily placed among three pipe supports. The lower edge of the sheet rests on small collars at A and B and its upper edge leans against pipe C. Neglecting friction at all surfaces, determine the reactions at A, B, and C.

SOLUTION

rG/B =

3.75 1.3919 i+ j+k 2 2

We have 5 unknowns and 6 Eqs. of equilibrium. Plywood sheet is free to move in z direction, but equilibrium is maintained (ΣFz = 0). ΣM B = 0: rA/B × ( Ax i + Ay j) + rC/B × ( −Ci ) + rG/B × ( − wj) = 0 i 0 Ax

j 0 Ay

k i j k i j k 4 + 3.75 1.3919 2 + 1.875 0.696 1 = 0 0 0 0 0 −C −34 0 −4 Ay i + 4 Ax j − 2Cj + 1.3919Ck + 34i − 63.75k = 0

Equating coefficients of unit vectors to zero: i:

− 4 Ay + 34 = 0

j:

− 2C + 4 Ax = 0

k : 1.3919C − 63.75 = 0

Ay = 8.5 lb Ax =

1 1 C = (45.80) = 22.9 lb 2 2

C = 45.80 lb

C = 45.8 lb

ΣFx = 0:

Ax + Bx − C = 0:

Bx = 45.8 − 22.9 = 22.9 lb

ΣFy = 0:

Ay + By − W = 0:

By = 34 − 8.5 = 25.5 lb

A = (22.9 lb)i + (8.5 lb) j B = (22.9 lb)i + (25.5 lb) j C = −(45.8 lb)i PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 444

PROBLEM 4.92 Two tape spools are attached to an axle supported by bearings at A and D. The radius of spool B is 30 mm and the radius of spool C is 40 mm. Knowing that TB = 80 N and that the system rotates at a constant rate, determine the reactions at A and D. Assume that the bearing at A does not exert any axial thrust and neglect the weights of the spools and axle.

SOLUTION Dimensions in mm

We have six unknowns and six Eqs. of equilibrium. ΣM A = 0: (90i + 30k ) × (−80 j) + (210i + 40 j) × (−TC k ) + (300i ) × ( Dx i + Dy j + Dz k ) = 0 −7200k + 2400i + 210TC j − 40TC i + 300 D y k − 300 Dz j = 0

Equate coefficients of unit vectors to zero: i: j:

2400 − 40TC = 0

TC = 60 N

210TC − 300 Dz = 0 (210)(60) − 300 Dz = 0

Dz = 42 N

k : −7200 + 300 Dy = 0

Dy = 24 N

ΣFx = 0:

Dx = 0

ΣFy = 0:

Ay + D y − 80 N = 0

Ay = 80 − 24 = 56 N

ΣFz = 0: Az + Dz − 60 N = 0

Az = 60 − 42 = 18 N

A = (56.0 N) j + (18.00 N)k

D = (24.0 N) j + (42.0 N)k

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PROBLEM 4.93 Solve Problem 4.92, assuming that the spool C is replaced by a spool of radius 50 mm. PROBLEM 4.92 Two tape spools are attached to an axle supported by bearings at A and D. The radius of spool B is 30 mm and the radius of spool C is 40 mm. Knowing that TB = 80 N and that the system rotates at a constant rate, determine the reactions at A and D. Assume that the bearing at A does not exert any axial thrust and neglect the weights of the spools and axle.

SOLUTION Dimensions in mm

We have six unknowns and six Eqs. of equilibrium. ΣM A = 0: (90i + 30k ) × (−80 j) + (210i + 50 j) × ( −TC k ) + (300i) × ( Dx i + D y j + Dz k ) = 0 −7200k + 2400i + 210TC j − 50TC i + 300 Dy k − 300 Dz j = 0

Equate coefficients of unit vectors to zero: i: j:

2400 − 50TC = 0

TC = 48 N

210TC − 300 Dz = 0 (210)(48) − 300 Dz = 0

Dz = 33.6 N

k : − 7200 + 300 Dy = 0

ΣFx = 0:

Dy = 24 N

Dx = 0

ΣFy = 0: Ay + Dy − 80 N = 0

Ay = 80 − 24 = 56 N

ΣFz = 0:

Az = 48 − 33.6 = 14.4 N

Az + Dz − 48 = 0

A = (56.0 N) j + (14.40 N)k

D = (24.0 N) j + (33.6 N)k

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 446

PROBLEM 4.94 Two transmission belts pass over sheaves welded to an axle supported by bearings at B and D. The sheave at A has a radius of 2.5 in., and the sheave at C has a radius of 2 in. Knowing that the system rotates at a constant rate, determine (a) the tension T, (b) the reactions at B and D. Assume that the bearing at D does not exert any axial thrust and neglect the weights of the sheaves and axle.

SOLUTION

Assume moment reactions at the bearing supports are zero. From f.b.d. of shaft (a) (b)

ΣM x -axis = 0: (24 lb − 18 lb)(5 in.) + (30 lb − T )(4 in.) = 0

T = 37.5 lb

ΣFx = 0: Bx = 0 ΣM D ( z -axis) = 0: (30 lb + 37.5 lb)(6 in.) − By (12 in.) = 0 By = 33.75 lb

ΣM D ( y -axis) = 0: (24 lb + 18 lb)(20 in.) + Bz (12 in.) = 0 Bz = −70.0 lb

or B = (33.8 lb) j − (70.0 lb)k ΣM B ( z -axis) = 0: − (30 lb + 37.5 lb)(6 in.) + D y (12 in.) = 0 Dy = 33.75 lb

ΣM B ( y -axis) = 0: (24 lb + 18 lb)(8 in.) + Dz (12 in.) = 0 Dz = −28.0 lb !

or D = (33.8 lb) j − (28.0 lb)k

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 447

PROBLEM 4.95 A 200-mm lever and a 240-mm-diameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N vertical load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.

SOLUTION Dimensions in mm

We have six unknowns and six Eqs. of equilibrium–OK ΣM C = 0: (−120k ) × ( Dx i + Dy j) + (120 j − 160k ) × Ti + (80k − 200i ) × (−720 j) = 0

−120 Dx j + 120 D y i − 120Tk − 160Tj + 57.6 × 103 i + 144 × 103 k = 0

Equating to zero the coefficients of the unit vectors: k:

−120T + 144 × 103 = 0

i:

120 Dy + 57.6 × 103 = 0

j: − 120 Dx − 160(1200 N) = 0

(b)

ΣFx = 0:

Cx + Dx + T = 0

ΣFy = 0:

C y + Dy − 720 = 0

ΣFz = 0:

Cz = 0

(a) T = 1200 N Dy = −480 N Dx = −1600 N Cx = 1600 − 1200 = 400 N C y = 480 + 720 = 1200 N

C = (400 N)i + (1200 N) j D = −(1600 N)i − (480 N) j

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 448

PROBLEM 4.96 Solve Problem 4.95, assuming that the axle has been rotated clockwise in its bearings by 30° and that the 720-N load remains vertical. PROBLEM 4.95 A 200-mm lever and a 240-mmdiameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N vertical load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.

SOLUTION Dimensions in mm

We have six unknowns and six Eqs. of equilibrium. ΣM C = 0: (−120k ) × ( Dx i + Dy j) + (120 j − 160k ) × T i + (80k − 173.21i ) × (−720 j) = 0 −120 Dx j + 120 D y i −120T k −160T j + 57.6 × 103 i + 124.71 × 103 k = 0

Equating to zero the coefficients of the unit vectors: k : − 120T + 124.71 × 103 = 0 i:

T = 1039 N

120 Dy + 57.6 × 103 = 0 Dy = −480 N

j: − 120 Dx − 160(1039.2)

(b)

T = 1039.2 N

ΣFx = 0:

Cx + Dx + T = 0

ΣFy = 0:

C y + Dy − 720 = 0

ΣFz = 0:

Cz = 0

Dx = −1385.6 N Cx = 1385.6 − 1039.2 = 346.4 C y = 480 + 720 = 1200 N

C = (346 N)i + (1200 N) j D = −(1386 N)i − (480 N) j

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 449

PROBLEM 4.97 An opening in a floor is covered by a 1 × 1.2-m sheet of plywood of mass 18 kg. The sheet is hinged at A and B and is maintained in a position slightly above the floor by a small block C. Determine the vertical component of the reaction (a) at A, (b) at B, (c) at C.

SOLUTION rB/A = 0.6i rC/A = 0.8i + 1.05k rG/A = 0.3i + 0.6k W = mg = (18 kg)9.81 W = 176.58 N ΣM A = 0: rB/A × Bj + rC/A × Cj + rG/A × (−Wj) = 0 (0.6i ) × Bj + (0.8i + 1.05k ) × Cj + (0.3i + 0.6k ) × ( −Wj) = 0 0.6 Bk + 0.8Ck − 1.05Ci − 0.3Wk + 0.6Wi = 0

Equate coefficients of unit vectors of zero: 0.6 ! i : 1.05C + 0.6W = 0 C = " #176.58 N = 100.90 N 1.05 $ % k : 0.6 B + 0.8C − 0.3W = 0 0.6 B + 0.8(100.90 N) − 0.3(176.58 N) = 0 B = −46.24 N ΣFy = 0: A + B + C − W = 0 A − 46.24 N + 100.90 N + 176.58 N = 0

A = 121.92 N

(a ) A = 121.9 N (b) B = −46.2 N (c) C = 100.9 N

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PROBLEM 4.98 Solve Problem 4.97, assuming that the small block C is moved and placed under edge DE at a point 0.15 m from corner E. PROBLEM 4.97 An opening in a floor is covered by a 1 × 1.2-m sheet of plywood of mass 18 kg. The sheet is hinged at A and B and is maintained in a position slightly above the floor by a small block C. Determine the vertical component of the reaction (a) at A, (b) at B, (c) at C.

SOLUTION rB/A = 0.6i rC/A = 0.65i + 1.2k rG/A = 0.3i + 0.6k W = mg = (18 kg) 9.81 m/s 2 W = 176.58 N

ΣM A = 0: rB/A × Bj + rC/A × Cj + rG/A × (−Wj) = 0 0.6i × Bj + (0.65i + 1.2k ) × Cj + (0.3i + 0.6k ) × (−Wj) = 0 0.6 Bk + 0.65Ck − 1.2Ci − 0.3Wk + 0.6Wi = 0

Equate coefficients of unit vectors to zero: 0.6 ! i : −1.2C + 0.6W = 0 C = " #176.58 N = 88.29 N $ 1.2 % k : 0.6 B + 0.65C − 0.3W = 0 0.6 B + 0.65(88.29 N) − 0.3(176.58 N) = 0 B = −7.36 N ΣFy = 0: A + B + C − W = 0 A − 7.36 N + 88.29 N − 176.58 N = 0

A = 95.648 N

(a ) A = 95.6 N (b) − 7.36 N (c) 88.3 N

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PROBLEM 4.99 The rectangular plate shown weighs 80 lb and is supported by three vertical wires. Determine the tension in each wire.

SOLUTION Free-Body Diagram:

ΣM B = 0: rA/B × TA j + rC/B × TC j + rG/B × (−80 lb) j = 0 (60 in.)k × TA j + [(60 in.)i + (15 in.)k ] × TC j + [(30 in.)i + (30 in.)k ] × ( −80 lb) j = 0 −60TAi + 60TC k − 15TC i − 2400k + 2400i = 0

Equating to zero the coefficients of the unit vectors: i:

60TA − 15(40) + 2400 = 0

TA = 30.0 lb

k:

60TC − 2400 = 0

TC = 40.0 lb

ΣFy = 0:

TA + TB + TC − 80 lb = 0

30 lb + TB + 40 lb − 80 lb = 0

TB = 10.00 lb

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PROBLEM 4.100 The rectangular plate shown weighs 80 lb and is supported by three vertical wires. Determine the weight and location of the lightest block that should be placed on the plate if the tensions in the three wires are to be equal.

SOLUTION Free-Body Diagram:

Let −Wb j be the weight of the block and x and z the block’s coordinates. Since tensions in wires are equal, let TA = TB = TC = T

ΣM 0 = 0: (rA × Tj) + (rB × Tj) + (rC × Tj) + rG × (−Wj) + ( xi + zk ) × ( −Wb j) = 0

or,

(75 k ) × Tj + (15 k ) × Tj + (60i + 30k ) × Tj + (30i + 45k ) × (−Wj) + ( xi + zk ) × (−Wb j) = 0

or,

−75Ti − 15T i + 60T k − 30T i − 30W k + 45W i − Wb × k + Wb zi = 0

Equate coefficients of unit vectors to zero: i:

−120T + 45W + Wb z = 0

(1)

k:

60T − 30W − Wb x = 0

(2)

ΣFy = 0:

3T − W − Wb = 0

(3)

Eq. (1) + 40 Eq. (3):

5W + ( z − 40)Wb = 0

(4)

Eq. (2) – 20 Eq. (3):

−10W − ( x − 20)Wb = 0

(5)

Also,

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PROBLEM 4.100 (Continued)

Solving (4) and (5) for Wb /W and recalling of 0 # x # 60 in., 0 # z # 90 in., (4):

Wb 5 5 = $ = 0.125 W 40 − z 40 − 0

(5):

Wb 10 10 = $ = 0.5 W 20 − x 20 − 0

Thus, (Wb ) min = 0.5W = 0.5(80) = 40 lb

(Wb ) min = 40.0 lb

Making Wb = 0.5W in (4) and (5): 5W + ( z − 40)(0.5W ) = 0

z = 30.0 in.

−10W − ( x − 20)(0.5W ) = 0

x = 0 in.

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PROBLEM 4.101 Two steel pipes AB and BC, each having a mass per unit length of 8 kg/m, are welded together at B and supported by three wires. Knowing that a = 0.4 m, determine the tension in each wire.

SOLUTION

W1 = 0.6m′g W2 = 1.2m′g

ΣM D = 0: rA/D × TA j + rE/D × (−W1 j) + rF/D × (−W2 j) + rC/D × TC j = 0 (−0.4i + 0.6k ) × TA j + (−0.4i + 0.3k ) × (−W1 j) + 0.2i × (−W2 j) + 0.8i × TC j = 0 −0.4TAk − 0.6TA i + 0.4W1k + 0.3W1i − 0.2W2 k + 0.8TC k = 0

Equate coefficients of unit vectors to zero: 1 1 i : − 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m′g = 0.3m′g 2 2

k : − 0.4TA + 0.4W1 − 0.2W2 + 0.8TC = 0 −0.4(0.3m′g ) + 0.4(0.6m′g ) − 0.2(1.2m′g ) + 0.8TC = 0 TC =

(0.12 − 0.24 − 0.24)m′g = 0.15m′g 0.8

ΣFy = 0: TA + TC + TD − W1 − W2 = 0 0.3m′g + 0.15m′g + TD − 0.6m′g − 1.2m′g = 0 TD = 1.35m′g 2

m′g = (8 kg/m)(9.81m/s ) = 78.48 N/m TA = 0.3m′g = 0.3 × 78.45

TA = 23.5 N

TB = 0.15m′g = 0.15 × 78.45

TB = 11.77 N

TC = 1.35m′g = 1.35 × 78.45

TC = 105.9 N

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PROBLEM 4.102 For the pipe assembly of Problem 4.101, determine (a) the largest permissible value of a if the assembly is not to tip, (b) the corresponding tension in each wire.

SOLUTION

W1 = 0.6m′g W2 = 1.2m′g

ΣM D = 0: rA/D × TA j + rE/D × (−W1 j) + rF/D × (−W2 j) + rC/D × TC j = 0 (− ai + 0.6k ) × TA j + (− ai + 0.3k ) × (−W1 j) + (0.6 − a)i × (−W2 j) + (1.2 − a)i × TC j = 0 −TA ak − 0.6TA i + W1ak + 0.3W1i − W2 (0.6 − a)k + TC (1.2 − a )k = 0

Equate coefficients of unit vectors to zero: 1 1 i : − 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m′g = 0.3m′g 2 2

k : − TA a + W1a − W2 (0.6 − a ) + TC (1.2 − a) = 0 −0.3m′ga + 0.6m′ga − 1.2m′g (0.6 − a ) + TC (1.2 − a) = 0 TC =

(a)

0.3a − 0.6a + 1.2(0.6 − a) 1.2 − a

For Max a and no tipping, TC = 0

−0.3a + 1.2(0.6 − a) = 0 −0.3a + 0.72 − 1.2a = 0 1.5a = 0.72

a = 0.480 m

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PROBLEM 4.102 (Continued)

(b)

Reactions:

m′g = (8 kg/m) 9.81 m/s 2 = 78.48 N/m TA = 0.3m′g = 0.3 × 78.48 = 23.544 N

TA = 23.5 N

ΣFy = 0: TA + TC + TD − W1 − W2 = 0 TA + 0 + TD − 0.6m′g − 1.2m′g = 0 TD = 1.8m′g − TA = 1.8 × 78.48 − 23.544 = 117.72

TD = 117.7 N

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PROBLEM 4.103 The 24-lb square plate shown is supported by three vertical wires. Determine (a) the tension in each wire when a = 10 in., (b) the value of a for which the tension in each wire is 8 lb.

SOLUTION rB/A = ai + 30k rC/A = 30i + ak rG/A = 15i + 15k

By symmetry: B = C ΣM A = 0: rB/A × Bj + rC × Cj + rG/A × (−Wj) = 0 (ai + 30k ) × Bj + (30i + ak ) × Bj + (15i + 15k ) × (−Wj) = 0 Bak − 30 Bi + 30 Bk − Bai − 15Wk + 15Wi = 0

Equate coefficient of unit vector i to zero: i : − 30 B − Ba + 15W = 0 B=

15W 30 + a

C=B=

15W 30 + a

(1)

ΣFy = 0: A + B + C − W = 0

& 15W ' A+ 2( ) − W = 0; * 30 + a +

(a)

For

a = 10 in.

Eq. (1)

C=B=

Eq. (2)

A=

A=

aW 30 + a

(2)

15(24 lb) = 9.00 lb 30 + 10

10(24 lb) = 6.00 lb 30 + 10

A = 6.00 lb B = C = 9.00 lb

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!

PROBLEM 4.103 (Continued)

(b)

For tension in each wire = 8 lb Eq. (1)

8 lb =

15(24 lb) 30 + a

30 in. + a = 45

a = 15.00 in.

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PROBLEM 4.104 The table shown weighs 30 lb and has a diameter of 4 ft. It is supported by three legs equally spaced around the edge. A vertical load P of magnitude 100 lb is applied to the top of the table at D. Determine the maximum value of a if the table is not to tip over. Show, on a sketch, the area of the table over which P can act without tipping the table.

SOLUTION r = 2 ft b = r sin 30° = 1 ft

We shall sum moments about AB. (b + r )C + (a − b) P − bW = 0 (1 + 2)C + (a − 1)100 − (1)30 = 0 1 C = [30 − (a − 1)100] 3

If table is not to tip, C $ 0 [30 − ( a − 1)100] $ 0 30 $ (a − 1)100 a − 1 # 0.3 a # 1.3 ft a = 1.300 ft

Only ⊥ distance from P to AB matters. Same condition must be satisfied for each leg. P must be located in shaded area for no tipping

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PROBLEM 4.105 A 10-ft boom is acted upon by the 840-lb force shown. Determine the tension in each cable and the reaction at the ball-and-socket joint at A.

SOLUTION We have five unknowns and six Eqs. of equilibrium but equilibrium is maintained (ΣM x = 0). Free-Body Diagram: ! BD = (−6 ft)i + (7 ft) j + (6 ft)k BD = 11 ft BE = (−6 ft)i + (7 ft) j − (6 ft)k BE = 11 ft ! BD TBD TBD = TBD (−6i + 7 j + 6k ) = BD 11 ! BE TBE TBE = TBE (−6i + 7 j − 6k ) = BE 11 ΣM A = 0: rB × TBD + rB × TBE + rC × ( −840 j) = 0 6i ×

TBD T (−6i + 7 j + 6k ) + 6i × BE (−6i + 7 j − 6k ) + 10i × (−840 j) = 0 11 11 42 36 42 36 TBD k − TBD j + TBE k + TBE j − 8400k 11 11 11 11

Equate coefficients of unit vectors to zero. i: − k:

36 36 TBD + TBE = 0 TBE = TBD 11 11

42 42 TBD + TBE − 8400 = 0 11 11 42 ! 2 " TBD # = 8400 $ 11 %

TBD = 1100 lb TBE = 1100 lb

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!

PROBLEM 4.105 (Continued)

ΣFx = 0: Ax −

6 6 (1100 lb) − (1100 lb) = 0 11 11 Ax = 1200 lb

ΣFy = 0: Ay +

7 7 (1100 lb) + (1100 lb) − 840 lb = 0 11 11 Ay = −560 lb

ΣFz = 0: Az +

6 6 (1100 lb) − (1100 lb) = 0 11 11 Az = 0

A = (1200 lb)i − (560 lb) j

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PROBLEM 4.106 A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE. Determine the tension in each cable and the reaction at C.

SOLUTION Free-Body Diagram:

Five Unknowns and six Eqs. of equilibrium. Equilibrium is maintained (ΣMAC = 0). rB = 1.2k

TAD TAE

rA = 2.4k ! AD = −0.8i + 0.6 j − 2.4k AD = 2.6 m ! AE = 0.8i + 1.2 j − 2.4k AE = 2.8 m ! AD TAD = = (−0.8i + 0.6 j − 2.4k ) AD 2.6 ! AE TAE = = (0.8i + 1.2 j − 2.4k ) AE 2.8

ΣM C = 0: rA × TAD + rA × TAE + rB × (−3 kN) j = 0 i j k i j k TAD T 0 0 2.4 0 2.4 AE + 1.2k × (−3.6 kN) j = 0 + 0 2.6 2.8 −0.8 0.6 −2.4 0.8 1.2 −2.4

Equate coefficients of unit vectors to zero. i : − 0.55385 TAD − 1.02857 TAE + 4.32 = 0

(1)

j : − 0.73846 TAD + 0.68671 TAE = 0 TAD = 0.92857 TAE

Eq. (1):

(2)

−0.55385(0.92857) TAE − 1.02857 TAE + 4.32 = 0 1.54286 TAE = 4.32 TAE = 2.800 kN

TAE = 2.80 kN

!

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PROBLEM 4.106 (Continued)

Eq. (2):

TAD = 0.92857(2.80) = 2.600 kN

TAD = 2.60 kN

0.8 0.8 (2.6 kN) + (2.8 kN) = 0 2.6 2.8 0.6 1.2 (2.6 kN) + (2.8 kN) − (3.6 kN) = 0 ΣFy = 0: C y + 2.6 2.8 2.4 2.4 (2.6 kN) − (2.8 kN) = 0 ΣFz = 0: C z − 2.6 2.8 ΣFx = 0: C x −

!

Cx = 0 C y = 1.800 kN C z = 4.80 kN

C = (1.800 kN) j + (4.80 kN)k

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PROBLEM 4.107 Solve Problem 4.106, assuming that the 3.6-kN load is applied at Point A. PROBLEM 4.106 A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE. Determine the tension in each cable and the reaction at C.

SOLUTION Free-Body Diagram:

Five unknowns and six Eqs. of equilibrium. Equilibrium is maintained (ΣMAC = 0). ! AD = −0.8i + 0.6 j − 2.4k AD = 2.6 m ! AE = 0.8i + 1.2 j − 2.4k AE = 2.8 m ! AD TAD (−0.8i + 0.6 j − 2.4k ) TAD = = AD 2.6 ! AE TAE (0.8i + 1.2 j − 2.4k ) TAE = = AE 2.8 ΣM C = 0: rA × TAD + rA × TAE + rA × (−3.6 kN) j

Factor rA : or: Coefficient of i:

rA × (TAD + TAE − (3.6 kN) j) TAD + TAE − (3 kN) j = 0

−

(Forces concurrent at A)

TAD T (0.8) + AE (0.8) = 0 2.6 2.8 TAD =

Coefficient of j:

2.6 TAE 2.8

(1)

TAD T (0.6) + AE (1.2) − 3.6 kN = 0 2.6 2.8 2.6 0.6 ! 1.2 TAE " TAE − 3.6 kN = 0 #+ 2.8 $ 2.6 % 2.8 TAE " $

0.6 + 1.2 ! # = 3.6 kN 2.8 % TAE = 5.600 kN

TAE = 5.60 kN

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PROBLEM 4.107 (Continued)

Eq. (1):

TAD =

2.6 (5.6) = 5.200 kN 2.8

0.8 0.8 (5.2 kN) + (5.6 kN) = 0; 2.6 2.8 0.6 1.2 (5.2 kN) + (5.6 kN) − 3.6 kN = 0 ΣFy = 0: C y + 2.6 2.8 2.4 2.4 (5.2 kN) − (5.6 kN) = 0 ΣFz = 0: Cz − 2.6 2.8 ΣFx = 0: C x −

TAD = 5.20 kN

!

C = (9.60 kN)k

!

Cx = 0 Cy = 0 Cz = 9.60 kN

Note: Since forces and reaction are concurrent at A, we could have used the methods of Chapter 2.

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PROBLEM 4.108 A 600-lb crate hangs from a cable that passes over a pulley B and is attached to a support at H. The 200-lb boom AB is supported by a ball-and-socket joint at A and by two cables DE and DF. The center of gravity of the boom is located at G. Determine (a) the tension in cables DE and DF, (b) the reaction at A.

SOLUTION Free-Body Diagram:

WC = 600 lb WG = 200 lb

We have five unknowns (TDE , TDF , Ax , Ay , Az ) and five equilibrium equations. The boom is free to spin about the AB axis, but equilibrium is maintained, since ΣM AB = 0. " We have BH = (30 ft)i − (22.5 ft) j BH = 37.5 ft " 8.8 DE = (13.8 ft)i − (22.5 ft) j + (6.6 ft)k 12 = (13.8 ft)i − (16.5 ft) j + (6.6 ft)k DE = 22.5 ft " DF = (13.8 ft)i − (16.5 ft) j − (6.6 ft)k DF = 22.5 ft

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PROBLEM 4.108 (Continued)

TBH = TBH

Thus:

TDE = TDE TDF = TDF

(a)

" BH 30i − 22.5 j = (600 lb) = (480 lb)i − (360 lb) j 37.5 BH " DE TDE (13.8i − 16.5 j + 6.6k ) = DE 22.5 " DF TDE (13.8i − 16.5 j − 6.6k ) = DF 22.5

ΣM A = 0: (rJ × WC ) + (rK × WG ) + (rH × TBH ) + (rE × TDE ) + (rF × TDF ) = 0 − (12i ) × (−600 j) − (6i ) × (−200 j) + (18i) × (480i − 360 j) i j k i j k TDE TDF 5 0 6.6 + 5 0 + −6.6 = 0 22.5 22.5 13.8 −16.5 6.6 13.8 −16.5 −6.6

7200k + 1200k − 6480k + 4.84(TDE − TDF )i

or,

+

58.08 82.5 (TDE − TDF ) j − (TDE + TDF )k = 0 22.5 22.5

Equating to zero the coefficients of the unit vectors: i or j:

TDE − TDF = 0

k : 7200 + 1200 − 6480 −

TDE = TDF *

82.5 (2TDE ) = 0 22.5

TDE = 261.82 lb TDE = TDF = 262 lb

(b)

13.8 ! ΣFx = 0: Ax + 480 + 2 " # (261.82) = 0 $ 22.5 % 16.5 ! ΣFy = 0: Ay − 600 − 200 − 360 − 2 " # (261.82) = 0 $ 22.5 % ΣFz = 0: Az = 0

Ax = −801.17 lb Ay = 1544.00 lb A = −(801 lb)i + (1544 lb) j

*Remark: The fact that TDE = TDF could have been noted at the outset from the symmetry of structure with respect to xy plane.

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PROBLEM 4.109 A 3-m pole is supported by a ball-and-socket joint at A and by the cables CD and CE. Knowing that the 5-kN force acts vertically downward (φ = 0), determine (a) the tension in cables CD and CE, (b) the reaction at A.

SOLUTION Free-Body Diagram: By symmetry with xy plane TCD = TCE = T ! CD = −3i + 1.5 j + 1.2k CD = 3.562 m

TCD TCE rB/A = 2i

" −3i + 1.5 j + 1.2k CD =T =T CD 3.562 −3i + 1.5 j − 1.2k =T 3.562

rC/A = 3i

ΣMA = 0: rC/A × TCD + rC/A × TCE + rB/A × (−5 kN) j = 0 i j k i j k i j k T T + 3 0 + 2 0 0 =0 3 0 0 0 3.562 3.562 −3 1.5 1.2 −3 1.5 −1.2 0 −5 0

Coefficient of k:

T ' & 2 (3 × 1.5 × ) − 10 = 0 T = 3.958 kN 3.562 * + ΣF = 0: A + TCD + TCE − 5 j = 0

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PROBLEM 4.109 (Continued)

Coefficient of k:

Az = 0

Coefficient of i:

Ax − 2[3.958 × 3/3.562] = 0

Ax = 6.67 kN

Coefficient of j:

Ay + 2[3.958 × 1.5/3.562] − 5 = 0

Ay = 1.667 kN

(a)

TCD = TCE = 3.96 kN A = (6.67 kN)i + (1.667 kN) j

(b)

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PROBLEM 4.110 A 3-m pole is supported by a ball-and-socket joint at A and by the cables CD and CE. Knowing that the line of action of the 5-kN force forms an angle φ = 30° with the vertical xy plane, determine (a) the tension in cables CD and CE, (b) the reaction at A.

SOLUTION Free-Body Diagram: Five unknowns and six Eqs. of equilibrium but equilibrium is maintained (ΣM AC = 0) rB/A = 2i rC/A = 3i

Load at B.

= −(5cos 30) j + (5sin 30)k = −4.33j + 2.5k ! CD = −3i + 1.5 j + 1.2k CD = 3.562 m ! CD T ( −3i + 1.5 j + 1.2k ) TCD = TCD = CD 3.562

Similarly,

TCE =

T (−3i + 1.5 j − 1.2k ) 3.562

ΣMA = 0: rC/A × TCD + rC/A × TCE + rB/A × ( −4.33j + 2.5k ) = 0 i j k i j k i j k TCD TCE + 3 +2 3 0 0 0 0 0 0 =0 3.562 3.562 −3 1.5 1.2 −3 1.5 −1.2 0 −4.33 2.5

Equate coefficients of unit vectors to zero. j: − 3.6

TCD T + 3.6 CE − 5 = 0 3.562 3.562

−3.6TCD + 3.6TCE − 17.810 = 0

(1)

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PROBLEM 4.110 (Continued)

k : 4.5

TCD T + 4.5 CE − 8.66 = 0 3.562 3.562 4.5TCD + 4.5TCE = 30.846

(2) + 1.25(1):

Eq. (1):

(2)

9TCE − 53.11 = 0 TCE = 5.901 kN −3.6TCD + 3.6(5.901) − 17.810 = 0 TCD = 0.954 kN ΣF = 0: A + TCD + TCE − 4.33j + 2.5k = 0 i : Ax +

0.954 5.901 (−3) + (−3) = 0 3.562 3.562 Ax = 5.77 kN

j: Ay +

0.954 5.901 (1.5) + (1.5) − 4.33 = 0 3.562 3.562 Ay = 1.443 kN

k : Az +

0.954 5.901 (1.2) + ( −1.2) + 2.5 = 0 3.562 3.562 Az = −0.833 kN

Answers: (a)

TCD = 0.954 kN

TCE = 5.90 kN A = (5.77 kN)i + (1.443 kN) j − (0.833 kN)k

(b)

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PROBLEM 4.111 A 48-in. boom is held by a ball-and-socket joint at C and by two cables BF and DAE; cable DAE passes around a frictionless pulley at A. For the loading shown, determine the tension in each cable and the reaction at C.

SOLUTION Free-Body Diagram: Five unknowns and six Eqs. of equilibrium but equilibrium is maintained (ΣMAC = 0). T = Tension in both parts of cable DAE. rB = 30k rA = 48k ! AD = −20i − 48k AD = 52 in. ! AE = 20 j − 48k AE = 52 in. ! BF = 16i − 30k BF = 34 in. ! AD T T TAD = T = (−20i − 48k ) = ( −5i − 12k ) AD 52 13 ! AE T T TAE = T = (20 j − 48k ) = (5 j − 12k ) AE 52 13 ! T BF TBF TBF = TBF = (16i − 30k ) = BF (8i − 15k ) BF 34 17

ΣM C = 0: rA × TAD + rA × TAE + rB × TBF + rB × (−320 lb) j = 0 i j k i j k i j k T T T + 0 0 48 + 0 0 30 BF + (30k ) × (−320 j) = 0 0 0 48 13 13 17 −5 0 −12 0 5 −12 8 0 −15

Coefficient of i:

−

240 T + 9600 = 0 13

T = 520 lb

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PROBLEM 4.111 (Continued)

Coefficient of j:

−

240 240 T+ TBD = 0 13 17 TBD =

17 17 T = (520) TBD = 680 lb 13 13

ΣF = 0: TAD + TAE + TBF − 320 j + C = 0

Coefficient of i:

−

20 8 (520) + (680) + Cx = 0 52 17

−200 + 320 + Cx = 0

Coefficient of j:

20 (520) − 320 + C y = 0 52 200 − 320 + C y = 0

Coefficient of k:

Cx = −120 lb

−

C y = 120 lb

48 48 30 (520) − (520) − (680) + Cz = 0 52 52 34 −480 − 480 − 600 + Cz = 0 Cz = 1560 lb

Answers: TDAE = T

TDAE = 520 lb

!

TBD = 680 lb

!

C = −(120.0 lb)i + (120.0 lb) j + (1560 lb)k

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PROBLEM 4.112 Solve Problem 4.111, assuming that the 320-lb load is applied at A. PROBLEM 4.111 A 48-in. boom is held by a ball-andsocket joint at C and by two cables BF and DAE; cable DAE passes around a frictionless pulley at A. For the loading shown, determine the tension in each cable and the reaction at C.

SOLUTION Free-Body Diagram: Five unknowns and six Eqs. of equilibrium but equilibrium is maintained (ΣMAC = 0). T = tension in both parts of cable DAE. rB = 30k rA = 48k ! AD = −20i − 48k AD = 52 in. ! AE = 20 j − 48k AE = 52 in. ! BF = 16i − 30k BF = 34 in. ! AD T T TAD = T = (−20i − 48k ) = ( −5i − 12k ) AD 52 13 ! AE T T TAE = T = (20 j − 48k ) = (5 j − 12k ) AE 52 13 ! T BF TBF TBF = TBF = (16i − 30k ) = BF (8i − 15k ) BF 34 17 ΣM C = 0: rA × TAD + rA × TAE + rB × TBF + rA × ( −320 lb) j = 0 i j k i j k i j k T T T + 0 0 48 + 0 0 30 BF + 48k × (−320 j) = 0 0 0 48 13 13 17 −5 0 −12 0 5 −12 8 0 −15

Coefficient of i:

−

240 T + 15360 = 0 13

T = 832 lb

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PROBLEM 4.112 (Continued)

Coefficient of j:

−

240 240 T+ TBD = 0 13 17 TBD =

17 17 T = (832) 13 13

TBD = 1088 lb

ΣF = 0: TAD + TAE + TBF − 320 j + C = 0 −

Coefficient of i:

20 8 (832) + (1088) + C x = 0 52 17

−320 + 512 + Cx = 0 20 (832) − 320 + C y = 0 52

Coefficient of j:

320 − 320 + C y = 0

Coefficient of k:

−

Cy = 0

48 48 30 (832) − (852) − (1088) + Cz = 0 52 52 34 −768 − 768 − 960 + Cz = 0

Answers:

Cx = −192 lb

TDAE = T

Cz = 2496 lb TDAE = 832 lb

!

TBD = 1088 lb

!

C = −(192.0 lb)i + (2496 lb)k

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PROBLEM 4.113 A 20-kg cover for a roof opening is hinged at corners A and B. The roof forms an angle of 30° with the horizontal, and the cover is maintained in a horizontal position by the brace CE. Determine (a) the magnitude of the force exerted by the brace, (b) the reactions at the hinges. Assume that the hinge at A does not exert any axial thrust.

SOLUTION Force exerted by CD F = F (sin 75°)i + F (cos 75°) j F = F (0.2588i + 0.9659 j) W = mg = 20 kg(9.81 m/s 2 ) = 196.2 N rA/B = 0.6k rC/B = 0.9i + 0.6k rG/B = 0.45i + 0.3k F = F (0.2588i + 0.9659 j) ΣM B = 0: rG/B × (−196.2 j) + rC/B × F + rA/B × A = 0 i j k i j k i 0.45 0 0.3 + 0.9 0 0.6 F + 0 −196.2 0 Ax 0 0.2588 +0.9659 0

j 0 Ay

k 0.6 = 0 0

Coefficient of i :

+58.86 − 0.5796 F − 0.6 Ay = 0

(1)

Coefficient of j:

+0.1553F + 0.6 Ax = 0

(2)

Coefficient of k:

−88.29 + 0.8693F = 0: F = 101.56 N

Eq. (2):

+58.86 − 0.5796(101.56) − 0.6 Ay = 0

Eq. (3):

+0.1553(101.56) + 0.6 Ax = 0

Ay = 0 Ax = −26.29 N

F = 101.6 N

A = −(26.3 N)i

ΣF : A + B + F − Wj = 0

Coefficient of i:

26.29 + Bx + 0.2588(101.56) = 0

Bx = 0

Coefficient of j:

By + 0.9659(101.56) − 196.2 = 0 By = 98.1 N Bz = 0

Coefficient of k:

B = (98.1 N) j

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 477

PROBLEM 4.114 The bent rod ABEF is supported by bearings at C and D and by wire AH. Knowing that portion AB of the rod is 250 mm long, determine (a) the tension in wire AH, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.

SOLUTION ∆ABH is equilateral

Free-Body Diagram: Dimensions in mm

rH/C = −50i + 250 j rD/C = 300i rF/C = 350i + 250k T = T (sin 30°) j − T (cos 30°)k = T (0.5 j − 0.866k ) ΣM C = 0: rH/C × T + rD × D + rF/C × (−400 j) = 0 i j k i j −50 250 0 T + 300 0 0 0.5 −0.866 0 Dy

Coefficient i:

k i j k 0 + 350 0 250 = 0 Dz 0 −400 0

−216.5T + 100 × 103 = 0 T = 461.9 N

Coefficient of j:

T = 462 N

−43.3T − 300 Dz = 0 −43.3(461.9) − 300 Dz = 0

Dz = −66.67 N

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PROBLEM 4.114 (Continued)

Coefficient of k:

−25T + 300 D y − 140 × 103 = 0 −25(461.9) + 300 Dy − 140 × 103 = 0

D y = 505.1 N D = (505 N) j − (66.7 N)k

ΣF = 0: C + D + T − 400 j = 0

Coefficient i:

Cx = 0

Cx = 0

Coefficient j:

C y + (461.9)0.5 + 505.1 − 400 = 0 C y = −336 N

Coefficient k:

C z − (461.9)0.866 − 66.67 = 0

C z = 467 N

C = −(336 N) j + (467 N)k

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PROBLEM 4.115 A 100-kg uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE that passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust.

SOLUTION rB/A (960 − 180)i = 780i

Dimensions in mm

960 ! 450 − 90 # i + rG/A = " k 2 $ 2 % = 390i + 225k rC/A = 600i + 450k T = Tension in cable DCE

! CD = −690i + 675 j − 450k ! CE = 270i + 675 j − 450k

CD = 1065 mm CE = 855 mm

T (−690i + 675 j − 450k ) 1065 T (270i + 675 j − 450k ) TCE = 855 W = −mgi = −(100 kg)(9.81 m/s 2 ) j = −(981 N) j

TCD =

ΣM A = 0: rC/A × TCD + rC/A × TCE + rG/A × (−Wj) + rB/A × B = 0 i j k i j k T T 600 0 450 450 + 600 0 + 1065 855 270 675 −450 −690 675 −450 i + 390 0

j 0 −981

k i 225 + 780 0

0

j 0

k 0 =0

By

Bz

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PROBLEM 4.115 (Continued)

Coefficient of i:

−(450)(675)

T T − (450)(675) + 220.725 × 103 = 0 1065 855 T = 344.6 N

Coefficient of j:

(−690 × 450 + 600 × 450)

T = 345 N

344.6 344.6 + (270 × 450 + 600 × 450) − 780 Bz = 0 1065 855 Bz = 185.49 N

Coefficient of k:

(600)(675)

344.6 344.6 + (600)(675) − 382.59 × 103 + 780 By 1065 855

By = 113.2N

B = (113.2 N) j + (185.5 N)k ΣF = 0: A + B + TCD + TCE + W = 0 690 270 (344.6) + (344.6) = 0 1065 855

Coefficient of i:

Ax −

Ax = 114.4 N

Coefficient of j:

Ay + 113.2 +

675 675 (344.6) + (344.6) − 981 = 0 1065 855

Ay = 377 N

Coefficient of k:

Az + 185.5 −

450 450 (344.6) − (344.6) = 0 1065 855

Az = 141.5 N A = (114.4 N)i + (377 N) j + (144.5 N)k

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!

PROBLEM 4.116 Solve Problem 4.115, assuming that cable DCE is replaced by a cable attached to Point E and hook C. PROBLEM 4.115 A 100-kg uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE that passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust.

SOLUTION See solution to Problem 4.115 for free-body diagram and analysis leading to the following: CD = 1065 mm CE = 855 mm

T (−690i + 675 j − 450k ) 1065 T (270i + 675 j − 450k ) TCE = 855 W = −mgi = −(100 kg)(9.81 m/s 2 ) j = −(981 N)j

TCD =

Now:

ΣM A = 0: rC/A × TCE + rG/A × ( −Wj) + rB/A × B = 0

i j k i j k i j T + 390 600 0 450 0 225 + 780 0 855 270 675 −450 0 −981 0 0 By

Coefficient of i:

−(450)(675)

k 0 =0 Bz

T + 220.725 × 103 = 0 855 T = 621.3 N

Coefficient of j: Coefficient of k:

(270 × 450 + 600 × 450) (600)(675)

T = 621 N

621.3 − 980 Bz = 0 Bz = 364.7 N 855

621.3 − 382.59 × 103 + 780 B y = 0 B y = 113.2 N 855

B = (113.2 N)j + (365 N)k

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PROBLEM 4.116 (Continued)

ΣF = 0: A + B + TCE + W = 0 270 (621.3) = 0 855

Coefficient of i:

Ax +

Coefficient of j:

Ay + 113.2 +

675 (621.3) − 981 = 0 855

Ay = 377.3 N

Coefficient of k:

Az + 364.7 −

450 (621.3) = 0 855

Az = −37.7 N !

!

Ax = −196.2 N

A = −(196.2 N)i + (377 N)j − (37.7 N)k

!

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!

PROBLEM 4.117 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B.

SOLUTION rB/A = (38 − 8)i = 30i rE/A = (30 − 4)i + 20k

rG/A

= 26i + 20k 38 = i + 10k 2 = 19i + 10k

! EF = 8i + 25 j − 20k EF = 33 in. ! AE T T =T = (8i + 25 j − 20k ) AE 33 ΣM A = 0: rE/A × T + rG/A × (−75 j) + rB/A × B = 0 i j k i j k i j T + 19 0 10 + 30 0 26 0 20 33 8 25 −20 0 −75 0 0 By −(25)(20)

Coefficient of i: Coefficient of j: Coefficient of k:

(160 + 520) (26)(25)

T + 750 = 0: 33

k 0 =0 Bz T = 49.5 lb

49.5 − 30 Bz = 0: Bz = 34 lb 33

49.5 − 1425 + 30 By = 0: By = 15 lb 33

B = (15 lb)j + (34 lb)k

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!

PROBLEM 4.117 (Continued)

!

ΣF = 0: A + B + T − (75 lb)j = 0 ! Ax +

Coefficient of i: Coefficient of j: Coefficient of k:

Ay + 15 +

8 (49.5) = 0 33

25 (49.5) − 75 = 0 33

Az + 34 −

20 (49.5) = 0 33

Ax = −12.00 lb Ay = 22.5 lb Az = −4.00 lb ! A = −(12.00 lb)i + (22.5 lb)j − (4.00 lb)k

!

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!

PROBLEM 4.118 Solve Problem 4.117, assuming that cable EF is replaced by a cable attached at points E and H. PROBLEM 4.117 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B.

SOLUTION rB/A = (38 − 8)i = 30i rE/A = (30 − 4)i + 20k = 26i + 20k

38 i + 10k 2 = 19i + 10k

rG/A =

! EH = −30i + 12 j − 20k EH = 38 in. ! EH T T=T = (−30i + 12 j − 20k ) EH 38 ΣM A = 0: rE/A × T + rG/A × (−75 j) + rB/A × B = 0 i j k i j k i j T 26 0 20 + 19 0 10 + 30 0 38 0 −75 0 0 By −30 12 −20 −(12)(20)

Coefficient of i: Coefficient of j: Coefficient of k:

(−600 + 520) (26)(12)

T + 750 = 0 38

T = 118.75

k 0 =0 Bz

T = 118.8lb

118.75 − 30 Bz = 0 Bz = −8.33lb 38

118.75 − 1425 + 30 By = 0 By = 15.00 lb 38

B = (15.00lb)j − (8.33 lb)k

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PROBLEM 4.118 (Continued)

ΣF = 0:

Coefficient of j: Coefficient of k:

30 (118.75) = 0 38

Ax = 93.75 lb

12 (118.75) − 75 = 0 38

Ay = 22.5 lb

Ax −

Coefficient of i: Ay + 15 +

A + B + T − (75 lb)j = 0

Az − 8.33 −

20 (118.75) = 0 38

Az = 70.83 lb A = (93.8 lb)i + (22.5 lb)j + (70.8 lb)k

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 487

!

PROBLEM 4.119 Solve Problem 4.114, assuming that the bearing at D is removed and that the bearing at C can exert couples about axes parallel to the y and z axes. PROBLEM 4.114 The bent rod ABEF is supported by bearings at C and D and by wire AH. Knowing that portion AB of the rod is 250 mm long, determine (a) the tension in wire AH, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.

SOLUTION Free-Body Diagram:

∆ABH is Equilateral

Dimensions in mm

rH/C = −50i + 250 j rF/C = 350i + 250k T = T (sin 30°) j − T (cos 30°)k = T (0.5 j − 0.866k ) ΣM C = 0: rF/C × (−400 j) + rH/C × T + ( M C ) y j + ( M C ) z k = 0

i j k i j k 350 0 250 + −50 250 0 T + (M C ) y j + (M C ) z k = 0 0 −400 0 0 0.5 −0.866

Coefficient of i: Coefficient of j:

+100 × 103 − 216.5T = 0 T = 461.9 N

T = 462 N

−43.3(461.9) + ( M C ) y = 0 ( M C ) y = 20 × 103 N ⋅ mm (M C ) y = 20.0 N ⋅ m

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PROBLEM 4.119 (Continued)

Coefficient of k:

−140 × 103 − 25(461.9) + ( M C ) z = 0 ( M C ) z = 151.54 × 103 N ⋅ mm (M C ) z = 151.5 N ⋅ m ΣF = 0: C + T − 400 j = 0

M C = (20.0 N ⋅ m)j + (151.5 N ⋅ m)k

Coefficient of i: Coefficient of j: Coefficient of k:

Cx = 0 C y + 0.5(461.9) − 400 = 0 C y = 169.1 N C z − 0.866(461.9) = 0 C z = 400 N

C = (169.1 N)j + (400 N)k

!

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PROBLEM 4.120 Solve Problem 4.117, assuming that the hinge at B is removed and that the hinge at A can exert couples about axes parallel to the y and z axes. PROBLEM 4.117 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B.

SOLUTION rE/A = (30 − 4)i + 20k = 26i + 20k rG/A = (0.5 × 38)i + 10k = 19i + 10k ! AE = 8i + 25 j − 20k AE = 33 in. ! AE T = (8i + 25 j − 20k ) T =T AE 33 ΣM A = 0: rE/A × T + rG/A × (−75 j) + ( M A ) y j + ( M A ) z k = 0

i j k i j k T 26 0 20 + 19 0 10 + ( M A ) y j + ( M A ) z k = 0 33 8 25 −20 0 −75 0 −(20)(25)

Coefficient of i: Coefficient of j: Coefficient of k:

(160 + 520) (26)(25)

T + 750 = 0 33

T = 49.5 lb

49.5 + ( M A ) y = 0 ( M A ) y = −1020 lb ⋅ in. 33

49.5 − 1425 + ( M A ) z = 0 33

( M A ) z = 450 lb ⋅ in.

ΣF = 0: A + T − 75 j = 0 Ax +

Coefficient of i:

8 (49.5) = 0 33

M A = −(1020 lb ⋅ in)j + (450 lb ⋅ in.)k

Ax = 12.00 lb

Coefficient of j:

Ay +

25 (49.5) − 75 = 0 33

Ay = 37.5 lb

Coefficient of k:

Az −

20 (49.5) 33

Az = 30.0 lb

A = −(12.00 lb)i + (37.5 lb)j + (30.0 lb)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 490

PROBLEM 4.121 The assembly shown is used to control the tension T in a tape that passes around a frictionless spool at E. Collar C is welded to rods ABC and CDE. It can rotate about shaft FG but its motion along the shaft is prevented by a washer S. For the loading shown, determine (a) the tension T in the tape, (b) the reaction at C.

SOLUTION Free-Body Diagram: rA/C = 4.2 j + 2k rE/C = 1.6i − 2.4 j ΣM C = 0: rA/C × (−6 j) + rE/C × T (i + k ) + ( M C ) y j + ( M C ) z k = 0 (4.2 j + 2k ) × (−6 j) + (1.6i − 2.4 j) × T (i + k ) + ( M C ) y j + ( M C ) z k = 0

Coefficient of i: Coefficient of j: Coefficient of k:

12 − 2.4T = 0

T = 5 lb

−1.6(5 lb) + (M C ) y = 0 ( M C ) y = 8 lb ⋅ in. 2.4(5 lb) + ( M C ) z = 0 ( M C ) z = −12 lb ⋅ in.

M C = (8 lb ⋅ in.)j − (12 lb ⋅ in.)k ΣF = 0: C x i + C y j + C z k − (6 lb)j + (5 lb)i + (5 lb)k = 0

Equate coefficients of unit vectors to zero. C x = −5 lb C y = 6 lb C z = −5 lb

C = −(5.00 lb)i + (6.00 lb)j − (5.00 lb)k

!

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PROBLEM 4.122 The assembly shown is welded to collar A that fits on the vertical pin shown. The pin can exert couples about the x and z axes but does not prevent motion about or along the y axis. For the loading shown, determine the tension in each cable and the reaction at A.

SOLUTION Free-Body Diagram: First note

TCF =

CF TCF

−(0.08 m)i + (0.06 m) j

=

(0.08) 2 + (0.06)2 m

TCF

= TCF (−0.8i + 0.6 j) TDE =

DE TDE

=

(0.12 m)j − (0.09 m)k (0.12) 2 + (0.09)2 m

TDE

= TDE (0.8 j − 0.6k )

(a)

From f.b.d. of assembly ΣFy = 0: 0.6TCF + 0.8TDE − 480 N = 0 0.6TCF + 0.8TDE = 480 N

or

(1)

ΣM y = 0: − (0.8TCF )(0.135 m) + (0.6TDE )(0.08 m) = 0 TDE = 2.25TCF

or

(2)

Substituting Equation (2) into Equation (1) 0.6TCF + 0.8[(2.25)TCF ] = 480 N TCF = 200.00 N TCF = 200 N

or !

and from Equation (2)

TDE = 2.25(200.00 N) = 450.00 TDE = 450 N

or

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PROBLEM 4.122 (Continued)

(b)

From f.b.d. of assembly ΣFz = 0: Az − (0.6)(450.00 N) = 0

Az = 270.00 N

ΣFx = 0: Ax − (0.8)(200.00 N) = 0

Ax = 160.000 N

or A = (160.0 N)i + (270 N)k ΣM x = 0:

MAx + (480 N)(0.135 m) − [(200.00 N)(0.6)](0.135 m) − [(450 N)(0.8)](0.09 m) = 0

M Ax = −16.2000 N ⋅ m ΣM z = 0:

MAz − (480 N)(0.08 m) + [(200.00 N)(0.6)](0.08 m) + [(450 N)(0.8)](0.08 m) = 0

M Az = 0

or M A = −(16.20 N ⋅ m)i

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PROBLEM 4.123 The rigid L-shaped member ABC is supported by a ball-and-socket joint at A and by three cables. If a 450-lb load is applied at F, determine the tension in each cable.

SOLUTION Free-Body Diagram:

In this problem: We have

Thus

a = 21 in. ! CD = (24 in.)j − (32 in.)k CD = 40 in. ! BD = −(42 in.)i + (24 in.)j − (32 in.)k BD = 58 in. ! BE = (42 in.)i − (32 in.)k BE = 52.802 in. TCD TBD TBE

! CD = TCD = TCD (0.6 j − 0.8k ) CD ! BD = TBD = TBD (−0.72414i + 0.41379 j − 0.55172k ) BD ! BE = TBE = TBE (0.79542i − 0.60604k ) BE

ΣM A = 0: (rC × TCD ) + (rB × TBD ) + (rB × TBE ) + (rW × W) = 0

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PROBLEM 4.123 (Continued)

rC = −(42 in.)i + (32 in.)k

Noting that

rB = (32 in.)k rW = − ai + (32 in.)k

and using determinants, we write i j k i j k 32 TCD + 0 0 32 −42 0 TBD 0 0.6 −0.8 −0.72414 0.41379 −0.55172 i j k i j k 0 0 32 0 32 = 0 + TBE + −a 0.79542 0 −0.60604 0 −450 0

Equating to zero the coefficients of the unit vectors: i: −19.2TCD − 13.241TBD + 14400 = 0

(1)

j: − 33.6TCD − 23.172TBD + 25.453TBE = 0

(2)

k:

−25.2TCD + 450a = 0

(3)

Recalling that a = 21 in., Eq. (3) yields TCD =

From (1):

450(21) = 375 lb 25.2

−19.2(375) − 13.241TBD + 14400 = 0 TBD = 543.77 lb

From (2):

TCD = 375 lb

TBD = 544 lb

−33.6(375) − 23.172(543.77) + 25.453TBE = 0 TBE = 990.07 lb !

!

TBE = 990 lb

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PROBLEM 4.124 Solve Problem 4.123, assuming that the 450-lb load is applied at C. PROBLEM 4.123 The rigid L-shaped member ABC is supported by a ball-and-socket joint at A and by three cables. If a 450-lb load is applied at F, determine the tension in each cable.

SOLUTION See solution of Problem 4.123 for free-body diagram and derivation of Eqs. (1), (2), and (3): −19.2TCD − 13.241TBD + 14400 = 0

(1)

−33.6TCD − 23.172TBD + 25.453TBE = 0

(2)

−25.2TCD + 450a = 0

(3)

In this problem, the 450-lb load is applied at C and we have a = 42 in. Carrying into (3) and solving for TCD , TCD = 750 lb

From (1): From (2):

−19.2(750) − 13.241TBD + 14400 = 0 −33.6(750) − 0 + 25.453TBE = 0 !

TCD = 750 lb TBD = 0 TBE = 990 lb

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PROBLEM 4.125 Frame ABCD is supported by a ball-and-socket joint at A and by three cables. For a = 150 mm, determine the tension in each cable and the reaction at A.

SOLUTION First note

TDG = λ DG TDG =

−(0.48 m)i + (0.14 m)j (0.48) 2 + (0.14) 2 m

TDG

−0.48i + 0.14 j TDG 0.50 T = DG (24i + 7 j) 25 =

TBE = λ BE TBE =

−(0.48 m)i + (0.2 m)k (0.48) 2 + (0.2) 2 m

TBE

−0.48i + 0.2k TBE 0.52 T = BE (−12 j + 5k ) 13 =

From f.b.d. of frame ABCD 7 ! ΣM x = 0: " TDG # (0.3 m) − (350 N)(0.15 m) = 0 25 $ % TDG = 625 N

or 24 5 ! ! ΣM y = 0: " × 625 N # (0.3 m) − " TBE # (0.48 m) = 0 $ 25 % $ 13 %

TBE = 975 N

or 7 ! ΣM z = 0: TCF (0.14 m) + " × 625 N # (0.48 m) − (350 N)(0.48 m) = 0 $ 25 %

TCF = 600 N

or

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PROBLEM 4.125 (Continued)

ΣFx = 0: Ax + TCF + (TBE ) x + (TDG ) x = 0 12 24 ! ! Ax − 600 N − " × 975 N # − " × 625 N # = 0 $ 13 % $ 25 % Ax = 2100 N ΣFy = 0: Ay + (TDG ) y − 350 N = 0 7 ! Ay + " × 625 N # − 350 N = 0 $ 25 % Ay = 175.0 N ΣFz = 0: Az + (TBE ) z = 0 5 ! Az + " × 975 N # = 0 $ 13 % Az = −375 N

A = (2100 N)i + (175.0 N) j − (375 N)k

Therefore

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PROBLEM 4.126 Frame ABCD is supported by a ball-and-socket joint at A and by three cables. Knowing that the 350-N load is applied at D (a = 300 mm), determine the tension in each cable and the reaction at A.

SOLUTION First note

TDG = λ DG TDG =

−(0.48 m)i + (0.14 m)j (0.48) 2 + (0.14)2 m

TDG

−0.48i + 0.14 j TDG 0.50 T = DG (24i + 7 j) 25 =

TBE = λ BE TBE =

−(0.48 m)i + (0.2 m)k (0.48) 2 + (0.2) 2 m

TBE

−0.48i + 0.2k TBE 0.52 T = BE (−12i + 5k ) 13 =

From f.b.d. of frame ABCD 7 ! ΣM x = 0: " TDG # (0.3 m) − (350 N)(0.3 m) = 0 $ 25 % TDG = 1250 N

or 24 5 ! ! ΣM y = 0: " × 1250 N # (0.3 m) − " TBE # (0.48 m) = 0 25 13 $ % $ %

TBE = 1950 N

or 7 ! ΣM z = 0: TCF (0.14 m) + " × 1250 N # (0.48 m) − (350 N)(0.48 m) = 0 $ 25 %

TCF = 0

or !

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PROBLEM 4.126 (Continued)

ΣFx = 0: Ax + TCF + (TBE ) x + (TDG ) x = 0 12 24 ! ! Ax + 0 − " × 1950 N # − " × 1250 N # = 0 ! $ 13 % $ 25 % Ax = 3000 N !

! ΣFy = 0: Ay + (TDG ) y − 350 N = 0 7 ! Ay + " × 1250 N # − 350 N = 0 $ 25 % Ay = 0 ΣFz = 0: Az + (TBE ) z = 0 5 ! Az + " × 1950 N # = 0 $ 13 % Az = −750 N

A = (3000 N)i − (750 N)k

Therefore

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PROBLEM 4.127 Three rods are welded together to form a “corner” that is supported by three eyebolts. Neglecting friction, determine the reactions at A, B, and C when P = 240 lb, a = 12 in., b = 8 in., and c = 10 in.

SOLUTION From f.b.d. of weldment ΣM O = 0: rA/O × A + rB/O × B + rC/O × C = 0

i 12 0

j 0 Ay

k i 0 + 0 Az Bx

j k i 8 0 + 0 0 Bz Cx

j k 0 10 = 0 Cy 0

(−12 Az j + 12 Ay k ) + (8 Bz i − 8 Bx k ) + (−10 C y i + 10 C x j) = 0

From i-coefficient

Bz = 1.25C y

or j-coefficient

k-coefficient

or!

(3)

ΣF = 0: A + B + C − P = 0 ! ( Bx + C x )i + ( Ay + C y − 240 lb) j + ( Az + Bz )k = 0 !

From i-coefficient

Bx + C x = 0 C x = − Bx

or j-coefficient or

(2)

12 Ay − 8 Bx = 0 Bx = 1.5 Ay

or

(1)

−12 Az + 10 C x = 0 C x = 1.2 Az

or

!

8 Bz − 10 C y = 0

(4)

Ay + C y − 240 lb = 0 Ay + C y = 240 lb

(5)

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PROBLEM 4.127 (Continued)

k-coefficient

Az + Bz = 0 Az = − Bz

or

(6)

Substituting C x from Equation (4) into Equation (2) − Bz = 1.2 Az

(7)

Using Equations (1), (6), and (7) Cy =

Bz − Az 1 Bx ! Bx = = = 1.25 1.25 1.25 "$ 1.2 #% 1.5

(8)

From Equations (3) and (8) Cy =

1.5 Ay 1.5

or C y = Ay

and substituting into Equation (5) 2 Ay = 240 lb Ay = C y = 120 lb

(9)

Using Equation (1) and Equation (9) Bz = 1.25(120 lb) = 150.0 lb

Using Equation (3) and Equation (9) Bx = 1.5(120 lb) = 180.0 lb

From Equation (4)

C x = −180.0 lb

From Equation (6)

Az = −150.0 lb

Therefore

A = (120.0 lb) j − (150.0 lb)k

!

!

B = (180.0 lb)i + (150.0 lb)k

!

!

C = −(180.0 lb)i + (120.0 lb) j

!

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PROBLEM 4.128 Solve Problem 4.127, assuming that the force P is removed and is replaced by a couple M = +(600 lb ⋅ in.)j acting at B. PROBLEM 4.127 Three rods are welded together to form a “corner” that is supported by three eyebolts. Neglecting friction, determine the reactions at A, B, and C when P = 240 lb, a = 12 in., b = 8 in., and c = 10 in.

SOLUTION From f.b.d. of weldment ΣM O = 0: rA/O × A + rB/O × B + rC/O × C + M = 0

i 12 0

j 0 Ay

k i 0 + 0 Az Bx

j k i 8 0 + 0 0 Bz Cx

j k 0 10 + (600 lb ⋅ in.) j = 0 Cy 0

(−12 Az j + 12 Ay k ) + (8 Bz j − 8 Bx k ) + ( −10C y i + 10C x j) + (600 lb ⋅ in.) j = 0

From i-coefficient

8 Bz − 10 C y = 0 C y = 0.8Bz

or j-coefficient

(1)

−12 Az + 10 C x + 600 = 0 C x = 1.2 Az − 60

or k-coefficient

(2)

12 Ay − 8 Bx = 0 Bx = 1.5 Ay

or

(3)

!

ΣF = 0: A + B + C = 0 !

!

( Bx + C x )i + ( Ay + C y ) j + ( Az + Bz )k = 0 !

From i-coefficient

C x = − Bx

(4)

j-coefficient

C y = − Ay

(5)

k-coefficient

Az = − Bz

(6)

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PROBLEM 4.128 (Continued)

Substituting C x from Equation (4) into Equation (2) B ! Az = 50 − " x # $ 1.2 %

(7)

2! C y = 0.8 Bz = − 0.8 Az = " # Bx − 40 $3%

(8)

Using Equations (1), (6), and (7)

From Equations (3) and (8) C y = Ay − 40

Substituting into Equation (5)

2 Ay = 40 Ay = 20.0 lb

From Equation (5)

C y = −20.0 lb

Equation (1)

Bz = −25.0 lb

Equation (3)

Bx = 30.0 lb

Equation (4)

C x = −30.0 lb

Equation (6)

Az = 25.0 lb

A = (20.0 lb) j + (25.0 lb)k

!

!

! B = (30.0 lb)i − (25.0 lb)k

!

!

C = − (30.0 lb)i − (20.0 lb) j

!

Therefore

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PROBLEM 4.129 In order to clean the clogged drainpipe AE, a plumber has disconnected both ends of the pipe and inserted a power snake through the opening at A. The cutting head of the snake is connected by a heavy cable to an electric motor that rotates at a constant speed as the plumber forces the cable into the pipe. The forces exerted by the plumber and the motor on the end of the cable can be represented by the wrench F = −(48 N)k , M = −(90 N ⋅ m)k. Determine the additional reactions at B, C, and D caused by the cleaning operation. Assume that the reaction at each support consists of two force components perpendicular to the pipe.

SOLUTION From f.b.d. of pipe assembly ABCD ΣFx = 0: Bx = 0 ΣM D ( x -axis) = 0: (48 N)(2.5 m) − Bz (2 m) = 0 Bz = 60.0 N

and B = (60.0 N)k ΣM D ( z -axis) = 0: C y (3 m) − 90 N ⋅ m = 0 C y = 30.0 N ΣM D ( y -axis) = 0: −C z (3 m) − (60.0 N)(4 m) + (48 N)(4 m) = 0 C z = −16.00 N

and C = (30.0 N) j − (16.00 N)k ΣFy = 0: Dy + 30.0 = 0 Dy = −30.0 N ΣFz = 0: Dz − 16.00 N + 60.0 N − 48 N = 0 Dz = 4.00 N

and D = − (30.0 N) j + (4.00 N)k

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 505

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PROBLEM 4.130 Solve Problem 4.129, assuming that the plumber exerts a force F = −(48 N)k and that the motor is turned off (M = 0). PROBLEM 4.129 In order to clean the clogged drainpipe AE, a plumber has disconnected both ends of the pipe and inserted a power snake through the opening at A. The cutting head of the snake is connected by a heavy cable to an electric motor that rotates at a constant speed as the plumber forces the cable into the pipe. The forces exerted by the plumber and the motor on the end of the cable can be represented by the wrench F = −(48 N)k , M = −(90 N ⋅ m)k. Determine the additional reactions at B, C, and D caused by the cleaning operation. Assume that the reaction at each support consists of two force components perpendicular to the pipe.

SOLUTION From f.b.d. of pipe assembly ABCD ΣFx = 0: Bx = 0 ΣM D ( x -axis) = 0: (48 N)(2.5 m) − Bz (2 m) = 0 Bz = 60.0 N

and B = (60.0 N)k

ΣM D ( z -axis) = 0: C y (3 m) − Bx (2 m) = 0 Cy = 0 ΣM D ( y -axis) = 0: C z (3 m) − (60.0 N)(4 m) + (48 N)(4 m) = 0 C z = −16.00 N

and C = − (16.00 N)k

ΣFy = 0: Dy + C y = 0 Dy = 0 ΣFz = 0: Dz + Bz + C z − F = 0 Dz + 60.0 N − 16.00 N − 48 N = 0 Dz = 4.00 N

and D = (4.00 N)k

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PROBLEM 4.131 The assembly shown consists of an 80-mm rod AF that is welded to a cross consisting of four 200-mm arms. The assembly is supported by a ball-and-socket joint at F and by three short links, each of which forms an angle of 45° with the vertical. For the loading shown, determine (a) the tension in each link, (b) the reaction at F.

SOLUTION

rE/F = −200 i + 80 j TB = TB (i − j) / 2

rB/F = 80 j − 200k

TC = TC (− j + k ) / 2 rC/F = 200i + 80 j TD = TD (− i + j) / 2

rD/E = 80 j + 200k

ΣM F = 0: rB/F × TB + rC/F × TC + rD/F × TD + rE/F × (− Pj) = 0 i j k i j k i j k i j k Tc TB TD 0 80 −200 + 200 80 0 + 0 80 200 + −200 80 0 = 0 2 2 2 1 −1 0 0 −1 1 −1 −1 0 0 −P 0

Equate coefficients of unit vectors to zero and multiply each equation by 2. i : − 200 TB + 80 TC + 200 TD = 0

(1)

j: − 200 TB − 200 TC − 200 TD = 0

(2)

k : − 80 TB − 200 TC + 80 TD + 200 2 P = 0 80 (2): 200

− 80 TB − 80 TC − 80 TD = 0

Eqs. (3) + (4):

−160TB − 280TC + 200 2 P = 0

Eqs. (1) + (2):

−400TB − 120TC = 0 TB = −

(3) (4) (5)

120 TC − 0.3TC 400

(6)

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 507

PROBLEM 4.131 (Continued)

Eqs. (6)

−160( −0.3TC ) − 280TC + 200 2 P = 0

(5):

−232TC + 200 2 P = 0 TC = 1.2191P

TC = 1.219 P

From Eq. (6):

TB = −0.3(1.2191P) = − 0.36574 = P

From Eq. (2):

− 200(− 0.3657 P) − 200(1.2191P) − 200Tθ D = 0

TB = −0.366 P

TD = − 0.8534 P ΣF = 0:

TD = − 0.853P

F + TB + TC + TD − Pj = 0

i : Fx +

(− 0.36574 P) 2

−

Fx = − 0.3448P j: Fy −

k : Fz +

=0

2

Fx = − 0.345P

(− 0.36574 P) 2

Fy = P

( − 0.8534 P)

−

(1.2191P) 2

−

(− 0.8534 P) 2

− 200 = 0

Fy = P

(1.2191P) 2

=0

Fz = − 0.8620 P Fz = − 0.862 P

F = − 0.345 Pi + Pj − 0.862 Pk

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 508

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PROBLEM 4.132 The uniform 10-kg rod AB is supported by a ball-and-socket joint at A and by the cord CG that is attached to the midpoint G of the rod. Knowing that the rod leans against a frictionless vertical wall at B, determine (a) the tension in the cord, (b) the reactions at A and B.

SOLUTION Free-Body Diagram: Five unknowns and six Eqs. of equilibrium. But equilibrium is maintained (ΣMAB = 0) W = mg = (10 kg) 9.81m/s 2 W = 98.1 N ! GC = − 300i + 200 j − 225k GC = 425 mm ! GC T = T =T (− 300i + 200 j − 225k ) GC 425 rB/ A = − 600i + 400 j + 150 mm rG/ A = − 300i + 200 j + 75 mm ΣMA = 0: rB/ A × B + rG/ A × T + rG/ A × (− Wj) = 0 i j k i j k i j k T − 600 400 150 + − 300 200 75 + − 300 200 75 425 B 0 0 − 300 200 − 225 0 − 98.1 0

Coefficient of i : (−105.88 − 35.29)T + 7357.5 = 0 T = 52.12 N

T = 52.1 N

! !

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 509

!

PROBLEM 4.132 (Continued)

Coefficient of j : 150 B − (300 × 75 + 300 × 225)

52.12 =0 425 B = 73.58 N

B = (73.6 N)i

ΣF = 0: A + B + T − Wj = 0

Coefficient of i : Ax + 73.58 − 52.15

300 =0 425

Ax = 36.8 N

Coefficient of j: Ay + 52.15

200 − 98.1 = 0 425

Ay = 73.6 N

Coefficient of k : Az − 52.15

225 =0 425

Az = 27.6 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 510

!

PROBLEM 4.133 The bent rod ABDE is supported by ball-and-socket joints at A and E and by the cable DF. If a 60-lb load is applied at C as shown, determine the tension in the cable.

SOLUTION Free-Body Diagram: ! DF = −16i + 11j − 8k DF = 21 in. ! DE T = (−16i + 11j − 8k ) T=T DF 21 rD/E = 16 i rC/E = 16i − 14k ! EA 7i − 24k = EA = 25 EA

ΣM EA = 0:

EA

⋅ (rB/E × T) +

EA

⋅ (rC/E ⋅ (− 60 j)) = 0

7 0 −24 7 0 −24 T 1 + 16 0 −14 =0 16 0 0 21 × 25 25 −16 11 −8 0 −60 0 −

−7 × 14 × 60 + 24 × 16 × 60 24 × 16 × 11 =0 T+ 21 × 25 25 201.14 T + 17,160 = 0 T = 85.314 lb

T = 85.3 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 511

!

PROBLEM 4.134 Solve Problem 4.133, assuming that cable DF is replaced by a cable connecting B and F.

SOLUTION Free-Body Diagram: rB/ A = 9i rC/ A = 9i + 10k ! BF = −16i + 11j + 16k BF = 25.16 in. ! BF T (−16i + 11j + 16k ) = T =T BF 25.16 ! AE 7i − 24k = AE = AE 25

ΣMAE = 0:

AF

⋅ (rB/ A × T) +

AE

⋅ (rC/ A ⋅ (− 60 j)) = 0

7 0 −24 7 0 −24 1 T 9 0 0 +9 0 10 =0 25 × 25.16 25 −16 11 16 0 −60 0 −

24 × 9 × 11 24 × 9 × 60 + 7 × 10 × 60 T+ 25 × 25.16 25 94.436 T − 17,160 = 0

T = 181.7 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 512

!

PROBLEM 4.135 The 50-kg plate ABCD is supported by hinges along edge AB and by wire CE. Knowing that the plate is uniform, determine the tension in the wire.

SOLUTION Free-Body Diagram: W = mg = (50 kg)(9.81 m/s 2 ) W = 490.50 N ! CE = − 240i + 600 j − 400k CE = 760 mm ! CE T (− 240i + 600 j − 400k ) T =T = CE 760 ! AB 480i − 200 j 1 = = = (12i − 5 j) AB 520 13 AB ΣMAB = 0:

AB

⋅ (rE/ A × T ) +

AB

⋅ (rG/ A × − Wj) = 0

rE/ A = 240i + 400 j; rG/ A = 240i − 100 j + 200k −5 −5 12 0 12 0 T 1 + 240 −100 200 =0 240 400 0 13 × 20 13 − 240 600 − 400 −W 0 0 (−12 × 400 × 400 − 5 × 240 × 400)

T + 12 × 200W = 0 760 T = 0.76W = 0.76(490.50 N)

T = 373 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 513

!

PROBLEM 4.136 Solve Problem 4.135, assuming that wire CE is replaced by a wire connecting E and D. PROBLEM 4.135 The 50-kg plate ABCD is supported by hinges along edge AB and by wire CE. Knowing that the plate is uniform, determine the tension in the wire.

SOLUTION Free-Body Diagram: Dimensions in mm

W = mg = (50 kg)(9.81 m/s 2 ) W = 490.50 N ! DE = − 240i + 400 j − 400k DE = 614.5 mm ! DE T (240i + 400 j − 400k ) T =T = DE 614.5 ! AB 480i − 200 j 1 = = (12i − 5 j) AB = 520 13 AB rE/ A = 240i + 400 j; rG/ A = 240i − 100 j + 200k 12 −5 0 12 5 0 T 1 + 240 −100 200 =0 240 400 0 13 × 614.5 13 −W 240 400 − 400 0 0 (−12 × 400 × 400 − 5 × 240 × 400)

T + 12 × 200 × W = 0 614.5 T = 0.6145W = 0.6145(490.50 N) T = 301 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 514

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PROBLEM 4.137 Two rectangular plates are welded together to form the assembly shown. The assembly is supported by ball-and-socket joints at B and D and by a ball on a horizontal surface at C. For the loading shown, determine the reaction at C.

SOLUTION λ BD =

First note

−(6 in.)i − (9 in.) j + (12 in.)k (6) 2 + (9) 2 + (12)2 in.

1 (− 6i − 9 j + 12k ) 16.1555 = − (6 in.)i =

rA/B

P = (80 lb)k rC/D = (8 in.)i C = (C ) j

From the f.b.d. of the plates ΣM BD = 0: λ BD ⋅ (rA/B × P ) + λ BD ⋅ ( rC/D × C ) = 0

−6 −9 12 −6 −9 12 & 6(80) ' & C (8) ' −1 0 0 ( + 1 0 0 ( =0 ) *16.1555 + * 16.1555 )+ 0 0 1 0 1 0 ( −9)(6)(80) + (12)(8)C = 0 C = 45.0 lb

or C = (45.0 lb) j

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!

PROBLEM 4.138 Two 2 × 4-ft plywood panels, each of weight 12 lb, are nailed together as shown. The panels are supported by ball-and-socket joints at A and F and by the wire BH. Determine (a) the location of H in the xy plane if the tension in the wire is to be minimum, (b) the corresponding minimum tension.

SOLUTION Free-Body Diagram:

! AF = 4i − 2 j − 4k AF

rG1/A

AF = 6 ft

1 = (2i − j − 2k ) 3 = 2i − j

rG2 /A = 4i − j − 2k rB/A = 4i

ΣM AF = 0:

AF

⋅ (rG/A × ( −12 j) +

AF

⋅ (rG 2/A × (−12 j)) +

2 −1 −2 2 −1 −2 1 1 + 4 −1 −2 + 2 −1 0 3 3 0 −12 0 0 −12 0 1 1 (2 × 2 × 12) + (−2 × 2 × 12 + 2 × 4 × 12) + 3 3 AF

AF

⋅ (rB/A × T ) = 0

AF

AF

⋅ (rB/A × T) = −32 or T ⋅ (

⋅ (rB/A × T) = 0

⋅ (rB/A × T) = 0 A/F

× rB/A ) = −32

(1)

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 516

PROBLEM 4.138 (Continued)

Projection of T on (

AF

× rB/A ) is constant. Thus, Tmin is parallel to AF

Corresponding unit vector is

1 5

1 1 × rB/A = (2i − j − 2k ) × 4i = (−8 j + 4k ) 3 3

(−2 j + k ) Tmin = T ( −2 j + k )

Eq. (1):

1

(2)

5

T &1 ' (−2 j + k ) ⋅ ( (2i − j − 2k ) × 4i ) = −32 3 5 * + T 1 (−2 j + k ) ⋅ ( −8 j + 4k ) = −32 3 5 T (16 + 4) = −32 3 5

T =−

3 5(32) = 4.8 5 20

T = 10.7331 lb

Eq. (2)

Tmin = T ( −2 j + k )

1 5

= 4.8 5( −2 j + k )

1

5 Tmin = −(9.6 lb)j + (4.8 lb k )

Since Tmin has no i component, wire BH is parallel to the yz plane, and x = 4 ft.

(a) (b)

x = 4.00 ft;

y = 8.00 ft

Tmin = 10.73 lb

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 517

!

PROBLEM 4.139 Solve Problem 4.138, subject to the restriction that H must lie on the y axis. PROBLEM 4.138 Two 2 × 4-ft plywood panels, each of weight 12 lb, are nailed together as shown. The panels are supported by ball-andsocket joints at A and F and by the wire BH. Determine (a) the location of H in the xy plane if the tension in the wire is to be minimum, (b) the corresponding minimum tension.

SOLUTION Free-Body Diagram: ! AF = 4i − 2 j − 4k AF

rG1/A

1 = (2i − j − 2k ) 3 = 2i − j

rG2 /A = 4i − j − 2k rB/A = 4i ΣMAF = 0:

AF

⋅ (rG/A × (−12 j) +

AF

⋅ (rG2 /A × (−12 j)) +

2 −1 2 2 −1 −2 1 1 2 −1 0 + 4 −1 −2 + 3 3 0 −12 0 0 −12 0 1 1 (2 × 2 × 12) + (−2 × 2 × 12 + 2 × 4 × 12) + 3 3 AF

AF

AF

AF

⋅ (rB/A × T ) = 0

⋅ (rB/A × T) = 0

⋅ (rB/A × T) = 0

⋅ (rB/A × T) = −32 ! BH = −4i + yj − 4k BH = (32 + y 2 )1/2 ! BH −4i + yj − 4k =T T=T BH (32 + y 2 )1/ 2

(1)

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 518

PROBLEM 4.139 (Continued)

Eq. (1):

AF

2 −1 −2 T ⋅ (rB/A × T ) = 4 0 0 = −32 3(32 + y 2 )1/2 −4 y −4

(−16 − 8 y )T = −3 × 32(32 + y 2 )1/2

T = 96

(32 + y 2 )1/2 8 y + 16

(2)

(8y +16) 12 (32 + y 2 ) −1/ 2 (2 y ) + (32 + y 2 )1/2 (8) dT = 0: 96 dy (8 y + 16) 2

Numerator = 0:

(8 y + 16) y = (32 + y 2 )8 8 y 2 + 16 y = 32 × 8 + 8 y 2

Eq. (2):

T = 96

(32 + 162 )1/ 2 = 11.3137 lb 8 × 16 + 16

y = 16.00 ft Tmin = 11.31 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 519

!

PROBLEM 4.140 The pipe ACDE is supported by ball-and-socket joints at A and E and by the wire DF. Determine the tension in the wire when a 640-N load is applied at B as shown.

SOLUTION Free-Body Diagram: Dimensions in mm

! AE = 480i + 160 j − 240k AE = 560 mm ! AE 480i + 160 j − 240k = = AE AE 560 6i + 2 j − 3k AE = 7 rB/A = 200i rD/A = 480i + 160 j ! DF = −480i + 330 j − 240k ; DF = 630 mm ! DF −480i + 330 j − 240k −16i + 11j − 8k = TDF = TDF TDF = TDF 630 21 DF ΣM AE =

AE

⋅ (rD/A × TDF ) +

AE

⋅ (rB/A × (−600 j)) = 0

6 2 −3 6 2 −3 TDE 1 480 160 0 0 0 =0 + 200 21 × 7 7 0 −640 0 −16 11 −8 3 × 200 × 640 −6 × 160 × 8 + 2 × 480 × 8 − 3 × 480 × 11 − 3 × 160 × 16 TDF + =0 21 × 7 7 −1120TDF + 384 × 103 = 0 TDF = 342.86 N

TDF = 343 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 520

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PROBLEM 4.141 Solve Problem 4.140, assuming that wire DF is replaced by a wire connecting C and F. PROBLEM 4.140 The pipe ACDE is supported by ball-and-socket joints at A and E and by the wire DF. Determine the tension in the wire when a 640-N load is applied at B as shown.

SOLUTION Free-Body Diagram: Dimensions in mm

! AE = 480i + 160 j − 240k AE = 560 mm ! AE 480i + 160 j − 240k = = AE 560 AE 6i + 2 j − 3k AE = 7 rB/A = 200i rC/A = 480i ! CF = −480i + 490 j − 240k ; CF = 726.70 mm ! CE −480i + 490 j − 240k TCF = TCF = CF 726.70 ΣMAE = 0:

AE

⋅ (rC/A × TCF ) +

AE

⋅ (rB/A × (−600 j)) = 0

6 2 6 2 −3 −3 TCF 1 480 0 0 0 0 =0 + 200 726.7 × 7 7 0 −640 0 −480 +490 −240 2 × 480 × 240 − 3 × 480 × 490 3 × 200 × 640 TCF + =0 726.7 × 7 7 −653.91TCF + 384 × 103 = 0

TCF = 587 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 521

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PROBLEM 4.142 A hand truck is used to move two kegs, each of mass 40 kg. Neglecting the mass of the hand truck, determine (a) the vertical force P that should be applied to the handle to maintain equilibrium when α = 35°, (b) the corresponding reaction at each of the two wheels.

SOLUTION Free-Body Diagram: W = mg = (40 kg)(9.81 m/s 2 ) = 392.40 N a1 = (300 mm)sinα − (80 mm)cosα a2 = (430 mm)cosα − (300 mm)sinα b = (930 mm)cosα

From free-body diagram of hand truck Dimensions in mm

ΣM B = 0: P(b) − W ( a2 ) + W (a1 ) = 0

(1)

ΣFy = 0: P − 2W + 2 B = 0

(2)

α = 35°

For

a1 = 300sin 35° − 80 cos 35° = 106.541 mm a2 = 430 cos 35° − 300sin 35° = 180.162 mm b = 930cos 35° = 761.81 mm

(a)

From Equation (1) P(761.81 mm) − 392.40 N(180.162 mm) + 392.40 N(106.54 mm) = 0 P = 37.921 N

(b)

or P = 37.9 N

From Equation (2) 37.921 N − 2(392.40 N) + 2 B = 0

or B = 373 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 522

PROBLEM 4.143 Determine the reactions at A and C when (a) α = 0, (b) α = 30°.

SOLUTION (a)

α = 0° From f.b.d. of member ABC ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − A(0.8 m) = 0 A = 225 N

or

A = 225 N

ΣFy = 0: C y + 225 N = 0 C y = −225 N or C y = 225 N

ΣFx = 0: 300 N + 300 N + C x = 0 C x = −600 N or C x = 600 N

Then

C = Cx2 + C y2 = (600) 2 + (225) 2 = 640.80 N

and

θ = tan −1 "

Cy ! −1 −225 ! # = tan " # = 20.556° $ −600 % $ Cx %

or (b)

C = 641 N

20.6°

α = 30° From f.b.d. of member ABC ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − ( A cos 30°)(0.8 m) + ( A sin 30°)(20 in.) = 0 A = 365.24 N

or

A = 365 N

60.0°

ΣFx = 0: 300 N + 300 N + (365.24 N) sin 30° + C x = 0 Cx = −782.62

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 523

PROBLEM 4.143 (Continued)

ΣFy = 0: C y + (365.24 N) cos 30° = 0 C y = −316.31 N or C y = 316 N

Then

C = Cx2 + C y2 = (782.62) 2 + (316.31) 2 = 884.12 N

and

θ = tan −1 "

Cy ! −1 −316.31 ! # = tan " # = 22.007° $ −782.62 % $ Cx % C = 884 N

or

22.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 524

PROBLEM 4.144 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 75-lb vertical force at B, determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION Free-Body Diagram: Geometry: x AC = (10 in.) cos 20° = 9.3969 in. y AC = (10 in.) sin 20° = 3.4202 in. , yDA = 12 in. − 3.4202 in. = 8.5798 in.

yDA ! −1 8.5798 ! # = tan " # = 42.397° $ 9.3969 % $ x AC %

α = tan −1 "

β = 90° − 20° − 42.397° = 27.603° Equilibrium for lever: ΣM C = 0: TAD cos 27.603°(10 in.) − (75 lb)[(15 in.)cos 20°] = 0

(a)

TAD = 119.293 lb

TAD = 119.3 lb

ΣFx = 0: C x + (119.293 lb) cos 42.397° = 0

(b)

Cx = −88.097 lb

ΣFy = 0: C y − 75 lb − (119.293 lb) sin 42.397° = 0 C y = 155.435

Thus:

C = Cx2 + C y2 = (−88.097) 2 + (155.435) 2 = 178.665 lb

and

θ = tan −1

Cy Cx

= tan −1

155.435 = 60.456° 88.097

C = 178.7 lb

60.5°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 525

PROBLEM 4.145 Neglecting friction and the radius of the pulley, determine (a) the tension in cable ADB, (b) the reaction at C.

SOLUTION Free-Body Diagram: Dimensions in mm

Geometry: Distance

AD = (0.36) 2 + (0.150) 2 = 0.39 m

Distance

BD = (0.2)2 + (0.15)2 = 0.25 m

Equilibrium for beam: ΣM C = 0:

(a)

0.15 ! 0.15 ! T # (0.36 m) − " T # (0.2 m) = 0 (120 N)(0.28 m) − " $ 0.39 % $ 0.25 % T = 130.000 N

or

T = 130.0 N

0.36 ! 0.2 ! ΣFx = 0: Cx + " (130.000 N) + " # # (130.000 N) = 0 $ 0.39 % $ 0.25 %

(b)

Cx = − 224.00 N

0.15 ! 0.15 ! ΣFy = 0: C y + " (130.00 N) + " # # (130.00 N) − 120 N = 0 $ 0.39 % $ 0.25 % C y = − 8.0000 N 2

Thus:

C = Cx2 + C y2 = (−224) 2 + ( − 8 ) = 224.14 N

and

θ = tan −1

Cy Cx

= tan −1

8 = 2.0454° 224

C = 224 N

2.05°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 526

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PROBLEM 4.146 The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C, D, and E when θ = 30°.

SOLUTION Free-Body Diagram: ΣFy = 0: E cos 30° − 20 − 40 = 0 E=

60 lb = 69.282 lb cos 30°

E = 69.3 lb

60.0°

ΣM D = 0: (20 lb)(4 in.) − (40 lb)(4 in.) − C (3 in.) + E sin 30°(3 in.) = 0 −80 − 3C + 69.282(0.5)(3) = 0 C = 7.9743 lb

C = 7.97 lb

ΣFx = 0: E sin 30° + C − D = 0 (69.282 lb)(0.5) + 7.9743 lb − D = 0 D = 42.615 lb

D = 42.6 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 527

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PROBLEM 4.147 The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine (a) the smallest value of θ for which the equilibrium of the bracket is maintained, (b) the corresponding reactions at C, D, and E.

SOLUTION Free-Body Diagram: ΣFy = 0: E cos θ − 20 − 40 = 0 E=

60 cos θ

(1)

ΣM D = 0: (20 lb)(4 in.) − (40 lb)(4 in.) − C (3 in.) ! 60 +" sin θ # 3 in. = 0 θ cos $ % 1 C = (180 tan θ − 80) 3

(a)

For C = 0,

180 tan θ = 80 tan θ =

Eq. (1)

E=

4 θ = 23.962° 9

θ = 24.0°

60 = 65.659 cos 23.962°

ΣFx = 0: −D + C + E sin θ = 0 D = (65.659) sin 23.962 = 26.666 lb

(b)

C = 0 D = 26.7 lb

E = 65.71 lb

66.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 528

PROBLEM 4.148 For the frame and loading shown, determine the reactions at A and C.

SOLUTION

Since member AB is acted upon by two forces, A and B, they must be colinear, have the same magnitude, and be opposite in direction for AB to be in equilibrium. The force B acting at B of member BCD will be equal in magnitude but opposite in direction to force B acting on member AB. Member BCD is a three-force body with member forces intersecting at E. The f.b.d.’s of members AB and BCD illustrate the above conditions. The force triangle for member BCD is also shown. The angle β is found from the member dimensions: 6 in. ! # = 30.964° 10 $ in. %

β = tan −1 "

Applying of the law of sines to the force triangle for member BCD, 30 lb B C = = sin(45° − β ) sin β sin135°

or

30 lb B C = = sin14.036° sin 30.964° sin135° A= B=

(30 lb)sin 30.964° = 63.641 lb sin14.036°

or and

C=

A = 63.6 lb

45.0°

C = 87.5 lb

59.0°

(30 lb) sin135° = 87.466 lb sin14.036°

or

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 529

PROBLEM 4.149 Determine the reactions at A and B when β = 50°.

SOLUTION Free-Body Diagram: (Three-force body) Reaction A must pass through Point D where 100-N force and B intersect In right ∆ BCD

α = 90° − 75° = 15° BD = 250 tan 75° = 933.01 mm Dimensions in mm

In right ∆ ABD tan γ =

150 mm AB = BD 933.01 mm

γ = 9.13 Force Triangle Law of sines 100 N A B = = sin 9.13° sin15° sin155.87° A = 163.1 N; B = 257.6 N A = 163.1 N

74.1° B = 258 N

65.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 530

PROBLEM 4.150 The 6-m pole ABC is acted upon by a 455-N force as shown. The pole is held by a ball-and-socket joint at A and by two cables BD and BE. For a = 3 m, determine the tension in each cable and the reaction at A.

SOLUTION Free-Body Diagram: Five unknowns and six Eqs. of equilibrium, but equilibrium is maintained (ΣM AC = 0) rB = 3j rC = 6 j ! CF = −3i − 6 j + 2k CF = 7 m ! BD = 1.5i − 3j − 3k BD = 4.5 m ! BE = 1.5i − 3j + 3k BE = 4.5 m ! CF P P=P = (−3i − 6 j + 2k ) CE 7 ! T BD TBD (1.5i − 3j − 3k ) = BD (i − 2 j − 2k ) TBD = TBD = BD 4.5 3 ! BE TBD TBE = TBE = = (i − 2 j + 2k ) BE 3

ΣM A = 0: rB × TBD + rB × TBE + rC × P = 0 i j k i j k i j k TBD TBE P + 0 3 0 + 0 6 0 =0 0 3 0 3 3 7 −3 −6 2 1 −2 −2 1 −2 2

Coefficient of i:

−2TBD + 2TBE +

12 P=0 7

(1)

Coefficient of k:

−TBD − TBF +

18 P=0 7

(2)

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 531

PROBLEM 4.150 (Continued)

Eq. (1) + 2 Eq. (2): Eq. (2):

− 4TBD + −

48 12 P = 0 TBD = P 7 7

12 18 6 P − TBE + P = 0 TBE = P 7 7 7 P = 445 N TBD =

Since TBE =

12 (455) 7

6 (455) 7

TBD = 780 N TBE = 390 N

ΣF = 0: TBD + TBE + P + A = 0 780 390 455 (3) + Ax = 0 + − 3 3 7

Coefficient of i:

260 + 130 − 195 + Ax = 0

Coefficient of j:

−

780 390 455 (2) − (2) − (6) + Ay = 0 3 3 7 −520 − 260 − 390 + Ay = 0

Coefficient of k:

−

Ax = 195.0 N

Ay = 1170 N

780 390 455 (2) + (2) + (2) + Az = 0 3 3 7 −520 + 260 + 130 + Az = 0

Az = +130.0 N A = −(195.0 N)i + (1170 N) j + (130.0 N)k

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 532

!

PROBLEM 4.151 Solve Problem 4.150 for a = 1.5 m. PROBLEM 4.150 The 6-m pole ABC is acted upon by a 455-N force as shown. The pole is held by a ball-and-socket joint at A and by two cables BD and BE. For a = 3 m, determine the tension in each cable and the reaction at A.

SOLUTION Free-Body Diagram: Five unknowns and six Eqs. of equilibrium but equilibrium is maintained (ΣM AC = 0) rB = 3j

ΣM A = 0: rB × TBD

rC = 6 j ! CF = −1.5i − 6 j + 2k CF = 6.5 m ! BD = 1.5i − 3j − 3k BD = 4.5 m ! BE = 1.5i − 3j + 3k BE = 4.5 m ! CF P P ( −1.5i − 6 j + 2k ) = (−3i − 12 j + 4k ) P=P = CE 6.5 13 ! T BD TBD (1.5i − 3j − 3k ) = BD (i − 2 j − 2k ) TBD = TBD = BD 4.5 3 ! BE TBD TBE = TBE = = (i − 2 j + 2k ) BE 3 + rB × TBE + rC × P = 0

i j k i j k i j k TBD TBE P + 0 3 0 + 0 =0 0 3 0 6 0 3 3 13 −3 −12 + 4 1 −2 −2 1 −2 2

Coefficient of i:

−2TBD + 2TBE +

24 P=0 13

(1)

Coefficient of k:

−TBD − TBE +

18 P=0 13

(2)

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 533

PROBLEM 4.151 (Continued)

Eq. (1) + 2 Eq. (2): Eq (2):

−4TBD + −

60 15 P = 0 TBD = P 13 13

15 18 3 P − TBE + P = 0 TBE = P 13 13 13 P = 445 N TBD =

Since

TBE =

15 (455) 13

3 (455) 13

TBD = 525 N

!

TBE = 105.0 N

!

ΣF = 0: TBD + TBE + P + A = 0 525 105 455 (3) + Ax = 0 + − 3 3 13

Coefficient of i:

175 + 35 − 105 + Ax = 0

Coefficient of j:

−

525 105 455 (2) − (2) − (12) + Ay = 0 3 3 13 −350 − 70 − 420 + Ay = 0

Coefficient of k:

Ax = 105.0 N

−

Ay = 840 N

525 105 455 (2) + (2) + (4) + Az = 0 3 3 13 −350 + 70 + 140 + Az = 0

Az = 140.0 N A = −(105.0 N)i + (840 N) j + (140.0 N)k

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 534

PROBLEM 4.152 The rigid L-shaped member ABF is supported by a ball-and-socket joint at A and by three cables. For the loading shown, determine the tension in each cable and the reaction at A.

SOLUTION Free-Body Diagram: rB/A = 12i rF/A = 12 j − 8k rD/A = 12i − 16k rE/A = 12i − 24k rF/A = 12i − 32k ! BG = −12i + 9k BG = 15 in. BG = −0.8i + 0.6k ! DH = −12i + 16 j; DH = 20 in.; λDH = −0.6i + 0.8 j ! FJ = −12i + 16 j; FJ = 20 in.; λFJ = −0.6i + 0.8 j

ΣM A = 0: rB/A × TBG λBG + rDH × TDH λDH + rF/A × TFJ λFJ +rF/A × (−24 j) + rE/A × ( −24 j) = 0 i j k i j k i j k 12 0 0 TBG + 12 0 −16 TDH + 12 0 −32 TFJ −0.8 0 0.6 −0.6 0.8 0 −0.6 0.8 0 i j k i j k + 12 0 −8 + 12 0 −24 = 0 0 −24 0 0 −24 0

Coefficient of i:

+12.8TDH + 25.6TFJ − 192 − 576 = 0

(1)

Coefficient of k:

+9.6TDH + 9.6TFJ − 288 − 288 = 0

(2)

3 4

Eq. (1) − Eq. (2):

9.6TFJ = 0

TFJ = 0

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 535

PROBLEM 4.152 (Continued)

Eq. (1):

12.8TDH − 268 = 0

Coefficient of j:

−7.2TBG + (16 × 0.6)(60.0 lb) = 0 ΣF = 0:

TDH = 60 lb TBG = 80.0 lb

A + TBG λ BG + TDH λ DH + TFJ − 24 j − 24 j = 0

Coefficient of i:

Ax + (80)( −0.8) + (60.0)(−0.6) = 0

Ax = 100.0 lb

Coefficient of j:

Ay + (60.0)(0.8) − 24 − 24 = 0

Ay = 0

Coefficient of k:

Az + (80.0)(+0.6) = 0

Az = −48.0 lb A = (100.0 lb)i − (48.0 lb) j

Note: The value Ay = 0 Can be confirmed by considering ΣM BF = 0

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 536

PROBLEM 4.153 A force P is applied to a bent rod ABC, which may be supported in four different ways as shown. In each case, if possible, determine the reactions at the supports.

SOLUTION (a)

ΣM A = 0: − Pa + (C sin 45°)2a + (cos 45°)a = 0 3

2 C =P C= P 3 2

2 ! 1 ΣFx = 0: Ax − " P " 3 ## 2 $ %

C = 0.471P

Ax =

2 ! 1 ΣFy = 0: Ay − P + " P " 3 ## 2 $ %

P 3

Ay =

2P 3 A = 0.745P

(b)

45°

63.4°

ΣM C = 0: +Pa − ( A cos 30°)2a + ( A sin 30°)a = 0 A(1.732 − 0.5) = P

A = 0.812 P A = 0.812P

60.0°

ΣFx = 0: (0.812 P)sin 30° + Cx = 0 C x = −0.406 P ΣFy = 0: (0.812 P) cos 30° − P + C y = 0 C y = −0.297 P C = 0.503P

36.2°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 537

PROBLEM 4.153 (Continued)

ΣM C = 0: + Pa − ( A cos 30°)2a + ( A sin 30°)a = 0

(c)

A(1.732 + 0.5) = P

A = 0.448P A = 0.448P

ΣFx = 0: − (0.448P) sin 30° + Cx = 0

60.0°

Cx = 0.224 P

ΣFy = 0: (0.448 P) cos 30° − P + C y = 0 C y = 0.612 P C = 0.652P

69.9°

! (d)

Force T exerted by wire and reactions A and C all intersect at Point D.

ΣM D = 0: Pa = 0

Equilibrium not maintained Rod is improperly constrained

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 538

!

CHAPTER 5

PROBLEM 5.1 Locate the centroid of the plane area shown.

SOLUTION Dimensions in mm

Then

A, mm 2

x , mm

y , mm

x A, mm3

y A, mm3

1

6300

105

15

0.66150 × 106

0.094500 × 106

2

9000

225

150

2.0250 × 106

1.35000 × 106

Σ

15300

2.6865 × 106

1.44450 × 106

X=

Σ x A 2.6865 × 106 = 15300 ΣA

Y =

Σ y A 1.44450 × 106 = ΣA 15300

X = 175.6 mm Y = 94.4 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 541

PROBLEM 5.2 Locate the centroid of the plane area shown.

SOLUTION Dimensions in mm

Then

A, mm 2

x , mm

y , mm

x A, mm3

y A, mm3

1

1200

10

30

12000

36000

2

540

30

36

16200

19440

Σ

1740

28200

55440

X=

Σ x A 28200 = 1740 ΣA

X = 16.21 mm

Y =

Σ y A 55440 = 1740 ΣA

Y = 31.9 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 542

PROBLEM 5.3 Locate the centroid of the plane area shown.

SOLUTION Dimensions in in.

Then

A, in.2

x , in.

y , in.

x A, in.3

y A, in.3

1

1 × 12 × 15 = 90 2

8

5

720

450

2

21 × 15 = 315

22.5

7.5

7087.5

2362.5

Σ

405.00

7807.5

2812.5

X=

Σ x A 7807.5 = Σ A 405.00

X = 19.28 in.

Y =

Σ y A 2812.5 = Σ A 405.00

Y = 6.94 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 543

PROBLEM 5.4 Locate the centroid of the plane area shown.

SOLUTION

Then

A, in.2

x , in.

y , in.

x A, in.3

y A, in.3

1

1 (12)(6) = 36 2

4

4

144

144

2

(6)(3) = 18

9

7.5

162

135

Σ

54

306

279

XA = Σ xA X (54) = 306

X = 5.67 in.

YA = Σ yA Y (54) = 279

Y = 5.17 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 544

PROBLEM 5.5 Locate the centroid of the plane area shown.

SOLUTION

Then

A, in.2

x , in.

y , in.

x A, in.3

y A, in.3

1

14 × 20 = 280

7

10

1960

2800

2

−π (4) 2 = −16π

6

12

–301.59

–603.19

Σ

229.73

1658.41

2196.8

X=

Σ xA 1658.41 = ΣA 229.73

X = 7.22 in.

Y =

Σ y A 2196.8 = Σ A 229.73

Y = 9.56 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 545

PROBLEM 5.6 Locate the centroid of the plane area shown.

SOLUTION

A, mm 2

x , mm

y , mm

x A, mm3

y A, mm3

1

1 (120)(75) = 4500 2

80

25

360 × 103

112.5 × 103

2

(75)(75) = 5625

157.5

37.5

885.94 × 103

210.94 × 103

163.169

43.169

−720.86 × 103

−190.716 × 103

525.08 × 103

132.724 × 103

3 Σ

Then

−

π 4

(75) 2 = −4417.9

5707.1

XA = Σx A

X (5707.1) = 525.08 × 103

X = 92.0 mm

YA = Σ y A

Y (5707.1) = 132.724 × 103

Y = 23.3 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 546

PROBLEM 5.7 Locate the centroid of the plane area shown.

SOLUTION

A, in.2

x , in.

y , in.

x A, in.3

y A, in.3

(38)2 = 2268.2

0

16.1277

0

36581

2

−20 × 16 = −320

−10

8

3200

−2560

Σ

1948.23

3200

34021

1

Then

π 2

X=

Σ xA 3200 = Σ A 1948.23

X = 1.643 in.

Y =

Σ y A 34021 = Σ A 1948.23

Y = 17.46 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 547

PROBLEM 5.8 Locate the centroid of the plane area shown.

SOLUTION

1 2 Σ

Then

−

A, in.2

x , in.

y , in.

x A, in.3

y A, in.3

30 × 50 = 1500

15

25

22500

37500

23.634

30

–8353.0

–10602.9

14147.0

26.897

π 2

(15) 2 = 353.43

1146.57

X=

Σ x A 14147.0 = Σ A 1146.57

X = 12.34 in.

Y =

Σ y A 26897 = Σ A 1146.57

Y = 23.5 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 548

PROBLEM 5.9 Locate the centroid of the plane area shown.

SOLUTION

1

Σ

Then

x , mm

y , mm

x A, mm3

y A, mm3

(60)(120) = 7200

–30

60

−216 × 103

432 × 103

(60) 2 = 2827.4

25.465

95.435

72.000 × 103

269.83 × 103

(60) 2 = −2827.4

–25.465

25.465

72.000 × 103

−72.000 × 103

−72.000 × 103

629.83 × 103

π

2

3

A, mm 2

4 −

π 4

7200

XA = Σ x A

X (7200) = −72.000 × 103

YA = Σ y A

Y (7200) = 629.83 × 103

X = −10.00 mm Y = 87.5 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 549

PROBLEM 5.10 Locate the centroid of the plane area shown.

SOLUTION Dimensions in mm

A, mm 2

1

Then

π 2

× 47 × 26 = 1919.51

2

1 × 94 × 70 = 3290 2

Σ

5209.5

x , mm

y , mm

x A, mm3

y A, mm3

0

11.0347

0

21181

−15.6667

−23.333

−51543

−76766

−51543

−55584

X=

Σ x A −51543 = ΣA 5209.5

X = −9.89 mm

Y =

Σ y A −55584 = ΣA 5209.5

Y = −10.67 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 550

PROBLEM 5.11 Locate the centroid of the plane area shown.

SOLUTION X =0

First note that symmetry implies

A, in.2

1

2

−

π (8) 2 2

π (12) 2 2

yA, in.3

= −100.531

3.3953

–341.33

= 226.19

5.0930

1151.99

125.659

Σ

Then

y , in.

Y =

!

810.66

Σ y A 810.66 in.3 = Σ A 125.66 in.2

or Y = 6.45 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 551

PROBLEM 5.12 Locate the centroid of the plane area shown.

SOLUTION

Then

A, mm 2

x , mm

y , mm

xA, mm3

yA, mm3

1

(15)(80) = 1200

40

7.5

48 × 103

9 × 103

2

1 (50)(80) = 1333.33 3

60

30

80 × 103

40 × 103

Σ

2533.3

128 × 103

49 × 103

X A = Σ xA X (2533.3) = 128 × 103

X = 50.5 mm

YA = Σ yA Y (2533.3) = 49 × 103

Y = 19.34 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 552

PROBLEM 5.13 Locate the centroid of the plane area shown.

SOLUTION

1

2 Σ

Then

A, mm 2

x , mm

y , mm

x A, mm3

y A, mm3

1 × 30 × 20 = 200 3

9

15

1800

3000

12.7324

32.7324

9000.0

23137

10800

26137

π 4

(30)2 = 706.86

906.86

X=

Σ x A 10800 = Σ A 906.86

X = 11.91 mm

Y =

Σ y A 26137 = Σ A 906.86

Y = 28.8 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 553

PROBLEM 5.14 Locate the centroid of the plane area shown.

SOLUTION Dimensions in in.

Then

A, in.2

x , in.

y , in.

x A, in 3

y A, in.3

1

2 800 × (20)(20) = 3 3

12

7.5

3200

2000

2

−1 −400 (20)(20) = 3 3

15

6.0

–2000

–800

Σ

400 3

1200

1200

X=

Y =

Σ x A 1200 = 400 ! ΣA " # $ 3 % Σ yA 1200 = 400 ! ΣA " # $ 3 %

X = 9.00 in.

Y = 9.00 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 554

PROBLEM 5.15 Locate the centroid of the plane area shown.

SOLUTION

Then

A, mm 2

x , mm

y , mm

x A, mm3

y A, mm3

1

2 (75)(120) = 6000 3

28.125

48

168750

288000

2

1 − (75)(60) = −2250 2

25

20

–56250

–45000

Σ

3750

112500

243000

X ΣA = ΣxA X (3750 mm 2 ) = 112500 mm3

and

or X = 30.0 mm

Y ΣA = Σ yA Y (3750 mm 2 ) = 243000

or Y = 64.8 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 555

PROBLEM 5.16 Determine the y coordinate of the centroid of the shaded area in terms of r1, r2, and α.

SOLUTION

First, determine the location of the centroid. y2 =

From Figure 5.8A:

Similarly

=

2 cos α r2 π 3 ( 2 −α )

y1 =

2 cos α r1 3 ( π2 − α )

Σ yA =

Then

π 2 sin ( 2 − α ) r2 π 3 ( 2 −α )

=

π ! A2 = " − α # r22 $2 %

π ! A1 = " − α # r12 2 $ %

2 cos α & π ! ' 2 cos α & π ! 2' r2 − α # r22 ) − r1 π ( (" − α # r1 ) 3 ( π2 − α ) *"$ 2 3 2 α − % + % + ( 2 ) *$ 2 3 3 r2 − r1 cos α 3

(

)

π π ! ! ΣA = " − α # r22 − " − α # r12 $2 % $2 %

and

π ! = " − α # r22 − r12 2 $ %

(

)

Y ΣA = Σ yA

Now

& π ' 2 ! Y (" − α # r22 − r12 ) = r23 − r13 cos α 2 % *$ + 3

(

)

(

)

Y =

2 r23 − r13 " 3 "$ r22 − r12

! 2 cos α ! ## " # % $ π − 2α %

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 556

PROBLEM 5.17 Show that as r1 approaches r2, the location of the centroid approaches that for an arc of circle of radius (r1 + r2 )/2.

SOLUTION

First, determine the location of the centroid. y2 =

From Figure 5.8A:

Similarly

=

2 cos α r2 π 3 ( 2 −α )

y1 =

2 cos α r1 3 ( π2 − α )

Σ yA =

Then

π 2 sin ( 2 − α ) r2 π 3 ( 2 −α )

=

π ! A2 = " − α # r22 $2 %

π ! A1 = " − α # r12 2 $ %

2 cos α & π ! ' 2 cos α & π ! 2' r2 − α # r22 ) − r1 π ( (" − α # r1 ) 3 ( π2 − α ) *"$ 2 3 2 − α % + % + ( 2 ) *$ 2 3 3 r2 − r1 cos α 3

(

)

π π ! ! ΣA = " − α # r22 − " − α # r12 $2 % $2 %

and

π ! = " − α # r22 − r12 2 $ %

(

)

Y ΣA = Σ yA

Now

& π ' 2 ! Y (" − α # r22 − r12 ) = r23 − r13 cos α 2 % *$ + 3

(

)

Y =

(

)

2 r23 − r13 " 3 "$ r22 − r12

! 2cos α ! ## " # % $ π − 2α %

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 557

PROBLEM 5.17 (Continued)

1 (r1 + r2 ) is 2

Using Figure 5.8B, Y of an arc of radius Y = =

Now

r23 − r13 r22

−

r12

= =

sin( π − α ) 1 (r1 + r2 ) π 2 2 ( 2 −α) 1 cos α (r1 + r2 ) π 2 ( 2 −α)

(

(r2 − r1 ) r22 + r1r2 + r12

(1)

)

(r2 − r1 )(r2 + r1 ) r22 + r1r2 + r12 r2 + r1

r2 = r + ∆

Let

r1 = r − ∆ r=

Then and

In the limit as ∆

r23 − r13 r22

−

r12

1 (r1 + r2 ) 2

=

(r + ∆) 2 + (r + ∆)(r − ∆)( r − ∆) 2 (r + ∆ ) + (r − ∆ )

=

3r 2 + ∆ 2 2r

0 (i.e., r1 = r2 ), then r23 − r13 r22 − r12

So that

=

3 r 2

=

3 1 × ( r1 + r2 ) 2 2

Y =

2 3 cos α × (r1 + r2 ) π −α 3 4 2

or Y = ( r1 + r2 )

cos α π − 2α

Which agrees with Equation (1).

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 558

PROBLEM 5.18 For the area shown, determine the ratio a/b for which x = y .

SOLUTION

Then

A

x

y

xA

yA

1

2 ab 3

3 a 8

3 b 5

a 2b 4

2ab2 5

2

1 − ab 2

1 a 3

2 b 3

Σ

1 ab 6

−

a 2b 6

a 2b 12

−

ab 2 3

ab 2 15

X Σ A = Σ xA 1 ! a 2b X " ab # = $ 6 % 12

or

1 a 2 Y ΣA = Σ y A X=

1 ! ab 2 Y " ab # = $ 6 % 15 2 b 5

or

Y =

Now

X =Y ,

1 2 a= b 2 5

or

a 4 = b 5

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 559

PROBLEM 5.19 For the semiannular area of Problem 5.11, determine the ratio r2/r1 so that y = 3r1/4.

SOLUTION

A −

1

Σ

2

Let

4r1 3π

2 − r13 3

r22

4r2 3π

2 3 r2 3

(r

2 2

− r12

2 3 3 r2 − r1 3

)

(

)

Y Σ A = Σ yA

Then or

r12

2

2

π

YA

π

π

2

Y

3 2 π r1 × r22 − r12 = r23 − r13 4 2 3 2 3 ' r ! 9π & r2 ! (" # − 1) = " 2 # − 1 16 ($ r1 % ) $ r1 % * +

(

)

p=

(

)

r2 r1

9π [( p + 1)( p − 1)] = ( p − 1)( p 2 + p + 1) 16

or

16 p 2 + (16 − 9π ) p + (16 − 9π ) = 0

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 560

PROBLEM 5.19 (Continued)

−(16 − 9π ) ± (16 − 9π ) 2 − 4(16)(16 − 9π ) 2(16)

Then

p=

or

p = −0.5726 p = 1.3397 r2 = 1.340 r1

Taking the positive root

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 561

PROBLEM 5.20 A composite beam is constructed by bolting four plates to four 60 × 60 × 12-mm angles as shown. The bolts are equally spaced along the beam, and the beam supports a vertical load. As proved in mechanics of materials, the shearing forces exerted on the bolts at A and B are proportional to the first moments with respect to the centroidal x axis of the red shaded areas shown, respectively, in Parts a and b of the figure. Knowing that the force exerted on the bolt at A is 280 N, determine the force exerted on the bolt at B.

SOLUTION

From the problem statement: F is proportional to Qx . (Qx ) B FA (Qx ) A

Therefore:

FA FB , or = (Qx ) A (Qx ) B

For the first moments:

12 ! (Qx ) A = " 225 + # (300 × 12) 2% $

FB =

= 831600 mm3 12 ! (Qx ) B = (Qx ) A + 2 " 225 − # (48 × 12) + 2(225 − 30)(12 × 60) 2% $ = 1364688 mm3

Then

FB =

1364688 (280 N) 831600

or FB = 459 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 562

PROBLEM 5.21 The horizontal x axis is drawn through the centroid C of the area shown, and it divides the area into two component areas A1 and A2 . Determine the first moment of each component area with respect to the x axis, and explain the results obtained.

SOLUTION

Note that Then

and

Qx = Σ y A 5 ! 1 ! (Qx )1 = " in. #" × 6 × 5 # in.2 $ 3 %$ 2 %

or

(Qx )1 = 25.0 in.3

2 ! 1 ! (Qx ) 2 = " − × 2.5 in. #" × 9 × 2.5 # in.2 $ 3 %$ 2 % 1 ! 1 ! + " − × 2.5 in. #" × 6 × 2.5 # in.2 3 2 $ %$ %

or (Qx ) 2 = −25.0 in.3 Now

Qx = (Qx )1 + (Qx )2 = 0

This result is expected since x is a centroidal axis (thus y = 0) and

Qx = Σ yA = Y ΣA( y = 0 , Qx = 0) !

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 563

PROBLEM 5.22 The horizontal x axis is drawn through the centroid C of the area shown, and it divides the area into two component areas A1 and A2 . Determine the first moment of each component area with respect to the x axis, and explain the results obtained.

SOLUTION First determine the location of the centroid C. We have

Then

A, in.2

y ′, in.

y ′A, in.3

I

1 ! 2 " × 2 × 1.5 # = 3 $2 %

0.5

1.5

II

1.5 × 5.5 = 8.25

2.75

22.6875

III

4.5 × 2 = 9

6.5

58.5

Σ

20.25

82.6875

Y ′ Σ A = Σ y ′A Y ′(20.25) = 82.6875

or

Y ′ = 4.0833 in.

Now

Qx = Σ y1 A

Then

&1 ' (Qx )1 = ( (5.5 − 4.0833)in.) [(1.5)(5.5 − 4.0833)]in.2 *2 + + [(6.5 − 4.0833)in.][(4.5)(2)]in.2

and

or

(Qx )1 = 23.3 in.3

&1 ' (Qx ) 2 = − ( (4.0833 in.) ) [(1.5)(4.0833)]in.2 *2 + & 1 ' ! − [(4.0833 − 0.5)in.] × 2 (" × 2 × 1.5 # in.2 ) % *$ 2 +

or (Qx ) 2 = −23.3 in.3

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 564

PROBLEM 5.22 (Continued)

Now

Qx = (Qx )1 + (Qx )2 = 0

This result is expected since x is a centroidal axis (thus Y = 0) and

Qx = Σ y A = Y Σ A (Y = 0 , Qx = 0) !

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 565

PROBLEM 5.23 The first moment of the shaded area with respect to the x axis is denoted by Qx. (a) Express Qx in terms of b, c, and the distance y from the base of the shaded area to the x axis. (b) For what value of y is Ox maximum, and what is that maximum value?

SOLUTION Shaded area: A = b (c − y ) Qx = yA = Qx =

(a) (b)

For Qmax : For y = 0:

1 (c + y )[b(c − y )] 2 1 b (c 2 − y 2 ) 2

dQ = 0 or dy (Qx ) =

1 b(−2 y ) = 0 2

1 2 bc 2

y=0 (Qx ) =

1 2 bc 2

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 566

PROBLEM 5.24 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed.

SOLUTION Perimeter of Figure 5.1 Dimensions in mm

L

x

y

xL, mm 2

yL, mm 2

I

30

0

15

0

0.45 × 103

II

210

105

30

22.05 × 103

6.3 × 103

III

270

210

165

56.7 × 103

44.55 × 103

IV

30

225

300

6.75 × 103

9 × 103

V

300

240

150

72 × 103

45 × 103

VII

240

120

0

28.8 × 103

0

Σ

1080

186.3 × 103

105.3 × 103

X ΣL = Σ x L X (1080 mm) = 186.3 × 103 mm 2

X = 172.5 mm

Y ΣL = Σ y L Y (1080 mm) = 105.3 × 103 mm 2

Y = 97.5 mm

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 567

PROBLEM 5.25 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed.

SOLUTION First note that because wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line.

Then

L, mm

x , mm

y , mm

xL, mm 2

yL, mm 2

1

20

10

0

200

0

2

24

20

12

480

288

3

30

35

24

1050

720

4

46.861

35

42

1640.14

1968.16

5

20

10

60

200

1200

6

60

0

30

0

1800

Σ

200.86

3570.1

5976.2

X ΣL = Σ x L

X (200.86) = 3570.1

X = 17.77 mm

Y ΣL = Σ y L

Y (200.86) = 5976.2

Y = 29.8 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 568

!

PROBLEM 5.26 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed.

SOLUTION First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line.

Then

L, in.

x , in.

y , in.

xL, in.2

yL, in.2

1

33

16.5

0

544.5

0

2

15

33

7.5

495

112.5

3

21

22.5

15

472.5

315

4

122 + 152 = 19.2093

6

7.5

115.256

144.070

Σ

88.209

1627.26

571.57

X ΣL = Σx L X (88.209) = 1627.26

and

or X = 18.45 in.

Y ΣL = Σ y L Y (88.209) = 571.57

or

Y = 6.48 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 569

!

PROBLEM 5.27 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed.

SOLUTION First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line.

Y6 =

Then

2

π

(38 in.)

L, in.

x , in.

y , in.

xL, in.2

yL, in.2

1

18

–29

0

–522

0

2

16

–20

8

–320

128

3

20

–10

16

–200

320

4

16

0

8

0

128

5

38

19

0

722

0

6

π (38) = 119.381

0

24.192

0

2888.1

Σ

227.38

–320

3464.1

X=

Σx L −320 = ΣL 227.38

X = −1.407 in.

Y =

Σ y L 3464.1 = ΣL 227.38

Y = 15.23 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 570

PROBLEM 5.28 A uniform circular rod of weight 8 lb and radius 10 in. is attached to a pin at C and to the cable AB. Determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION For quarter circle (a)

r=

2r

π

2r ! ΣM C = 0: W " # − Tr = 0 $π % 2! 2! T = W " # = (8 lb) " # $π % $π %

(b)

T = 5.09 lb

ΣFx = 0: T − C x = 0 5.09 lb − Cx = 0

C x = 5.09 lb

ΣFy = 0: C y − W = 0

C y = 8 lb

C y − 8 lb = 0

C = 9.48 lb

57.5° !

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 571

PROBLEM 5.29 Member ABCDE is a component of a mobile and is formed from a single piece of aluminum tubing. Knowing that the member is supported at C and that l = 2 m, determine the distance d so that portion BCD of the member is horizontal.

SOLUTION First note that for equilibrium, the center of gravity of the component must lie on a vertical line through C. Further, because the tubing is uniform, the center of gravity of the component will coincide with the centroid of the corresponding line. Thus, X = 0 So that

Σx L = 0

Then 0.75 ! cos 55° # m × (0.75 m) −" d − 2 $ % + (0.75 − d )m × (1.5 m) & 1 !' + ((1.5 − d )m − " × 2 m × cos 55° # ) × (2 m) = 0 $2 %+ *

or

&1 ' (0.75 + 1.5 + 2)d = ( (0.75) 2 − 2 ) cos 55° + (0.75)(1.5) + 3 2 * +

or d = 0.739 m

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 572

PROBLEM 5.30 Member ABCDE is a component of a mobile and is formed from a single piece of aluminum tubing. Knowing that the member is supported at C and that d is 0.50 m, determine the length l of arm DE so that this portion of the member is horizontal.

SOLUTION First note that for equilibrium, the center of gravity of the component must lie on a vertical line through C. Further, because the tubing is uniform, the center of gravity of the component will coincide with the centroid of the corresponding line. Thus, X =0

So that or

Σx L = 0 0.75 ! −" sin 20° + 0.5sin 35° # m × (0.75 m) $ 2 % + (0.25 m × sin 35°) × (1.5 m) l! + "1.0 × sin 35° − # m × (l m) = 0 2 $ %

or

+ ( sin 35° − 2l ) l = 0 −0.096193 !!!!!!!!" !!!!!!" ( xL)DE ( xL) AB +( xL)BD

The equation implies that the center of gravity of DE must be to the right of C. Then

l 2 − 1.14715l + 0.192386 = 0

or

l=

or

l = 0.204 m

1.14715 ± (−1.14715)2 − 4(0.192386) 2 or l = 0.943 m

Note that sin 35° − 12 l . 0 for both values of l so both values are acceptable.!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 573

PROBLEM 5.31 The homogeneous wire ABC is bent into a semicircular arc and a straight section as shown and is attached to a hinge at A. Determine the value of θ for which the wire is in equilibrium for the indicated position.

SOLUTION First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through A. Further, because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. Thus, X =0

So that

Σx L = 0

Then

1 2r ! ! " − 2 r cos θ # (r ) + " π − r cos θ # (π r ) = 0 $ % $ %

or

cos θ =

4 1 + 2π = 0.54921

or θ = 56.7°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 574

!

PROBLEM 5.32 Determine the distance h for which the centroid of the shaded area is as far above line BB as possible when (a) k = 0.10, (b) k = 0.80.

SOLUTION

A

y

yA

1

1 ba 2

1 a 3

1 2 a b 6

2

1 − (kb)h 2

1 h 3

1 − kbh 2 6

Σ

b (a − kh) 2

b 2 ( a − kh 2 ) 6

Y Σ A = Σ yA

Then

&b ' b Y ( ( a − kh) ) = (a 2 − kh 2 ) *2 + 6

Y =

or

(1)

dY 1 −2kh(a − kh) − (a 2 − kh 2 )(− k ) = =0 dh 3 ( a − kh) 2

and or

a 2 − kh 2 3(a − kh)

2h(a − kh) − a 2 + kh 2 = 0

(2)

Simplifying Eq. (2) yields kh 2 − 2ah + a 2 = 0

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 575

PROBLEM 5.32 (Continued)

Then

2a ± (−2a) 2 − 4( k )(a 2 ) 2k a = &*1 ± 1 − k '+ k

h=

Note that only the negative root is acceptable since h , a. Then (a)

k = 0.10 h=

(b)

a & 1 − 1 − 0.10 ' + 0.10 *

or h = 0.513a

k = 0.80 h=

a & 1 − 1 − 0.80 ' + 0.80 *

or h = 0.691a

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 576

PROBLEM 5.33 Knowing that the distance h has been selected to maximize the distance y from line BB to the centroid of the shaded area, show that y = 2h/3.

SOLUTION See solution to Problem 5.32 for analysis leading to the following equations: Y =

a 2 − kh 2 3(a − kh)

(1)

2h(a − kh) − a 2 + kh 2 = 0

(2)

Rearranging Eq. (2) (which defines the value of h which maximizes Y ) yields a 2 − kh 2 = 2h(a − kh)

Then substituting into Eq. (1) (which defines Y ) Y =

1 × 2h(a − kh) 3( a − kh)

or Y =

2 h 3

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 577

PROBLEM 5.34 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h.

SOLUTION y h = x a h y= x a xEL = x 1 yEL = y 2 dA = ydx A=

-x -

EL dA

-

a 0

ydx =

-

= xydx =

yEL dA =

-

a 0

1 "2 $

-

a 0

a 0

1 h ! " a x # dx = 2 ah $ % a

h ! h & x3 ' 1 x " x # dx = ( ) = ha 2 a * 3 +0 3 $a %

1 ! y # ydx = 2 %

-

a 0

2

h ! 1 h2 " a x # dx = 2 2 b $ %

a

& x3 ' 1 2 ( ) = h a 3 * +0 6

1 ! 1 xA = xEL dA: x " ah # = ha 2 $2 % 3

x=

1 ! 1 yA = yEL dA: y " ah # = h 2 a $2 % 6

1 y= h 3

-

-

2 a 3

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 578

PROBLEM 5.35 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h.

SOLUTION y1 : h = ka 2

At (a, h)

k=

or

h a2

y2 : h = ma m=

or

xEL = x

Now

yEL =

1 ( y1 + y2 ) 2

h ' &h dA = ( y2 − y1 )dx = ( x − 2 x 2 ) dx a a * + h = 2 (ax − x 2 ) dx a

and

-

A = dA =

Then

and

h a

-

-

a 0

a

1 ' 1 h h &a (ax − x 2 )dx = 2 ( x 2 − x3 ) = ah 3 +0 6 a2 a *2 a

h &a 1 ' 1 &h x 2 (ax − x 2 ) dx = 2 ( x3 − x 4 ) = a 2 h 0 ( 4 + 0 12 a *3 *a 1 1 2 ( y1 + y2 )[( y2 − y1 ) dx] = yEL dA = y2 − y12 dx 2 2 xEL dA =

-

a

-

=

1 2

- (

-

a 0

)

h 2 2 h2 4 ! "" 2 x − 4 x ##dx a $a % a

1 h2 & a 2 3 1 5 ' = ( x − x ) 2 a4 * 3 5 +0 1 = ah 2 15

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 579

PROBLEM 5.35 (Continued)

1 ! 1 xA = xEL dA: x " ah # = a 2 h $ 6 % 12

x=

1 a 2

1 ! 1 yA = yEL dA: y " ah # = ah 2 $ 6 % 15

y=

2 h 5

-

-

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 580

!

PROBLEM 5.36 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h.

SOLUTION For the element (EL) shown At

x = a,

Then

x=

xEL yEL

h a3

a 1/3 y h1/3

a 1/3 y dy h1/3 1 1 a 1/3 = x= y 2 2 h1/ 3 =y

-

A = dA =

Then

Hence

or k =

dA = xdy =

Now

and

y = h : h = ka3

-

xEL dA =

-

yEL dA =

-

h

0 h

0

-

h

0

a 1/3 3 a y dy = y 4/3 1/3 4 h1/3 h

( )

h

= 0

3 ah 4 h

1 a 1/3 a 1/3 ! 1 a 3 5/3 ! 3 y " 1/3 y dy # = y # = a2h 2/3 " 2 h1/3 2 5 10 h h $ % $ %0 h

a a 3 3 ! ! y " 1/3 y1/3 dy # = 1/3 " y 7/3 # = ah 2 $h % h $7 %0 7

-

3 ! 3 x " ah # = a 2 h $ 4 % 10

x=

2 a 5

!

-

3 ! 3 y " ah # = ah2 $4 % 7

y=

4 h 7

!

xA = xEL dA: yA = yEL dA:

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 581

PROBLEM 5.37 Determine by direct integration the centroid of the area shown.

SOLUTION y=

For the element (EL) shown

b 2 a − x2 a

dA = (b − y ) dx

and

xEL

)

(

b a − a 2 − x 2 dx a =x =

1 ( y + b) 2 b a + a 2 − x2 = 2a

yEL =

(

-

A = dA =

Then

-

To integrate, let

x = a sin θ :

Then

A= =

-

π /2 0

a 0

) )

(

b a − a 2 − x 2 dx a

a 2 − x 2 = a cos θ , dx = a cos θ dθ

b (a − a cos θ )(a cos θ dθ ) a

b& 2 2θ 2 θ ( a sin θ − a " + sin a* 2 4 $

π /2

!' #) %+ 0

π! = ab "1 − # 4% $ and

-x

EL dA

=

-

a

0

&b ' x ( a − a 2 − x 2 dx ) a * +

)

(

π /2

b& a 1 !' = (" x 2 + (a 2 − x 2 )3/2 # ) a *$ 2 3 %+ 0 =

1 3 ab 6

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 582

PROBLEM 5.37 (Continued)

-y

EL dA

=

-

a 0

b &b ' a + a 2 − x 2 ( a − a 2 − x 2 dx ) 2a a * +

b2 = 2 2a =

) (

(

)

a

-

a 0

b 2 x3 ! ( x ) dx = 2 "" ## 2a $ 3 % 0 2

1 2 ab 6

-

& π !' 1 x ( ab "1 − # ) = a 2 b 4 %+ 6 * $

or x =

2a 3(4 − π )

-

& π !' 1 y ( ab "1 − # ) = ab 2 4 %+ 6 * $

or y =

2b 3(4 − π )

xA = xEL dA:

yA = yEL dA:

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 583

!

PROBLEM 5.38 Determine by direct integration the centroid of the area shown.

SOLUTION x =0

First note that symmetry implies For the element (EL) shown yEL =

2r

(Figure 5.8B)

π dA = π rdr

-

A = dA =

Then

and

-y

EL dA

=

-

r2 r1

-

r2 r1

r2 π r d r = π "" $ 2

r2

! π 2 2 r2 − r1 ## = % r1 2

(

r2

1 ! 2 (π rdr ) = 2 " r 3 # = r23 − r13 π $3 %r 3

2r

(

)

)

1

So

&π ' 2 yA = yEL dA: y ( r22 − r12 ) = r23 − r13 2 * + 3

-

(

)

(

)

or y =

4 r23 − r13 3π r22 − r12

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 584

PROBLEM 5.39 Determine by direct integration the centroid of the area shown.

SOLUTION y =0

First note that symmetry implies

xEL

-

A = dA =

Then

-

= a[ x]0a −

and

-

xEL dA =

1 a( adφ ) 2 2 = a cos φ 3

dA =

dA = adx xEL = x

-

a

0

a

0

adx −

α

1

- α 2 a dφ 2

−

a2 α [φ ]α = a 2 (1 − α ) 2

x(adx) −

α

2

1

!

- α 3 a cos φ "$ 2 a dφ #% 2

−

a

& x2 ' 1 = a ( ) − a3 [sin φ ]α−α 2 * +0 3 1 2 ! = a3 " − sin α # $2 3 % 1 2 ! xA = xEL dA: x [a 2 (1 − α )] = a3 " − sin α # $2 3 %

-

or x =

3 − 4sin α a 6(1 − α )

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 585

!

PROBLEM 5.40 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b.

SOLUTION x = 0,

At

y =b

b = k (0 − a) 2 y=

Now

xEL = x

-

A = dA =

Then and

b ( x − a) 2 dx a2

-

a

0

a 1 b b & 3' 2 ( ) − = − = ab x a dx x a ( ) 2 2 * + 0 3 3a a

&b ' b a 3 x ( 2 ( x − a) 2 dx ) = 2 x − 2ax 2 + a 2 x dx 0 *a 0 + a 4 2 2 b x a 2! 1 2 = 2 "" − ax3 + x ## = a b 2 a $ 4 3 % 12

-

xEL dA =

-

yEL dA = =

Hence

y b = 2 ( x − a)2 2 2a

dA = ydx =

and

b a2

b ( x − a )2 a2

Then

yEL =

or k =

-

a

-

a 0

- (

)

a

2 b &b ' b &1 ' ( x − a) 2 ( 2 ( x − a) 2 dx ) = 4 ( ( x − a)5 ) 2 5 2a *a + 2a * +0

1 2 ab 10

1 ! 1 xA = xEL dA: x " ab # = a 2 b $ 3 % 12

-

1 ! 1 yA = yEL dA: y " ab # = ab 2 $ 3 % 10

-

x=

y=

1 a 4

3 b 10

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 586

PROBLEM 5.41 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b.

SOLUTION b 2 x a2 b y2 = k2 x 4 but b = k2 a 4 y2 = 4 x 4 a b x4 ! dA = ( y2 − y1 )dx = 2 "" x 2 − 2 ## dx a $ a % xEL = x y1 = k1 x 2

yEL = =

but b = k1a 2

y1 =

1 ( y1 + y2 ) 2 b 2a 2

x4 2 "" x + 2 a $

,

A = dA =

b a2

,

a 0

! ## % x4 2 "" x − 2 a $

! ## dx %

a

b & x3 x5 ' = 2 ( − 2) a * 3 5a + 0 2 = ba 15

,x

EL dA

! ## dx %

×

b x4 2 x − 2 " a "$ a2

b a2

,

x5 ! 3 "" x − 2 ## dx 0 a % $

=

b a2

& x4 x6 ' − ( 2) * 4 6a + 0

=

1 2 a b 12

=

,

=

a 0

a

a

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 587

PROBLEM 5.41 (Continued)

,

yEL dA = =

,

a 0

b x4 ! b x4 ! 2 2 + − x x " # " # dx 2a 2 "$ a 2 #% a 2 "$ a 2 #%

b2 2a 4

,

a 0

x8 4 "" x − 4 a $

! ## dx % a

2 2 b 2 & x5 x9 ' = 4 ( − 4) = ab 2a * 5 9a + 0 45 2 ! 1 xA = xEL dA: x " ba # = a 2b 15 $ % 12

5 x= a 8

2 ! 2 2 yA = yEL dA: y " ba # = ab ! 15 $ % 45

1 y= b 3

,

!

,

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 588

!

PROBLEM 5.42 Determine by direct integration the centroid of the area shown.

SOLUTION xEL = x

We have

! ## % 2 ! x x dA = ydx = a ""1 − + 2 ## dx L L % $

yEL =

1 a x x2 y = ""1 − + 2 2 2$ L L

,

A = dA =

Then

,

2L 0

2L

& x x2 ! x2 x3 ' a ""1 − + 2 ## dx = a ( x − + 2) L L % 2 L 3L + 0 $ *

8 = aL 3

and

,

xEL dA = =

,

yEL dA = =

,

2L 0

& x x2 x ( a ""1 − + 2 L L (* $

2L

! ' & x 2 x3 x4 ' + 2) ## dx ) = a ( − % )+ * 2 3L 4 L + 0

10 2 aL 3

,

2L 0

a2 2

a x x2 ! & x x2 ! ' ""1 − + 2 ## ( a ""1 − + 2 ## dx ) 2$ L L % *( $ L L % +)

,

EL 0

x x2 x3 x 4 ! ""1 − 2 + 3 2 − 2 3 + 4 ## dx L L L L % $ 2L

a2 & x 2 x3 x4 x5 ' = + 2 − 3 + 4) (x − 2 * L L 2L 5L + 0 11 = a2 L 5

Hence

8 ! 10 xA = xEL dA: x " aL # = aL2 $3 % 3

,

1 ! 11 yA = yEL dA: y " a # = a 2 $8 % 5

,

x=

y=

5 L 4

33 a 40

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 589

!

PROBLEM 5.43 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b.

SOLUTION x = a, y = b : a = kb2

For y2 at

b

Then

y2 =

Now

xEL = x

and for

a 0# x# : 2

yEL =

a

y2 b x1/ 2 = 2 2 a x1/2 a

dx

1 a b x 1 x1/2 ! # x # a : yEL = ( y1 + y2 ) = "" − + # 2 2 2$ a 2 a #% x1/2

dA = ( y2 − y1 )dx = b "" $

Then

a b2

x1/2

dA = y2 dx = b

For

or k =

,

A = dA =

,

a/2 0

b

a

x1/2 a

−

dx +

,

x 1! + # dx a 2 #% x1/ 2 x 1 ! b "" − + ## dx a/2 $ a a 2% a

a

a/ 2 & 2 x3/2 x 2 1 ' b & 2 3/2 ' = + − + x) x b ( ) a (* 3 +0 * 3 a 2a 2 + a/2

=

3/ 2 3/ 2 a! ' 2 b & a! (" # + (a)3/ 2 − " # ) 3 a *($ 2 % $ 2 % +)

2 a! ' 1& a ! ' /. /- 1 & 2 (a ) − $ % ! + (a) − $ % ! # + b "− & 2 ' !) 2 ( & 2 ' ) *, *+ 2a ( 13 ab = 24

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 590

PROBLEM 5.43 (Continued)

and

1

xEL dA =

1

a/2 0

/ x1/ 2 0 x $$ b dx %% + a & '

1

- / x1/ 2 x 1 0 . x b $$ − + %% ! dx a/2 ( & a a 2 ' )! a

a

a/ 2 - 2 x5/2 x3 x 4 . b - 2 5/2 . = − + ! x ! +b a (5 )0 ( 5 a 3a 4 ) a/2 5/2 5/2 2 b -/ a 0 /a0 . 5/2 + − ( ) a $ % $ % ! 5 a (& 2 ' & 2 ' )! 3 2 *2 1 - 3 / a 0 . 1 - 2 / a 0 . *3 + b "− ( a) − $ % ! + (a) − $ % ! # 3a ( & 2 ' !) 4 ( & 2 ' !) ,* +* 71 2 a b = 240

=

1

yEL dA =

1

a/2 0

+

1

b x1/ 2 - x1/ 2 . b dx ! 2 a ( a )

b / x 1 x1/2 0 - / x1/ 2 x 1 0 . − + % dx ! $ − + % b$ a/ 2 2 $ a 2 a %' ( $& a a 2 %' )! & a

a

a/ 2 3 b2 - 1 2 . b 2 -/ x 2 1 / x 1 0 0 . $ x ! + = − $ − % %! 2a ( 2 ) 0 2 $& 2a 3a & a 2 ' %' ! ( ) a/2

=

b -/ a 0 /a0 2 $ % + (a ) − $ % 4 a (& 2 ' &2'

=

11 2 ab 48

2

Hence

2.

b2 / a 1 0 !− $ − % !) 6a & 2 2 '

/ 13 0 71 2 xA = xEL dA: x $ ab % = a b & 24 ' 240

1

/ 13 0 11 2 yA = yEL dA: y $ ab % = ab & 24 ' 48

1

3

x=

17 a = 0.546a 130

y=

11 b = 0.423b 26

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 591

!

PROBLEM 5.44 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b.

SOLUTION x = a,

For y1 at

y = 2b 2b = ka 2

or k =

2b a2

2b 2 x a2

Then

y1 =

By observation

b x0 / y2 = − ( x + 2b) = b $ 2 − % a a' &

Now

xEL = x

and for 0 # x # a :

yEL =

1 b y1 = 2 x 2 2 a

For a # x # 2a :

yEL =

1 b/ x0 x0 / y2 = $ 2 − % and dA = y2 dx = b $ 2 − % dx 2 2& a' a' &

1

A = dA =

Then

1

a 0

and dA = y1dx =

2b 2 x dx + a2

1

2a a

2b 2 x dx a2

x0 / b $ 2 − % dx a' & 2a

a

2 - a/ 2b - x3 . 7 x0 . = 2 ! + b − $ 2 − % ! = ab 6 a ' )! a ( 3 )0 ( 2& 0

and

1x

EL dA

a

/ 2b 0 x $ 2 x 2 dx % + &a '

- / x0 . x b $ 2 − % dx ! a ' ) ( &

1

=

- 2 x3 . 2b - x 4 . ! +b x − ! 2 3a ) 0 a ( 4 )0 (

0

1

2a

=

a

a

2a

1 2 1 - 2 2 3 (2a ) − (a )3 .) # a b + b " -((2a) 2 − (a) 2 .) + ( 2 3a + , 7 2 = a b 6 =

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 592

PROBLEM 5.44 (Continued)

1y

EL dA

=

1

a 0

b 2 - 2b 2 . x x dx ! + a2 ( a2 )

1

2a 0

b/ x 0- / x0 . $ 2 − % b $ 2 − % dx ! 2& a '( & a' ) 2a

a

3 2b 2 - x5 . b2 - a / x0 . = 4 − $2 − % ! ! + a ' )! a ( 5 )0 2 ( 3 & a 17 2 = ab 30

Hence

/7 0 7 xA = xEL dA: x $ ab % = a 2b &6 ' 6

1

/ 7 0 17 2 yA = yEL dA: y $ ab % = ab & 6 ' 30

1

x =a

y=

17 b 35

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 593

!

PROBLEM 5.45 A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid.

SOLUTION First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line xEL = a cos3 θ

Now

and dL = dx 2 + dy 2

x = a cos3 θ : dx = −3a cos 2 θ sin θ dθ

Where

y = a sin 3 θ : dy = 3a sin 2 θ cos θ dθ dL = [(−3a cos 2 θ sin θ dθ )2 + (3a sin 2 θ cos θ dθ )2 ]1/ 2

Then

= 3a cos θ sin θ (cos 2 θ + sin 2 θ )1/ 2 dθ = 3a cos θ sin θ dθ

1

L = dL =

and

1x

EL dL

=

3 a 2

=

1

π /2 0

1

π /2 0

π /2

3a cos θ sin θ dθ = 3a

a cos3θ (3a cos θ sin θ dθ ) π /2

- 1 . = 3a 2 − cos5 θ ! ( 5 )0

Hence

-1 2 . sin θ ! (2 )0

3 = a2 5

/3 0 3 xL = xEL dL : x $ a % = a 2 &2 ' 5

1

x=

2 a 5

Alternative Solution / x0 x = a cos3 θ 4 cos 2 θ = $ % &a' / y0 y = a sin 3 θ 4 sin 2 θ = $ % &a' /x0 $a% & '

2/3

/ y0 +$ % &a'

2/3

2/3

2/3

= 1 or

y = (a 2/3 − x 2/3 )3/2

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 594

PROBLEM 5.45 (Continued)

dy = (a 2/3 − x 2/3 )1/ 2 (− x −1/3 ) dx

Then Now

xEL = x

and

/ dy 0 dL = 1 + $ % & dx '

2

{

dx = 1 + -(( a 2/3 − x 2/3 )1/2 (− x −1/3 ) .)

1

L = dL =

Then

and

Hence

1

xEL dL =

1

a 0

1

a 0

2

1/2

}

dx a

a1/3 3 - 3 2/3 . dx = a1/3 x ! = a 1/3 x (2 )0 2

a / a1/3 0 3 -3 . x $$ 1/3 dx %% = a1/3 x5/3 ! = a 2 (5 )0 5 &x '

/3 0 3 xL = xEL dL : x $ a % = a 2 &2 ' 5

1

x=

2 a 5

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 595

PROBLEM 5.46 A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid.

SOLUTION First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line Now

xEL = r cos θ

Then

L = dL =

and

1

1

xEL dL =

7π / 4

1π

/4

and dL = rdθ 7π /4

1π

/4

3 rdθ = r[θ ]π7π/ 4/ 4 = π r 2

r cos θ ( rdθ )

= r 2 [sin θ ]π7π/4/ 4 / 1 1 0 = r2 $ − − % 2 2' & = −r 2 2

Thus

/3 0 xL = xdL : x $ π r % = −r 2 2 &2 '

1

x =−

2 2 r 3π

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 596

PROBLEM 5.47* A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid. Express your answer in terms of a.

SOLUTION First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. x = a,

We have at

y=a

a = ka 3/2 1

y=

Then

a

or k =

1 a

x3/ 2

dy 3 1/2 = x dx 2 a

and Now

xEL = x

and

/ dy 0 dL = 1 + $ % dx & dx '

2

1/2

2 - / 3 0 . x1/2 % ! = 1+ $ ' !) ( &2 a 1 4a + 9 x dx = 2 a

1

L = dL =

Then

1

a

1

0

2 a

dx

4a + 9 x dx a

1 -2 1 . × (4a + 9 x)3/ 2 ! 3 9 2 a( )0 a = [(13)3/2 − 8] 27 = 1.43971a =

and

1x

EL dL

=

1

a 0

x

- 1 (2 a

. 4a + 9 x dx ! )

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 597

PROBLEM 5.47* (Continued)

Use integration by parts with u=x du = dx

Then

1

xEL dL =

dv = 4a + 9 x dx 2 v= (4a + 9 x)3/ 2 27

a 1 2* 2 3/2 . × + x a x (4 9 ) " ! − 2 a *+ ( 27 )0

1

a 0

3* 2 (4a + 9 x)3/ 2 dx # 27 *, a

=

(13)3/2 2 1 -2 . a − (4a + 9 x)5/2 ! 27 45 27 a ( )0

=

a2 2 2 3 3/ 2 5/2 "(13) − [(13) − 32]# 27 + 45 ,

= 0.78566a 2

1

xL = xEL dL : x (1.43971a ) = 0.78566a 2

or x = 0.546a

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 598

PROBLEM 5.48* Determine by direct integration the centroid of the area shown.

SOLUTION We have

xEL = x 1 πx a yEL = y = cos 2 2 2L

and

dA = ydx = a cos

1

A = dA =

Then

1

πx

dx

2L

L/2

a cos

0

πx 2L

dx

L/ 2

πx. - 2L sin =a 2 L !) 0 (π 2

=

and

1x

EL dA

π

aL

πx 0 / = x $ a cos dx 2 L %' &

1

u=x

Use integration by parts with

dv = cos

du = dx

Then

πx

1 x cos 2L dx =

1

2L

v=

× sin

2L

π

πx

−

πx

dx

2L sin

πx 2L

2L

1π

πx

dx 2L 2L π x 2L πx0 2L / cos x sin = + π $& 2L π 2 L %'

π

xEL dA = a =a

2L -

π (

x sin

πx 2L

+

sin

2L

π

cos

π x.

L/ 2

2 L !) 0

2 L -/ L 2 0 2L . L %− + $$ % π ! π (& 2 2 π ' )!

= 0.106374aL2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 599

PROBLEM 5.48* (Continued)

Also

1y

EL

dA =

1

L/ 2 0

a2 = 2

πx/ πx 0 a dx % cos $ a cos 2 2L & 2L '

-x

L/ 2

sin π x . + 2π L ! L (2 )! 0

=

a −2 / L L 0 $ + % 2 & 4 2π '

= 0.20458a 2L / 2 0 xA = xEL dA: x $ aL % = 0.106374aL2 $ π % & '

or x = 0.236 L

/ 2 0 yA = yEL dA: y $ aL % = 0.20458a 2L $ π % & '

or y = 0.454a

1

1

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 600

PROBLEM 5.49* Determine by direct integration the centroid of the area shown.

SOLUTION 2 2 r cos θ = aeθ cos θ 3 3 2 2 θ = r sin θ = ae sin θ 3 3

xEL =

We have

yEL

dA =

and

1 1 ( r )(rdθ ) = a 2 e2θ dθ 2 2

1

A = dA =

Then

1

π 0

π

1 2 2θ 1 -1 . a e dθ = a 2 e 2θ ! 2 2 (2 )0

1 2 2π a (e − 1) 4 = 133.623a 2 =

and

1x

EL dA =

1

π 0

2 θ /1 0 ae cos θ $ a 2 e2θ dθ % 3 &2 '

1 = a3 3

1

π 0

e3θ cos θ dθ

To proceed, use integration by parts, with u = e3θ

du = 3e3θ dθ

and

dv = cos θ dθ

Then

1e

3θ

1

then

du = 3e3θ dθ

dv = sin θ dθ , then

Then

1e

3θ

v = sin θ

cos θ dθ = e3θ sin θ − sin θ (3e3θ dθ ) u = e3θ

Now let

and

v = − cos θ

sin θ dθ = e3θ sin θ − 3 - −e3θ cos θ − (− cos θ )(3e3θ dθ ) .! ( )

1

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 601

PROBLEM 5.49* (Continued)

So that

1

e3θ cos θ dθ =

e3θ (sin θ + 3cos θ ) 10 π

1

. 1 - e3θ (sin θ + 3cos θ ) ! xEL dA = a 3 3 ( 10 )0

1y

Also

EL dA

=

a3 (−3e3π − 3) = −1239.26a3 30

=

1

π 0

2 θ /1 0 ae sin θ $ a 2 e 2θ dθ % 3 &2 '

1 = a3 3

1

π 0

e3θ sin θ dθ

Using integration by parts, as above, with u = e3θ

and

1

dv = sin θ dθ

Then

1e

So that

1

3θ

du = 3e3θ dθ

and

v = − cos θ

1

sin θ dθ = −e3θ cos θ − (− cos θ )(3e3θ dθ )

e3θ sin θ dθ =

e3θ ( − cos θ + 3sin θ ) 10 π

1

. 1 - e3θ (− cos θ + 3sin θ ) ! yEL dA = a3 3 ( 10 )0 =

Hence

a 3 3π (e + 1) = 413.09a3 30

1

or x = −9.27a

1

or

xA = xEL dA: x (133.623a 2 ) = −1239.26a 3 yA = yEL dA: y (133.623a 2 ) = 413.09a3

y = 3.09a

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 602

PROBLEM 5.50 Determine the centroid of the area shown when a = 2 in.

SOLUTION xEL = x

We have

yEL =

/ 10 dA = ydx = $1 − % dx x' &

and

1

A = dA =

Then

and

1 1/ 10 y = $1 − % 2 2& x'

=

1

a

EL dA =

1

a

1x 1y

EL dA

1

1

1

a/

1

1 0 dx 2 a $1 − x % 2 = [ x − ln x]1 = (a − ln a − 1) in. & ' a

. / a2 -/ 1 0 . - x 2 10 − x ! = $$ − a + %% in.3 x $1 − % dx ! = x' ) ( 2 2' (& )1 & 2 1 / 1 0 -/ 1 0 . 1 1 − % $1 − % dx ! = 2 $& x ' (& x' ) 2

1

a/

1

2 1 0 $ 1 − x + 2 % dx x ' &

a

=

11. 1/ 10 x − 2ln x − ! = $ a − 2ln a − % in.3 2( x )1 2 & a'

1

xA = xEL dA: x =

1

yA = yEL dA: y =

a2 2

−a+

1 2

in. a − ln a − 1 a − 2ln a − 1a 2( a − ln a − 1)

in.

Find: x and y when a = 2 in. We have

x=

and

y=

1 2

(2) 2 − 2 +

1 2

2 − ln 2 − 1 2 − 2ln 2 − 12 2(2 − ln 2 − 1)

or

x = 1.629 in.

or y = 0.1853 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 603

PROBLEM 5.51 Determine the value of a for which the ratio x / y is 9.

SOLUTION xEL = x

We have

yEL =

1 1/ 10 y = $1 − % 2 2& x'

/ 10 dA = ydx = $1 − % dx x' &

and

1

A = dA =

Then

1

a/

1

1 0 dx = [ x − ln x]1a $1 − % x' 2 &

= (a − ln a − 1) in.2

and

1

xEL dA =

1

a

1

a

. -/ 1 0 . - x 2 − x! x $1 − % dx ! = x' ) ( 2 (& )1

/ a2 10 = $$ − a + %% in.3 2' & 2

1

yEL dA =

1

a

1

1 / 1 0 -/ 1 0 . 1 1 − % $1 − % dx ! = x ' (& x' ) 2 2 $&

1

a/

1

2 1 0 $ 1 − x + 2 % dx x ' &

a

=

11. x − 2ln x − ! 2( x )1

=

1/ 10 a − 2ln a − % in.3 $ 2& a'

1

xA = xEL dA: x =

1

yA = yEL dA: y =

a2 2

−a+

1 2

in. a − ln a − 1 a − 2ln a − 1a 2( a − ln a − 1)

in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 604

PROBLEM 5.51 (Continued)

Find: a so that

x =9 y

x xA = = y yA

We have

Then or

1 2 1 2

a2 − a +

1 2

( a − 2 ln a − 1a )

1x 1y

EL dA EL dA

=9

a 3 − 11a 2 + a + 18a ln a + 9 = 0

Using trial and error or numerical methods and ignoring the trivial solution a = 1 in., find a = 1.901 in. and a = 3.74 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 605

PROBLEM 5.52 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.1 about (a) the line x = 240 mm, (b) the y axis. PROBLEM 5.1 Locate the centroid of the plane area shown.

SOLUTION From the solution to Problem 5.1 we have A = 15.3 × 103 mm 2 Σ xA = 2.6865 × 106 mm3 Σ yA = 1.4445 × 106 mm3

Applying the theorems of Pappus-Guldinus we have (a)

Rotation about the line

x = 240 mm Volume = 2π (240 − x ) A

= 2π (240 A − Σ xA) = 2π [240(15.3 × 103 ) − 2.6865 × 106 ]

Volume = 6.19 × 106 mm3

Area = 2π X line L = 2π Σ( xline ) L

= 2π ( x1 L1 + x3 L3 + x4 L4 + x5 L5 + x6 L6 )

Where x1 ,

, x6 are measured with respect to line x = 240 mm. Area = 2π [(120)(240) + (15)(30) + (30)(270) + (135)(210) + (240)(30)]

(b)

Area = 458 × 103 mm 2

Rotation about the y axis Volume = 2π X area A = 2π (Σ xA)

= 2π (2.6865 × 106 mm3 )

Volume = 16.88 × 106 mm3

Area = 2π X line L = 2π Σ( xline ) L

= 2π ( x1 L1 + x2 L2 + x3 L3 + x4 L4 + x5 L5 ) = 2π [(120)(240) + (240)(300) + (225)(30) + (210)(270) + (105)(210)]

Area = 1.171 × 106 mm 2

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!

PROBLEM 5.53 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.2 about (a) the line y = 60 mm, (b) the y axis. PROBLEM 5.2 Locate the centroid of the plane area shown.

SOLUTION From the solution to Problem 5.2 we have A = 1740 mm 2 Σ xA = 28200 mm3 Σ yA = 55440 mm3

Applying the theorems of Pappus-Guldinus we have (a)

y = 60 mm

Rotation about the line

Volume = 2π (60 − y ) A = 2π (60 A − Σ yA) = 2π [60(1740) − 55440]

Volume = 308 × 103 mm3

Area = 2π Yline

= 2π Σ( yline ) L = 2π ( y1 L1 + y2 L2 + y3 L3 + y4 L4 + y6 L6 )

Where y1 ,

, y6 are measured with respect to line y = 60 mm. Area = 2π -(60)(20) + (48)(24) + (36)(30) + (18) (30) 2 + (36) 2 + (30)(60) . !) ( Area = 38.2 × 103 mm 2

(b)

Rotation about the y axis Volume = 2π X area A = 2π (Σ xA) = 2π (28200 mm3 )

Volume = 177.2 × 106 mm3

Area = 2π X line L = 2π Σ( xline ) L = 2π ( x1 L1 + x2 L2 + x3 L3 + x4 L4 + x5 L5 ) = 2π -(10)(20) + (20)(24) + (35)(30) + (35) (30) 2 + (36) 2 + (10)(20) . !) ( Area = 22.4 × 103 mm 2

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 607

PROBLEM 5.54 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.8 about (a) the x axis, (b) the y axis. PROBLEM 5.8 Locate the centroid of the plane area shown.

SOLUTION From the solution to Problem 5.8 we have A = 1146.57 in.2 ΣxA = 14147.0 in.3 ΣyA = 26897 in.3

Applying the theorems of Pappus-Guldinus we have (a)

Rotation about the x axis: Volume = 2π Yarea A = 2π Σ y A = 2π (26897 in.3 )

or Volume = 169.0 × 103 in.3

Area = 2π Yline A = 2π Σ( yline ) A = 2π ( y2 L2 + y3 L3 + y4 L4 + y5 L5 + y6 L6 ) = 2π [(7.5)(15) + (30)(π × 15) + (47.5)(5) + (50)(30) + (25)(50)]

(b)

or Area = 28.4 × 103 in.2

Rotation about the y axis Volume = 2π X area A = 2π Σ x A = 2π (14147.0 in.3 )

or Volume = 88.9 × 103 in.3

Area = 2π X line L = 2π Σ( xline ) L = 2π ( x1 L1 + x2 L2 + x3 L3 + x4 L4 + x5 L5 ) . 2 × 15 0 / = 2π (15)(30) + (30)(15) + $ 30 − (π × 15) + (30)(5) + (15)(30) ! % π ' & ( ) Area = 15.48 × 103 in.2

or

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 608

PROBLEM 5.55 Determine the volume of the solid generated by rotating the parabolic area shown about (a) the x axis, (b) the axis AA′.

SOLUTION First, from Figure 5.8a we have

4 ah 3 2 y= h 5

A=

Applying the second theorem of Pappus-Guldinus we have (a)

Rotation about the x axis: Volume = 2π yA / 2 0/ 4 0 = 2π $ h %$ ah % & 5 '& 3 '

(b)

or Volume =

16 π ah 2 15

or Volume =

16 2 πa h 3

Rotation about the line AA′: Volume = 2π (2a) A /4 0 = 2π (2a) $ ah % &3 '

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 609

PROBLEM 5.56 Determine the volume and the surface area of the chain link shown, which is made from a 6-mm-diameter bar, if R = 10 mm and L = 30 mm.

SOLUTION The area A and circumference C of the cross section of the bar are A=

π 4

d 2 and C = π d .

Also, the semicircular ends of the link can be obtained by rotating the cross section through a horizontal semicircular arc of radius R. Now, applying the theorems of Pappus-Guldinus, we have for the volume V: V = 2(Vside ) + 2(Vend ) = 2( AL) + 2(π RA) = 2( L + π R) A

or

V = 2[30 mm + π (10 mm)]

-π . (6 mm) 2 ! (4 )

= 3470 mm3

For the area A:

or V = 3470 mm3

A = 2( Aside ) + 2( Aend ) = 2(CL) + 2(π RC ) = 2( L + π R)C

or

A = 2[30 mm + π (10 mm)][π (6 mm)] = 2320 mm 2

or A = 2320 mm 2

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PROBLEM 5.57 Verify that the expressions for the volumes of the first four shapes in Figure 5.21 on Page 253 are correct.

SOLUTION Following the second theorem of Pappus-Guldinus, in each case a specific generating area A will be rotated about the x axis to produce the given shape. Values of y are from Figure 5.8a. (1)

Hemisphere: the generating area is a quarter circle We have

(2)

2 V = π a3 3

/ 4a 0/ π 0 V = 2π y A = 2π $ %$ ha % & 3π '& 4 '

2 or V = π a 2 h 3

Paraboloid of revolution: the generating area is a quarter parabola We have

(4)

or

Semiellipsoid of revolution: the generating area is a quarter ellipse We have

(3)

/ 4a 0/ π 2 0 V = 2π y A = 2π $ %$ a % & 3π '& 4 '

/ 3 0/ 2 0 V = 2π y A = 2π $ a %$ ah % & 8 '& 3 '

1 or V = π a 2 h 2

Cone: the generating area is a triangle We have

/ a 0/ 1 0 V = 2π y A = 2π $ %$ ha % & 3 '& 2 '

1 or V = π a 2 h 3

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 611

PROBLEM 5.58 A 34 -in.-diameter hole is drilled in a piece of 1-in.-thick steel; the hole is then countersunk as shown. Determine the volume of steel removed during the countersinking process.

SOLUTION The required volume can be generated by rotating the area shown about the y axis. Applying the second theorem of Pappus-Guldinus, we have

V = 2π x A = 2π

-3 1 / 1 0 . -1 1 1 . + $ % in.! × × in. × in.! 8 3 4 2 4 4 ) & ' ) ( (

V = 0.0900 in.3

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 612

PROBLEM 5.59 Determine the capacity, in liters, of the punch bowl shown if R = 250 mm.

SOLUTION The volume can be generated by rotating the triangle and circular sector shown about the y axis. Applying the second theorem of Pappus-Guldinus and using Figure 5.8a, we have V = 2π x A = 2π Σ x A = 2π ( x1 A1 + x2 A2 ) -/ 1 1 0 / 1 1 3 0 / 2 R sin 30° 0 / π 2 0 . = 2π $ × R % $ × R × R% + $ R %! % $ 2 '% &$ 3 × π6 '% &$ 6 ' !) (& 3 2 ' & 2 2 / R3 R3 0 = 2π $$ + %% & 16 3 2 3 ' 3 3 π R3 8 3 3 = π (0.25 m)3 8 = 0.031883 m3 =

Since

103 l = 1 m3 V = 0.031883 m3 ×

103 l 1 m3

V = 31.9 l

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PROBLEM 5.60 Three different drive belt profiles are to be studied. If at any given time each belt makes contact with one-half of the circumference of its pulley, determine the contact area between the belt and the pulley for each design.

SOLUTION

SOLUTION Applying the first theorem of Pappus-Guldinus, the contact area AC of a belt is given by: AC = π yL = π Σ yL

where the individual lengths are the lengths of the belt cross section that are in contact with the pulley. AC = π [2( y1 L1 ) + y2 L2 ]

(a)

0.125 0 . - 0.125 in. . *2 -/ *3 = π "2 $ 3 − + [(3 − 0.125)in.](0.625 in.) # in.! % ! 2 ' ) ( cos 20° ) +* (& ,* AC = 8.10 in.2

or AC = π [2( y1 L1 )]

(b)

-/ 0.375 0 . / 0.375 in. 0 in.! = 2π $ 3 − 0.08 − 2 '% ) &$ cos 20° '% (& AC = 6.85 in.2

or AC = π [2( y1 L1 )]

(c)

-/ 2(0.25) 0 . in. [π (0.25 in.)] = π $3 − π %' !) (& AC = 7.01 in.2

or

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 614

PROBLEM 5.61 The aluminum shade for the small high-intensity lamp shown has a uniform thickness of 1 mm. Knowing that the density of aluminum is 2800 kg/m3, determine the mass of the shade.

SOLUTION

The mass of the lamp shade is given by m = ρV = ρ At

Where A is the surface area and t is the thickness of the shade. The area can be generated by rotating the line shown about the x axis. Applying the first theorem of Pappus Guldinus we have A = 2π yL = 2π Σ yL = 2π ( y1 L1 + y2 L2 + y3 L3 + y4 L4 )

or

A = 2π

-13 mm / 13 + 16 0 2 2 (13 mm) + $ % mm × (32 mm) + (3 mm) & 2 ' ( 2

/ 16 + 28 0 +$ mm × (8 mm) 2 + (12 mm) 2 % & 2 ' / 28 + 33 0 2 2. +$ % mm × (28 mm) + (5 mm) ! 2 & ' ) = 2π (84.5 + 466.03 + 317.29 + 867.51) = 10903.4 mm 2

Then

m = ρ At = (2800 kg/m3 )(10.9034 × 10−3 m 2 )(0.001 m) m = 0.0305 kg

or

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 615

!

PROBLEM 5.62 The escutcheon (a decorative plate placed on a pipe where the pipe exits from a wall) shown is cast from brass. Knowing that the density of brass is 8470 kg/m3, determine the mass of the escutcheon.

SOLUTION The mass of the escutcheon is given by m = (density)V , where V is the volume. V can be generated by rotating the area A about the x-axis. From the figure:

L1 = 752 − 12.52 = 73.9510 m L2 =

37.5 = 76.8864 mm tan 26°

a = L2 − L1 = 2.9324 mm 12.5 = 9.5941° 75 26° − 9.5941° α= = 8.2030° = 0.143168 rad 2

φ = sin −1

Area A can be obtained by combining the following four areas:

Applying the second theorem of Pappus-Guldinus and using Figure 5.8a, we have V = 2π yA = 2π Σ yA

Seg.

A, mm 2

y , mm

y A , mm3

1

1 (76.886)(37.5) = 1441.61 2

1 (37.5) = 12.5 3

18020.1

2

−α (75) 2 = −805.32

2(75)sin α sin (α + φ ) = 15.2303 3α

−12265.3

3

1 − (73.951)(12.5) = −462.19 2

1 (12.5) = 4.1667 3

−1925.81

4

−(2.9354)(12.5) = −36.693

1 (12.5) = 6.25 2

−229.33

Σ

3599.7

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 616

PROBLEM 5.62 (Continued)

Then

V = 2π Σ yA

= 2π (3599.7 mm3 ) = 22618 mm3 m = (density)V = (8470 kg/m3 )(22.618 × 10−6 m3 ) = 0.191574 kg

or m = 0.1916 kg

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 617

PROBLEM 5.63 A manufacturer is planning to produce 20,000 wooden pegs having the shape shown. Determine how many gallons of paint should be ordered, knowing that each peg will be given two coats of paint and that one gallon of paint covers 100 ft2.

SOLUTION The number of gallons of paint needed is given by / 1 gallon 0 Number of gallons = (Number of pegs)(Surface area of 1 peg) $ (2 coats) 2 % & 100 ft '

Number of gallons = 400 As

or

( As ! ft 2 )

where As is the surface area of one peg. As can be generated by rotating the line shown about the x axis. Using the first theorem of Pappus-Guldinus and Figures 5.8b, We have

R = 0.875 in. sin 2α =

0.5 0.875

2α = 34.850° α = 17.425°

or

As = 2π Y L = 2π Σ y L yL, in.2

1

0.25

y , in. " 0.125

2

0.5

0.25

0.125

3

0.0625

0.25 + 0.3125 = 0.28125 2

0.0175781

4

3 − 0.875(1 − cos 34.850) − 0.1875 = 2.6556

0.3125

0.82988

L, in.

5

6

π 2

0.5 −

× 0.1875 = 0.29452

2 × 0.1875

π

0.875sin17.425°

2α (0.875)

α

0.03125

= 0.38063

0.112103

× sin17.425°

0.137314

Σ y L = 1.25312 in.2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 618

PROBLEM 5.63 (Continued)

Then

As = 2π (1.25312 in.2 ) ×

1 ft 2 144 in.2

= 0.054678 ft 2

Finally

Number of gallons = 400 × 0.054678

= 21.87 gallons

Order 22 gallons

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 619

!

PROBLEM 5.64 The wooden peg shown is turned from a dowel 1 in. in diameter and 4 in. long. Determine the percentage of the initial volume of the dowel that becomes waste.

SOLUTION To obtain the solution it is first necessary to determine the volume of the peg. That volume can be generated by rotating the area shown about the x axis.

The generating area is next divided into six components as indicated

sin 2α =

0.5 0.875

2α = 34.850°

or

α = 17.425°

Applying the second theorem of Pappus-Guldinus and then using Figure 5.8a, we have VPEG = 2π YA = 2π Σ yA A, in.2

yA, in.3

y , in. " 0.125

0.015625

1

0.5 × 0.25 × 0.125

2

[3 − 0.875(1 − cos 34.850°) − 0.1875] × (0.3125) = 0.82987

0.15625

0.129667

3

0.1875 × 0.5 × 0.9375

0.25

0.023438

4

−

π 4

(0.1875)2 = −0.027612

0.5 −

4 × 0.1875 = 0.42042 3π

−0.011609

5

α (0.875)2

2 × 0.875sin17.425° × sin17.425° 3α

0.04005

6

1 − (0.875cos 34.850°)(0.5) = −0.179517 2

1 (0.5) = 0.166667 3

−0.029920

Σ y L = 0.167252 in.3

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 620

PROBLEM 5.64 (Continued)

Then

V peg = 2π (0.167252 in.3 ) = 1.05088 in.3

Now

Vdowel = =

π 4

π

(diameter)2 (length) (1 in.)2 (4in.)

4 = 3.14159 in.3

Then

% Waste = =

Vwaste × 100% Vdowel Vdowel − V peg Vdowel

× 100%

/ 1.05088 0 = $1 − % × 100% & 3.14159 '

or % Waste = 66.5%

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 621

!

PROBLEM 5.65* The shade for a wall-mounted light is formed from a thin sheet of translucent plastic. Determine the surface area of the outside of the shade, knowing that it has the parabolic cross section shown.

SOLUTION First note that the required surface area A can be generated by rotating the parabolic cross section through π radians about the y axis. Applying the first theorem of Pappus-Guldinus we have A = π xL

Now at

x = 100 mm, 250 = k (100)

and

2

y = 250 mm or k = 0.025 mm −1

xEL = x 2

/ dy 0 dL = 1 + $ % dx & dx '

where

dy = 2kx dx

Then

dL = 1 + 4k 2 x 2 dx

We have

xL = xEL dL =

1

1

100

x

0

( 1 + 4k x dx ) 2 2

100

-1 1 . (1 + 4k 2 x 2 )3/ 2 ! 2 ( 3 4k )0 1 1 [1 + 4(0.025)2 (100) 2 ]3/2 − (1)3/2 = 2 12 (0.025)

xL =

{

}

= 17543.3 mm 2

Finally

A = π (17543.3 mm 2 )

or A = 55.1 × 103 mm 2

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 622

PROBLEM 5.66 For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports.

SOLUTION

1 (150 lb/ft) (9 ft) = 675 lb 2 1 RII = (120 lb/ft) (9 ft) = 540 lb 2 R = RI + RII = 675 + 540 = 1215 lb RI =

XR = ΣXR :

X (1215) = (3)675) + (6)(540)

X = 4.3333 ft R = 1215 lb

(a) (b)

Reactions:

X = 4.33 ft

Σ M A = 0: B (9 ft) − (1215 lb) (4.3333 ft) = 0 B = 585.00 lb

B = 585 lb

Σ Fy = 0: A + 585.00 lb − 1215 lb = 0 A = 630.00 lb

A = 630 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 623

PROBLEM 5.67 For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports.

SOLUTION

(a)

RI = (4 m) ( 200 N/m ) = 800 N

We have

RII =

Then or and

Σ Fy :

2 (4 m) (600 N/m) = 1600 N 3

− R = − RI − RII R = 800 + 1600 = 2400 N

Σ M A : − X (2400) = −2(800) − 2.5(1600) X=

or

7 m 3 R = 2400 N

(b)

Reactions

X = 2.33 m

Σ Fx = 0: Ax = 0

/7 0 Σ M A = 0: (4 m) By − $ m % (2400 N) = 0 &3 ' By = 1400 N

or

ΣFy = 0: Ay + 1400 N − 2400 N = 0 Ay = 1000 N

or

A = 1000 N

B = 1400 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 624

PROBLEM 5.68 Determine the reactions at the beam supports for the given loading.

SOLUTION 1 (4 kN/m) (6 m) 2 = 12 kN RII = (2 kN/m) (10 m) RI =

= 20 kN ΣFy = 0: A − 12 kN − 20 kN = 0 A = 32.0 kN

ΣM A = 0: M A − (12 kN) (2 m) − (20 kN) (5 m) = 0 M A = 124.0 kN ⋅ m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 625

PROBLEM 5.69 Determine the reactions at the beam supports for the given loading.

SOLUTION We have

Then

1 (3ft) ( 480 lb/ft ) = 720 lb 2 1 R II = (6 ft) ( 600 lb/ft ) = 1800 lb 2 R III = (2 ft) ( 600 lb/ft ) = 1200 lb R I=

ΣFx = 0: Bx = 0 ΣM B = 0: (2 ft)(720 lb) − (4 ft) (1800 lb) + (6 ft)C y − (7 ft) (1200 lb) = 0

or

C y = 2360 lb

C = 2360 lb

ΣFy = 0: −720 lb + By − 1800 lb + 2360 lb − 1200 lb = 0

or

By = 1360 lb

B = 1360 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 626

PROBLEM 5.70 Determine the reactions at the beam supports for the given loading.

SOLUTION R I = (200 lb/ft) (15 ft) R I = 3000 lb 1 (200 lb/ft) (6 ft) 2 R II = 600 lb R II =

ΣM A = 0: − (3000 lb) (1.5 ft) − (600 lb) (9 ft + 2 ft) + B(15 ft) = 0 B = 740 lb

B = 740 lb

ΣFy = 0: A + 740 lb − 3000 lb − 600 lb = 0 A = 2860 lb

A = 2860 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 627

PROBLEM 5.71 Determine the reactions at the beam supports for the given loading.

SOLUTION First replace the given loading with the loading shown below. The two loadings are equivalent because both are defined by a linear relation between load and distance and the values at the end points are the same.

We have

Then

1 (3.6 m)(2200 N/m) = 3960 N 2 RII = (3.6 m)(1200 N/m) = 4320 N RI =

ΣFx = 0: Bx = 0

ΣM B = 0: − (3.6 m) Ay + (2.4 m)(3960 N) −(1.8 m)(4320 N) = 0

or

Ay = 480 N

A = 480 N

ΣFy = 0: 480 N − 3960 N + 4320 + By = 0 By = −840 N

or

B = 840 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 628

PROBLEM 5.72 Determine the reactions at the beam supports for the given loading.

SOLUTION

We have

Then

1 RI = (12 ft)(200 lb/ft) = 800 lb 3 1 RII = (6 ft)(100 lb/ft) = 200 lb 3 ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 800 lb − 200 lb = 0

or

Ay = 1000 lb

A = 1000 lb

ΣM A = 0: M A − (3 ft)(800 lb) − (16.5 ft)(200 lb) = 0

or

M A = 5700 lb ⋅ ft

M A = 5700 lb ⋅ ft

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 629

PROBLEM 5.73 Determine the reactions at the beam supports for the given loading.

SOLUTION First replace the given loading with the loading shown below. The two loadings are equivalent because both are defined by a parabolic relation between load and distance and the values at the end points are the same.

We have

RI = (6 m)(300 N/m) = 1800 N RII =

Then

2 (6 m)(1200 N/m) = 4800 N 3

ΣFx = 0: Ax = 0 ΣFy = 0: Ay + 1800 N − 4800 N = 0

or

Ay = 3000 N

A = 3000 N

15 ! ΣM A = 0: M A + (3 m)(1800 N) − " m # (4800 N) = 0 $ 4 %

or

M A = 12.6 kN ⋅ m

M A = 12.6 kN ⋅ m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 630

PROBLEM 5.74 Determine (a) the distance a so that the vertical reactions at supports A and B are equal, (b) the corresponding reactions at the supports.

SOLUTION (a)

We have

Then

1 ( a m )(1800 N/m ) = 900a N 2 1 RII = &(( 4 − a ) m ') ( 600 N/m ) = 300 ( 4 − a ) N 2 RI =

ΣFy = 0: Ay − 900a − 300(4 − a ) + B y = 0

or

Ay + By = 1200 + 600a

Now

Ay = By * Ay = By = 600 + 300a (N)

Also

(1)

& a! ' ΣM B = 0: − (4 m) Ay + +" 4 − # m , &(( 900a ) N ') 3% ) ($ &1 ' + + (4 − a ) m , (&300 ( 4 − a ) N )' = 0 (3 )

or

Ay = 400 + 700a − 50a 2

Equating Eqs. (1) and (2)

600 + 300a = 400 + 700a − 50a 2

or

a 2 − 8a + 4 = 0

(2)

8 ± (−8) 2 − 4(1)(4) 2

Then

a=

or

a = 0.53590 m

Now

a#4m*

a = 7.4641 m a = 0.536 m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 631

PROBLEM 5.74 (Continued)

(b)

We have Eq. (1)

ΣFx = 0: Ax = 0 Ay = By = 600 + 300(0.53590) = 761 N

A = B = 761 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 632

PROBLEM 5.75 Determine (a) the distance a so that the reaction at support B is minimum, (b) the corresponding reactions at the supports.

SOLUTION (a)

We have

Then or Then (b)

Eq. (1) and

1 (a m)(1800 N/m) = 900a N 2 1 RII = [(4 − a )m](600 N/m) = 300(4 − a) N 2 RI =

a ! 8+a ! ΣM A = 0: − " m # (900a N) − " m # &(300 ( 4 − a ) N ') + ( 4 m ) By = 0 $3 % $ 3 % By = 50a 2 − 100a + 800 dBy da

= 100a − 100 = 0

By = 50(1)2 − 100(1) + 800 = 750 N

(1) or a = 1.000 m B = 750 N

ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 900(1)N − 300(4 − 1)N + 750 N = 0

or

Ay = 1050 N

A = 1050 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 633

PROBLEM 5.76 Determine the reactions at the beam supports for the given loading when ω 0 = 150 lb/ft.

SOLUTION

We have

Then

1 (18 ft)(450 lb/ft) = 4050 lb 2 1 RII = (18 ft)(150 lb/ft) = 1350 lb 2 RI =

ΣFx = 0: Cx = 0 ΣM B = 0: − (44,100 kip ⋅ ft) − (2 ft) − (4050 lb) − (8 ft)(1350 lb) + (12 ft)C y = 0 C y = 5250 lb

or

C = 5250 lb

ΣFy = 0: By − 4050 lb − 1350 lb + 5250 lb = 0 By = 150 lb

or

B = 150.0 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 634

PROBLEM 5.77 Determine (a) the distributed load ω 0 at the end D of the beam ABCD for which the reaction at B is zero, (b) the corresponding reaction at C.

SOLUTION (a)

We have

Then

1 (18 ft)(450 lb/ft) = 4050 lb 2 1 RII = (18 ft)(ω0 lb/ft) = 9 ω0 lb 2 RI =

ΣM C = 0: − (44,100 lb ⋅ ft) + (10 ft)(4050 lb) + (4 ft)(9ω0 lb) = 0

ω0 = 100.0 lb/ft

or ΣFx = 0: C x = 0

(b)

ΣFy = 0: − 4050 lb − (9 × 100) lb + C y = 0 C y = 4950 lb

or

C = 4950 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 635

PROBLEM 5.78 The beam AB supports two concentrated loads and rests on soil that exerts a linearly distributed upward load as shown. Determine the values of ω A and ω B corresponding to equilibrium.

SOLUTION

1 RI = ω A (1.8 m) = 0.9ω A 2 1 RII = ωB (1.8 m) = 0.9ωB 2 ΣM D = 0: (24 kN)(1.2 − a) − (30 kN)(0.3 m) − (0.9ω A )(0.6 m) = 0

For

(1)

a = 0.6 m: 24(1.2 − 0.6) − (30)(0.3) − 0.54 ωa = 0 14.4 − 9 − 0.54ω A = 0 ΣFy = 0: − 24 kN − 30 kN + 0.9(10 kN/m) + 0.9ωB = 0

ω A = 10.00 kN/m

ωB = 50.0 kN/m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 636

PROBLEM 5.79 For the beam and loading of Problem 5.78, determine (a) the distance a for which ω A = 20 kN/m, (b) the corresponding value of ω B . PROBLEM 5.78 The beam AB supports two concentrated loads and rests on soil that exerts a linearly distributed upward load as shown. Determine the values of ω A and ω B corresponding to equilibrium.

SOLUTION

We have

1 (1.8 m)(20 kN/m) = 18 kN 2 1 RII = (1.8 m)(ω B kN/m) = 0.9ωB kN 2 RI =

ΣM C = 0: (1.2 − a)m × 24 kN − 0.6 m × 18 kN − 0.3 m × 30 kN = 0

(a)

a = 0.375 m

or ΣFy = 0: − 24 kN + 18 kN + (0.9ωB ) kN − 30 kN = 0

(b)

ω B = 40.0 kN/m

or

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 637

PROBLEM 5.80 The cross section of a concrete dam is as shown. For a 1-m-wide dam section determine (a) the resultant of the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of Part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam.

SOLUTION (a)

Consider free body made of dam and triangular section of water shown. (Thickness = 1 m)

p = (7.2 m)(103 kg/m3 )(9.81m/s 2 )

2 (4.8 m)(7.2 m)(1 m)(2.4 × 103 kg/m3 )(9.81 m/s 2 ) 3 = 542.5 kN 1 W2 = (2.4 m)(7.2 m)(1 m)(2.4 × 103 kg/m3 )(9.81 m/s 2 ) 2 = 203.4 kN 1 W3 = (2.4 m)(7.2 m)(1 m)(103 kg/m3 )(9.81 m/s 2 ) 2 = 84.8 kN 1 1 P = Ap = (7.2 m)(1 m)(7.2 m)(103 kg/m3 )(9.81 m/s 2 ) 2 2 = 254.3 kN W1 =

ΣFx = 0: H − 254.3 kN = 0

H = 254 kN

ΣFy = 0: V − 542.5 − 203.4 − 84.8 = 0 V = 830.7 kN

V = 831 kN

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 638

PROBLEM 5.80 (Continued)

5 x1 = (4.8 m) = 3 m 8 1 x2 = 4.8 + (2.4) = 5.6 m 3 2 x3 = 4.8 + (2.4) = 6.4 m 3

(b)

ΣM A = 0: xV − Σ xW + P(2.4 m) = 0

x(830.7 kN) − (3 m)(542.5 kN) − (5.6 m)(203.4 kN) − (6.4 m)(84.8 kN) + (2.4 m)(254.3 kN) = 0 x(830.7) − 1627.5 − 1139.0 − 542.7 + 610.3 = 0 x(830.7) − 2698.9 = 0 x = 3.25 m (To right of A)

(c)

Resultant on face BC Direct computation: P = ρ gh = (103 kg/m3 )(9.81 m/s 2 )(7.2 m) P = 70.63 kN/m 2 BC = (2.4) 2 + (7.2) 2 = 7.589 m θ = 18.43° 1 PA 2 1 = (70.63 kN/m 2 )(7.589 m)(1 m) 2

R=

R = 268 kN

18.43°

− R = 268 kN

18.43°

R = 268 kN

18.43°

Alternate computation: Use free body of water section BCD.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 639

PROBLEM 5.81 The cross section of a concrete dam is as shown. For a 1-ft-wide dam section determine (a) the resultant of the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of Part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam.

SOLUTION The free body shown consists of a 1-ft thick section of the dam and the quarter circular section of water above the dam.

Note:

4 × 21 ! x1 = " 21 − # ft 3π % $ = 12.0873 ft x2 = (21 + 4) ft = 25 ft 4 × 21 ! x4 = " 50 − ft 3π #% $ = 41.087 ft

For area 3 first note.

I II

Then

−

a

x

r2

1 r 2

π 4

r2

(

r−

& 1 (21)(21) 2 + ( 21 − 4×21 ) − π × 212 2 3π 4 x3 = 29 ft + ( ( (21) 2 − π4 (21) 2 * = (29 + 4.6907)ft = 33.691 ft

4r 3π

) ') ft ) +

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 640

PROBLEM 5.81 (Continued)

(a)

Now

W = γV

So that

&π ' W1 = (150 lb/ft 3 ) ( (21 ft)2 (1 ft) ) = 51,954 lb 4 * +

W2 = (150 lb/ft 3 )[(8 ft)(21 ft)(1 ft)] = 25, 200 lb & ' π ! W3 = (150 lb/ft 3 ) (" 212 − × 212 # ft 2 × (1 ft) ) = 14,196 lb 4 % *$ + &π ' W4 = (62.4 lb/ft 3 ) ( (21 ft) 2 (1 ft) ) = 21, 613 lb *4 +

Also Then

P=

1 1 Ap = [(21 ft)(1 ft)][(62.4 lb/ft 3 )(21 ft)] = 13,759 lb 2 2

ΣFx = 0: H − 13,759 lb = 0

or H = 13.76 kips

ΣFy = 0: V − 51,954 lb − 25,200 lb − 14,196 lb − 21,613 lb = 0

or (b)

We have

V = 112,963 lb

V = 113.0 kips

ΣM A = 0: x(112,963 lb) − (12.0873 ft)(51,954 lb) − (25 ft)(25,200 lb) −(33.691 ft)(14,196 lb) − (41.087 ft)(21,613 lb) +(7 ft)(13,759 lb) = 0

or

112,963x − 627,980 − 630, 000 − 478, 280 − 888, 010 + 96,313 = 0

x = 22.4 ft

or (c)

Consider water section BCD as the free body We have

Then

ΣF = 0

− R = 25.6 kips

57.5°

or R = 25.6 kips

57.5°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 641

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PROBLEM 5.82 The 3 × 4-m side AB of a tank is hinged at its bottom A and is held in place by a thin rod BC. The maximum tensile force the rod can withstand without breaking is 200 kN, and the design specifications require the force in the rod not to exceed 20 percent of this value. If the tank is slowly filled with water, determine the maximum allowable depth of water d in the tank.

SOLUTION Consider the free-body diagram of the side. We have Now Where

P=

1 1 Ap = A( ρ gd ) 2 2

ΣM A = 0: hT −

d P=0 3

h=3m

Then for d max . (3 m)(0.2 × 200 × 103 N) −

d max 3

&1 ' 3 3 2 ( 2 (4 m × d max ) × (10 kg/m × 9.81 m/s × d max ) ) = 0 * + 3 120 N ⋅ m − 6.54d max N/m 2 = 0

or

d max = 2.64 m

or

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!

PROBLEM 5.83 The 3 × 4-m side of an open tank is hinged at its bottom A and is held in place by a thin rod BC. The tank is to be filled with glycerine, whose density is 1263 kg/m3. Determine the force T in the rod and the reactions at the hinge after the tank is filled to a depth of 2.9 m.

SOLUTION Consider the free-body diagram of the side. We have

Then

1 1 Ap = A( ρ gd ) 2 2 1 = [(2.9 m)(4 m)] [(1263 kg/m3 )(9.81 m/s 2 )(2.9 m)] 2 = 208.40 kN

P=

ΣFy = 0: Ay = 0 2.9 ! ΣM A = 0: (3 m)T − " m # (208.4 kN) = 0 $ 3 %

T = 67.151 kN

or

T = 67.2 kN

ΣFx = 0: Ax + 208.40 kN − 67.151 kN = 0

Ax = −141.249 kN

or

A = 141.2 kN

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 643

!

PROBLEM 5.84 The friction force between a 6 × 6-ft square sluice gate AB and its guides is equal to 10 percent of the resultant of the pressure forces exerted by the water on the face of the gate. Determine the initial force needed to lift the gate if it weighs 1000 lb.

SOLUTION Consider the free-body diagram of the gate. Now

1 1 ApI = [(6 × 6) ft 2 ][(62.4 lb/ft 3 )(9 ft)] 2 2 = 10,108.8 lb

PI =

1 1 ApII = [(6 × 6) ft 2 ][(62.4 lb/ft 3 )(15 ft)] 2 2 = 16848 lb

PII =

Then

F = 0.1P = 0.1( PI + PII ) = 0.1(10108.8 + 16848)lb = 2695.7 lb

Finally

ΣFy = 0: T − 2695.7 lb − 1000 lb = 0

or T = 3.70 kips

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PROBLEM 5.85 A freshwater marsh is drained to the ocean through an automatic tide gate that is 4 ft wide and 3 ft high. The gate is held by hinges located along its top edge at A and bears on a sill at B. If the water level in the marsh is h = 6 ft, determine the ocean level d for which the gate will open. (Specific weight of salt water = 64 lb/ft 3 .)

SOLUTION

Since gate is 4 ft wide Thus:

w = (4 ft)p = 4γ (depth) w1 = 4γ (h − 3) w2 = 4γ h w1′ = 4γ ′(d − 3) w2′ = 4γ ′d 1 (3 ft)(w1′ − w1 ) 2 1 = (3 ft)[4γ ′(d − 3) − 4λ (h − 3)] = 6γ ′(d − 3) − 6γ (h − 3) 2

PI′ − PI =

1 (3 ft)(w2′ − w2 ) 2 1 = (3 ft)[4γ ′d − 4γ h] = 6γ ′d − 6γ h 2

PII′ − PII =

ΣM A = 0: (3 ft)B − (1 ft)(PI′ − PI ) − (2 ft)(PII′ − PII ) = 0 1 2 B = ( PI′ − PI ) − ( PII′ − PII ) 3 3 1 2 = [6γ ′(d − 3) − 6γ (h − 3)] − [6γ ′d − 6γ h] 3 3 = 2γ (d − 3) − 2γ (h − 3) + 4γ ′d − 4γ h B = 6γ ′(d − 1) − 6γ ( h − 1)

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PROBLEM 5.85 (Continued)

With

0 = 6γ ′(d − 1) − 6γ ( h − 1)

B = 0 and h = 6 ft:

d −1 = 5

Data:

γ γ′

γ ′ = 64 lb/ft 3 γ = 62.4 lb/ft 3 62.4 lb/ft 3 64 lb/ft 3 = 4.875 ft

d −1 = 5

d = 5.88 ft

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 646

PROBLEM 5.86 The dam for a lake is designed to withstand the additional force caused by silt that has settled on the lake bottom. Assuming that silt is equivalent to a liquid of density ρ s = 1.76 × 103 kg/m3 and considering a 1-m-wide section of dam, determine the percentage increase in the force acting on the dam face for a silt accumulation of depth 2 m.

SOLUTION First, determine the force on the dam face without the silt. We have

1 1 Apw = A( ρ gh) 2 2 1 = [(6.6 m)(1 m)][(103 kg/m3 )(9.81 m/s 2 )(6.6 m)] 2 = 213.66 kN

Pw =

Next, determine the force on the dam face with silt We have

1 Pw′ = [(4.6 m)(1 m)][(103 kg/m3 )(9.81 m/s2 )(4.6 m)] 2 = 103.790 kN ( Ps ) I = [(2.0 m)(1 m)][(103 kg/m3 )(9.81 m/s 2 )(4.6 m)] = 90.252 kN 1 ( Ps ) II = [(2.0 m)(1 m)][(1.76 × 103 kg/m3 )(9.81 m/s 2 )(2.0 m)] 2 = 34.531 kN

Then

P′ = Pw′ + ( Ps ) I + ( Ps ) II = 228.57 kN

The percentage increase, % inc., is then given by % inc. =

P′ − Pw × 100% Pw

(228.57 − 213.66) × 100% 213.66 = 6.9874% =

% inc. = 6.98%

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 647

PROBLEM 5.87 The base of a dam for a lake is designed to resist up to 120 percent of the horizontal force of the water. After construction, it is found that silt (that is equivalent to a liquid of density ρ s = 1.76 × 103 kg/m3 ) is settling on the lake bottom at the rate of 12 mm/year. Considering a 1-m-wide section of dam, determine the number of years until the dam becomes unsafe.

SOLUTION From Problem 5.86, the force on the dam face before the silt is deposited, is Pw = 213.66 kN. The maximum allowable force Pallow on the dam is then: Pallow = 1.2 Pw = (1.5)(213.66 kN) = 256.39 kN

Next determine the force P′ on the dam face after a depth d of silt has settled

We have

1 Pw′ = [(6.6 − d )m × (1 m)][(103 kg/m3 )(9.81 m/s 2 )(6.6 − d )m] 2 = 4.905(6.6 − d ) 2 kN ( Ps ) I = [d (1 m)][(103 kg/m3 )(9.81 m/s 2 )(6.6 − d )m] = 9.81 (6.6d − d 2 )kN 1 ( Ps ) II = [d (1 m)][(1.76 × 103 kg/m3 )(9.81 m/s 2 )(d )m] 2 = 8.6328d 2 kN P′ = Pw′ + ( Ps ) I + ( Ps ) II = [4.905(43.560 − 13.2000d + d 2 ) + 9.81(6.6d − d 2 ) + 8.6328d 2 ]kN = [3.7278d 2 + 213.66]kN

Now required that P′ = Pallow to determine the maximum value of d. (3.7278d 2 + 213.66)kN = 256.39 kN

or Finally

d = 3.3856 m 3.3856 m = 12 × 10−3

m ×N year

or N = 282 years

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 648

PROBLEM 5.88 A 0.5 × 0.8-m gate AB is located at the bottom of a tank filled with water. The gate is hinged along its top edge A and rests on a frictionless stop at B. Determine the reactions at A and B when cable BCD is slack.

SOLUTION First consider the force of the water on the gate. We have

P=

1 1 Ap = A( ρ gh) 2 2

1 PI = [(0.5 m)(0.8 m)] × [(103 kg/m3 )(9.81 m/s 2 )(0.45 m)] 2 = 882.9 N

so that

1 PII = [(0.5 m)(0.8 m)] × [(103 kg/m3 )(9.81 m/s 2 )(0.93 m)] 2 = 1824.66 N

Reactions at A and B when T = 0 We have

ΣM A = 0:

1 2 (0.8 m)(882.9 N) + (0.8 m)(1824.66 N) − (0.8 m)B = 0 3 3 B = 1510.74 N

or or

B = 1511 N

53.1°

A = 1197 N

53.1°

ΣF = 0: A + 1510.74 N − 882.9 N − 1824.66 N = 0

or

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 649

!

PROBLEM 5.89 A 0.5 × 0.8-m gate AB is located at the bottom of a tank filled with water. The gate is hinged along its top edge A and rests on a frictionless stop at B. Determine the minimum tension required in cable BCD to open the gate.

SOLUTION First consider the force of the water on the gate. We have so that

P=

1 1 Ap = A( ρ gh) 2 2

1 PI = [(0.5 m)(0.8 m)] × [(103 kg/m3 )(9.81 m/s 2 )(0.45 m)] 2 = 882.9 N 1 PII = [(0.5 m)(0.8 m)] × [(103 kg/m3 )(9.81 m/s 2 )(0.93 m)] 2 = 1824.66 N

T to open gate First note that when the gate begins to open, the reaction at B Then

or

ΣM A = 0:

0.

1 2 (0.8 m)(882.9 N) + (0.8 m)(1824.66 N) 3 3 8 ! − (0.45 + 0.27)m × " T # = 0 $ 17 %

235.44 + 973.152 − 0.33882 T = 0 T = 3570 N

or

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PROBLEM 5.90 A long trough is supported by a continuous hinge along its lower edge and by a series of horizontal cables attached to its upper edge. Determine the tension in each of the cables, at a time when the trough is completely full of water.

SOLUTION Consider free body consisting of 20-in. length of the trough and water l = 20-in. length of free body &π ' W = γ v = γ ( r 2l ) 4 * + PA = γ r P=

1 1 1 PA rl = (γ r )rl = γ r 2l 2 2 2

1 ! ΣM A = 0: Tr − Wr − P " r # = 0 $3 %

π ! 4r Tr − " γ r 2 l #" $ 4 %$ 3π

1 2 ! 1 ! ! # − " 2 γ r l #" 3 r # = 0 % $ %$ %

1 1 1 T = γ r 2l + γ r 2l = γ r 2l 3 6 2

Data:

γ = 62.4 lb/ft 3 r =

Then

T=

24 20 ft = 2ft l = ft 12 12

1 20 ! (62.4 lb/ft 3 )(2 ft) 2 " ft # 2 12 $ %

= 208.00 lb

T = 208 lb

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 651

PROBLEM 5.91 A 4 × 2-ft gate is hinged at A and is held in position by rod CD. End D rests against a spring whose constant is 828 lb/ft. The spring is undeformed when the gate is vertical. Assuming that the force exerted by rod CD on the gate remains horizontal, determine the minimum depth of water d for which the bottom B of the gate will move to the end of the cylindrical portion of the floor.

SOLUTION First determine the forces exerted on the gate by the spring and the water when B is at the end of the cylindrical portion of the floor sin θ =

We have

2 θ = 30° 4

Then

xSP = (3 ft) tan 30°

and

FSP = kxSP = 828 lb/ft × 3 ft × tan30° = 1434.14 lb

Assume

d $ 4 ft

We have

P=

1 1 Ap = A(γ h) 2 2

1 PI = [(4 ft)(2 ft)] × [(62.4 lb/ft 3 )(d − 4)ft] 2 = 249.6( d − 4)lb

Then

1 PII = [(4 ft)(2 ft)] × [(62.4 lb/ft 3 )(d − 4 + 4cos 30°)] 2 = 249.6(d − 0.53590°)lb W =0

For d min so that gate opens,

Using the above free-body diagrams of the gate, we have 4 ! 8 ! ΣM A = 0: " ft # [249.6(d − 4)lb] + " ft # [249.6( d − 0.53590)lb] 3 $ % $3 % −(3 ft)(1434.14 lb) = 0

or

(332.8d − 1331.2) + (665.6d − 356.70) − 4302.4 = 0 d = 6.00 ft

or d $ 4 ft , assumption correct

d = 6.00 ft

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 652

!

PROBLEM 5.92 Solve Problem 5.91 if the gate weighs 1000 lb. PROBLEM 5.91 A 4 × 2-ft gate is hinged at A and is held in position by rod CD. End D rests against a spring whose constant is 828 lb/ft. The spring is undeformed when the gate is vertical. Assuming that the force exerted by rod CD on the gate remains horizontal, determine the minimum depth of water d for which the bottom B of the gate will move to the end of the cylindrical portion of the floor.

SOLUTION First determine the forces exerted on the gate by the spring and the water when B is at the end of the cylindrical portion of the floor sin θ =

We have

2 θ = 30° 4

Then

xSP = (3 ft) tan 30°

and

FSP = kxSP = 828 lb/ft × 3 ft × tan30° = 1434.14 lb

Assume

d $ 4 ft

We have

P=

1 1 Ap = A(γ h) 2 2

1 PI = [(4 ft)(2 ft)] × [(62.4 lb/ft 3 )(d − 4)ft] 2 = 249.6(d − 4)lb 1 PII = [(4 ft)(2 ft)] × [(62.4 lb/ft 3 )(d − 4 + 4cos 30°)] 2 = 249.6(d − 0.53590°)lb

Then

For d min so that gate opens,

W = 1000 lb

Using the above free-body diagrams of the gate, we have 4 ! 8 ! ΣM A = 0: " ft # [249.6(d − 4) lb] + " ft # [249.6(d − 0.53590) lb] 3 $ % $3 % − (3 ft)(1434.14 lb) − (1 ft)(1000 lb) = 0

or

(332.8d − 1331.2) + (665.6d − 356.70) − 4302.4 − 1000 = 0 d = 7.00 ft

or d $ 4 ft , assumption correct

d = 7.00 ft

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 653

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PROBLEM 5.93 A prismatically shaped gate placed at the end of a freshwater channel is supported by a pin and bracket at A and rests on a frictionless support at B. The pin is located at a distance h = 0.10 m below the center of gravity C of the gate. Determine the depth of water d for which the gate will open.

SOLUTION First note that when the gate is about to open (clockwise rotation is impending), By 0 and the line of action of the resultant P of the pressure forces passes through the pin at A. In addition, if it is assumed that the gate is homogeneous, then its center of gravity C coincides with the centroid of the triangular area. Then

a=

d − (0.25 − h) 3

and

b=

2 8 d! (0.4) − " # 3 15 $ 3 %

Now

a 8 = b 15

so that

d − (0.25 − h) 3 2 (0.4) − 158 d3 3

( )

=

8 15

Simplifying yields 289 70.6 d + 15h = 45 12

(1)

Alternative solution Consider a free body consisting of a 1-m thick section of the gate and the triangular section BDE of water above the gate. Now

1 1 Ap′ = (d × 1 m)( ρ gd ) 2 2 1 = ρ gd 2 (N) 2 1 8 ! W ′ = ρ gV = ρ g " × d × d × 1 m # 2 15 $ % 4 = ρ gd 2 (N) 15

P′ =

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PROBLEM 5.93 (Continued)

Then with By = 0 (as explained above), we have &2 1 8 !' 4 ! &d ' 1 ! ΣM A = 0: ( (0.4) − " d # ) " ρ gd 2 # − ( − (0.25 − h) ) " ρ gd 2 # = 0 3 3 15 15 3 2 $ %+ $ % * +$ % *

Simplifying yields

289 70.6 d + 15h = 45 12

as above. Find d , Substituting into Eq. (1)

h = 0.10 m 289 70.6 d + 15(0.10) = 45 12

or d = 0.683 m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 655

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PROBLEM 5.94 A prismatically shaped gate placed at the end of a freshwater channel is supported by a pin and bracket at A and rests on a frictionless support at B. Determine the distance h if the gate is to open when d = 0.75 m.

SOLUTION First note that when the gate is about to open (clockwise rotation is impending), By 0 and the line of action of the resultant P of the pressure forces passes through the pin at A. In addition, if it is assumed that the gate is homogeneous, then its center of gravity C coincides with the centroid of the triangular area. Then

a=

d − (0.25 − h) 3

and

b=

2 8 d! (0.4) − " # 3 15 $ 3 %

Now

a 8 = b 15

so that

d − (0.25 − h) 3 2 (0.4) − 158 d3 3

( )

=

8 15

Simplifying yields 289 70.6 d + 15h = 45 12

(1)

Alternative solution Consider a free body consisting of a 1-m thick section of the gate and the triangular section BDE of water above the gate. Now

1 1 Ap′ = (d × 1 m)( ρ gd ) 2 2 1 2 (N) = ρ gd 2 1 8 ! W ′ = ρ gV = ρ g " × d × d × 1 m # 2 15 $ % 4 2 (N) = ρ gd 15

P′ =

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 656

PROBLEM 5.94 (Continued)

Then with By = 0 (as explained above), we have &2 1 8 !' 4 ! &d ' 1 ! ΣM A = 0: ( (0.4) − " d # ) " ρ gd 2 # − ( − (0.25 − h) ) " ρ gd 2 # = 0 3 $ 15 % + $ 15 % *3 +$ 2 % *3

Simplifying yields

289 70.6 d + 15h = 45 12

as above. Find h, Substituting into Eq. (1)

d = 0.75 m 289 70.6 (0.75) + 15h = 45 12

or h = 0.0711 m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 657

!

PROBLEM 5.95 A 55-gallon 23-in.-diameter drum is placed on its side to act as a dam in a 30-in.-wide freshwater channel. Knowing that the drum is anchored to the sides of the channel, determine the resultant of the pressure forces acting on the drum.

SOLUTION Consider the elements of water shown. The resultant of the weights of water above each section of the drum and the resultants of the pressure forces acting on the vertical surfaces of the elements is equal to the resultant hydrostatic force acting on the drum. Then 1 1 ApI = A(γ h) 2 2 1 & 30 ! 23 ! ' & 23 ! ' = (" # ft × " # ft ) × ( (62.4 lb/ft 3 ) " ft # ) 2 *$ 12 % $ 12 % + * $ 12 % +

PI =

= 286.542 lb 1 1 ApII = A(γ h) 2 2 1 & 30 ! 11.5 ! ' & 11.5 ! ' 3 = (" # ft × " ft # ) # ft ) × ( (62.4 lb/ft ) " 2 *$ 12 % 12 $ % + * $ 12 % +

PII =

= 71.635 lb & 11.5 !2 2 π 11.5 ! 2 2 ' 30 ! W1 = γ V1 = (62.4 lb/ft 3 ) (" # ft − " # ft ) " ft # = 30.746 lb 4 $ 12 % (*$ 12 % )+ $ 12 % & 11.5 !2 2 π 11.5 !2 2 ' 30 ! W2 = γ V2 = (62.4 lb/ft 3 ) (" # ft + " # ft ) " ft # = 255.80 lb 4 $ 12 % (*$ 12 % )+ $ 12 % & π 11.5 !2 2 ' 30 ! ft # = 112.525 lb W3 = γ V3 = (62.4 lb/ft 3 ) ( " # ft ) " *( 4 $ 12 % +) $ 12 %

Then

ΣFx : Rx = (286.542 − 71.635) lb = 214.91 lb ΣFy : Ry = (−30.746 + 255.80 + 112.525) lb = 337.58 lb

Finally

R = Rx2 + Ry2 = 400.18 1b tan θ =

Ry Rx

θ = 57.5°

R = 400 lb

57.5°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 658

!

PROBLEM 5.96 Determine the location of the centroid of the composite body shown when (a) h = 2b, (b) h = 2.5b.

SOLUTION

V

x

xV

Cylinder I

π a 2b

1 b 2

1 2 2 πa b 2

Cone II

1 2 πa h 3

b+

1 h 4

1 2 1 ! π a h"b + h# 3 4 % $

1 ! V = π a2 " b + h # 3 % $ 1 1 1 ! Σ xV = π a 2 " b 2 + hb + h 2 # 3 12 % $2

(a)

For h = 2b :

1 & ' 5 V = π a 2 (b + (2b) ) = π a 2 b 3 * + 3 1 1 &1 ' Σ xV = π a 2 ( b 2 + (2b)b + (2b) 2 ) 2 3 12 * + 1 2 1 3 & ' = π a 2b 2 ( + + ) = π a 2 b2 * 2 3 3+ 2

5 ! 3 XV = Σ xV : X " π a 2b # = π a 2 b 2 $3 % 2

X=

9 b 10

Centroid is

1 b 10

to left of base of cone

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 659

PROBLEM 5.96 (Continued)

(b)

For h = 2.5b :

1 & ' V = π a 2 (b + (2.5b) ) = 1.8333π a 2 b 3 * + 1 1 &1 ' Σ xV = π a 2 ( b 2 + (2.5b)b + (2.5b) 2 ) 3 12 *2 + = π a 2b 2 [0.5 + 0.8333 + 0.52083] = 1.85416π a 2b 2

XV = Σ xV : X (1.8333π a 2 b) = 1.85416π a 2b 2

X = 1.01136b

Centroid is 0.01136b to right of base of cone Note: Centroid is at base of cone for h = 6b = 2.449b !

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 660

PROBLEM 5.97 Determine the y coordinate of the centroid of the body shown.

SOLUTION First note that the values of Y will be the same for the given body and the body shown below. Then

V

y

Cone

1 2 πa h 3

1 − h 4

2

Cylinder

a! 1 −π " # b = − π a 2b 4 $2%

1 − b 2

Σ

We have Then

π 12

a 2 (4h − 3b)

yV −

1 π a 2 h2 12 1 2 2 πa b 8

−

π 24

a 2 (2h 2 − 3b 2 )

Y ΣV = Σ yV

π &π ' Y ( a 2 (4h − 3b) ) = − a 2 (2h 2 − 3b 2 ) 24 *12 +

or Y = −

2h 2 − 3b 2 2(4h − 3b)

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 661

!

PROBLEM 5.98 Determine the z coordinate of the centroid of the body shown. (Hint: Use the result of Sample Problem 5.13.)

SOLUTION First note that the body can be formed by removing a “half-cylinder” from a “half-cone,” as shown.

V 1 2 πa h 6

Half-Cone

Half-Cylinder

Σ

−

π a!

2

−

π

b = − a 2b 2 "$ 2 #% 8

π 24

zV

z

−

a

π

1 − a3 h 6

*

4 a! 2a =− " # 3π $ 2 % 3π

a 2 (4h − 3b)

1 3 ab 12

−

1 3 a (2h − b) 12

From Sample Problem 5.13 We have Then

Z ΣV = Σ zV 1 &π ' Z ( a 2 (4h − 3b) ) = − a3 (2h − b) 12 * 24 +

or Z = −

a 4h − 2b ! π "$ 4h − 3b #%

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 662

!

PROBLEM 5.99 The composite body shown is formed by removing a semiellipsoid of revolution of semimajor axis h and semiminor axis a/2 from a hemisphere of radius a. Determine (a) the y coordinate of the centroid when h = a/2, (b) the ratio h/a for which y = −0.4a.

SOLUTION

V

y

yV

Hemisphere

2 3 πa 3

3 − a 8

1 − π a4 4

2

Semiellipsoid

2 a! 1 − π " # h = − π a2 h 3 $2% 6

3 − h 8

ΣV =

Then

π 6

Σ yV = −

+

1 π a 2 h2 16

a 2 (4a − h)

π 16

a 2 (4a 2 − h 2 )

Y ΣV = Σ yV

Now So that

π &π ' Y ( a 2 (4a − h) ) = − a 2 (4a 2 − h 2 ) 16 *6 + 2 h! h! ' 3 & Y " 4 − # = − a (4 − " # ) a% 8 (* $ a % )+ $

or

Y = ? when h =

(a) Substituting

(1)

a 2

h 1 = into Eq. (1) a 2 2 1! 3 & 1! ' Y " 4 − # = − a (4 − " # ) 2% 8 *( $ 2 % +) $

or

Y =−

45 a 112

Y = −0.402a

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 663

PROBLEM 5.99 (Continued)

h = ? when Y = −0.4a a

(b) Substituting into Eq. (1)

2 h! h! ' 3 & (−0.4a) " 4 − # = − a ( 4 − " # ) a% 8 (* $ a % )+ $ 2

or

Then

h! h! 3 " # − 3.2 " # + 0.8 = 0 $a% $a% 2 h 3.2 ± (−3.2) − 4(3)(0.8) = a 2(3)

=

3.2 ± 0.8 6

or

h 2 = a 5

and

h 2 = a 3

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 664

!

PROBLEM 5.100 For the stop bracket shown, locate the x coordinate of the center of gravity.

SOLUTION Assume that the bracket is homogeneous so that its center of gravity coincides with the centroid of the volume.

Then

V , mm3

x , mm

xV , mm 4

1

(100)(88)(12) = 105600

50

5280000

2

(100)(12)(88) = 105600

50

5280000

3

1 (62)(51)(10) = 15810 2

39

616590

4

1 − (66)(45)(12) = −17820 2

Σ

209190

X =

34 +

2 (66) = 78 3

−1389960

9786600

Σ xV 9786600 mm = 209190 ΣV

or

X = 46.8 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 665

PROBLEM 5.101 For the stop bracket shown, locate the z coordinate of the center of gravity.

SOLUTION Assume that the bracket is homogeneous so that it center of gravity coincides with the centroid of the volume.

V , mm3

Then

z , mm

zV , mm 4

6

633600

1

(100)(88)(12) = 105600

2

(100)(12)(88) = 105600

12 +

1 (88) = 56 2

5913600

3

1 (62)(51)(10) = 15810 2

1 12 + (51) = 29 3

458490

4

1 − (66)(45)(12) = −17820 2

Σ

209190

Z =

55 +

2 (45) = 85 3

−1514700

5491000

Σ zV 5491000 mm = 209190 ΣV

or Z = 26.2 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 666

!

PROBLEM 5.102 For the machine element shown, locate the y coordinate of the center of gravity.

SOLUTION First assume that the machine element is homogeneous so that its center of gravity will coincide with the centroid of the corresponding volume.

V , mm3

x , mm

y , mm

xV, mm 4

yV, mm 4

I

(100)(18)(90) = 162000

50

9

8100000

1458000

II

(16)(60)(50) = 48000

92

48

4416000

2304000

III

π (12) 2 (10) = 4523.9

105

54

475010

244290

IV

−π (13) 2 (18) = −9556.7

28

9

–267590

–86010

Σ

204967.2

12723420

3920280

We have

Y ΣV = Σ yV Y (204967.2 mm3 ) = 3920280 mm 4

or Y = 19.13 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 667

!

PROBLEM 5.103 For the machine element shown, locate the y coordinate of the center of gravity.

SOLUTION For half cylindrical hole: r = 1.25 in. 4(1.25) yIII = 2 − 3π = 1.470 in.

For half cylindrical plate:

r = 2 in. 4(2) zIV = 7 + = 7.85 in. 3π V, in.3

y , in.

z , in.

yV , in.4

z V , in.4

I

Rectangular plate

(7)(4)(0.75) = 21.0

–0.375

3.5

–7.875

73.50

II

Rectangular plate

(4)(2)(1) = 8.0

1.0

2

8.000

16.00

III

–(Half cylinder)

(1.25) 2 (1) = 2.454

1.470

2

–3.607

–4.908

IV

Half cylinder

(2) 2 (0.75) = 4.712

–0.375

–7.85

–1.767

36.99

V

–(Cylinder)

−π (1.25) 2 (0.75) = −3.682

–0.375

7

1.381

–25.77

Σ

27.58

–3.868

95.81

−

π 2

π 2

Y ΣV = Σ yV Y (27.58 in.3 ) = −3.868 in.4

Y = −0.1403 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 668

!

PROBLEM 5.104 For the machine element shown, locate the z coordinate of the center of gravity.

SOLUTION For half cylindrical hole: r = 1.25 in. 4(1.25) yIII = 2 − 3π = 1.470 in.

For half cylindrical plate:

r = 2 in. 4(2) zIV = 7 + = 7.85 in. 3π V, in.3

Now

y , in.

z , in.

yV , in.4

z V , in.4

I

Rectangular plate

(7)(4)(0.75) = 21.0

–0.375

3.5

–7.875

73.50

II

Rectangular plate

(4)(2)(1) = 8.0

1.0

2

8.000

16.00

III

–(Half cylinder)

(1.25) 2 (1) = 2.454

1.470

2

–3.607

–4.908

IV

Half cylinder

(2) 2 (0.75) = 4.712

–0.375

–7.85

–1.767

36.99

V

–(Cylinder)

−π (1.25) 2 (0.75) = −3.682

–0.375

7

1.381

–25.77

Σ

27.58

–3.868

95.81

−

π 2

π 2

Z ΣV = zV Z (27.58 in.3 ) = 95.81 in.4

Z = 3.47 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 669

!

PROBLEM 5.105 For the machine element shown, locate the x coordinate of the center of gravity.

SOLUTION First assume that the machine element is homogeneous so that its center of gravity will coincide with the centroid of the corresponding volume.

V , mm3

x , mm

y , mm

xV, mm 4

yV, mm 4

I

(100)(18)(90) = 162000

50

9

8100000

1458000

II

(16)(60)(50) = 48000

92

48

4416000

2304000

III

π (12) 2 (10) = 4523.9

105

54

475010

244290

IV

−π (13) 2 (18) = −9556.7

28

9

–267590

–86010

Σ

204967.2

12723420

3920280

We have

X ΣV = Σ xV X (204967.2 mm3 ) = 12723420 mm 4

X = 62.1 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 670

!

PROBLEM 5.106 Locate the center of gravity of the sheet-metal form shown.

SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with the centroid of the corresponding area. 1 yI = − (1.2) = −0.4 m 3 1 zI = (3.6) = 1.2 m 3 4(1.8) 2.4 m xIII = − =− 3π π

A, m 2

x, m

y, m

z, m

xA, m3

yA, m3

zA, m3

I

1 (3.6)(1.2) = 2.16 2

1.5

−0.4

1.2

3.24

−0.864

2.592

II

(3.6)(1.7) = 6.12

0.75

0.4

1.8

4.59

2.448

11.016

0.8

1.8

−3.888

4.0715

9.1609

3.942

5.6555

22.769

III Σ

We have

π 2

(1.8)2 = 5.0894

−

2.4

π

13.3694

X ΣV = Σ xV : X (13.3694 m 2 ) = 3.942 m3

or X = 0.295 m

Y ΣV = Σ yV : Y (13.3694 m 2 ) = 5.6555 m3

or Y = 0.423 m

Z ΣV = Σ zV : Z (13.3694 m 2 ) = 22.769 m3

or Z = 1.703 m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 671

PROBLEM 5.107 Locate the center of gravity of the sheet-metal form shown.

SOLUTION First assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with the centroid of the corresponding area. Now note that symmetry implies X = 125 mm

yII = 150 +

2 × 80

π

= 200.93 mm 2 × 80 zII =

π

yIII

I II

III Σ

We have

A, mm 2

y , mm

(250)(170) = 42500

75

π 2

(80)(250) = 31416

π 2

(125)2 = 24544

= 50.930 mm 4 × 125 = 230 + 3π = 283.05 mm

z , mm

yA, mm3

zA, mm3

40

3187500

1700000

200.93

50930

6312400

1600000

283.05

0

6947200

0

16447100

3300000

98460

Y Σ A = Σ y A: Y (98460 mm 2 ) = 16447100 mm3

or Y = 1670 mm

Z Σ A = Σ z A: Z (98460 mm 2 ) = 3.300 × 106 mm3

or Z = 33.5 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 672

PROBLEM 5.108 A wastebasket, designed to fit in the corner of a room, is 16 in. high and has a base in the shape of a quarter circle of radius 10 in. Locate the center of gravity of the wastebasket, knowing that it is made of sheet metal of uniform thickness.

SOLUTION By symmetry:

X =Z

For III (Cylindrical surface)

x= A=

For IV (Quarter-circle bottom)

x= A=

2r

π π 2

=

2(10)

rh =

π π 2

= 6.3662 in.

(10)(16) = 251.33 in.2

4r 4(10) = = 4.2441 in. 3π 3π

π 4

r2 =

π 4

(10) 2 = 78.540 in.2

A, in.2

x , in.

x , in.

xA, in.3

yA, in.3

I

(10)(16) = 160

5

8

800

1280

II

(10)(16) = 160

0

8

0

1280

III

251.33

6.3662

8

1600.0

2010.6

IV

78.540

4.2441

0

333.33

0

Σ

649.87

2733.3

4570.6

X Σ A = Σ x A:

X (649.87 in.2 ) = 2733.3 in.3 X = 4.2059 in.

Y Σ A = Σ y A:

X = Z = 4.21 in.

Y (649.87 in.2 ) = 4570.6 in.3 Y = 7.0331 in.

Y = 7.03 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 673

!

PROBLEM 5.109 A mounting bracket for electronic components is formed from sheet metal of uniform thickness. Locate the center of gravity of the bracket.

SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the bracket coincides with the centroid of the corresponding area. Then (see diagram) 4(0.625) 3π = 1.98474 in.

zV = 2.25 −

π

(0.625) 2 2 = −0.61359 in.2

AV = −

A, in.2

x , in.

y , in.

z , in.

xA, in.3

yA, in.3

zA, in.3

I

(2.5)(6) = 15

1.25

0

3

18.75

0

45

II

(1.25)(6) = 7.5

2.5

–0.625

3

18.75

–4.6875

22.5

III

(0.75)(6) = 4.5

2.875

–1.25

3

12.9375

–5.625

13.5

IV

5! − " # (3) = − 3.75 $4%

1.0

0

3.75

3.75

0

–14.0625

V

− 0.61359

1.0

0

1.98474

0.61359

0

–1.21782

Σ

22.6364

46.0739

10.3125

65.7197

We have

X Σ A = Σx A X (22.6364 in.2 ) = 46.0739 in.3

or

X = 2.04 in.

Y ΣA = Σy A Y (22.6364 in.2 ) = −10.3125 in.3

or Y = − 0.456 in.

Z Σ A = Σz A Z (22.6364 in.2 ) = 65.7197 in.3

or

Z = 2.90 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 674

PROBLEM 5.110 A thin sheet of plastic of uniform thickness is bent to form a desk organizer. Locate the center of gravity of the organizer.

SOLUTION First assume that the plastic is homogeneous so that the center of gravity of the organizer will coincide with the centroid of the corresponding area. Now note that symmetry implies Z = 30.0 mm

x2 = 6 −

2× 6

x4 = 36 + x8 = 58 −

π

= 2.1803 mm

2×6

π 2× 6

π

x10 = 133 +

= 39.820 mm = 54.180 mm

2×6

π

= 136.820 mm

y2 = y4 = y8 = y10 = 6 − y6 = 75 +

2×5

π

2×6

π

= 2.1803 mm

= 78.183 mm

π

× 6 × 60 = 565.49 mm 2 2 A6 = π × 5 × 60 = 942.48 mm 2

A2 = A4 = A8 = A10 =

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 675

PROBLEM 5.110 (Continued)

We have

A, mm 2

x , mm

y , mm

xA, mm3

yA, mm3

1

(74)(60) = 4440

0

43

0

190920

2

565.49

2.1803

2.1803

1233

1233

3

(30)(60) = 1800

21

0

37800

0

4

565.49

39.820

2.1803

22518

1233

5

(69)(60) = 4140

42

40.5

173880

167670

6

942.48

47

78.183

44297

73686

7

(69)(60) = 4140

52

40.5

215280

167670

8

565.49

54.180

2.1803

30638

1233

9

(75)(60) = 4500

95.5

0

429750

0

10

565.49

136.820

2.1803

77370

1233

Σ

22224.44

1032766

604878

X Σ A = Σ x A: X (22224.44 mm 2 ) = 1032766 mm3

or X = 46.5 mm

Y Σ A = Σ y A: Y (22224.44 mm 2 ) = 604878 mm3

or Y = 27.2 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 676

PROBLEM 5.111 A window awning is fabricated from sheet metal of uniform thickness. Locate the center of gravity of the awning.

SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the awning coincides with the centroid of the corresponding area. (4)(25) = 14.6103 in. 3π (4)(25) 100 zII = zVI = = in. 3π 3π (2)(25) yIV = 4 + = 19.9155 in. yII = yVI = 4 +

π

zIV =

(2)(25)

π

AII = AVI = AIV =

π 2

π 4

=

50

π

in.

(25) 2 = 490.87 in.2

(25)(34) = 1335.18 in.2

A, in.2

y , in.

z , in.

yA, in.3

zA, in.3

I

(4)(25) = 100

2

12.5

200

1250

II

490.87

14.6103

7171.8

5208.3

III

(4)(34) = 136

2

272

3400

IV

1335.18

19.9155

26591

21250

V

(4)(25) = 100

2

200

1250

VI

490.87

14.6103

7171.8

5208.3

Σ

2652.9

41607

37567

100 3π 25 50

π 12.5 100 3π

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 677

PROBLEM 5.111 (Continued)

X = 17.00 in.

Now, symmetry implies and

Y Σ A = Σ y A: Y (2652.9 in.2 ) = 41607 in.3

or Y = 15.68 in.

Z Σ A = Σ z A: Z (2652.9 in.2 ) = 37567

or Z = 14.16 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 678

PROBLEM 5.112 An elbow for the duct of a ventilating system is made of sheet metal of uniform thickness. Locate the center of gravity of the elbow.

SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides with the centroid of the corresponding area. Also, note that the shape of the duct implies Y = 38.0 mm ! Note that

xI = zI = 400 − xII = 400 − zII = 300 −

2

π 2

π

2

π

(400) = 145.352 mm

(200) = 272.68 mm (200) = 172.676 mm

xIV = zIV = 400 −

4 (400) = 230.23 mm 3π

4 (200) = 315.12 mm 3π 4 zV = 300 − (200) = 215.12 mm 3π

xV = 400 −

Also note that the corresponding top and bottom areas will contribute equally when determining x and z . Thus

x , mm

z , mm

xA, mm3

zA, mm3

(400)(76) = 47752

145.352

145.352

6940850

6940850

(200)(76) = 23876

272.68

172.676

6510510

4122810

A, mm 2

I II

π 2

π 2

III

100(76) = 7600

200

350

1520000

2660000

IV

π! 2 " # (400) 2 = 251327 $4%

230.23

230.23

57863020

57863020

V

π! −2 " # (200)2 = −62832 $4%

315.12

215.12

–19799620

–13516420

VI

−2(100)(200) = −40000

300

350

–12000000

–14000000

Σ

227723

41034760

44070260

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 679

PROBLEM 5.112 (Continued)

We have

X Σ A = Σ x A: X (227723 mm 2 ) = 41034760 mm3

or X = 180.2 mm

Z Σ A = Σ z A: Z (227723 mm 2 ) = 44070260 mm3

or Z = 193.5 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 680

PROBLEM 5.113 An 8-in.-diameter cylindrical duct and a 4 × 8-in. rectangular duct are to be joined as indicated. Knowing that the ducts were fabricated from the same sheet metal, which is of uniform thickness, locate the center of gravity of the assembly.

SOLUTION Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the area.

A, in.2

x , in.

y , in.

xA, in.3

yA, in.3

π (8)(12) = 96π

0

6

0

576π

10

−128

−160π

4(4) 16 =− 3π 3π

12

− 42.667

96π

6

12

576

1152

6

8

576

768

4(4) 16 = 3π 3π

8

− 42.667

− 64π

7

(4) 2 = − 8π 2 (4)(12) = 48

6

10

288

480

8

(4)(12) = 48

6

10

288

480

Σ

539.33

1514.6

4287.4

1 2

−

π 2

π

3

Then

(4) 2 = 8π

4

2 (8)(12) = 96

5

(8)(12) = 96

6

−

2(4)

(8)(4) = −16π

π

π −

=

8

π

X =

Σ x A 1514.67 = in. ΣA 539.33

or X = 2.81 in.

Y =

Σ y A 4287.4 = in. ΣA 539.33

or Y = 7.95 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 681

PROBLEM 5.114 A thin steel wire of uniform cross section is bent into the shape shown. Locate its center of gravity.

SOLUTION First assume that the wire is homogeneous so that its center of gravity will coincide with the centroid of the corresponding line. x1 = 0.3sin 60° = 0.15 3 m z1 = 0.3cos 60° = 0.15 m 0.6sin 30° ! 0.9 m ## sin 30° = π π 6 % 0.6sin 30° ! 0.9 3m z2 = " ## cos 30° = π " π 6 $ %

x2 = " " $

π! L2 = " # (0.6) = (0.2π ) m $3% L, m

x, m

y, m

z, m

xL, m 2

yL, m 2

zL, m 2

1

1.0

0.15 3

0.4

0.15

0.25981

0.4

0.15

2

0.2π

0.18

0

0.31177

3

0.8

0

0.4

0.6

0

0.32

0.48

4

0.6

0

0.8

0.3

0

0.48

0.18

Σ

3.0283

0.43981

1.20

1.12177

We have

0.9

π

0.9 3

0

π

X Σ L = Σ x L: X (3.0283 m) = 0.43981 m 2

or X = 0.1452 m

Y Σ L = Σ yL:

Y (3.0283 m) = 1.20 m 2

or

Y = 0.396 m

Z Σ L = Σ z L:

Z (3.0283 m) = 1.12177 m 2

or

Z = 0.370 m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 682

PROBLEM 5.115 Locate the center of gravity of the figure shown, knowing that it is made of thin brass rods of uniform diameter.

SOLUTION Uniform rod

AB 2 = (1 m) 2 + (0.6 m) 2 + (1.5 m) 2 AB = 1.9 m

L, m

x, m

y, m

z, m

xL, m 2

yL, m 2

ΣL , m 2

AB

1.9

0.5

0.75

0.3

0.95

1.425

0.57

BD

0.6

1.0

0

0.3

0.60

0

0.18

DO

1.0

0.5

0

0

0.50

0

0

OA

1.5

0

0.75

0

0

1.125

0

Σ

5.0

2.05

2.550

0.75

X Σ L = Σ x L : X (5.0 m) = 2.05 m 2

X = 0.410 m

Y Σ L = Σ y L : Y (5.0 m) = 2.55 m 2

Y = 0.510 m

Z Σ L = Σ z L : Z (5.0 m) = 0.75 m 2

Z = 0.1500 m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 683

!

PROBLEM 5.116 Locate the center of gravity of the figure shown, knowing that it is made of thin brass rods of uniform diameter.

SOLUTION X =0

By symmetry:

L, in.

y , in.

z , in.

yL, in.2

zL, in.2

AB

302 + 162 = 34

15

0

510

0

AD

302 + 162 = 34

15

8

510

272

AE

302 + 162 = 34

15

0

510

0

BDE

π (16) = 50.265

0

0

512

Σ

152.265

1530

784

2(16)

π

= 10.186

Y ΣL = Σ y L : Y (152.265 in.) = 1530 in.2 Y = 10.048 in.

!

Y = 10.05 in.

!

Z = 5.15 in.

!

Z ΣL = Σ z L : Z (152.265 in.) = 784 in.2 ! Z = 5.149 in. !

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 684

PROBLEM 5.117 The frame of a greenhouse is constructed from uniform aluminum channels. Locate the center of gravity of the portion of the frame shown.

SOLUTION First assume that the channels are homogeneous so that the center of gravity of the frame will coincide with the centroid of the corresponding line. x8 = x9 =

2×3

π

y8 = y9 = 5 +

6

π

2×3

π

ft = 6.9099 ft

L, ft

x , ft

y , ft

z , ft

xL, ft 2

yL, ft 2

zL, ft 2

1

2

3

0

1

6

0

2

2

3

1.5

0

2

4.5

0

6

3

5

3

2.5

0

15

12.5

0

4

5

3

2.5

2

15

12.5

10

5

8

0

4

2

0

32

16

6

2

3

5

1

6

10

2

7

3

1.5

5

2

4.5

15

6

6.9099

0

9

32.562

0

6.9099

2

9

32.562

9.4248

8

1

0

16

2

69

163.124

53.4248

8 9

π 2

π 2

× 3 = 4.7124 × 3 = 4.7124

10

2

Σ

39.4248

We have

=

6

π 6

π 0

X Σ L = Σ x L : X (39.4248 ft) = 69 ft 2

or X = 1.750 ft

Y Σ L = Σ y L : Y (39.4248 ft) = 163.124 ft 2

or

Y = 4.14 ft

!

Z Σ L = Σ z L : Z (39.4248 ft) = 53.4248 ft 2

or Z = 1.355 ft

!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 685

PROBLEM 5.118 A scratch awl has a plastic handle and a steel blade and shank. Knowing that the density of plastic is 1030 kg/m3 and of steel is 7860 kg/m3, locate the center of gravity of the awl.

SOLUTION Y =Z =0

First, note that symmetry implies

5 xI = (12.5 mm) = 7.8125 mm 8 2π ! 3 WI = (1030 kg/m3 ) " # (0.0125 m) $ 3 % = 4.2133 × 10−3 kg xII = 52.5 mm

π! WII = (1030 kg/m3 ) " # (0.025 m) 2 (0.08 m) $4% −3 = 40.448 × 10 kg xIII = 92.5 mm − 25 mm = 67.5 mm π! WIII = −(1030 kg/m3 ) " # (0.0035 m)2 (.05 m) $4% −3 = −0.49549 × 10 kg xIV = 182.5 mm − 70 mm = 112.5 mm

π! WIV = (7860 kg/m3 ) " # (0.0035 m) 2 (0.14 m) 2 = 10.5871 × 10−3 kg $4% 1 xV = 182.5 mm + (10 mm) = 185 mm 4 π! WV = (7860 kg/m3 ) " # (0.00175 m) 2 (0.01 m) = 0.25207 × 10−3 kg $3%

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 686

!

PROBLEM 5.118 (Continued)

We have

W, kg

x, mm

xW, kg ⋅ mm

I

4.123 × 10−3

7.8125

32.916 × 10−3

II

40.948 × 10−3

52.5

2123.5 × 10−3

III

−0.49549 × 10−3

67.5

−33.447 × 10−3

IV

10.5871 × 10−3

112.5

1191.05 × 10−3

V

0.25207 × 10−3

185

46.633 × 10−3

Σ

55.005 × 10−3

3360.7 × 10−3

X ΣW = Σ xW : X (55.005 × 10−3 kg) = 3360.7 × 10−3 kg ⋅ mm X = 61.1 mm

or

(From the end of the handle)

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 687

!

PROBLEM 5.119 A bronze bushing is mounted inside a steel sleeve. Knowing that the specific weight of bronze is 0.318 lb/in.3 and of steel is 0.284 lb/in.3, determine the location of the center of gravity of the assembly.

SOLUTION X =Z =0

First, note that symmetry implies

Now

W = ( pg )V

& π! ' yI = 0.20 in. WI = (0.284 lb/in.3 ) *" # ( 1.82 − 0.752 in.2 ) ( 0.4 in.) + = 0.23889 lb .$ 4 % , /

(

)

& π! ' yII = 0.90 in. WII = (0.284 lb/in.3 ) *" # ( 1.1252 − 0.752 in.2 ) (1 in.) + = 0.156834 lb .$ 4 % , /

(

)

& π! ' yIII = 0.70 in. WIII = (0.318 lb/in.3 ) *" # ( 0.752 − 0.52 in.2 ) (1.4 in.) + = 0.109269 lb , 4 .$ % /

(

We have

)

Y ΣW = Σ yW Y =

(0.20 in.)(0.23889 lb) + (0.90 in.)(0.156834 lb) + (0.70 in.)(0.109269 lb) 0.23889 lb + 0.156834 lb + 0.109269 lb Y = 0.526 in.

or

(above base)!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 688

PROBLEM 5.120 A brass collar, of length 2.5 in., is mounted on an aluminum rod of length 4 in. Locate the center of gravity of the composite body. (Specific weights: brass = 0.306 lb/in.3, aluminum = 0.101 lb/in.3)

SOLUTION

Aluminum rod:

W = γV (π ) = (0.101 lb/in.3 ) 0 (1.6 in.) 2 (4 in.) 1 ,4 = 0.81229 lb

Brass collar:

W = γV

π

= (0.306 lb/in.3 ) [(3 in.)2 − (1.6 in.)2 ](2.5 in.) 4 = 3.8693 lb

Component

W(lb)

y (in.)

yW (lb ⋅ in.)

Rod

0.81229

2

1.62458

Collar

3.8693

1.25

4.8366

Σ

4.6816

6.4612

Y ΣW = Σ y W : Y (4.6816 lb) = 6.4612 lb ⋅ in. Y = 1.38013 in.

Y = 1.380 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 689

!

PROBLEM 5.121 The three legs of a small glass-topped table are equally spaced and are made of steel tubing, which has an outside diameter of 24 mm and a cross-sectional area of 150 mm 2 . The diameter and the thickness of the table top are 600 mm and 10 mm, respectively. Knowing that the density of steel is 7860 kg/m3 and of glass is 2190 kg/m3, locate the center of gravity of the table.

SOLUTION X =Z =0

First note that symmetry implies Also, to account for the three legs, the masses of components I and II will each bex multiplied by three yI = 12 + 180 −

2 × 180

mI = ρ ST VI = 7860 kg/m3 × (150 × 10−6 m 2 ) ×

π 2 × 280

mII = ρ ST VII = 7860 kg/m3 × (150 × 10−6 m 2 ) ×

π

mIII = ρGLVIII = 2190 kg/m3 ×

yIII = 24 + 180 + 280 + 5

= 489 mm

or

(0.180 m)

π 2

(0.280 m)

= 0.51855 kg

= 370.25 mm

We have

2

= 0.33335 kg

= 77.408 mm yII = 12 + 180 +

π

π 4

(0.6 m) 2 × (0.010 m)

= 6.1921 kg m, kg

y , mm

ym, kg ⋅ mm

I

3(0.33335)

77.408

77.412

II

3(0.51855)

370.25

515.98

III

6.1921

489

3027.9

Σ

8.7478

3681.3

Y Σm = Σ ym : Y (8.7478 kg) = 3681.3 kg ⋅ mm Y = 420.8 mm

The center of gravity is 421 mm (above the floor)!

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 690

PROBLEM 5.122 Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A hemisphere

SOLUTION Choose as the element of volume a disk of radius r and thickness dx. Then dV = π r 2 dx, xEL = x

The equation of the generating curve is x 2 + y 2 = a 2 so that r 2 = a 2 − x 2 and then dV = π ( a 2 − x 2 ) dx

V1 =

Component 1

and

2x 1

EL dV

2

a/2 0

a/2

( x3 ) π (a − x )dx = π 0 a 2 x − 1 3 -0 , 2

=

11 3 πa 24

=

2

a/2 0

2

x (,π (a 2 − x 2 ) dx )a/2

( x2 x4 ) = π 0a2 − 1 2 4 -0 , 7 = π a4 64

Now

x1V1 =

2x 1

EL dV :

11 ! 7 x1 " π a 3 # = π a 4 $ 24 % 64

or

x1 =

21 a 88

Component 2 V2 =

2

a a /2

(

π (a 2 − x 2 )dx = π 0 a 2 x − ,

a

x3 ) 1 3 - a/2

& ( a 3 )' a3 ) 0 2 a ! ( 2 ) 1 3 3( 2 = π * 0 a (a ) − 1 − a " # − + 3 - 0 $2% 3 13 3, , -/ . 5 = π a3 24

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 691

PROBLEM 5.122 (Continued)

and

2

a

( x 2 x4 ) − 1 xEL dV = x (,π ( a 2 − x 2 )dx )- = π 0 a 2 a/2 2 2 4 - a/2 ,

2

a

2 4 )' & ( a a 2 ( (a ) 4 ) 0 2 ( 2 ) 3 ( 2 (a) 2 ) 13 = π *0a − − + 1− a 2 4 - 0 2 4 13 3, , -/ . 9 = π a4 64

Now

x2V2 =

2

5 ! 9 xEL dV : x2 " π a3 # = π a 4 2 $ 24 % 64

or

x2 =

27 a 40

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 692

!

PROBLEM 5.123 Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A semiellipsoid of revolution

SOLUTION Choose as the element of volume a disk of radius r and thickness dx. Then dV = π r 2 dx, xEL = x x2 y2 + = 1 so that h2 a 2

The equation of the generating curve is r2 =

a2 2 (h − x 2 )dx h2

dV = π

and then

V1 =

Component 1

=

and

a2 2 (h − x 2 ) h2

2

1

xEL dV =

2

h/2 0

h/2

a2 a2 ( x3 ) π 2 (h 2 − x 2 )dx = π 2 0 h 2 x − 1 3 -0 h h ,

11 2 πa h 24

2

h/2

0

( a2 ) x 0π 2 (h 2 − x 2 ) dx 1 , h h/2

a2 ( x2 x4 ) = π 2 0 h2 − 1 4 -0 h , 2 7 = π a 2 h2 64

Now

Component 2

x1V1 =

2x 1

EL dV :

11 ! 7 x1 " π a 2 h # = π a 2 h 2 $ 24 % 64

or

x1 =

21 h 88

h

a2 a2 ( x3 ) π 2 (h 2 − x 2 )dx = π 2 0 h 2 x − 1 V2 = h/2 h 3 - h/2 h ,

2

h

a2 =π 2 h =

& ( h 3 )' ( h )3 ) 0 2 h ! ( 2 ) 1 3 3( 2 *0h ( h ) − + 1− h " #− 3 - 0 $2% 3 13 3, , -/ .

5 π a2h 24

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 693

PROBLEM 5.123 (Continued)

and

2

2

xEL dV =

h

a2 =π 2 h

( 2 x2 x4 ) − 1 0h 4 - h/2 , 2

a2 =π 2 h

& ( h 2 h 4 )' 2 ( ( h) 4 ) 0 2 ( 2 ) 3 ( 2 ( h) 2 ) 13 − − *0h + 1− h 2 4 - 0 2 4 13 3, , -/ .

=

Now

2

( a2 ) x 0π 2 (h 2 − x 2 ) dx 1 h/2 , h h

x2V2 =

9 π a 2 h2 64

2

5 ! 9 xEL dV : x2 " π a 2 h # = π a 2 h 2 2 $ 24 % 64

or

x2 =

27 h 40

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 694

!

PROBLEM 5.124 Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A paraboloid of revolution

SOLUTION Choose as the element of volume a disk of radius r and thickness dx. Then dV = π r 2 dx, xEL = x

The equation of the generating curve is x = h −

h 2 a2 2 y so that (h − x) r = h a2 a2 (h − x)dx h

dV = π

and then

V1 =

Component 1

2

h/2

0

π

a2 ( h − x) dx h h/2

a2 ( x2 ) =π 0 hx − 1 h , 2 -0 3 = π a2 h 8

2

and

1

xEL dV =

2

h/2

0

( a2 ) x 0π (h − x)dx 1 , h h/2

a 2 ( x 2 x3 ) 1 2 2 =π 0h − 1 = π a h h , 2 3 -0 12 x1V1 =

Now

2x 1

EL dV :

3 ! 1 x1 " π a 2 h # = π a 2 h 2 8 $ % 12

or

x1 =

2 h 9

Component 2 V2 =

2

h h/2

h

π

a2 a2 ( x2 ) (h − x)dx = π 0 hx − 1 h h , 2 - h/2

& ( h 2 )' ( h) 2 ) 0 h ! ( 2 ) 1 3 a2 3( =π * 0 h( h ) − + 1 − h" # − h 3, 2 - 0 $2% 2 13 , -/ . 1 2 = πa h 8

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PROBLEM 5.124 (Continued)

and

2

2

a2 =π h =

Now

h

( a2 ) a 2 ( x 2 x3 ) xEL dV = x 0π ( h − x)dx 1 = π 0h − 1 h/2 2 h , 2 3 - h/2 , h h

x2V2 =

& ( h 2 ( h )3 ) ' 2 ( h )3 ) 0 ( 2 ) 3 ( ( h) 3 − − 2 1+ *0h 1− h 0 1 2 3 2 3 - , 3, - 3/ .

1 π a 2 h2 12

2

2

1 ! 1 xEL dV : x2 " π a 2 h # = π a 2 h 2 $8 % 12

or

x2 =

2 h 3

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!

PROBLEM 5.125 Locate the centroid of the volume obtained by rotating the shaded area about the x axis.

SOLUTION First note that symmetry implies

y =0

and

z =0

We have

y = k ( X − h) 2

at

x = 0,

or

k=

y = a : a = k ( − h) 2

a h2

Choose as the element of volume a disk of radius r and thickness dx. Then dV = π r 2 dx, X EL = x r=

Now

a ( x − h) 2 h2

so that

dV = π

Then

V=

2

a2 ( x − h) 4 dx 4 h h 0

π

a2 π a2 ( 5 h 4 ( ) x h dx x − h) ) − = 4 4 ,( -0 5h h

1 = π a2h 5

and

2x

EL dV

( a2 ) x 0π 4 ( x − h) 4 dx 1 0 , h 2 h a =π 4 ( x5 − 4hx 4 + 6h 2 x3 − 4h3 x 2 + h4 x)dx 0 h =

2

h

2

2

=π =

Now

a h4

h

(1 6 4 5 3 2 4 4 3 3 1 4 2 ) 0 6 x − 5 hx + 2 h x − 3 h x + 2 h x 1 , -0

1 π a 2 h2 30

π ! π 2 2 xV = xEL dV : x " a 2 h # = a h 5 $ % 30

2

or

x=

1 h 6

!

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PROBLEM 5.126 Locate the centroid of the volume obtained by rotating the shaded area about the x axis.

SOLUTION y =0

First note that symmetry implies

z =0

Choose as the element of volume a disk of radius r and thickness dx. Then dV = π r 2 dx, xEL = x r = kx1/3

Now

dV = π k 2 x 2/3 dx

so that at x = h, y = a :

a = kh1/3

or

k=

a h1/3 a 2 2/3 x dx h 2/3

dV = π

Then

V=

and

2

h 0

π

a 2 2/3 x dx h 2/3

a2 h 2/3

=π

h

( 3 5/3 ) 05 x 1 , -0

3 = π a2 h 5

Also

2

xEL dV =

2

h 0

! a2 a2 x "" π 2/3 x 2/3 dx ## = π 2/3 h $ h %

h

( 3 8/3 ) 08 x 1 , -0

3 = π a 2 h2 8

Now

2

xV = xdV :

3 ! 3 x " π a 2 h # = π a 2 h2 5 $ % 8

5 or x = h 8

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 698

PROBLEM 5.127 Locate the centroid of the volume obtained by rotating the shaded area about the line x = h.

SOLUTION x =h

First, note that symmetry implies

z =0

Choose as the element of volume a disk of radius r and thickness dx. Then dV = π r 2 dy, yEL = y

Now x 2 =

Then

and Let Then

h 2 h2 2 a − y2 (a − y 2 ) so that r = h − 2 a a dV = π

V=

2

h2 a − a2 − y 2 a2

(

a 0

π

2

) dy

h2 a − a2 − y2 2 a

(

2

) dy

y = a sin θ 4 dy = a cos θ dθ h2 a2 h2 =π 2 a

V =π

= π ah 2

2 2

π /2 0

π /2 0

2

π /2 0

(

a − a 2 − a 2 sin 2 θ

2

) a cosθ dθ

( a 2 − 2a (a cos θ ) + (a 2 − a 2 sin 2 θ ) ) a cos θ dθ , -

(2cos θ − 2 cos 2 θ − sin 2 θ cos θ ) dθ

( θ sin 2θ = π ah 2 0 2sin θ − 2 " + 4 $2 ,

π /2

! 1 3 ) # − 3 sin θ 1 % -0

π ! ( 1) = π ah 2 0 2 − 2 "" 2 ## − 1 0, $ 2 % 3 1-

= 0.095870π ah 2

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PROBLEM 5.127 (Continued)

and

2

yEL dV =

2

=π

a 0

( h2 y 0π 2 a − a 2 − y 2 , a

h2 a2

(

2

a 0

( 2a y − 2ay 2

) dy )12

)

a 2 − y 2 − y 3 dy a

2 1 ) h2 ( = π 2 0 a 2 y 2 + a (a 2 − y 2 )3/2 − y 4 1 3 4 -0 a , =π =

Now

h2 a2

&( 2 2 1 4 ) ( 2 2 3/2 ) ' * 0 a (a ) − a 1 − 0 a( a ) 1 + 4 - ,3 -/ .,

1 π a 2 h2 12

2

y (0.095870π ah2 ) =

yV = yEL dV :

1 π a 2 h2 12

or y = 0.869a

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!

PROBLEM 5.128* Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the x axis.

SOLUTION y =0

First, note that symmetry implies

z =0

Choose as the element of volume a disk of radius r and thickness dx. dV = π r 2 dx, xEL = x

Then

r = b sin

Now

πx 2a

dV = π b 2 sin 2

so that

V=

Then

&

2a

a

πx 2a

dx

π b 2 sin 2

πx 2a

dx 2a

x sin π x ! = π b 2 " − π2 a # 2 a %# $" 2 a = π b 2 $( 22a ) − ( a2 ) !% 1 = π ab 2 2

and

&x

EL dV

=

&

2a

a

πx ( ' x ) π b 2 sin 2 dx 2a *, +

Use integration by parts with dV = sin 2

u=x du = dx

V=

x − 2

πx 2a sin πax 2π a

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PROBLEM 5.128* (Continued)

Then

&

2a - ' . 2a ' x sin π x ( / x sin πax ( ! / xEL dV = π b 0 " x ) − 2π * # − − 2π a * dx 1 ) * * a )2 " )2 a a , %# a + , 3/ 2/ $ + 2a 1 2 a2 π x ! ./ ' 2a ( ' a (! 2/ = π b 0 " 2a ) * − a ) * # − " x + 2 cos # 1 a %a / 2π + 2 , + 2 ,% $ 4 2/ $ 3

&

2

-/' 3 ( 1 a2 a 2 ! ./ 1 = π b 2 0) a 2 * − " (2a) 2 + 2 − (a) 2 + 2 # 1 4 2π 2π % /3 /2+ 2 , $ 4 '3 1 ( = π a 2b 2 ) − 2 * +4 π ,

= 0.64868π a 2b 2

Now

'1 ( xV = xEL dV : x ) π ab 2 * = 0.64868π a 2b 2 +2 ,

&

or x = 1.297a

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!

PROBLEM 5.129* Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the y axis. (Hint: Use a thin cylindrical shell of radius r and thickness dr as the element of volume.)

SOLUTION x =0

First note that symmetry implies

z =0

Choose as the element of volume a cylindrical shell of radius r and thickness dr. Then Now so that Then

dV = (2π r )( y )(dr ), y = b sin

&

2a a

1 y 2

πr 2a

dV = 2π br sin V=

yEL =

πr 2a

dr

2π br sin

πr 2a

dr

Use integration by parts with u = rd

dv = sin

dr 2a 2a πr v = − cos 2a π

du = dr

Then

πr

2a -/ π r (! ' 2a V = 2π b 0 "( r ) ) − cos * # − 2a , % a /2 $ + π

&

2a π r ( ./ ) π cos 2a * dr 1 + , 3/

2a ' a

2a - 2a π r ! ./ 4a 2 / = 2π b 0 − [ (2a)(−1) ] + " 2 sin # 1 2a % a / π $π 2/ 3

' 4a 2 4 a 2 V = 2π b )) − 2 π + π 1( ' = 8a 2 b ) 1 − * + π,

( ** ,

= 5.4535a 2b

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PROBLEM 5.129* (Continued)

Also

&y

EL dV

π r (' πr ( 1 dr * ) 2 b sin *) 2π br sin 2 a 2a , + ,+ 2a πr r sin 2 dr = π b2 a 2a =

&

2a '

a

&

Use integration by parts with dv = sin 2

u=r du = dr

Then

&

v=

r − 2

πr

dr 2a sin πar 2π a

2a . 2a ' r ' r sin πar ( ! sin π r ( / / yEL dV = π b 0 "( r ) ) − 2π * # − − 2π a * dr 1 ) * * a )2 " )2 a a , %# a + , 3/ 2/ $ + 2a π r ! ./ r2 a2 ' 2a ( ' a (! 2/ = π b 0 "(2a) ) * − ( a) ) * # − " + 2 cos # 1 a %a / + 2 , + 2 , % $ 4 2π 2/ $ 3

&

2

(2a) 2 (a) 2 a2 a 2 ! /. /- 3 = π b2 0 a 2 − " + 2− + 2 #1 4 2π 2π % 3/ $ 4 2/ 2 '3 1 ( = π a 2b 2 ) − 2 * +4 π ,

= 2.0379a 2 b 2

Now

&

yV = yEL dV : y (5.4535a 2b) = 2.0379a 2b 2

or y = 0.374b

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PROBLEM 5.130* Show that for a regular pyramid of height h and n sides (n = 3, 4, is located at a distance h/4 above the base.

) the centroid of the volume of the pyramid

SOLUTION Choose as the element of a horizontal slice of thickness dy. For any number N of sides, the area of the base of the pyramid is given by Abase = kb 2

where k = k ( N ); see note below. Using similar triangles, have s h− y = b h s=

or

b (h − y ) h

dV = Aslice dy = ks 2 dy = k

Then

V=

and

h

&

0

b2 (h − y ) 2 dy h2 h

b2 b2 1 ! k 2 ( h − y ) 2 dy = k 2 " − (h − y )3 # h h $ 3 %0

1 = kb 2 h 3 yEL = y

Also So then

&

yEL dV =

&

h

0

=k

Now

Note

! b2 b2 y " k 2 ( h − y ) 2 dy # = k 2 h $ h %

b2 h2

&

h

0

( h 2 y − 2hy 2 + y 3 ) dy

h

1 2 2 2 3 1 4! 1 2 2 " 2 h y − 3 hy + 4 y # = 12 kb h $ %0

'1 ( 1 yV = yEL dV : y ) kb 2 h * = kb 2 h 2 3 + , 12

&

'1 Abase = N )) × b × +2 N = b2 4 tan πN

b 2 tan πN

or y =

1 h Q.E.D. 4

( ** ,

= k ( N )b 2

!

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PROBLEM 5.131 Determine by direct integration the location of the centroid of one-half of a thin, uniform hemispherical shell of radius R.

SOLUTION x =0

First note that symmetry implies

The element of area dA of the shell shown is obtained by cutting the shell with two planes parallel to the xy plane. Now dA = (π r )( Rdθ ) 2r

yEL = −

π

r = R sin θ

Where

dA = π R 2 sin θ dθ 2R sin θ yEL = −

so that

π

A=

Then

π /2

&

0

π R 2 sin θ dθ = π R 2 [− cos θ ]π0 /2

= π R2

and

&y

EL dA

=

&

π /2' 0

2R ( sin θ * (π R 2 sin θ dθ ) )− π + , π /2

sin 2θ ! = −2 R " − 4 #% 0 $2 3

=−

Now Symmetry implies

π 2

θ

R3

&

yA = yEL dA: y (π R 2 ) = −

π 2

R3

1 or y = − R 2 1 z =− R 2

z=y

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PROBLEM 5.132 The sides and the base of a punch bowl are of uniform thickness t. If t ,, R and R = 250 mm, determine the location of the center of gravity of (a) the bowl, (b) the punch.

SOLUTION (a)

Bowl x =0

First note that symmetry implies

z =0

for the coordinate axes shown below. Now assume that the bowl may be treated as a shell; the center of gravity of the bowl will coincide with the centroid of the shell. For the walls of the bowl, an element of area is obtained by rotating the arc ds about the y axis. Then dAwall = (2π R sin θ )( Rdθ )

and Then

( yEL ) wall = − R cos θ

Awall =

π /2

&π

/6

2π R 2 sin θ dθ π /2

= 2π R 2 [− cos θ ]π /6 = π 3R 2

and

& π =& π

ywall Awall = ( yEL ) wall dA /2

/6

(− R cos θ )(2π R 2 sin θ dθ )

= π R3 [cos 2 θ ]ππ /2 /6 3 = − π R3 4

π

R2 ,

By observation

Abase =

Now

y ΣA = Σ yA

or or

4

ybase = −

3 R 2

' 3 3 ( π π ' ( y ) π 3R 2 + R 2 * = − π R3 + R 2 ) − R* ) * 4 4 4 + , + 2 , y = − 0.48763R R = 250 mm

y = −121.9 mm

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PROBLEM 5.132 (Continued)

(b)

Punch x =0

First note that symmetry implies

z =0

and that because the punch is homogeneous, its center of gravity will coincide with the centroid of the corresponding volume. Choose as the element of volume a disk of radius x and thickness dy. Then dV = π x 2 dy,

Now

yEL = y

x2 + y 2 = R2 dV = π ( R 2 − y 2 )dy

so that

V=

Then

&

0 − 3/2 R

π ( R 2 − y 2 ) dy 0

1 ! = π " R y − y3 # 3 %− $ 2

3/2 R

3 ' ( 1' ( ! 3 3 3 R* − )− R * # = π 3R3 = −π " R ) − " )+ 2 *, 3 )+ 2 *, # 8 $ % 2

and

&

yEL dV =

&

0 − 3/2 R

( y ) π R 2 − y 2 dy ! $ %

(

)

0

1 1 ! = π " R2 y 2 − y4 # 2 4 %− $

3/2 R

4! 1 2' 3 ( 1' 3 ( # 15 " R )− R* − )− R * = − π R4 = −π ) * ) * "2 + 2 , 4+ 2 , # 64 $ % 2

Now

15 '3 ( yV = yEL dV : y ) π 3 R3 * = − π R 4 8 64 + ,

&

y =−

or

5 8 3

R R = 250 mm

y = − 90.2 mm

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PROBLEM 5.133 After grading a lot, a builder places four stakes to designate the corners of the slab for a house. To provide a firm, level base for the slab, the builder places a minimum of 3 in. of gravel beneath the slab. Determine the volume of gravel needed and the x coordinate of the centroid of the volume of the gravel. (Hint: The bottom surface of the gravel is an oblique plane, which can be represented by the equation y = a + bx + cz.)

SOLUTION The centroid can be found by integration. The equation for the bottom of the gravel is: y = a + bx + cz , where the constants a, b, and c can be determined as follows:

For x = 0, and z = 0:

y = − 3 in., and therefore −

3 1 ft = a, or a = − ft 12 4

For x = 30 ft, and z = 0: y = − 5 in., and therefore −

5 1 1 ft = − ft + b(30 ft), or b = − 12 4 180

For x = 0, and z = 50 ft: y = − 6 in., and therefore −

6 1 1 ft = − ft + c(50 ft), or c = − 12 4 200

Therefore:

1 1 1 y = − ft − x− z 4 180 200

Now

x =

&x

EL dV

V

A volume element can be chosen as: dV = | y| dxdz 1' 1 1 ( 1+ x+ z dx dz ) 4+ 45 50 *,

or

dV =

and

xEL = x

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PROBLEM 5.133 (Continued)

Then

&

xEL dV =

50

30

0

0

& &

x' 1 1 ( 1+ x+ z dx dz 4 )+ 45 50 *, 30

x2 z 2! 1 3 x + x # dz " + 100 % 0 $ 2 135

1 = 4

&

=

1 4

&

=

1 9 ! 650 z + z 2 # " 4$ 2 %0

50 0 50 0

(650 + 9 z )dz 50

= 10937.5 ft 4

The volume is:

&

V dV =

50

30

0

0

& &

1' 1 1 ( 1+ x+ z dx dz ) 4+ 45 50 *, 30

1 2 z ! " x + 90 x + 50 x # dz $ %0

=

1 4

=

1 4

=

1 3 ! 40 z + z 2 # " 4$ 10 % 0

& &

50 0

50 ' 0

3 ( ) 40 + z * dz 5 , + 50

= 687.50 ft 3

Then

x =

&x

EL dV

V

=

10937.5ft 4 = 15.9091 ft 687.5 ft 3 V = 688 ft 3

Therefore:

x = 15.91 ft

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!

PROBLEM 5.134 Determine by direct integration the location of the centroid of the volume between the xz plane and the portion shown of the surface y = 16h(ax − x2)(bz − z2)/a2b2.

SOLUTION

First note that symmetry implies

x=

a 2

z=

b 2

Choose as the element of volume a filament of base dx × dz and height y. Then dV = ydxdz ,

or Then

dV =

yEL =

1 y 2

16h (ax − x 2 )(bz − z 2 ) dx dz 2 2 a b b

a

0

0

V=

&&

V=

16h a 2b2

16h (ax − x 2 )(bz − z 2 ) dx dz 2 2 a b

&

a

a 1 ! (bz − z 2 ) " x 2 − x3 # dz 0 z 3 %0 $ b

b

=

16h a 2 1 3 ! b 2 1 3 ! ( a ) − (a) # " z − z # 3 3 %0 a 2 b 2 "$ 2 % $2

8ah b 2 1 3 ! (b) − (b) # 3 3b 2 "$ 2 % 4 = abh 9 =

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PROBLEM 5.134 (Continued)

and

&y

EL dV

b

a

0

0

=

&&

=

128h 2 a 4b 4

=

128h 2 a 2b 4

1 16h ! 16h ! ( ax − x 2 )(bz − z 2 ) # " 2 2 (ax − x 2 )(bz − z 2 ) dx dz # " 2 2 2 $a b %$a b % b

a

0

0

&& &

(a 2 x 2 − 2ax3 + x 4 )(b 2 z 2 − 2bz 3 + z 4 )dx dz a

a2 a 1 ! (b 2 z 2 − 2bz 3 + z 4 ) " x3 − x 4 + x5 # dz 0 2 5 %0 $3 b

b

! b2 128h 2 a 2 a 1 b 1 ! = 4 4 " ( a )3 − ( a ) 4 + ( a )5 # " Z 3 − Z 4 + Z 5 # 2 5 Z 5 %0 a b $3 %$ 3 =

Now

64ah 2 b3 3 b 4 1 5 ! 32 abh 2 (b) − (b) + (b) # = 4 " 2 5 15b $ 3 % 225

'4 ( 32 yV = yEL dV : y ) abh * = abh 2 +9 , 225

&

8 or y = 25 h

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!

PROBLEM 5.135 Locate the centroid of the section shown, which was cut from a thin circular pipe by two oblique planes.

SOLUTION x =0

First note that symmetry implies

Assume that the pipe has a uniform wall thickness t and choose as the element of volume A vertical strip of width adθ and height ( y2 − y1 ). Then

dV = ( y2 − y1 )tadθ ,

Now

y1 = =

and

h 3

2a

z+

h 6

yEL =

1 ( y1 + y2 ) z EL = z 2 y2 = −

h ( z + a) 6a

=

2h 3

2a

z+

2 h 3

h ( − z + 2a ) 3a

z = a cos θ h h (−a cos θ + 2a) − (a cos θ + a ) 3a 6a h = (1 − cos θ ) 2

Then

( y2 − y1 ) =

and

( y1 + y2 ) =

h h (a cos θ + a) + (− a cos θ + 2a) 6a 3a h = (5 − cos θ ) 6 aht h dV = (1 − cos θ )dθ yEL = (5 − cos θ ), z EL = a cos θ 2 12

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 713

PROBLEM 5.135 (Continued)

and

π

aht (1 − cos θ )dθ = aht[θ − sin θ ]π0 2 = π aht

Then

&y

V =2

&

=2

&

EL dV

0

h aht ! (5 − cos θ ) " (1 − cos θ ) dθ # 12 $ 2 %

π 0 2

π

=

ah t 12

=

ah 2 t θ sin 2θ ! 5θ − 6sin θ + + " 12 $ 2 4 #% 0

=

11 π ah2 t 24

&

0

(5 − 6 cos θ + cos 2 θ ) dθ π

&z

EL dV

=2

&

π 0

aht ! a cos θ " (1 − cos θ ) dθ # $ 2 %

θ

= a 2 ht "sin θ − − 2 $ 1 = − π a 2 ht 2

&

π

sin 2θ ! 4 #% 0

11 π ah 2t 24

Now

yV = yEL dV : y (π aht ) =

and

1 zV = zEL dV : z (π aht ) = − π a 2 ht 2

&

or y =

11 h 24

1 or z = − a 2

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 714

!

PROBLEM 5.136* Locate the centroid of the section shown, which was cut from an elliptical cylinder by an oblique plane.

SOLUTION x=0

First note that symmetry implies

Choose as the element of volume a vertical slice of width zx, thickness dz, and height y. Then dV = 2 xydz , x=

and

y=−

Then

V=

Then

&

h/2 h h z+ = (b − z ) b 2 2b b

' a 2 ! 2 ( h ) 2 b b − z * " 2b (b − z ) # dz ,$ %

−b +

z = b sin θ V=

1 , z EL = z 24

a 2 b − z2 b

Now

Let

yEL =

ah b2

π /2

&π

= abh

/2

dz = b cos θ dθ (b cos θ )[b(1 − sin θ )] b cos θ dθ

π /2

&π

− /2

θ

(cos 2 θ − sin θ cos 2 θ ) dθ

= abh " + $2

π /2

sin 2θ 1 ! + cos3 θ # 4 3 % −π /2

1 V = π abh 2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 715

PROBLEM 5.136* (Continued)

and

&y

EL dV

Now So that Then

&

=

1 ah 2 4 b3

h ! -' a 2 ( h ! . b − z 2 * " (b − z ) # dz 1 (b − z ) # 0) 2 b b 2b 2 % 2+ ,$ % 3

×

−b " $2

&

b −b

z = b sin θ

Let Then

1

b

=

&

dz = b cos θ dθ

1 ah 2 4 b3 1 = abh 2 4

sin 2 θ cos2 θ =

&y

EL dV

=

π /2

&π

yEL dV =

sin 2 θ =

(b − z ) 2 b 2 − z 2 dz

− /2

[b(1 − sin θ )]2 (b cos θ ) × (b cos θ dθ )

π /2

&π

− /2

(cos 2 θ − 2sin θ cos 2 θ + sin 2 θ cos 2 θ ) dθ

1 1 (1 − cos 2θ ) cos 2 θ = (1 + cos 2θ ) 2 2 1 (1 − cos 2 2θ ) 4 1 abh 2 4

1 ! cos 2 θ − 2sin θ cos 2 θ + (1 − cos 2 2θ ) # dθ " /2 $ 4 %

π /2

&π −

π /2

1 1 1 ' θ sin 4θ ( ! ' θ sin 2θ ( 1 = abh 2 ") + + cos3 θ + θ − ) + # * 4 4 , 3 4 4+ 2 8 *, % −π /2 $+ 2 =

Also

&z

EL dV

5 π abh2 32

- a 2 h ! . z 02 a − z 2 " (b − z ) # dz 1 −b 2 b $ 2b % 3 ah b z (b − z ) b 2 − z 2 dz = 2 b −b =

&

b

&

z = b sin θ

Let Then

&z

EL dV

=

ah b2

π /2

&π

− /2

= ab 2 h

Using

sin 2 θ cos2 θ =

&z

EL dV

dz = b cos θ dθ (b sin θ )[b(1 − sin θ )](b cos θ ) × (b cos θ dθ )

π /2

&π

− /2

(sin θ cos 2 θ − sin 2 θ cos 2 θ ) dθ

1 (1 − cos 2 2θ ) from above 4

= ab 2 h

1 ! sin θ cos 2 θ − (1 − cos 2 2θ ) # dθ " /2 $ 4 %

π /2

&π −

π /2

1 1 1 ' θ sin 4θ ( ! 1 = ab h " − cos3 θ − θ + ) + = − π ab 2 h # * 4 4+ 2 8 , % −π /2 8 $ 3 2

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 716

PROBLEM 5.136* (Continued)

Now

'1 ( 5 yV = yEL dV : y ) π abh * = π abh 2 2 + , 32

or y =

and

1 '1 ( z V = z EL dV : z ) π abh * = − π ab 2 h 2 8 + ,

1 or z = − b 4

&

&

5 h 16

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 717

!

PROBLEM 5.137 Locate the centroid of the plane area shown.

SOLUTION

Then

A, mm 2

x , mm

y , mm

xA, mm3

yA, mm3

1

126 × 54 = 6804

9

27

61236

183708

2

1 × 126 × 30 = 1890 2

30

64

56700

120960

3

1 × 72 × 48 = 1728 2

48

−16

82944

−27648

Σ

10422

200880

277020

X ΣA = Σ xA X (10422 m 2 ) = 200880 mm 2

and

or X = 19.27 mm

Y Σ A = Σ yA Y (10422 m 2 ) = 270020 mm3

or

Y = 26.6 mm

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 718

!

PROBLEM 5.138 Locate the centroid of the plane area shown.

SOLUTION

Dimensions in in.

Then

A, in.2

x , in.

y , in.

xA, in.3

yA, in.3

1

2 (4)(8) = 21.333 3

4.8

1.5

102.398

32.000

2

1 − (4)(8) = −16.0000 2

5.3333

1.33333

85.333

−21.333

Σ

5.3333

17.0650

10.6670

X ΣA = Σ xA X (5.3333 in.2 ) = 17.0650 in.3

and

or X = 3.20 in.

Y Σ A = Σ yA Y (5.3333 in.2 ) = 10.6670 in.3

or

Y = 2.00 in.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 719

!

PROBLEM 5.139 The frame for a sign is fabricated from thin, flat steel bar stock of mass per unit length 4.73 kg/m. The frame is supported by a pin at C and by a cable AB. Determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION First note that because the frame is fabricates from uniform bar stock, its center of gravity will coincide with the centroid of the corresponding line.

L, m

x, m

xL, m 2

1

1.35

0.675

0.91125

2

0.6

0.3

0.18

3

0.75

0

0

4

0.75

0.2

0.15

1.07746

1.26936

5 Σ

Then

π 2

(0.75) = 1.17810

4.62810

2.5106

X ΣL = Σx L X (4.62810) = 2.5106

or

X = 0.54247 m

The free-body diagram of the frame is then Where

W = (m′Σ L) g = 4.73 kg/m × 4.62810 m × 9.81 m/s 2 = 214.75 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 720

PROBLEM 5.139 (Continued)

Equilibrium then requires '3 ( ΣM C = 0: (1.55 m) ) TBA * − (0.54247 m)(214.75 N) = 0 +5 ,

(a)

TBA = 125.264 N

or

or TBA = 125.3 N

3 ΣFx = 0: C x − (125.264 N) = 0 5

(b)

C x = 75.158 N

or ΣFy = 0: C y +

4 (125.264 N) − (214.75 N) = 0 5 C y = 114.539 N

or

C = 137.0 N

Then

56.7°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 721

!

PROBLEM 5.140 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h.

SOLUTION y1 = −

By observation

h x( ' x + h = h )1 − * a + a,

For y2 : at

x = 0,

y = h : h = k (1 − 0 )

at

x = a,

y = 0: 0 = h(1 − ca 2 ) or C =

or

Then

' x2 y2 = h ))1 − 2 + a

Now

dA = ( y2 − y1 )dx

k=h 1 a2

( ** ,

' x2 = h "))1 − 2 $"+ a

( ' x (! ** − )1 − * # dx , + a , %#

' x x2 ( = h )) − 2 ** dx +a a ,

xEL = x yEL =

Then

1 ( y1 − y2 ) 2

=

h ' x( ' x2 ")1 − * + ))1 − 2 2 "$+ a , + a

=

h' x x2 ( )) 2 − − 2 ** 2+ a a ,

&

A = dA = =

&

a

0

' x x2 h )) − 2 +a a

(! ** # , %#

a

( x2 x3 ! ** dx = h " − 2 # , $ 2a 3a % 0

1 ah 6

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 722

PROBLEM 5.140 (Continued)

and

&x

&

EL dA

a

' x x2 ( ! ' x3 x4 − 2 x " h )) − 2 ** dx # = h ")) $" + a a , %# $"+ 3a 4a

=

&

=

1 2 a h 12

0

( ' x x2 ** " h )) − 2 0 , "$ + a a h2 a ' x x2 x4 ( = )) 2 − 3 2 + 4 ** dx 2 0+ a a a ,

yEL dA =

&

a

h' x x2 )) 2 − − 2 2+ a a

a

(! ** # , %# 0

( ! ** dx # , #%

&

a

1 2 h 2 x 2 x3 x5 ! = " − 2 + 4 # = ah 2 $a a 5a % 0 10 '1 ( 1 xA = xEL dA: x ) ah * = a 2 h + 6 , 12

x=

'1 ( 1 yA = yEL dA: y ) ah * = a 2 h + 6 , 10

3 y= h 5

&

&

1 a 2

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 723

!

PROBLEM 5.141 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b.

SOLUTION y =b

First note that symmetry implies x = a,

at

y=b

y1 : b = ka 2 y1 =

Then

k=

or

b a2

b 2 x a2

y2 : b = 2b − ca 2 c=

or

' x2 y2 = b )) 2 − 2 a +

Then

( ** ,

' x2 dA = ( y2 − y1 )dx2 = "b )) 2 − 2 a $" + ' x2 ( = 2b ))1 − 2 ** dx + a ,

Now

( b 2! ** − 2 x # dx #% , a

xEL = x

and

& &

A = dA

Then

and

b a2

&

a

0

' x2 2b ))1 − 2 + a

a

( 4 x3 ! 2 dx b x = − = ab " 2# ** 3 a , $ %0 3 a

' x2 ( ! x2 x4 ! 1 xEL dA = x " 2b ))1 − 2 ** dx # = 2b " − 2 # = a 2b 0 "$ + a , #% $ 2 4a % 0 2 '4 ( 1 xA = xEL dA: x ) ab * = a 2b +3 , 2

&

a

&

3 x= a 8

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 724

PROBLEM 5.142 Knowing that two equal caps have been removed from a 10-in.-diameter wooden sphere, determine the total surface area of the remaining portion.

SOLUTION The surface area can be generated by rotating the line shown about the y axis. Applying the first theorem of Pappus-Guldinus, we have A = 2π X L = 2π Σ x L = 2π (2 x1 L1 + x2 L2 )

Now

tan α =

4 3

or

α = 53.130°

Then

x2 =

5 in. × sin 53.130° π 53.130° × 180 °

= 4.3136 in.

and

π ( ' L2 = 2 ) 53.130° × * (5 in.) 180 °, + = 9.2729 in. ! '3 ( A = 2π " 2 ) in. * (3 in.) + (4.3136 in.)(9.2729 in.) # $ +2 , % A = 308 in.2

or

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 725

PROBLEM 5.143 Determine the reactions at the beam supports for the given loading.

SOLUTION RI = (3 m)(900 N/m) = 2700 N 1 RII = (1 m)(900 N/m) 2 = 450 N

Now

ΣFx = 0: Ax = 0 '1 ( ΣM B = 0: − (3 m) Ay + (1.5 m)(2700 N) − ) m * (450 N) = 0 +3 , Ay = 1300 N

or

A = 1300 N

ΣFy = 0: 1300 N − 2700 N + By − 450 N = 0

or

By = 1850 N

B = 1850 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 726

PROBLEM 5.144 A beam is subjected to a linearly distributed downward load and rests on two wide supports BC and DE, which exert uniformly distributed upward loads as shown. Determine the values of wBC and wDE corresponding to equilibrium when wA = 600 N/m.

SOLUTION

We have

1 (6 m)(600 N/m) = 1800 N 2 1 RII = (6 m)(1200 N/m) = 3600 N 2 RBC = (0.8 m) (WBC N/m) = (0.8 WBC )N RI =

RDE = (1.0 m) (WDE N/m) = (WDE ) N

Then

ΣM G = 0: −(1 m)(1800 N) − (3 m)(3600 N) + (4 m)(WDE N) = 0 WDE = 3150 N/m

or and or

ΣFy = 0: (0.8WBC ) N − 1800 N − 3600 N + 3150 N = 0 WBC = 2812.5 N/m

WBC = 2810 N/m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 727

PROBLEM 5.145 The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water d = 3.5 ft, determine the force exerted on the gate by the shear pin.

SOLUTION First consider the force of the water on the gate. We have 1 Ap 2 1 = A(γ h) 2

P=

Then

1 (1.8 ft) 2 (62.4 lb/ft 3 )(1.7 ft) 2 = 171.850 lb

PI =

1 (1.8 ft) 2 (62.4 lb/ft 3 ) × (1.7 + 1.8cos 30°) ft 2 = 329.43 lb

PII =

Now or

'1 ( '2 ( ΣM A = 0: ) LAB * PI + ) LAB * PII − LAB FB = 0 3 3 + , + , 1 2 (171.850 lb) + (329.43 lb) − FB = 0 3 3 FB = 276.90 lb

or

FB = 277 lb

30.0°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 728

PROBLEM 5.146 Consider the composite body shown. Determine (a) the value of x when h = L/2, (b) the ratio h/L for which x = L.

SOLUTION

V

x

xV

Rectangular prism

Lab

1 L 2

1 2 L ab 2

Pyramid

1 'b( a h 3 )+ 2 *,

1 1 ( ' abh ) L + h * 6 4 , +

1 h 4

1 ( ' ΣV = ab ) L + h * 6 , +

Then

Σ xV =

1 1 (! ' ab "3L2 + h ) L + h * # 6 $ 4 ,% +

X ΣV = Σ xV

Now so that

1 (! 1 ' 1 ( ' X " ab ) L + h * # = ab ) 3L2 + hL + h 2 * 6 6 4 , ,% + $ + h 1 h2 ( ' 1 h( 1 ' = + + X )1 + L 3 ) * * L 4 L2 *, + 6 L , 6 )+

or

(a)

L+

X = ? when h =

Substituting

1 L 2

h 1 = into Eq. (1) L 2 1 ' 1 (! 1 '1( 1'1( X "1 + ) * # = L "3 + ) * + ) * $ 6 + 2 , % 6 "$ + 2 , 4 + 2 ,

or

(1)

X=

57 L 104

2!

# #% X = 0.548L

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 729

PROBLEM 5.146 (Continued)

(b)

h = ? when X = L L

Substituting into Eq. (1)

or or

h 1 h2 ( ' 1 h( 1 ' = + + L )1 + L 3 ) * * L 4 L2 *, + 6 L , 6 )+ 1+

1 h 1 1 h 1 h2 = + + 6 L 2 6 L 24 L2 h2 = 12 L2

h =2 3 L

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 730

PROBLEM 5.147 Locate the center of gravity of the sheet-metal form shown.

SOLUTION First assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with the centroid of the corresponding area. 1 yI = 0.18 + (0.12) = 0.22 m 3 1 zI = (0.2 m) 3 2 × 0.18 0.36 m xII = yII = =

π

xIV

I II

IV Σ

A, m 2

x, m

y, m

z, m

x A, m3

yA, m3

z A, m3

1 (0.2)(0.12) = 0.012 2

0

0.22

0.2 3

0

0.00264

0.0008

0.36

0.36

π

π

0.1

0.00648

0.00648

0.005655

0.26

0

0.1

0.00832

0

0.0032

0.31878

0

0.1

–0.001258

0

–0.000393

0.013542

0.00912

0.009262

π

(0.18)(0.2) = 0.018π

2

(0.16)(0.2) = 0.032

III −

π

4 × 0.05 = 0.34 − 3π = 0.31878 m

π 2

(0.05) 2 = −0.00125π

0.096622

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 731

PROBLEM 5.147 (Continued)

We have

X ΣV = Σ xV : X (0.096622 m 2 ) = 0.013542 m3

or X = 0.1402 m

Y ΣV = Σ yV : Y (0.096622 m 2 ) = 0.00912 m3

or Y = 0.0944 m

Z ΣV = Σ zV : Z (0.096622 m 2 ) = 0.009262 m3

or Z = 0.0959 m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 732

PROBLEM 5.148 Locate the centroid of the volume obtained by rotating the shaded area about the x axis.

SOLUTION y =0

First, note that symmetry implies

z =0

Choose as the element of volume a disk of radius r and thickness dx. dV = π r 2 dx, xEL = x

Then Now r = 1 −

2

' 1( dV = π )1 − * dx x, + ' 2 1 ( = π )1 − + 2 * dx x x , +

1 so that x

V=

Then

3

&

1

' +

π )1 −

3

2 1 ( 1! + 2 * dx = π " x − 2 ln x − # x x , x %1 $

1( ' 1 (! ' = π ") 3 − 2ln 3 − * − ) 1 − 2 ln1 − * # 3, + 1 ,% $+ = (0.46944π ) m3

and

&x

EL dV

=

&

3

1

3

! x2 ' 2 1 ( ! x "π )1 − + 2 * dx # = π " − 2 x + ln x # x x , % $ + $2 %1

-/ 32 ! 13 ! ./ = π 0 " − 2(3) + ln 3# − " − 2(1) + ln1# 1 /2 $ 2 % $2 % /3 = (1.09861π ) m

Now

&

xV = xEL dV : x (0.46944π m3 ) = 1.09861π m 4

or x = 2.34 m

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