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231

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Ans:

(MF1)A = 3.00 kip # ft (Clockwise)



*4–4.

Determine the moment about point A of each of the threeforces acting on the beam.

SOLUTION

a

(Clockwise) Ans.

a

(Clockwise) Ans.

a

(Clockwise) Ans.= -2593 lb # ft = 2.59 kip # ft

+ 1MF32A = -1601cos 30°21192 + 160 sin 30°10.52

= -5600 lb # ft = 5.60 kip # ft

+ 1MF22A = -500a4

5b1142

= -3000 lb # ft = 3.00 kip # ft

+ 1MF12A = -375182

= 500 lbF2

= 160 lbF3

4

3

5

= 375 lbF1

8 ft 6 ft

0.5 ft

30˚

5 ft

BA

232


Ans:(MF1

)B = 4.125 kip # ftd(MF2

)B = 2.00 kip # ftd(MF3

)B = 40.0 lb # ftd

4–5.

SOLUTION

a

(Counterclockwise) Ans.

a

(Counterclockwise) Ans.

a

(Counterclockwise) Ans.= 40.0 lb # ft

+1MF32B = 160 sin 30°10.52 - 160 cos 30°102

= 2000 lb # ft = 2.00 kip # ft

+1MF22B = 500a4

5b152

= 4125 lb # ft = 4.125 kip # ft

+1MF12B = 3751112

Determine the moment about point B of each of the threeforces acting on the beam.

= 500 lbF2

= 160 lbF3

4

3

5

= 375 lbF1

8 ft 6 ft

0.5 ft

30˚

5 ft

BA

235


Ans:MB = 150 N # mdMB = 600 N # mdMB = 0

*4–8.

Determine the moment of each of the three forces about point B.

SOLUTION

The forces are resolved into horizontal and vertical component as shown in Fig. a.For F1,

a

d Ans.

For F2,

a

d Ans.

Since the line of action of F3 passes through B, its moment arm about point B iszero. Thus

Ans. MB = 0

= 600 N # m

+ MB = 300 sin 60° (0) + 300 cos 60° (4)

= 149.51 N # m = 150 N # m

+ MB = 250 cos 30° (3) - 250 sin 30° (4)

2 m 3 m

4 m

60

30F1 250 N

B

F2 300 N

F3 500 N

A

43

5

236


Ans:MB = 90.6 lb # ftbMC = 141 lb # ftd

4–9.

20

2.5 ft

AFB

FC

0.75 ft

30B

C

25

Determine the moment of each force about the bolt locatedat A. Take FB = 40 lb, FC = 50 lb.

SOLUTION

a d Ans.

a d Ans.+MC = 50 cos 30°(3.25) = 141 lb # ft

+MB = 40 cos 25°(2.5) = 90.6 lb # ft

240


SolutionForce Vector And Position Vector. Referring to Fig. a,

F = 20 (sin 60°i - cos 60°j) = {17.32i - 10j} N

rBA = {-0.01i + 0.2j} m

Moment of Force F about point B.

MB = rBA * F

= †i j k

-0.01 0.2 017.32 -10 0

†

= {-3.3641 k} N # m

= {-3.36 k} N # m Ans.

Here the unit vector for MB is u = -k. Thus, the coordinate direction angles of MB are

a = cos-1 0 = 90° Ans.

b = cos-1 0 = 90° Ans.

g = cos-1 (-1) = 108° Ans.

4–13.

The 20-N horizontal force acts on the handle of the socket wrench. What is the moment of this force about point B. Specify the coordinate direction angles a, b, g of the moment axis.

O

x

z

B A

y

50 mm

200 mm

10 mm

20 N

60�

Ans:MB = {-3.36 k} N # ma = 90°b = 90°g = 180°

242


Ans:(MA)C = 768 lb # ftb(MA)B = 636 lb # ftdClockwise

4–15.

Two men exert forces of and on theropes. Determine the moment of each force about A. Whichway will the pole rotate, clockwise or counterclockwise?

P = 50 lbF = 80 lb

A

P

F

B

C

6 ft

45

12 ft3

4

5

SOLUTION

c b Ans.

a d Ans.

Since

Clockwise Ans.

(MA)C 7 (MA)B

+ (MA)B = 50 (cos 45°)(18) = 636 lb # ft

+ (MA)C = 80a45b(12) = 768 lb # ft

248


Ans:F = 27.6 lb

4–21.

SOLUTIONResolving 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.

259


Ans:MA = {-110i + 70j - 20k} N # m

SolutionPosition Vector. The coordinates of point C are C (0.5, 0.7, -0.3) m. Thus

rAC = {0.5i + 0.7 j - 0.3k} m

Moment of Force F About Point A.

MA = rAC * F

= †i j k

0.5 0.7 -0.3600 800 -500

†

= {-110i + 70j - 20k} N # m Ans.

*4–32.

The pipe assembly is subjected to the force of F = {600i + 800j - 500k} N. Determine the moment of this force about point A.

y

0.5 m

0.4 m

0.3 m

0.3 m

x

z

F

B

C

A

264


Ans:MO = {163i - 346j - 360k} N # m

SolutionPosition And Force Vectors. The coordinates of point A are A (0.4, 0.5, -0.3) m. Thus

rOA = {0.4i + 0.5j - 0.3k} m

F = FuF = 800 (cos 60°i + cos 120°j + cos 45°k)

= {400i - 400j + 565.69k} N

Moment of F About Point O.

MO = rOA * F

= †i j k

0.4 0.5 -0.3400 -400 565.69

†

= {162.84 i - 346.27j - 360 k} N # m

= {163i - 346j - 360k} N # m Ans.

4–37.

Determine the moment of force F about point O. The force has a magnitude of 800 N and coordinate direction angles of a = 60°, b = 120°, g = 45°. Express the result as a Cartesian vector.

x

F

0.4 m

A

z

O y

0.5 m0.3 m

267


Ans:MC = {-35.4i - 128j - 222k} lb # ft

*4–40.

The curved rod has a radius of 5 ft. If a force of 60 lb acts atits end as shown, determine the moment of this force aboutpoint C.

SOLUTION

Position Vector and Force Vector:

Moment of Force About Point C: Applying Eq. 4–7, we have

Ans.= -35.4i - 128j - 222k lb # ft

=i j k0 4.330 -2.50

51.231 22.797 -21.346

MC = rCA * FAB

FAB

= 551.231i + 22.797j - 21.346k6 lb

FAB = 60¢16 - 02i + 17 - 5 sin 60°2j + 10 - 5 cos 60°2k

216 - 022 + 17 - 5 sin 60°22 + 10 - 5 cos 60°22≤ lb

= 54.330j - 2.50k6 m

rCA = 515 sin 60° - 02j + 15 cos 60° - 52k6 m

5 ft

5 ft

60°

z

x

y

6 ft60 lb

A

C

B

7 ft

269


Ans: MO = 4.27 N # ma = 95.2°b = 110°g = 20.6°

4–42.

A 20-N horizontal force is applied perpendicular to thehandle of the socket wrench. Determine the magnitude andthe coordinate direction angles of the moment created bythis force about point O.

SOLUTION

Ans.

Ans.

Ans.

Ans.g = cos -1 a 44.272

b = 20.6°

b = cos -1 a -1.4494.272

b = 110°

a = cos -1 a -0.38824.272

b = 95.2°

MO = 4.272 = 4.27 N # m= {-0.3882 i - 1.449 j + 4.00k} N # m

MO = rA * F = 3i j k

0.05176 0.1932 0.075-19.32 5.176 0

3

= -19.32 i + 5.176 j

F = -20 cos 15°i + 20 sin 15°j

= 0.05176 i + 0.1932 j + 0.075k

rA = 0.2 sin 15°i + 0.2 cos 15°j + 0.075k 15�

200 mm

75 mm

20 N

A

O

x

y

z

270


Ans:MA = {-5.39i + 13.1j + 11.4k} N # m

4–43.

SOLUTIONPosition Vector And Force Vector:

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

Ans.= {-5.39i + 13.1j + 11.4k} N # m

= 3i j k

0.55 0.4 -0.244.53 53.07 -40.0

3

MA = rAC * F

= (44.53i + 53.07j - 40.0k} N

F = 80(cos 30° sin 40°i + cos 30° cos 40°j - sin 30°k) N

= {0.55i + 0.4j - 0.2k} m

rAC = {(0.55 - 0)i + (0.4 - 0)j + (-0.2 - 0)k} m

The pipe assembly is subjected to the 80-N force. Determinethe moment of this force about point A.

400 mm

y300 mm

200 mm

250 mm200 mm

x

z

30

40

F 80 N

B

C

A

274


Ans:MA = {-16.0i - 32.1k} N # m

4–47.

SOLUTION

Ans.

Also,

Ans.MA = rC * F =i j k

-0.4 0 0.2-53.5 80.2 26.7

= -16.0 i - 32.1 k N # m

MA = rB * F = 3i j k0 -0.6 0

-53.5 80.2 26.7

3 = 5-16.0 i - 32.1 k6 N # m

F = 5-53.5 i + 80.2 j + 26.7 k6 N

F = 100 a-0.4 i + 0.6 j + 0.2 k

0.7483b

A force F having a magnitude of acts along thediagonal of the parallelepiped. Determine the moment of Fabout point A, using and MA = rC : F.MA = rB : F

F = 100 N

F

F

z

y

x

B

A

C

200 mm

400 mm

600 mm

rC

rB

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