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Solution Manual Of Engineering Mechanics Statics 15th edition by Hibbeler - Updated 2024 $33.54   Add to cart

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Solution Manual Of Engineering Mechanics Statics 15th edition by Hibbeler - Updated 2024

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Solution Manual Of Engineering Mechanics Statics 15th edition by Hibbeler - Updated 2024 Complete Solution Manual

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  • October 26, 2024
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© 2022 by R. C. Hibbeler. Published by Pearson Education, Inc., Hoboken, NJ. 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.

Determine the magnitude of the resultant force y
FR = F1 + F2 and its orientation u, measured
counterclockwise from the positive x axis.
F1 5 260 lb




12 13

5

SOLUTION x

Sine Law: 458
sin 67.62° sin (45° + a)
= a = 3.728°
320 260
u = 180° - a = 176° Ans. F 5 310 lb
2

12
tan-1 = 67.38°
5
67.38° + 45° = 112.38°
360° - a (112.38°)
= 67.62°
a

Cosine Law:

FR = 23102 + 2602 - 2(310)(260) cos 67.62° = 320 lb Ans.




Ans:
FR = 320 lb
u = 176°

22

,© 2022 by R. C. Hibbeler. Published by Pearson Education, Inc., Hoboken, NJ. 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.



2–2.

Determine the magnitude of the resultant force y
FR′ = F1 - F2 and its orientation u, measured
counterclockwise from the positive x axis.
F1 5 260 lb




12 13

5

SOLUTION x

12 458
tan-1 = 67.38°
5
67.38° + 45° = 112.38°

F2 5 310 lb
Cosine Law:

FR = 23102 + 2602 - 2(310)(260) cos 112.38° = 474 lb Ans.

Sine Law:
sin 112.38° sin (u - 45°)
= u = 75.4° Ans.
474 260




Ans:
FR = 474 lb
u = 75.4°

23

,© 2022 by R. C. Hibbeler. Published by Pearson Education, Inc., Hoboken, NJ. 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.



2–3.

Two forces are applied at the end of a screw eye in order to y
remove the post. Determine the angle u 10° … u … 90°2 F
and the magnitude of force F so that the resultant force
30°
acting on the post is directed vertically upward and has a 500 N θ
magnitude of 750 N.

x


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

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

sin f sin 30°
=
750 500

sin f = 0.750

f = 131.41° 1By observation, f 7 80°2

Thus,

u = 180° - 30° - 131.41° = 18.59° = 18.6° Ans.

F 500
=
sin 18.59° sin 30°

F = 319 N Ans.




Ans:
u = 18.6°
F = 319 N

24

, © 2022 by R. C. Hibbeler. Published by Pearson Education, Inc., Hoboken, NJ. 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.



*2–4.
Determine the magnitudes
The vertical force of the
F acts downward at two components
on the of F
two-membered
along
frame.members
Determine and
ABthe AC. Set F of
magnitudes 500two
= the N. components of
B
F directed along the axes of and . Set 500 N.




SOLUTION A
Parallelogram Law:
Parallelogram Law: The parallelogram law of addition is shown in Fig. a.
Trigonometry:
Trigonometry: Using the law of sines (Fig. b), we have
F
500 C
sin 60° sin 75°

448 N Ans.

500
sin 45° sin 75°

366 N Ans.




Ans:
FAB = 448 N
FAC = 366 N


25

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