Engineering Mechanics - Statics Episode 1 Part 10 potx

Engineering Mechanics - Statics Chapter 5
Solution:
When tipping occurs, R
c
= 0
Σ
M
D
= 0;
W
2
− fW
1
c+ W
B
bc+()+ 0=
W
B
W
2
fW
1
c−
bc+
=
W
B
78.6 lb=
Problem 5-28
The articulated crane boom has a weight W and mass center at G. If it supports a load L,
determine the force acting at the pin A and the compression in the hydraulic cylinder BC when

the boom is in the position shown.
Units Used:
kip 10
3
lb=
Given:
W 125 lb=
361
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
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Engineering Mechanics - Statics Chapter 5
L 600 lb=
a 4ft=
b 1ft=
c 1ft=
d 8ft=
θ
40 deg=
Solution:
Guesses A
x
1lb= A
y
1lb= F
B
1lb=
Given A
x
− F

B
cos
θ
()
+ 0= A
y
− F
B
sin
θ
()
+ W− L− 0=
F
B
cos
θ
()
bF
B
sin
θ
()
c+ Wa− Ld c+()− 0=
A
x
A
y
F
B
⎛

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
Find A
x
A
y
, F
B
,
()
= F
B
4.19 kip=
A
x
A
y
⎛
⎜
⎝
⎞
⎟
⎠

3.208
1.967
⎛
⎜
⎝
⎞
⎟
⎠
kip=
Problem 5-29
The device is used to hold an elevator
door open. If the spring has stiffness k
and it is compressed a distnace
δ
,
determine the horizontal and vertical
components of reaction at the pin A and
the resultant force at the wheel bearing B.
Given:
k 40
N
m
= b 125 mm=
δ
0.2 m= c 100 mm=
a 150 mm=
θ
30 deg=
362
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
Solution:
F
s
k
δ
=
Σ
M
A
= 0;
F
s
− aF
B
cos
θ
()
ab+()+ F
B
sin
θ
()
c− 0=
F
B
F
s

a
cos
θ
()
ab+( ) sin
θ
()
c−
=
F
B
6.378 N=
+
→
Σ
F
x
= 0;
A
x
F
B
sin
θ
()
− 0=
A
x
F
B

sin
θ
()
=
A
x
3.189 N=
+
↑
Σ
F
y
= 0;
A
y
F
s
− F
B
cos
θ
()
+ 0=
A
y
F
s
F
B
cos

θ
()
−=
A
y
2.477 N=
Problem 5-30
Determine the reactions on the bent rod which is supported by a smooth surface at B and by a
collar at A, which is fixed to the rod and is free to slide over the fixed inclined rod.
Given:
F 100 lb=
M 200 lb ft=
a 3ft=
b 3ft=
c 2ft=
363
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
d 3=
e 4=
f 12=
g 5=
Solution:
Initial Guesses:
N
A
20 lb= N
B

10 lb= M
A
30 lb ft=
Given
Σ
M
A
=
0;
M
A
Fa− M− N
B
f
f
2
g
2
+
⎛
⎜
⎝
⎞
⎟
⎠
ab+()+ N
B
g
f
2

g
2
+
⎛
⎜
⎝
⎞
⎟
⎠
c− 0=
Σ
F
x
= 0;
N
A
e
e
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
N
B

g
f
2
g
2
+
⎛
⎜
⎝
⎞
⎟
⎠
− 0=
Σ
F
y
= 0;
N
A
d
e
2
d
2
+
⎛
⎜
⎝
⎞
⎟

⎠
N
B
f
f
2
g
2
+
⎛
⎜
⎝
⎞
⎟
⎠
+ F− 0=
N
A
N
B
M
A
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟

⎟
⎠
Find N
A
N
B
, M
A
,
()
=
N
A
N
B
⎛
⎜
⎝
⎞
⎟
⎠
39.7
82.5
⎛
⎜
⎝
⎞
⎟
⎠
lb= M

A
106lb ft⋅=
Problem 5-31
The cantilevered jib crane is used to support the load F. If the trolley T can be placed anywhere in
the range
x
1
x≤ x
2
≤
, determine the maximum magnitude of reaction at the supports A and B.
Note that the supports are collars that allow the crane to rotate freely about the vertical axis. The
collar at B supports a force in the vertical direction, whereas the one at A does not.
Units Used:
kip 1000 lb=
Given:
F 780 lb=
364
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
a 4ft=
b 8ft=
x
1
1.5 ft=
x
2
7.5 ft=

Solution:
The maximum occurs when x = x
2
Σ
M
A
= 0;
F− x
2
B
x
a+ 0=
B
x
F
x
2
a
=
B
x
1.462 10
3
× lb=
+
→
Σ
F
x
= 0;

A
x
B
x
− 0= A
x
B
x
= A
x
1.462 10
3
× lb=
+
↑
Σ
F
y
= 0;
B
y
F− 0= B
y
F= B
y
780lb=
F
B
B
x

2
B
y
2
+= F
B
1.657 kip=
Problem 5-32
The uniform rod AB has weight W. Determine the force in the cable when the rod is in the
position shown.
Given:
W 15 lb=
L 5ft=
365
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
θ
1
30 deg=
θ
2
10 deg=
Solution:
Σ
M
A
=
0;

N
B
Lsin
θ
1
θ
2
+
()
W
L
2
⎛
⎜
⎝
⎞
⎟
⎠
cos
θ
1
θ
2
+
()
− 0=
N
B
W cos
θ

1
θ
2
+
()
2 sin
θ
1
θ
2
+
()
=
N
B
8.938 lb=
Σ
F
x
= 0;
Tcos
θ
2
()
N
B
−
T
N
B

cos
θ
2
()
=
T 9.08 lb=
Problem 5-33
The power pole supports the three lines, each line exerting a vertical force on the pole due to its
weight as shown. Determine the reactions at the fixed support D. If it is possible for wind or ice
to snap the lines, determine which line(s) when removed create(s) a condition for the greatest
moment reaction at D.
Units Used:
kip 10
3
lb=
366
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
Given:
W
1
800 lb=
W
2
450 lb=
W
3
400 lb=

a 2ft=
b 4ft=
c 3ft=
Solution:
+
→
Σ
F
x
= 0;
D
x
0=
+
↑
Σ
F
y
= 0;
D
y
W
1
W
2
+ W
3
+
()
− 0=

D
y
W
1
W
2
+ W
3
+= D
y
1.65 kip=
Σ
M
D
= 0;
W
2
− bW
3
c− W
1
a+ M
D
+ 0=
M
D
W
2
bW
3

c+ W
1
a−= M
D
1.4 kip ft⋅=
Examine all cases. For these numbers we require line 1 to snap.
M
Dmax
W
2
bW
3
c+= M
Dmax
3kip ft⋅=
Problem 5-34
The picnic table has a weight W
T
and a center of gravity at G
T
. If a man weighing W
M
has a
center of gravity at G
M
and sits down in the centered position shown, determine the vertical
reaction at each of the two legs at B.Neglect the thickness of the legs. What can you conclude
from the results?
Given:
W

T
50 lb=
367
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
W
M
225 lb=
a 6in=
b 20 in=
c 20 in=
Solution:
Σ
M
A
=
0;
2 N
B
bc+()W
M
a+ W
T
b− 0=
N
B
W
T

bW
M
a−
2 bc+()
=
N
B
4.37− lb=
Since N
B
has a negative sign, the table will tip over.
Problem 5-35
If the wheelbarrow and its contents have a mass of M and center of mass at G, determine the
magnitude of the resultant force which the man must exert on each of the two handles in order to
hold the wheelbarrow in equilibrium.
Given:
M 60 kg=
a 0.6 m=
b 0.5 m=
c 0.9 m=
d 0.5 m=
g 9.81
m
s
2
=
368
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics Chapter 5
Solution:
Σ
M
B
= 0;
A
y
− bc+()Mgc+ 0=
A
y
Mgc
bc+
=
A
y
378.386 N=
+
→
Σ
F
x
= 0;
B
x
0N= B
x
0=
+
↑

A
y
Mg− 2 B
y
+ 0=
Σ
F
y
= 0;
B
y
Mg A
y
−
2
= B
y
105.107 N=
Problem 5-36
The man has weight W and stands at the center of the plank. If the planes at A and B are
smooth, determine the tension in the cord in terms of W and
θ
.
Solution:
Σ
M
B
=
0;
W

L
2
cos
φ
()
N
A
Lcos
φ
()
− 0=
N
A
W
2
=
Σ
F
x
= 0;
Tcos
θ
()
N
B
sin
θ
()
− 0=
(1)

369
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
Σ
F
y
= 0;
Tsin
θ
(
)
N
B
cos
θ
()
+ N
A
+ W− 0=
(2)
Solving Eqs. (1) and (2) yields:
T
W
2
sin
θ
()
=

N
B
W
2
cos
θ
()
=
Problem 5-37
When no force is applied to the brake pedal of the lightweight truck, the retainer spring AB
keeps the pedal in contact with the smooth brake light switch at C. If the force on the switch is
F
, determine the unstretched length of the spring if the stiffness of the spring is k.
Given:
F 3N=
k 80
N
m
=
a 100 mm=
b 50 mm=
c 40 mm=
d 10 mm=
θ
30 deg=
Solution:
Σ
M
D
=

0;
F
s
bFcos
θ
()
c− F sin
θ
()
d− 0=
F
s
F
cos
θ
()
c sin
θ
()
d+
b
= F
s
2.378 N=
F
s
kx= x
F
s
k

= x 29.73mm=
L
0
ax−= L
0
70.3 mm=
370
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
Problem 5-38
The telephone pole of negligible thickness is subjected to the force
F
directed as shown. It is
supported by the cable BCD and can be assumed pinned at its base A. In order to provide
clearance for a sidewalk right of way, where D is located, the strut CE is attached at C, as
shown by the dashed lines (cable segment CD is removed). If the tension in CD' is to be twice
the tension in BCD, determine the height h for placement of the strut CE.
Given:
F 80 lb=
θ
30 deg=
a 30 ft=
b 10 ft=
Solution:
+
Σ
M
A

= 0;
F− cos
θ
()
a
b
a
2
b
2
+
⎛
⎜
⎝
⎞
⎟
⎠
T
BCD
a+ 0=
T
BCD
F cos
θ
()
a
2
b
2
+

b
= T
BCD
219.089 lb=
Require
T
CD'
2 T
BCD
= T
CD'
438.178 lb=
+
Σ
M
A
= 0;
T
CD'
dFcos
θ
()
a− 0=
dFa
cos
θ
()
T
CD'
⎛

⎜
⎝
⎞
⎟
⎠
= d 4.7434ft=
Geometry:
ah−
d
a
b
= haa
d
b
⎛
⎜
⎝
⎞
⎟
⎠
−= h 15.8ft=
371
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
Problem 5-39
The worker uses the hand truck to move material down the ramp. If the truck and its contents
are held in the position shown and have weight W with center of gravity at G, determine the
resultant normal force of both wheels on the ground A and the magnitude of the force required

at the grip B.
Given:
W 100 lb= e 1.5 ft=
a 1ft= f 0.5 ft=
b 1.5 ft=
θ
60 deg=
c 2ft=
φ
30 deg=
d 1.75 ft=
Solution:
Σ
M
B
= 0;
N
A
cos
θφ
−
()
bc+ d+()N
A
sin
θφ
−
()
af−()W cos
θ

()
bc+()−+ W sin
θ
()
ea+()− 0=
N
A
W cos
θ
()
bc+()W sin
θ
()
ea+()+
cos
θφ
−
()
bc+ d+( ) sin
θφ
−
()
af−()+
= N
A
81.621 lb=
Σ
F
x

= 0;
B
x
− N
A
sin
φ
()
+ 0= B
x
N
A
sin
φ
()
= B
x
40.811 lb=
Σ
F
y

= 0;
B
y
N
A
cos
φ
()

W− 0=
()
+ B
y
WN
A
cos
φ
()
−= B
y
29.314 lb=
F
B
B
x
2
B
y
2
+= F
B
50.2 lb=
372
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
Σ
F

y
= 0;
Tsin
θ
(
)
N
B
cos
θ
()
+ N
A
+ W− 0=
(2)
Solving Eqs. (1) and (2) yields:
T
W
2
sin
θ
()
=
N
B
W
2
cos
θ
()

=
Problem 5-37
When no force is applied to the brake pedal of the lightweight truck, the retainer spring AB
keeps the pedal in contact with the smooth brake light switch at C. If the force on the switch is
F
, determine the unstretched length of the spring if the stiffness of the spring is k.
Given:
F 3N=
k 80
N
m
=
a 100 mm=
b 50 mm=
c 40 mm=
d 10 mm=
θ
30 deg=
Solution:
Σ
M
D
=
0;
F
s
bFcos
θ
()
c− F sin

θ
()
d− 0=
F
s
F
cos
θ
()
c sin
θ
()
d+
b
= F
s
2.378 N=
F
s
kx= x
F
s
k
= x 29.73mm=
L
0
ax−= L
0
70.3 mm=
373

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
The shelf supports the electric motor which has mass m
1
and mass center at G
m
. The platform
upon which it rests has mass m
2
and mass center at G
p
.

Assuming that a single bolt B holds the
shelf up and the bracket bears against the smooth wall at A, determine this normal force at A and
the horizontal and vertical components of reaction of the bolt B on the bracket.
Given:
m
1
15 kg= c 50 mm=
m
2
4kg= d 200 mm=
a 60 mm= e 150 mm=
b 40 mm=
g 9.81
m
s

2
=
Solution:
Σ
M
A
= 0;
B
x
am
2
gd− m
1
gd e+()− 0=
B
x
g
m
2
dm
1
de+()+
a
= B
x
989 N=
+
→
Σ
F

x
= 0;
A
x
B
x
− 0=
A
x
B
x
= A
x
989 N=
+
↑
Σ
F
y
= 0;
B
y
m
2
g− m
1
g− 0=
B
y
m

2
gm
1
g+= B
y
186 N=
Problem 5-42
A cantilever beam, having an extended length L, is subjected to a vertical force
F
. Assuming that
the wall resists this load with linearly varying distributed loads over the length a of the beam
portion inside the wall, determine the intensities w
1
and w
2
for equilibrium.
374
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
Units Used:
kN 10
3
N=
Given:
F 500 N=
a 0.15 m=
L 3m=
Solution:

The initial guesses
w
1
1
kN
m
= w
2
1
kN
m
=
Given
+
↑
Σ
F
y
= 0;
1
2
w
1
a
1
2
w
2
a− F− 0=
Σ

M
A
= 0;
F− L
1
2
w
1
a
a
3
⎛
⎜
⎝
⎞
⎟
⎠
−
1
2
w
2
a
2 a
3
⎛
⎜
⎝
⎞
⎟

⎠
+ 0=
w
1
w
2
⎛
⎜
⎝
⎞
⎟
⎠
Find w
1
w
2
,
()
=
w
1
w
2
⎛
⎜
⎝
⎞
⎟
⎠
413

407
⎛
⎜
⎝
⎞
⎟
⎠
kN
m
=
Problem 5-43
The upper portion of the crane boom consists of the jib AB, which is supported by the pin at A,
the guy line BC, and the backstay CD, each cable being separately attached to the mast at C. If
the load F is supported by the hoist line, which passes over the pulley at B, determine the
magnitude of the resultant force the pin exerts on the jib at A for equilibrium, the tension in the
guy line BC, and the tension T in the hoist line. Neglect the weight of the jib. The pulley at B has
a radius of r.
375
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
Units Used:
kN 10
3
N=
Given:
F 5kN=
r 0.1 m=
ar=

b 1.5 m=
c 5m=
Solution:
From pulley, tension in the hoist line is
Σ
M
B
= 0
;
Ta() Fr()− 0=
TF
r
a
=
T 5kN=
From the jib,
Σ
M
A
= 0
;
F− c() T
BC
ba+
c
2
ba+()
2
+
c+ 0=

T
BC
F
c
2
ba+()
2
+
ba+
= T
BC
16.406 kN=
+
↑
Σ
F
y
= 0;
A
y
− T
BC
ba+
c
2
ba+()
2
+
⎡
⎢

⎣
⎤
⎥
⎦
+ F− 0=
A
y
T
BC
ba+
c
2
ba+()
2
+
⎡
⎢
⎣
⎤
⎥
⎦
F−= A
y
0kN=
+
→
Σ
F
x
= 0;

A
x
T
BC
c
c
2
ba+()
2
+
⎡
⎢
⎣
⎤
⎥
⎦
− F− 0=
376
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Engineering Mechanics - Statics Chapter 5
A
x
T
BC
c
c
2
ba+()

2
+
F+= A
x
20.6 kN=
F
A
A
x
2
A
y
2
+= F
A
20.6 kN=
Problem 5-44
The mobile crane has weight W
1
and center of gravity at G
1
; the boom has weight W
2
and center
of gravity at G
2
. Determine the smallest angle of tilt
θ
of the boom, without causing the crane to
overturn if the suspended load has weight W. Neglect the thickness of the tracks at A and B.

Given:
W
1
120000 lb=
W
2
30000 lb=
W 40000 lb=
a 4ft=
b 6ft=
c 3ft=
d 12 ft=
e 15 ft=
Solution:
When tipping occurs, R
A
= 0
Σ
M
B
= 0;
W
2
− dcos
θ
()
c−
()
Wd e+( )cos
θ

()
c−
⎡
⎣
⎤
⎦
− W
1
bc+()+ 0=
θ
acos
W
2
cWc+ W
1
bc+()+
W
2
dWde+()+
⎡
⎢
⎣
⎤
⎥
⎦
=
θ
26.4 deg=
377
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
Problem 5-45
The mobile crane has weight W
1
and center of gravity at G
1
; the boom has weight W
2
and center
of gravity at G
2
. If the suspended load has weight W determine the normal reactions at the tracks
A and B. For the calculation, neglect the thickness of the tracks.
Units Used:
kip 10
3
lb=
Given:
W
1
120000 lb= a 4ft=
W
2
30000 lb= b 6ft=
W 16000 lb= c 3ft=
θ
30 deg= d 12 ft=
e 15 ft=

Solution:
Σ
M
B
= 0
;
W
2
− dcos
θ
()
c−
()
Wd e+( )cos
θ
()
c−
⎡
⎣
⎤
⎦
− R
A
ab+ c+()− W
1
bc+()+ 0=
R
A
W
2

− d cos
θ
()
c−
()
Wd e+( )cos
θ
()
c−
⎡
⎣
⎤
⎦
− W
1
bc+()+
ab+ c+
= R
A
40.9 kip=
+
↑
Σ
F
y
= 0;
R
A
R
B

+ W
1
− W
2
− W− 0=
R
B
R
A
− W
1
+ W
2
+ W+= R
B
125kip=
Problem 5-46
The man attempts to support the load of boards having a weight W and a center of gravity at G.
If he is standing on a smooth floor, determine the smallest angle
θ
at which he can hold them up
in the position shown. Neglect his weight
.
378
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
Given:
a 0.5 ft=

b 3ft=
c 4ft=
d 4ft=
Solution:
Σ
M
B

=
0;
N
A
− ab+()Wb ccos
θ
()
−
()
+ 0=
As
θ
becomes smaller, N
A
goes to 0 so that,
cos
θ
()
b
c
=
θ

acos
b
c
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
41.4 deg=
Problem 5-47
The motor has a weight W. Determine the force that each of the chains exerts on the
supporting hooks at A, B, and C. Neglect the size of the hooks and the thickness of the beam.
379
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
Given:
W 850 lb=
a 0.5 ft=
b 1ft=
c 1.5 ft=
θ
1
10 deg=
θ
2

30 deg=
θ
3
10 deg=
Solution:
Guesses
F
A
1lb= F
B
1lb= F
C
1lb=
Given
Σ
M
B
=
0;
F
A
cos
θ
3
()
bWa+ F
C
cos
θ
1

()
ac+()− 0=
Σ
F
x
= 0;
F
C
sin
θ
1
()
F
B
sin
θ
2
()
− F
A
sin
θ
3
()
− 0=
Σ
F
y
= 0;
WF

A
cos
θ
3
()
− F
B
cos
θ
2
()
− F
C
cos
θ
1
()
− 0=
F
A
F
B
F
C
⎛
⎜
⎜
⎜
⎝
⎞

⎟
⎟
⎟
⎠
Find F
A
F
B
, F
C
,
()
=
F
A
F
B
F
C
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
432

0−
432
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
380
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
The boom supports the two vertical loads. Neglect the size of the collars at D and B and the
thickness of the boom, and compute the horizontal and vertical components of force at the pin
A and the force in cable CB.
Given:
F
1
800 N=
F
2
350 N=
a 1.5 m=
b 1m=
c 3=
d 4=

θ
30 deg=
Solution:
Σ
M
A
= 0;
F
1
− acos
θ
()
F
2
ab+( ) cos
θ
()
−
d
c
2
d
2
+
F
CB
ab+( ) sin
θ
()
+

c
c
2
d
2
+
F
CB
ab+( ) cos
θ
()
+ 0=
F
CB
F
1
aF
2
ab+()+
⎡
⎣
⎤
⎦
cos
θ
()
c
2
d
2

+
dsin
θ
()
ab+()ccos
θ
()
ab+()+
= F
CB
782 N=
+
→
+
→
ΣΣ
F
x
= 0;
A
x
d
c
2
d
2
+
F
CB
− 0=

A
x
d
c
2
d
2
+
F
CB
= A
x
625 N=
+
↑
Σ
F
y
= 0;
A
y
F
1
− F
2
−
c
c
2
d

2
+
F
CB
+ 0=
A
y
F
1
F
2
+
c
c
2
d
2
+
F
CB
−= A
y
681 N=
Problem 5-49
The boom is intended to support two vertical loads
F
1
and
F
2

. If the cable CB can sustain a
381
Problem 5-48
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
maximum load T
max
before it fails, determine the critical loads if F
1
= 2F
2
. Also, what is the
magnitude of the maximum reaction at pin A?
Units Used:
kN 10
3
N=
Given:
T
max
1500 N=
a 1.5 m=
b 1m=
c 3=
d 4=
θ
30deg=
Solution:

Σ
M
A
= 0
;
F
1
2 F
2
=
2− F
2
acos
θ
()
F
2
ab+( ) cos
θ
()
−
d
c
2
d
2
+
T
max
ab+( ) sin

θ
()
+
c
c
2
d
2
+
T
max
ab+( ) cos
θ
()
+
0
=
F
2
ab+()T
max
dsin
θ
()
ccos
θ
()
+
()
c

2
d
2
+ cos
θ
()
3 ab+()
= F
2
724N=
F
1
2 F
2
= F
1
1.448 kN=
+
→
Σ
F
x
= 0;
A
x
d
c
2
d
2

+
T
max
− 0=
A
x
d
c
2
d
2
+
T
max
= A
x
1.20 kN=
+
↑
Σ
F
y
= 0;
A
y
F
2
− F
1
−

c
c
2
d
2
+
T
max
+ 0=
382
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
A
y
F
2
F
1
+
c
c
2
d
2
+
T
max
−= A

y
1.27 kN=
F
A
A
x
2
A
y
2
+= F
A
1.749 kN=
Problem 5-50
The uniform rod of length L and weight W is supported on the smooth planes. Determine its
position
θ
for equilibrium. Neglect the thickness of the rod.
Solution:
Σ
M
B
=
0;
W−
L
2
cos
θ
()

N
A
cos
φθ
−
()
L+ 0= N
A
W cos
θ
()
2 cos
φθ
−
()
=
Σ
M
A
=
0;
W
L
2
cos
θ
()
N
B
cos

ψθ
+
()
L− 0= N
B
W cos
θ
()
2 cos
ψθ
+
()
=
Σ
F
x
= 0;
N
B
sin
ψ
()
N
A
sin
φ
()
− 0=
W cos
θ

()
2 cos
ψθ
+
()
sin
ψ
()
W cos
θ
()
2 cos
φθ
−
()
sin
φ
()
− 0=
sin
ψ
()
cos
φθ
−
()
sin
φ
()
cos

ψθ
+
()
− 0=
sin
ψ
()
cos
φ
()
cos
θ
()
sin
φ
()
sin
θ
()
+
()
sin
φ
()
cos
ψ
()
cos
θ
()

sin
ψ
()
sin
θ
()
−
()
− 0=
2 sin
ψ
()
sin
φ
()
sin
θ
()
sin
φ
()
cos
ψ
()
sin
ψ
()
cos
φ
()

−
()
cos
θ
()
=
tan
θ
()
sin
φ
()
cos
ψ
()
sin
ψ
()
cos
φ
()
−
2 sin
ψ
()
sin
φ
()
=
cot

ψ
()
cot
φ
()
−
2
=
θ
atan
cot
ψ
()
cot
φ
()
−
2
⎛
⎜
⎝
⎞
⎟
⎠
=
383
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5

Problem 5-51
The toggle switch consists of a cocking lever that is pinned to a fixed frame at A and held in
place by the spring which has unstretched length
δ
. Determine the magnitude of the resultant
force at A and the normal force on the peg at B when the lever is in the position shown.
Given:
δ
200 mm=
k 5
N
m
=
a 100 mm=
b 300 mm=
c 300 mm=
θ
30 deg=
Solution:
Using the law of cosines and the law of sines
lc
2
ab+()
2
+ 2 ca b+( ) cos 180 deg
θ
−
()
−=
sin

φ
()
c
sin 180 deg
θ
−
()
l
=
φ
asin c
sin 180 deg
θ
−
()
l
⎛
⎜
⎝
⎞
⎟
⎠
=
φ
12.808 deg=
F
s
ks= kl
δ
−

()
= F
s
kl
δ
−
()
= F
s
2.3832 N=
Σ
M
A
= 0;
F
s
− sin
φ
()
ab+()N
B
a+ 0= N
B
F
s
sin
φ
()
ab+
a

= N
B
2.11 N=
Σ
F
x
= 0;
A
x
F
s
cos
φ
()
− 0= A
x
F
s
cos
φ
()
= A
x
2.3239 N=
Σ
F
y
= 0;
A
y

N
B
+ F
s
sin
φ
()
− 0= A
y
F
s
sin
φ
()
N
B
−= A
y
1.5850− N=
F
A
A
x
2
A
y
2
+= F
A
2.813 N=

384
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
Problem 5-52
The rigid beam of negligible weight is supported horizontally by two springs and a pin. If the
springs are uncompressed when the load is removed, determine the force in each spring when the
load
P
is applied. Also, compute the vertical deflection of end C. Assume the spring stiffness k is
large enough so that only small deflections occur. Hint: The beam rotates about A so the
deflections in the springs can be related.
Solution:
Σ
M
A
= 0;
F
B
LF
C
2 L+ P
3
2
L− 0=
F
B
2 F
C

+ 1.5 P=
Δ
C
2 Δ
B
=
F
C
k
2 F
B
k
=
F
C
2F
B
=
5 F
B
1.5 P=
F
B
0.3 P=
F
c
0.6 P=
Δ
C
0.6 P

k
=
Problem 5-53
The rod supports a weight W and is pinned at its end A. If it is also subjected to a couple
moment of M, determine the angle
θ
for equilibrium.The spring has an unstretched length
δ
and
a stiffness k.
385
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be reproduced, in any form or by any means, without permission in writing from the publisher.