978-0133915426 Chapter 5 Part 2

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Ans:

Pmin =271 N

SOLUTION

Pmin =271.13 N =271 N

5–30.

Determine the magnitude and direction

u

of the minimum

force P needed to pull the 50-kg roller over the smooth step.

A

B

P

300 mm

50 mm

u

5–31.

The operation of the fuel pump for an automobile depends

on the reciprocating action of the rocker arm ABC, which

is pinned at Band is spring loaded at Aand D.When the

smooth cam Cis in the position shown, determine the

horizontal and vertical components of force at the pin and

the force along the spring DF for equilibrium.The vertical

force acting on the rocker arm at Ais , and at C

it is .FC=125 N

FA=60 N

30°

F

A

= 60 N

C

D

B

A

F

E

F

C

= 125 N

SOLUTION

*5–32.

Determine the magnitude of force at the pin and in the

cable needed to support the 500-lb load. Neglect the

weight of the boom .

SOLUTION

summing moments about point .

Ans.

Thus,Ans.F

A=Ax=2060.9 lb =2.06 kip

Ay=0

Ay+1820.7 sin 13° –500 cos 35° =0a+©F

y=0;

Ax=2060.9 lb

Ax–1820.7 cos 13° –500 sin 35° =0

+

Q©F

x=0;

F

BC =1820.7 lb =1.82 kip

F

BC sin 13°(8) –500 cos 35°(8) =0a+©MA=0;

A

AB

BC

A

35

22

8 ft

C

B

A

5–33.

SOLUTION

Ans.

Ans.Bx=25.4 kN

:

+©Fx=0; Bx–25.4 =0

By=22.8 kN

+c©Fy=0; By–800 (9.81) –15 000 =0

Ax=25.4 kN

The dimensions of a jib crane,which is manufactured by the

Basick Co., are given in the figure. If the crane has a mass of

800 kg and a center of mass at G,and the maximum rated

force at its end is F15 kN,determine the reactions at its

bearings. The bearing at Ais a journal bearing and supports

only a horizontal force,whereas the bearing at Bisathrust

bearing that supports both horizontal and vertical components.

F

G

A

3m

2m

0.75 m

421

5–34.

The dimensions of a jib crane,which is manufactured by the

Basick Co., are given in the figure. The crane has a mass of

800 kg and a center of mass at G.The bearing at Ais a journal

bearing and can support a horizontal force,whereas the

bearing at Bis a thrust bearing that supports both horizontal

andvertical components.Determine the maximum load Fthat

can be suspended from its end if the selected bearings at A

and Bcan sustain a maximum resultant load of 24 kN and

34 kN,respectively.

SOLUTION

Assume .

Solving,

Ans.

OKFB=(24)2+(21.9)2=32.5 kN 634 kN

F=14.0 kN

By=21.9 kN

Bx=24 kN

Ax=24 000 N

:

+©Fx=0; Bx–Ax=0

+c©Fy=0; By–800 (9.81) –F=0

F

G

A

3m

2m

B

0.75 m

422

Ans:

NA=173 N

NC=416 N

NB=69.2 N

NA=173.21 N =173 N

NB=69.22 N =69.2 N

5–37.

The cantilevered jib crane is used to support the load of

780 lb.If ,determine the reactions at the supports.

Note that the supports are collars that allow the crane to

rotate freely about the vertical axis.The collar at Bsupports a

force in the vertical direction, whereas the one at Adoes not.

x=5 ft

8 ft

4 ft

780 lb

x

T

B

SOLUTION

5–38.

T

h

e cant

il

evere

d

jib

crane

i

s use

d

to support t

h

e

l

oa

d

of

780 lb.If the trolley Tcan be placed anywhere between

determine the maximum magnitude of

reaction

at the supports Aand B.Note that the supports

are collars that allow the crane to rotate freely about the

vertical axis.The collar at Bsupports a force in the vertical

direction, whereas the one at Adoes not.

1.5 ft …x…7.5 ft,

SOLUTION

a

Ans.

=

1657.5 lb

=

1.66 ki

p

FB=2(1462.5)2+(780)2

By=780 lb

+c©Fy=0; By–780 =0

Ax=1462.5 =1462 lb

:

+©Fx=0; Ax–1462.5 =0

Bx=1462.5 lb

+©MA=0; –780(7.5) +Bx(4) =0

8ft

4ft

780 lb

x

T

B

A

SOLUTION

u

5–39.

The bar of negligible weight is supported by two springs,

each having a stiffness k

=

100 N

>

m. If the springs are

originally unstretched, and the force is vertical as shown,

determine the angle

u

the bar makes with the horizontal,

when the 30-N force is applied to the bar. 2 m

1 m

B

C

30 N

k

5–41.

The bulk head AD is subjected to both water and soil–

backfill pressures.Assuming AD is “pinned” to the

ground at A,determine the horizontal and vertical

reactions there and also the required tension in the

ground anchor BC necessary for equilibrium. The bulk

head has a mass of 800 kg.

6m

4m

310 kN/m118 kN/m

0.5 m

CF

A

B

D

SOLUTION

5–42.

SOLUTION

a

Ans.

Ans.Ay=681 N

+c©Fy=0; Ay–800 –350 +3

5(781.6) =0

Ax=625 N

:

+©Fx=0; Ax–4

5(781.6) =0

FCB =781.6 =782 N

+4

5FCB (2.5 sin 30°) +3

5FCB(2.5 cos 30°) =0

+©MA=0;

–800(1.5 cos 30°) –350(2.5 cos 30°)

Theboom supports the two vertical loads.Neglect the size

of

the collars at Dand Band the thickness of the boom,

and

compute the horizontal and vertical components of

force

at the pin Aand the force in cable CB.Set

and F

2=350 N.

F

1=800 N

1.5 m

30

3

C

B

F

1

F

2

D

A

4

5

1m

5–43.

The boom is intended to support two vertical loads,and

If the cable CB can sustain a maximum load of 1500 N before

it fails, determine the critical loads if Also, what is

the magnitude of the maximum reaction at pin A?

F

1=2F

2.

F

2.

F

1

SOLUTION

3

C

4

5

432

Ans:

P=660 N

NA=442 N

u=48.0°

b

NA=442.07 N =442 N

5–46.

Three uniform books, each having a weight Wand length a,

are stacked as shown. Determine the maximum distance d

that the top book can extend out from the bottom one so

the stack does not topple over.

SOLUTION

Ans.d=3a

4

ad

5–47.

Determine the reactions at the pin A and the tension in cord

BC. Set F = 40 kN. Neglect the thickness of the beam.

SOLUTION

3

C

A

F

26 kN

13 12

5

53

4

B

4 m

2 m

*5–48.

If rope BC will fail when the tension becomes 50 kN,

determine the greatest vertical load F that can be applied to

the beam at B. What is the magnitude of the reaction at A

for this loading? Neglect the thickness of thebeam.

SOLUTION

3

C

A

F

26 kN

13 12

5

53

4

B

4 m

2 m

5–49.

The rigid metal strip of negligible weight is used as part of an

electromagnetic switch. If the stiffness of the springs at A

and Bis and the strip is originally horizontal

when the springs are unstretched, determine the smallest

force needed to close the contact gap at C.

k=5N>m,

SOLUTION

Set , then

From Eq. (2),

Ans.F

C=

F

A=

ky

A=

(5)(0.002)

=

10 mN

yB=4mm

yA=2mm

x=16.67

yA

=100 –x

10

x

©Fy=0; FB=2F

©MB=0; FA=FC=F

50 mm 50 mm

10 mm

A

B

C

k