978-0133915426 Chapter 3 Part 2

3–21.

Determine the unstretched length of spring AC if a force

causes the angle for equilibrium. Cord

AB is 2 ft long.Take k=50 lb>ft.

u=60°P=80 lb

SOLUTION

Solving for ,

Ans.l=212 –40

50 =2.66 ft40 =50(212 –l¿)

F

s=kx

F

s=40 lb

Fs

:

+©Fx=0; –Tcos 60° +Fscos 30° =0

+c©Fy=0; Tsin 60° +Fssin 30° –80 =0

f=sin–1

¢

2sin 60°

212

≤

=30°

212

sin 60° =2

sin f

l=212

l=242+22–2(2)(4) cos 60°

2 ft

k

2 ft

A

BC

u

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3–22.

SOLUTION

Ans.F=158 N

T=ks =500(232+(1.5)2–3) =177.05 N

+©F

The springs BA and BC each have a stiffness of 500 N

>

m and an

unstretched length of 3 m. Determine the horizontal force F

applied to the cord which is attached to the small ring B so

that the displacement of the ring from the wall is d = 1.5 m.

F

B

A

k 500 N/m

6 m

3–23.

The springs

BA and BC each have a stiffness of 500 N

>

m and an

unstretched length of 3 m. Determine the displacement

d of the

cord from the wall when a force F = 175 N is applied to the cord.

SOLUTION

By trial and error

:

Ans.d=1.56 m

da1–3

29+d2b=0.175

T=ks =500(232+d2–3)

TCd

232+d2S=87.5

:

+©F

x=0; 175 =2Tsin u

F

B

A

k 500 N/m

6 m

SOLUTION

x

*3–24.

Determine the distances x and y for equilibrium if F1 = 800 N

and F2 = 1000 N.

B

CDF1

F2

2 m

y

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

Ans:

y=2 m

F1=833 N

SOLUTION

y=2 m

F1=833.33 N =833 N

3–25.

Determine the magnitude of F1 and the distance y if x = 1.5 m

and F2 = 1000 N.

B

CDF1

F2

2 m

y

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

3–26.

SOLUTION

The 30-kg pipe is supported at A by a system of ﬁvecords.

Determine the force in each cord for equilibrium.

A

H

E

B

C

D

60°

3

4

5

187

3–27.

Each cord can sustain a maximum tension of 500 N.

Determine the largest mass of pipe that can be supported.

SOLUTION

At A:

At B:

By comparison, cord BC carries the largest load. Thus

Ans.m=261.69

9.81 =26.7 kg

W=261.69 N

500 =1.9107 W

FBC =1.9107 W

:

+©Fx=0; –FBC +1.667 Wa4

5b+1.1547 sin 30° =0

FBD =1.667 W

+c©Fy=0; FBD a3

5b–(1.1547 cos 30°)W=0

FAE =0.5774 W

:

+©Fx=0; FAE –(1.1547 W) cos 60° =0

FAB =1.1547 W

+c©Fy=0; FAB sin 60° –W=0

A

E

B

C

D

60°

3

4

5

*3–28.

The street-lights at A and B are suspended from the two

poles as shown. If each light has a weight of 50 lb, determine

the tension in each of the three supporting cables and the

required height h of the pole DE so that cable AB is

horizontal.

SOLUTION

D

A

h

B

C

3–29.

SOLUTION

Ans.

Using the result FCD = 339.83 N and applying the equations of equilibrium along the

xand yaxes to the free-body diagram of joint Dshown in Fig. b, we have

(1)

(2)

Solving Eqs. (1) and (2), yields

Ans.F

CA =243 NF

CB =275 N

F

CA a4

5b–F

CB sin 45° =0+c©F

y=0;

339.83 –F

CA a3

5b–F

CD cos 45° =0©F

x=0;

:

+

F

CD =339.83 N =340 N392.4 cos 30° –F

CD =0+c©F

y=0;

F

DE =392.4 N =392 NF

DE sin 30°–20(9.81) =0©F

x=0;

:

Determine the tension developed in each cord required for

equilibrium of the 20-kg lamp.

A

B

D

E

C

45°

30°

3

45

3–30.

SOLUTION

Using the result FCD = 16.99mand applying the equations of equilibrium along the

xand yaxes to the free-body diagram of joint Dshown in Fig. b, we have

(1)

(2)

Solving Eqs. (1) and (2), yields

Notice that cord DE is subjected to the greatest tensile force, and so it will achieve

the maximum allowable tensile force first. Thus

Ans.

m=20.4 kg

F

DE =400 =19.62m

F

CA =12.14mF

CB =13.73m

F

CA a4

5b–F

CB sin 45° =0+c©F

y=0;

16.99m–F

CA a3

5b–F

CD cos 45° =0©F

x=0;

:

+

F

CD =16.99m19.62m cos 30° –F

CD =0©F

x=0;

:

+

Determine the maximum mass of the lamp that the cord

system can support so that no single cord develops a tension

exceeding 400 N.

A

B

D

E

C

45°

30°

3

45

195

3–35.

Thus,

Solving by trial and error,

Ans.

Also,

Ans.

Ans.l1.40 ft

2

sin 66.75° =l

sin 40°

FAB =82.4 lb

u=66.75°

l=2.34 ft

2

sin 33.25° =l

sin 40°

FAB =175 lb

u=33.25°

sin u+0.8391 cos u=1.25

+c©Fy=0; 160 sin u+FAB sin 40° –200 =0

:

+©Fx=0; FAB cos 40° –160 cos u=0

The ring of negligible size is subjected to a vertical force of

200 lb. Determine the required length lof cord AC such that

the tension acting in AC is 160 lb. Also, what is the force in

cord AB?Hint: Use the equilibrium condition to determine

the required angle for attachment, then determine lusing

trigonometry applied to triangle ABC.

u

40°

θ

BC

A

l2ft

200 lb

SOLUTION

*3–36.

SOLUTION

Geometry: Referring to Fig. b, we can write

Also,

Ans.

Substituting into Eq. (1),yields

Ans.T=687 N

f=45.57°

x=1.38 m

x tan 45.57° +0.75 =(3.5 –x) tan 45.57°

f=cos–1 a3.5

5b=45.57°

3.5 –x

cos f+x

cos f=5

Cable ABC has a length of 5 m. Determine the position x

and the tension developed in ABC required for equilibrium

of the 100-kg sack.Neglect the size of the pulley at B.

A

B

C

x

3.5 m

0.75 m

3–37.

SOLUTION

Geometry

:The angle which the surface make with the horizontal is to be

determined first.

F

ree Body Diagram:The tension in the cord is the same throughout the cord and is

equal to the weight of block

B,.

Equations of Equilibrium

:

[1]

[2]

Solving Eqs

. [1] and [2] yields

Ans.mB=3.58 kg

N=19.7 N

8.4957mB+0.4472N=39.24

+c©F

y=0; mB(9.81) sin 60° +Ncos 63.43° –39.24 =0

N=5.4840mB

:

+©Fx=0; mB(9.81) cos 60° –Nsin 63.43° =0

W

B=mB(9.81)

u=63.43°

tan u

`

x=0.4 m

=dy

dx

`

x=0.4m

=5.0x

`

x=0.4 m

=2.00

u

A4–kg sphere rests on the smooth parabolic surface.

Determine the normal force it exerts on the surface and the

mass of block Bneeded to hold it in the equilibrium

position shown.

mB

B

A

y

x

0.4m

60fi

y2.5x

2

3–38.

Determine the forces in cables AC and AB needed to hold

the 20-kg ball Din equilibrium.Take F= 300 N and d= 1 m.

SOLUTION

(1)

(2)

Solving Eqs. (1) and (2) yields

Ans.F

AB =98.6 N F

AC =267 N

0.7809F

AB +0.4472F

AC =196.2

+c©F

y=0; F

AB a5

241 b+F

AC a1

25b–196.2 =0

06247F

AB +0.8944F

AC =300

:

+©F

x=0; 300 –F

AB a4

241 b–F

AC a2

25b=0

A

C

B

D

1.5 m

d

3–39.

The ball Dhas a mass of 20 kg.If a force of F= 100 N is

applied horizontally to the ring at A,determine the largest

dimension dso that the force in cable AC is zero.

SOLUTION

(1)

(2)

Solving Eqs. (1) and (2) yields

From the geometry,

Ans.d=2.42 m

d+1.5 =2 tan 62.99°

u=62.99° F

AB =220.21 N

+c©F

y=0; F

AB sin u–196.2 =0 F

AB sin u=196.2

:

+©F

x=0; 100 –F

AB cos u=0 F

AB cos u=100

A

C

B

D

1.5 m

d

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*3–40.

T

h

e 200-

lb

un

i

form tan

k

i

s suspen

d

e

d

b

y means of a

6

-ft-

long cable, which is attached to the sides of the tank and

passes over the small pulley located at O.If the cable can be

attached at either points Aand B,or Cand D,determine

which attachment produces the least amount of tension in

the cable. What is this tension?

SOLUTION

Free-Body Diagram: By observation, the force F has to support the entire weight

of the tank. Thus, F = 200 lb. The tension in cable AOB or COD is the same

throughout the cable.

Equations of Equilibrium:

(1)

F

rom the function obtained above, one realizes that in order to produce the least

amount of tension in the cable, hence must be as great as possible. Since the

attachment of the cable to point Cand Dproduces a greater

as compared to the attachment of the cable to points Aand ,

the attachment of the cable to point C and D will produce the least amount

of tension in the cable. Ans.

Thus,

Ans.T=100

sin 70.53° =106 lb

B

A

u=cos–12

3=48.19°

B

u

A

u=cos–11

3=70.53°

B

usin u

+c©F

y=0; 200 –2Tsin u=0T=100

sin u

:

A

O

C

1 ft

B

2 ft

F

D

2 ft