Beer Vector Mechanics for Engineers STATICS 10th solutions PDF

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www.elsolucionario.net ww.elsolucionario. SOLUTION MANUAL www.elsolucionario.net www.elsolucionario.net www.elsolucionario.net CHAPTER 2 www.elsolucionario.net www.elsolucionario.net www.elsolucionario.net PROBLEM 2.1 Two forces are applied at point B of beam AB. Determine graphically the magnitude ...

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ww.elsolucionario. SOLUTION MANUAL

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CHAPTER 2

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www.elsolucionario.net PROBLEM 2.1 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.

(a)

Parallelogram law:

(b)

Triangle rule:

We measure:

R = 3.30 kN, α = 66.6°

R = 3.30 kN

66.6° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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

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SOLUTION

www.elsolucionario.net PROBLEM 2.2 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.

We measure: (a)

Parallelogram law:

(b)

Triangle rule:

We measure:

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SOLUTION

α = 51.3° β = 59.0°

R = 139.1 lb,

γ = 67.0°

R = 139.1 lb

67.0° 



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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

www.elsolucionario.net PROBLEM 2.3 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 10 kN and Q = 15 kN, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.

(a)

Parallelogram law:

(b)

Triangle rule:

We measure:

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SOLUTION

α = 21.2°

R = 20.1 kN,

R = 20.1 kN

21.2° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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

www.elsolucionario.net PROBLEM 2.4 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 6 kips and Q = 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.

(a)

Parallelogram law:

(b)

Triangle rule:

We measure:

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SOLUTION

R = 8.03 kips, α = 3.8°

R = 8.03 kips

3.8° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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

www.elsolucionario.net PROBLEM 2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant.

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

120 N P = sin 30° sin 25°

P = 101.4 N 

30° + β + 25° = 180°

β = 180° − 25° − 30° = 125° 120 N R = sin 30° sin125°

R = 196.6 N 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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

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SOLUTION

www.elsolucionario.net PROBLEM 2.6 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?

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. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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

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SOLUTION

www.elsolucionario.net PROBLEM 2.7 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.

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. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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

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SOLUTION

www.elsolucionario.net PROBLEM 2.8 A telephone cable is clamped at A to the pole AB. Knowing that the tension in the left-hand portion of the cable is T1 = 800 lb, determine by trigonometry (a) the required tension T2 in the right-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R.

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

75° + 40° + α = 180°

α = 180° − 75° − 40° = 65°

(b)

T2 800 lb = sin 65° sin 75°

T2 = 853 lb 

800 lb R = sin 65° sin 40°

R = 567 lb 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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

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SOLUTION

www.elsolucionario.net PROBLEM 2.9 A telephone cable is clamped at A to the pole AB. Knowing that the tension in the right-hand portion of the cable is T2 = 1000 lb, determine by trigonometry (a) the required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R.

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

75° + 40° + β = 180°

β = 180° − 75° − 40° = 65°

(b)

T1 1000 lb = sin 75° sin 65°

T1 = 938 lb 

1000 lb R = sin 75° sin 40°

R = 665 lb 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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

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SOLUTION

www.elsolucionario.net PROBLEM 2.10 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: 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° = 117.862° R 35 N = sin117.862° sin 25°

R = 73.2 N 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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

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(a)

www.elsolucionario.net 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.

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. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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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SOLUTION

www.elsolucionario.net 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.

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

=

500 lb

90° − α = 47.402°

(b)

R 500 lb = sin (42.598° + 30°) sin 60°

α = 42.6°  R = 551 lb 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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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SOLUTION

www.elsolucionario.net PROBLEM 2.13 A steel tank is to be positioned in an excavation. 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.

SOLUTION

(a)

P = (425 lb) cos 30°

(b)

R = (425 lb)sin 30°

P = 368 lb



R = 213 lb 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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

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The smallest force P will be perpendicular to R.

www.elsolucionario.net PROBLEM 2.14 For the hook support of Prob. 2.10, 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

(a)

P = (50 N)sin 25°

(b)

R = (50 N) cos 25°

P = 21.1 N



R = 45.3 N 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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

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The smallest force P will be perpendicular to R.

www.elsolucionario.net PROBLEM 2.15 Solve Problem 2.2 by trigonometry. PROBLEM 2.2 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.

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SOLUTION

8 10 α = 38.66°

tan α =

6 10 β = 30.96°

tan β =

Using the triangle rule:

Using the law of cosines:

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

Using the law of sines:

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. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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

www.elsolucionario.net PROBLEM 2.16 Solve Problem 2.4 by trigonometry. PROBLEM 2.4 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 6 kips and Q = 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.

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

γ = 180° − (50° + 25°)

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We have:

= 105°

Then

R 2 = (4 kips) 2 + (6 kips)2 − 2(4 kips)(6 kips) cos105°

= 64.423 kips 2 R = 8.0264 kips

And

4 kips 8.0264 kips = sin(25° + α ) sin105° sin(25° + α ) = 0.48137 25° + α = 28.775° α = 3.775° R = 8.03 kips

3.8° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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

www.elsolucionario.net PROBLEM 2.17 For the stake of Prob. 2.5, knowing that the tension in one rope is 120 N, determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 160 N. PROBLEM 2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant.

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SOLUTION

Using the laws of cosines and sines: P 2 = (120 N) 2 + (160 N)2 − 2(120 N)(160 N) cos 25° P = 72.096 N

And

sin α sin 25° = 120 N 72.096 N sin α = 0.70343 α = 44.703° P = 72.1 N

44.7° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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

www.elsolucionario.net PROBLEM 2.18 For the hook support of Prob. 2.10, 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 2.10 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°)

Then

R 2 = (75 N) 2 + (50 N) 2 − 2(75 N)(50 N) cos 105° R = 10, 066.1 N 2 R = 100.330 N 2

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<...