978-0133915426 Chapter 3 Part 4

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

3–55.

SOLUTION

Equations of Equilibrium: Equilibrium requires

Equating the i,j,and kcomponents yields

(1)

(2)

(3)

Assuming that cable AD achieves maximum tension first, substituting

into Eqs. (2) and (3),and solving Eqs. (1) through (3) yields

Ans.

Since the above assumption is correct.F

AB =70 lb 6250 lb and F

AC =105 lb,

W

C=265 lb

F

AB =70 lb

F

AC =105 lb

F

AD =250 lb

2

7 F

AB +3

7 F

AC +4

5 F

AD –W

C=0

–6

7 F

AB –6

7 F

AC +3

5 F

AD =0

–3

7 F

AB +2

7 F

AC =0

¢

–3

7 F

AB +2

7 F

AC

≤

i+

¢

–6

7 F

AB –6

7 F

AC +3

5 F

AD

≤

j+

¢

2

7 F

AB +3

7 F

AC +4

5 F

AD –W

C

≤

k=0

¢

–3

7 F

AB i–6

7 F

AB j+2

7 F

AB k

≤

+

¢

2

7 F

AC i–6

7 F

AC j+3

7 F

AC k

≤

+

¢

3

5 F

AD j+4

5 F

AD k

≤

+(–W

C k)=0

gF=0;

FAB +FAC +FAD +W=0

W=-W

C k

FAD =F

AD C(0 –0)i+(3 –0)j+(4 –0)k

2(0 –0)2+(3 –0)2+(4 –0)2S=3

5 F

AD j+4

5 F

AD k

FAC =F

AC C(2 –0)i+(–6–0)j+(3 –0)k

2(2 –0)2+(–6–0)2+(3 –0)2S=2

7 F

AC i –6

7 F

AC j+3

7 F

AC k

2(–3–0)2+(–6–0)2+(2 –0)2S=-

7 F

Determine the maximum weight of the crate that can be

suspended from cables AB,AC,and AD so that the tension

developed in any one of the cables does not exceed 250 lb.

y

A

B

C

D

x

z

6 ft

3 ft

2 ft 2 ft

3 ft

4 ft

*3–56.

The 25-kg flowerpot is supported at Aby the three cords.

Determine the force acting in each cord for equilibrium.

SOLUTION

Thus,

[1]

[2]

[3]

Solving Eqs. [1], [2], and [3] yields:

Ans.F

AD =F

A

C

=104 N F

AB =220 N

©Fz=0; 0.4330FAD +0.7071FAB +0.4330FAC –245.25 =0

©F

y=0; –0.75F

AD +0.7071F

AB –0.75F

AC =0

©F

x=0; 0.5F

AD –0.5F

AC =0

+(0.4330F

AD +0.7071F

AB +0.4330F

AC –245.25) k=0

(0.5F

AD –0.5F

AC)i+(–0.75F

AD +0.7071F

AB –0.75F

AC)j

+(–0.5F

ACi–0.75F

ACj+0.4330F

ACk) +(–245.25k)=0

(0.5F

AD i–0.75F

AD j) +0.4330F

AD k+(0.7071F

AB j+0.7071F

AB k)

©F=0;FAD +FAB +FAC +F=0

F=-25(9.81)k={–245.25k}N

FAB =FAB(sin 45°j+cos 45°k)=0.7071FAB j+0.7071FAB k

=-0.5FAC i–0.75FACj+0.4330FACk

FAC =FAC (–sin 30°i–cos 30° sin 60°j+cos 30° cos 60°k)

=0.5FADi–0.75FADj+0.4330FAD k

FAD =FAD (sin 30°i–cos 30° sin 60°j+cos 30° cos 60°k)

30fi

60fi

45fi

x

A

z

B

y

D

C

3–57.

SOLUTION

(1)

(2)

Assume

then

Assume

.Then

T

hus,

Ans.W

=

2.366(23.57)

=

55.767

=

55.8 N

F

AC =0.7071(50)

1.5 =23.57 N 650 N(O. K!)

F

AB =50 N

F

AB =1.5(50)

0.7071 =106.07 N 750 N(N.G!)

F

AC =50 N

2.366F

AC =W

0.8660F

AC +1.5F

AC –W=0

©F

z=0; 0.4330F

AD +0.7071F

AB +0.4330F

AC –W=0

0.7071F

AB =1.5F

AC

©F

y=0; –0.75F

AD +0.7071F

AB –0.75F

AC =0

F

AD =F

AC

©F

x=0; 0.5F

AD –0.5F

AC =0

W

=-Wk

F

AB =FAB (sin 45° j+cos 45° k)=0.7071FAB j+0.7071FAB k

=-0.5FAC i–0.75FAC j+0.4330FAC k

FAC =FAC (–sin 30°i–cos 30° sin 60° j+cos 30° cos 60° k)

=0.5FAD i–0.75FAD j+0.4330FAD k

F

AD =FAD (sin 30°i–cos 30° sin 60°j+cos 30° cos 60°k)

If each cord can sustain a maximum tension of 50 N before

it

fails, determine the greatest weight of the flowerpot the

cords can support.

30fi

60fi

45fi

x

A

z

B

y

D

C

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

3–58.

Determine the tension developed in the three cables

required to support the traffic light, which has a mass of

15 kg. Take h = 4 m.

SOLUTION

D

A

hB

C

4 m

6 m

3 m

z

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

3–59.

Determine the tension developed in the three cables

required to support the traffic light, which has a mass of

20 kg. Take h = 3.5 m.

SOLUTION

D

A

hB

C

4 m

6 m

3 m

4 m

z

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

*3–60.

SOLUTION

T

h

e 800-

lb

cy

li

n

d

er

i

ssupporte

db

yt

h

reec

h

a

i

ns as s

h

own.

Determinethe forcein eachchain for equilibrium. Take

d=1ft.

90

135

1ft

D

B

C

z

y

*3–64.

SOLUTION

Equations of Equilibrium: Equilibrium requires

Equating the i,j,and kcomponents yields

(1)

(2)

(3)

Solving Eqs. (1) through (3) yields

Ans.

Ans.F

AE =2354 lb =2.35 kip

F

AC =538 lb

F

AB =808 lb

–6

7 F

AB –6

7 F

AC +F

AE –1200 =0

–3

7 F

AB –2

7 F

AC +500 =0

2

7 F

AB –3

7 F

AC =0

¢

2

7 F

AB –3

7 F

AC

≤

i+

¢

–3

7 F

AB –2

7 F

AC +500

≤

j+

¢

–6

7 F

AB –6

7 F

AC +F

AE –1200

≤

k=0

¢

2

7 F

AB i–3

7 F

AB j–6

7 F

AB k

≤

+

¢

–3

7 F

AC i–2

7 F

AC j–6

7 F

AC k

≤

+(500j–1200k)+F

AE k=0

gF=0;

FAB +FAC +FAD +FAE =0

FAE =F

AE k

FAD =F

AD C(0 –0)i+(12.5 –0)j+(–30 –0)k

2(0 –0)2+(12.5–0)2+(–30 –0)2S={500j–1200k} lb

FAC =F

AC C(–15 –0)i+(–10 –0)j+(–30 –0)k

2(–15 –0)2+(–10 –0)2+(–30 –0)2S=-

3

7 F

AC i –2

7 F

AC j–6

7 F

AC k

2(10 –0)2+(–15 –0)2+(–30 –0)2S=2

7 F

If cable AD is tightened by a turnbuckle and develops a

tension of 1300 lb,determine the tension developed in

cables AB and AC and the force developed along the

antenna tower AE at point A.

15 ft 15 ft

10 ft

z

BE

D

C

A

30 ft

3–65.

SOLUTION

If the tension developed in either cable AB or AC cannot

exceed 1000 lb,determine the maximum tension that can

be developed in cable AD when it is tightened by the

turnbuckle.Also, what is the force developed along the

antenna tower at point A?

10 ft

z

BE

C

A

30 ft

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

3–66.

Determine the tension developed in cables ,,and

required for equilibrium of the 300-lb crate.

ADACAB

A

C

1 ft

3 ft

2 ft 1 ft 2 ft

2 ft

y

z

2 ft

B

SOLUTION

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

3–67.

SOLUTION

F

orce Vectors: We can express each of the forces on the free-body diagram shown in

F

ig.(a) in Cartesian vector form as

Determine the maximum weight of the crate so that the

tension developed in any cable does not exceed 450 lb

.

A

C

1ft

3ft

2ft1ft2ft

2ft

y

z

2ft

B

FAD =FAD i

FAC =FAC C(–2–0)i+(–2–0)j+(1 –0)k

2(–2–0)2+(–2–0)2+(1 –0)2S=- 2

3FAC i–2

3FAC j+1

3FAC k

FAB =FAB C(–2–0)i+(1 –0)j+(2 –0)k

2(–2–0)2+(1 –0)2+(2 –0)2S=- 2

3FAB i+1

3FAB j+2

3FAB k