Statics Mechanics of Materials 4th Edition Hibbeler Solutions Manual

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Statics Mechanics of Materials 4th Edition Hibbeler Solutions Manual

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2–2.

If and ,determine the magnitude of the resultant force and its direction,measured counterclockwise from the positive x axis.

F = 450 N u = 60°

SOLUTION

The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively.

Applying the law of consines to Fig. b,

FR = 27002 + 4502 - 2(700)(450) cos 45°

= 497.01 N = 497 N

This yields

sin a 700 = sin 45° 497.01 a = 95.19°

Ans.

Thus,the direction of angle of measured counterclockwise from the positive axis,is

f = a + 60° = 95.19° + 60° = 155°

Ans.

x FR
f
x y 700 N F u 15 18
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2–3.

If the magnitude of the resultant force is to be 500 N, directed along the positive y axis,determine the magnitude of force F and its direction . u

SOLUTION

The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively.

Applying the law of cosines to Fig. b,

F = 25002 + 7002 - 2(500)(700) cos 105°

= 959.78 N = 960 N

Applying the law of sines to Fig. b,and using this result,yields

sin (90° + u) 700 = sin 105° 959.78

Ans.

Ans. u = 45.2°

x y 700 N F u 15 19
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Determine the magnitude of the resultant force and its direction,measured clockwise from the positive

axis.

2–4.
u
FR = F1 + F2 SOLUTION Ans. Ans. f = 55.40° + 30° = 85.4° u = 55.40° 605.1 sin 95° = 500 sin u FR = 2(300)2 + (500)2 - 2(300)(500) cos 95° = 605.1 = 605 N u v 70 30 45 F1 300N F2 500N 20 © Pearson Education South Asia Pte Ltd 2014. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

© Pearson Education South Asia Pte Ltd 2014. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Resolvetheforceintocomponentsactingalongthe u and v axesanddeterminethemagnitudesofthecomponents.

SOLUTION

2–5.
Ans. Ans. F1v = 160N F1v sin 30° = 300 sin 110° F1u = 205N F1u sin 40° = 300 sin 110°
F1 u v 70 30 45 F1 300N F2 500N 21

© Pearson Education South Asia Pte Ltd 2014. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Resolvethe force intocomponentsactingalongthe u and v axesanddeterminethe magnitudesof thecomponents

2–6.
F2 SOLUTION Ans. Ans. F2v = 482 N F2v sin 65° = 500 sin 70° F2u = 376 N F2u sin 45° = 500 sin 70° u v 70 30 45 F1 300 N F2 500 N 22

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2–7. Ifand the resultant force acts along the positive u axis,determine the magnitude of the resultant force and the angle . u

FB = 2 kN y x u B FA 3 kN FB A u 30 23

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2–8. If the resultant force is required to act along the positive u axis and have a magnitude of 5 kN,determine the required magnitude of FB and its direction . u

y x u B FA 3 kN FB A u 30 24

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Resolve F1 into components along the u and axes and determine the magnitudes of these components v

2–9.
Ans. Ans. F1u sin 45° = 250 sin 105° F1u = 183N F1v sin 30° = 250 sin 105° F1v = 129N F1 250 N F2 150N u v 30 30 105 25
SOLUTION Sine law:

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Resolve F2 into components along the u and axes and determine the magnitudes of these components. v

SOLUTION

Sine law:

2–10.
Ans. Ans. F2u sin 75° = 150 sin 75° F2u = 150N F2v sin 30° = 150 sin 75° F2v = 77.6N
F1 250 N F2 150N u v 30 30 105 26
. . 27 © Pearson Education South Asia Pte Ltd 2014. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
. . . 28 © Pearson Education South Asia Pte Ltd 2014. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

© Pearson Education South Asia Pte Ltd 2014. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2–1. The device is used for surgical replacement of the knee joint If the force acting along the leg is 360 N, determine its components along the x and y axes ¿

x x¿
60 360 N 10 y x y¿ x¿ 3 . . 29

© Pearson Education South Asia Pte Ltd 2014. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2–1 . The device is used for surgical replacement of the knee joint.If the force acting along the leg is 360 N, determine its components along the x and y axes. ¿

60 360 N 10 y x y¿ x¿ 4 . . 30

u = 60°

The plate is subjected to the two forces at A and B as shown. If ,determine the magnitude of the resultant of these two forces and its direction measured clockwise from the horizontal.

SOLUTION

Parallelogram Law: The parallelogram law of addition is shown in Fig. a

Trigonometry: Using law of cosines (Fig. b),we have

FR = 282 + 62 - 2(8)(6) cos 100°

= 10.80kN = 10.8kN

Ans.

The angle can be determined using law of sines (Fig. b).

u

sin u 6 = sin 100° 10.80

sin u = 0.5470

u = 33.16°

f

Thus,the direction of FR measured from the x axis is

f = 33.16° - 30° = 3.16°

Ans.

2–15.
A B
40
31
FA 8kN FB 6kN
u
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2–16.

Determine the angle of for connecting member A to the plate so that the resultant force of FA and FB isdirected horizontally to the right.Also,what is the magnitude of the resultant force?

SOLUTION

Parallelogram Law: The parallelogram law of addition is shown in Fig. a.

Trigonometry: Using law of sines (Fig .b),we have

sin (90° - u) 6 = sin 50° 8

sin (90° - u) = 0.5745

u = 54.93° = 54.9°

Ans.

From the triangle,.Thus,using law of cosines,the magnitude of FR is

f = 180° - (90° - 54.93°) - 50° = 94.93°

FR = 282 + 62 - 2(8)(6) cos 94.93°

Ans. = 10.4kN

u
A B FA 8kN FB 6kN 40 u 32 ©
Pearson Education South Asia Pte Ltd 2014. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
. . 33 © Pearson Education South Asia Pte Ltd 2014. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
. . . 34 © Pearson Education South Asia Pte Ltd 2014. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

© Pearson Education South Asia Pte Ltd 2014. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Determinethemagnitudeanddirectionoftheresultant ofthethreeforcesbyfirstfindingthe resultantandthenforming

2–19. SOLUTION Ans. Ans. 19.18 sin 1.47° = 30.85 sin u ; u = 2.37° FR = 2(30.85)2 + (50)2 - 2(30.85)(50) cos 1.47° = 19.18 = 19.2N 30.85 sin 73.13° = 30 sin (70° - u ¿) ; u ¿ = 1.47° F ¿ = 2(20)2 + (30)2 - 2(20)(30) cos 73.13° = 30.85 N
FR = F ¿+ F3. F ¿= F1 + F2 FR = F1 + F2 + F3 y x F2 20N F1 30N 20 3 5 4 F3 50N 35

© Pearson Education South Asia Pte Ltd 2014. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Statics Mechanics of Materials 4th Edition Hibbeler Solutions Manual Full Download: http://testbanktip.com/download/statics-mechanics-of-materials-4th-edition-hibbeler-solutions-manual/

Determinethemagnitudeanddirectionoftheresultant ofthethreeforcesbyfirstfindingthe resultantandthenforming

2–20.
FR = F ¿+ F1 F ¿= F2 + F3 FR = F1 + F2 + F3 SOLUTION Ans. Ans. u = 23.53° - 21.15° = 2.37° 19.18 sin 13.34° = 30 sin f ; f = 21.15° FR = 2(47.07)2 + (30)2 - 2(47.07)(30) cos 13.34° = 19.18 = 19.2N 20 sin u ¿ = 47.07 sin 70° ; u ¿ = 23.53° F ¿ = 2(20)2 + (50)2 - 2(20)(50) cos 70° = 47.07N y x F2 20N F1 30N 20 3 5 4 F3 50N 36
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