Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 04

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If A, B, and D are given vectors, prove the

distributive law for the vector cross product, i.e.,

A : (B + D) = (A : B) + (A : D)

SOLUTION

Consider the three vectors; with A vertical.

Note obd is perpendicular to A.

Also, these three cross products all lie in the plane obd since they are all

perpendicular to A As noted the magnitude of each cross product is proportional to

the length of each side of the triangle

The three vector cross products also form a closed triangle which is similar to

triangle obd Thus from the figure,

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Given the three nonzero vectors A, B, and C, show that if

, the three vectors must lie in the same

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Determine the moment about point A of each of the three

forces acting on the beam

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Determine the moment about point B of each of the three

forces acting on the beam

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The crane can be adjusted for any angle and

any extension For a suspended mass of

120 kg, determine the moment developed at A as a function

of x and What values of both x and develop the

maximum possible moment at A? Compute this moment.

Neglect the size of the pulley at B.

uu

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Determine the moment of each of the three forces about

point A.

SOLUTION

The moment arm measured perpendicular to each force from point A is

Using each force where we have

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202.5 ft

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202.5 ft

moment about the bolt located at A.

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The railway crossing gate consists of the 100-kg gate arm

having a center of mass at Ga and the 250-kg counterweight

having a center of mass at GW Determine the magnitude

and directional sense of the resultant moment produced by

the weights about point A.

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The railway crossing gate consists of the 100-kg gate arm

having a center of mass at Ga and the 250-kg counterweight

having a center of mass at GW Determine the magnitude

and directional sense of the resultant moment produced by

the weights about point B.

1 m0.5 m

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The two boys push on the gate with forces of ,

and , as shown Determine the moment of each

force about C Which way will the gate rotate, clockwise or

counterclockwise? Neglect the thickness of the gate

A

FB

3 4 5

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Two boys push on the gate as shown If the boy at B exerts

a force of , determine the magnitude of the

force the boy at A must exert in order to prevent the

gate from turning Neglect the thickness of the gate

A

FB

3 4 5

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The Achilles tendon force of is mobilized when

the man tries to stand on his toes As this is done, each of his

feet is subjected to a reactive force of

Determine the resultant moment of and about the

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The Achilles tendon force is mobilized when the man tries

to stand on his toes As this is done, each of his feet is

subjected to a reactive force of If the resultant

moment produced by forces and about the ankle joint

A is required to be zero, determine the magnitude of Ff

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The total hip replacement is subjected to a force of

Determine the moment of this force about the

neck at A and the stem at B.

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The tower crane is used to hoist the 2-Mg load upward at

constant velocity The 1.5-Mg jib BD, 0.5-Mg jib BC, and

6-Mg counterweight C have centers of mass at G1, G2, and

G3, respectively Determine the resultant moment produced

by the load and the weights of the tower crane jibs about

point A and about point B.

SOLUTION

Since the moment arms of the weights and the load measured to points A and B are

the same, the resultant moments produced by the load and the weight about points

A and B are the same.

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The tower crane is used to hoist a 2-Mg load upward at

con-stant velocity The 1.5-Mg jib BD and 0.5-Mg jib BC have

centers of mass at G1 and G2, respectively Determine the

required mass of the counterweight C so that the resultant

moment produced by the load and the weight of the tower

crane jibs about point A is zero The center of mass for the

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The handle of the hammer is subjected to the force of

Determine the moment of this force about the

point A.

F = 20 lb

SOLUTION

Resolving the 20-lb force into components parallel and perpendicular to the

hammer, Fig a, and applying the principle of moments,

18 in

5 in

30

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SOLUTION

Resolving force F into components parallel and perpendicular to the hammer, Fig a,

and applying the principle of moments,

a

Ans.

F = 27.6 lb+MA = -500 = -F cos 30°(18) - F sin 30°(5)

F

B A

18 in

5 in

30

In order to pull out the nail at B, the force F exerted on the

handle of the hammer must produce a clockwise moment of

500 lb# in about point A Determine the required magnitude

of force F.

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The tool at A is used to hold a power lawnmower blade

stationary while the nut is being loosened with the wrench

If a force of 50 N is applied to the wrench at B in the direction

shown, determine the moment it creates about the nut at C.

What is the magnitude of force F at A so that it creates the

opposite moment about C?

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The towline exerts a force of at the end of the

20-m-long crane boom If determine the

placement x of the hook at A so that this force creates a

maximum moment about point O What is this moment?

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The towline exerts a force of P = 4 k N at the end of the

20-m-long crane boom If x = 25 m, determine the position u

of the boom so that this force creates a maximum moment

about point O What is this moment?

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If the 1500-lb boom AB, the 200-lb cage BCD, and the 175-lb

man have centers of gravity located at points G1, G2 and G3,

respectively, determine the resultant moment produced by

each weight about point A.

Moment of the weight of the man about point A:

a+ MA = -175(30cos75° + 4.25) = - 2102.55 lb#ft = 2.10 kip#ft(Clockwise) Ans.

= 2.05 kip#ft+ MA = -200(30cos75° + 2.5) = - 2052.91 lb#ft

+ MA = -1500(10cos75°) = - 3882.29 lb#ft = 3.88 kip#ft

75⬚

B C D

A

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If the 1500-lb boom AB, the 200-lb cage BCD, and the

175-lb man have centers of gravity located at points G1, G2

and G3, respectively, determine the resultant moment

pro-duced by all the weights about point A.

75⬚

B C D

A

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The connected bar BC is used to increase the lever arm of the

crescent wrench as shown If the applied force is F = 200 N and

d = 300 mm, determine the moment produced by this force

about the bolt at A.

300 mm

30⬚

15⬚

d C

B

A

F

SOLUTION

By resolving the 200-N force into components parallel and perpendicular to the box

wrench BC, Fig a, the moment can be obtained by adding algebraically the moments

of these two components about point A in accordance with the principle of

moments

a

= - 100.38 N#m = 100 N#m+(MR)A = ©Fd; MA = 200 sin 15°(0.3sin30°) - 200cos15°(0.3cos30° + 0.3)

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By resolving the 200-N force into components parallel and perpendicular to the box

wrench BC, Fig a, the moment can be obtained by adding algebraically the moments

of these two components about point A in accordance with the principle of moments.

a

Ans.

d = 0.4016 m = 402 mm+(MR)A = ©Fd; -120 = 200sin15°(0.3sin30°) - 200cos15°(0.3cos30° + d)

300 mm

30⬚

15⬚

d C

B

A

F

*4–28.

The connected bar BC is used to increase the lever arm of

the crescent wrench as shown If a clockwise moment of

is needed to tighten the bolt at A and the

force F = 200 N, determine the required extension d in

order to develop this moment

MA = 120 N#m

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The connected bar BC is used to increase the lever arm of

the crescent wrench as shown If a clockwise moment of

is needed to tighten the nut at A and the

extension d = 300 mm, determine the required force F in

order to develop this moment

SOLUTION

By resolving force F into components parallel and perpendicular to the box wrench

BC, Fig a, the moment of F can be obtained by adding algebraically the moments

of these two components about point A in accordance with the principle of

moments

a

Ans.

F = 239 N+(MR)A = ©Fd; -120 = Fsin15°(0.3sin30°) - Fcos15°(0.3cos30° + 0.3)

B

A

F

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A force F having a magnitude of acts along the

diagonal of the parallelepiped Determine the moment of F

about point A, using and MA = rB : F MA = rC : F.

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the beam Determine the moment of the force about

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Determine the moment produced by force about point O.

Express the result as a Cartesian vector

FB

SOLUTION

Position Vector and Force Vectors: Either position vector or can be used to

determine the moment of about point O.

The force vector is given by

Vector Cross Product: The moment of about point O is given by

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Determine the moment produced by force about point O.

Express the result as a Cartesian vector

FC

y x

Position Vector and Force Vectors: Either position vector or can be used to

determine the moment of about point O.

The force vector FCis given by

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Determine the resultant moment produced by forces

and about point O Express the result as a Cartesian

vector

FC

FB

y x

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Using a ring collar the 75-N force can act in the vertical

plane at various angles Determine the magnitude of the

moment it produces about point A, plot the result of M

(ordinate) versus (abscissa) for and

specify the angles that give the maximum and minimum

moment

0° … u … 180°,u

u

1.5 m

75 Nθ

dMA

du = 12112 656.25 sin2u + 22 5002- 1

112 656.25212 sin u cos u2 = 0

MA= 21112.5 sin u22 + 1-150 sin u22 + 1150 cos u22 = 212 656.25 sin2u + 22 500

= 112.5 sin u i - 150 sin u j + 150 cos u k

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3.742(80) - 2

3.742(80)3

rAC = 2(1)2 + (-3)2 + (-2)2 = 3.742 m

rAC = {1i - 3j - 2k} m

The curved rod lies in the x–y plane and has a radius of 3 m.

If a force of acts at its end as shown, determine

the moment of this force about point O.

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The curved rod lies in the x–y plane and has a radius of 3 m.

If a force of acts at its end as shown, determine

the moment of this force about point B.

3.742(80) - 3

3.742(80) - 2

3.742(80)3

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Force F acts perpendicular to the inclined plane Determine

the moment produced by F about point A Express the

result as a Cartesian vector

SOLUTION

Force Vector: Since force F is perpendicular to the inclined plane, its unit vector

is equal to the unit vector of the cross product, , Fig a Here

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Force F acts perpendicular to the inclined plane Determine

the moment produced by F about point B Express the

result as a Cartesian vector

Force Vector: Since force F is perpendicular to the inclined plane, its unit vector

is equal to the unit vector of the cross product, , Fig a Here

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SOLUTION

Position Vector And Force Vector:

Moment of Force F About Point A: Applying Eq 4–7, we have

The pipe assembly is subjected to the 80-N force Determine

the moment of this force about point A.

F 80 N

B

C

A

© 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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