Engineering Mechanics - Statics Chapter 5
Given:
M 5Mg= d 1.50 m=
a 0.3 m= e 0.6 m=
b 0.75 m=
θ
1
20 deg=
c 1m=
θ
2
30 deg=
g 9.81
m
s
2
=
Solution:
Guesses F 1kN= N
A
1kN= N
B
1kN=
Given
N
A
N
B
+ F sin
θ
2

()
+ Mgcos
θ
1
()
− 0=
F− cos
θ
2
()
Mgsin
θ
1
()
+ 0=
F cos
θ
2
()
aFsin
θ
2
()
b− Mgcos
θ
1
()
c− Mgsin
θ
1

()
e− N
B
cd+()+ 0=
F
N
A
N
B
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
Find FN
A
, N
B
,
()
=
F
N
A
N

B
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
19.37
13.05
23.36
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
Problem 5-17
The uniform bar has mass M and
center of mass at G. The supports
A, B, and C are smooth. Determine
the reactions at the points of contact
at A, B, and C.
Given:

M 100 kg=
a 1.75 m=
b 1.25 m=
351
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Engineering Mechanics - Statics Chapter 5
c 0.5 m=
d 0.2 m=
θ
30 deg=
g 9.81
m
s
2
=
Solution:
The initial guesses:
N
A
20 N=
N
B
30 N=
N
C
40 N=
Given
Σ

M
A
= 0;
M− g cos
θ
()
aMgsin
θ
()
d
2
− N
B
sin
θ
()
d+ N
C
ab+()+ 0=
+
↑
Σ
F
y
= 0;
N
B
Mg− N
C
cos

θ
()
+ 0=
+
↑
N
A
N
C
sin
θ
()
− 0=
Σ
F
y
= 0;
N
C
N
B
N
A
⎛
⎜
⎜
⎜
⎝
⎞
⎟

⎟
⎟
⎠
Find N
C
N
B
, N
A
,
()
=
N
C
N
B
N
A
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
493
554

247
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
Problem 5-18
The beam is pin-connected at A
and rocker-supported at B.
Determine the reactions at the pin A
and at the roller at B.
Given:
F 500 N=
M 800 N m⋅=
a 8m=
352
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Engineering Mechanics - Statics Chapter 5
b 4m=
c 5m=
Solution:
α
atan
c

ab+
⎛
⎜
⎝
⎞
⎟
⎠
=
Σ
M
A
= 0;
F−
a
cos
α
()
M− B
y
a+ 0=
B
y
Fa Mcos
α
()
+
cos
α
()
a

=
B
y
642 N=
+
→
Σ
F
x
= 0;
A
x
− F sin
α
()
+ 0= A
x
F sin
α
()
= A
x
192 N=
+
↑
Σ
F
y
= 0;
A

y
− F cos
α
()
− B
y
+ 0= A
y
F− cos
α
()
B
y
+= A
y
180 N=
Problem 5-19
Determine the magnitude of the reactions on the beam at A and B. Neglect the thickness of the beam.
Given:
F
1
600 N=
F
2
400 N=
θ
15 deg=
a 4m=
b 8m=
c 3=

d 4=
Solution:
Σ
M
A
= 0;
B
y
ab+()F
2
cos
θ
()
ab+()− F
1
a− 0=
353
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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Engineering Mechanics - Statics Chapter 5
B
y
F
2
cos
θ
(
)
ab+()F

1
a+
ab+
= B
y
586 N=
+
→
Σ
F
x
= 0;
A
x
F
2
sin
θ
()
− 0=
A
x
F
2
sin
θ
()
= A
x
104 N=

+
↑
Σ
F
y
= 0;
A
y
F
2
cos
θ
()
− B
y
+ F
1
− 0=
A
y
F
2
cos
θ
()
B
y
− F
1
+= A

y
400 N=
F
A
A
x
2
A
y
2
+= F
A
413 N=
Problem 5-20
Determine the reactions at the supports.
Given:
w 250
lb
ft
=
a 6ft=
b 6ft=
c 6ft=
Solution:
Guesses
A
x
1lb= A
y
1lb=

B
y
1lb=
Given
A
x
0= A
y
B
y
+ wa b+()−
1
2
wc− 0=
wa
a
2
⎛
⎜
⎝
⎞
⎟
⎠
wb
b
2
⎛
⎜
⎝
⎞

⎟
⎠
−
1
2
wc b
c
3
+
⎛
⎜
⎝
⎞
⎟
⎠
− B
y
b+ 0=
354
© 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
x
A
y
B
y
⎛

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

⎟
⎟
⎟
⎠
0
2750
1000
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
Problem 5-21
When holding the stone of weight W in equilibrium, the humerus H, assumed to be smooth, exerts
normal forces
F
C
and
F
A
on the radius C and ulna A as shown. Determine these forces and the
force
F
B
that the biceps B exerts on the radius for equilibrium. The stone has a center of mass at
G. Neglect the weight of the arm.

Given:
W 5lb=
θ
75 deg=
a 2in=
b 0.8 in=
c 14 in=
Solution:
Σ
M
B
= 0;
W− ca−()F
A
a+ 0=
F
A
W
ca−
a
⎛
⎜
⎝
⎞
⎟
⎠
=
F
A
30lb=

+
↑
Σ
F
y
= 0;
F
B
sin
θ
()
W− F
A
− 0=
F
B
WF
A
+
sin
θ
()
=
F
B
36.2 lb=
+
→
Σ
F

x
= 0;
F
C
F
B
cos
θ
()
− 0=
355
© 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
F
C
F
B
cos
θ
(
)
=
F
C
9.378 lb=
Problem 5-22
The uniform door has a weight W and a center of gravity at G. Determine the reactions at the
hinges if the hinge at A supports only a horizontal reaction on the door, whereas the hinge at B

exerts both horizontal and vertical reactions.
Given:
W 100 lb=
a 3ft=
b 3ft=
c 0.5 ft=
d 2ft=
Solution:
Σ
M
B
= 0;
Wd A
x
ab+()− 0=
A
x
W
d
ab+
⎛
⎜
⎝
⎞
⎟
⎠
= A
x
33.3 lb=
Σ

F
x
= 0;
B
x
A
x
= B
x
33.3 lb=
Σ
F
y
= 0;
B
y
W= B
y
100lb=
Problem 5-23
The ramp of a ship has weight W and center of gravity at G. Determine the cable force in CD
needed to just start lifting the ramp, (i.e., so the reaction at B becomes zero). Also, determine the
horizontal and vertical components of force at the hinge (pin) at A.
356
© 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 200 lb= a 4ft=

θ
30 deg= b 3ft=
φ
20 deg= c 6ft=
Solution:
Σ
M
A
= 0;
F
CD
− cos
θ
()
bc+( ) cos
φ
()
F
CD
sin
θ
()
bc+( ) sin
φ
()
+ Wccos
φ
()
+ 0=
F

CD
Wccos
φ
()
bc+( ) cos
θ
()
cos
φ
()
sin
θ
()
sin
φ
()
−
()
= F
CD
195lb=
+
→
Σ
F
x
= 0;
F
CD
sin

θ
()
A
x
− 0=
A
x
F
CD
sin
θ
()
= A
x
97.5 lb=
+
↑
Σ
F
y
= 0;
A
y
W− F
CD
cos
θ
()
+ 0=
A

y
WF
CD
cos
θ
()
−= A
y
31.2 lb=
Problem 5-24
The drainpipe of mass M is held in the tines of the fork lift. Determine the normal forces at A
and B as functions of the blade angle
θ
and plot the results of force (ordinate) versus
θ
(abscissa)
for
0
θ
≤ 90 deg≤
.
357
© 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
Units used:
Mg 10
3
kg=

Given:
M 1.4 Mg=
a 0.4 m=
g 9.81
m
s
2
=
Solution:
θ
090 =
N
A
θ
()
Mgsin
θ
deg
()
10
3
=
N
B
θ
()
Mgcos
θ
deg
()

10
3
=
0 20406080100
0
5
10
15
Angle in Degrees
Force in kN
N
A
θ
()
N
B
θ
()
θ
Problem 5-25
While slowly walking, a man having a total mass M places all his weight on one foot. Assuming
that the normal force
N
C

of the ground acts on his foot at C, determine the resultant vertical
compressive force
F
B


which the tibia T exerts on the astragalus B, and the vertical tension
F
A

in
the achilles tendon A at the instant shown.
358
© 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
Units Used:
kN 10
3
N=
Given:
M 80 kg=
a 15 mm=
b 5mm=
c 20 mm=
d 100 mm=
Solution:
N
C
Mg=
N
C
785 N=
Σ
M

A
=
0;
F
B
− cN
C
cd+()+ 0=
F
B
N
C
cd+
c
⎛
⎜
⎝
⎞
⎟
⎠
=
F
B
4.71 kN=
Σ
F
y
= 0;
F
A

F
B
− N
C
+ 0=
F
A
F
B
N
C
−=
F
A
3.92 kN=
Problem 5-26
Determine the reactions at the roller A and pin B.
359
© 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
M 800 lbft= c 3ft=
F 390 lb= d 5=
a 8ft= e 12=
b 4ft=
θ
30 deg=
Solution:
Guesses R

A
1lb= B
x
1lb= B
y
1lb=
Given
R
A
sin
θ
()
B
x
+
d
e
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F− 0=
R
A

cos
θ
()
B
y
+
e
e
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F− 0=
MR
A
cos
θ
()
ab+()− B
x
c+ 0=
R
A
B

x
B
y
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
Find R
A
B
x
, B
y
,
()
= R
A
105.1 lb=
B
x
B
y
⎛
⎜

⎝
⎞
⎟
⎠
97.4
269
⎛
⎜
⎝
⎞
⎟
⎠
lb=
Problem 5-27
The platform assembly has weight W
1
and center of gravity at G
1
. If it is intended to support a
maximum load W
2
placed at point G
2,
,determine the smallest counterweight W that should be
placed at B in order to prevent the platform from tipping over.
Given:
W
1
250 lb= a 1ft= c 1ft= e 6ft=
W

2
400 lb= b 6ft= d 8ft= f 2ft=
360
Given:
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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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
be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
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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
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
© 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
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
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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 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.
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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 9.08lb=
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 29.73 mm=
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 4.7434 ft=
Geometry:
ah−
d
a
b
= haa
d
b
⎛
⎜
⎝
⎞
⎟
⎠
−= h 15.8 ft=
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 29.73 mm=
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.