Engineering Mechanics - Statics Chapter 8
of the belt so that the collar rotates clockwise with a constant angular velocity. Assume that
the belt does not slip on the collar; rather, the collar slips on the shaft. Neglect the weight and
thickness of the belt and collar. The radius, measured from the center of the collar to the mean
thickness of the belt is R.
Given:
r 2in=
μ
k
0.3=
R 2.25 in=
F 20 lb=
Solution:
φ
k
atan
μ
k
()
=
φ
k
16.699 deg=
r
f
rsin
φ
k
()
= r
f

y
2
+= P
2
F
2
+=
Guess P 1lb=
Given P
2
F
2
+
()
r
f
FR+ PR− 0= P Find P()= P 29.00 lb=
Problem 8-127
The connecting rod is attached to the piston by a pin at B of diameter d
1
and to the crank shaft by a
bearing A of diameter d
2
. If the piston is moving downwards, and the coefficient of static friction at
these points is
μ
s
, determine the radius of the friction circle at each connection.
Given:
d

1
μ
s
= r
fB
0.075 in=
Problem 8-128
The connecting rod is attached to the piston by a pin at B of diameter d
1
and to the crank shaft
by a bearing A of diameter d
2
. If the piston is moving upwards, and the coefficient of static
friction at these points is
μ
s
, determine the radius of the friction circle at each connection.
Given:
d
1
20 mm=
d
2
50 mm=
μ
s
0.3=
Solution:
r
fA

needed to push the
roller at constant speed. Neglect friction developed at the axle, A, and assume that the
resultant force
P
acting on the handle is applied along arm BA.
Given:
M 80 kg=
θ
30 deg=
a 250 mm=
r 25 mm=
Solution:
θ
1
asin
r
a
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
1
5.74 deg=
Σ
M
0

P
that must be applied to the handle to overcome the rolling resistance. The
coefficient of rolling resistance is
μ
. Neglect the weight of the cart.
883
Problem 8-129
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Engineering Mechanics - Statics Chapter 8
Given:
W 1500 lb=
D 6in=
a 0.04 in=
c 3=
b 4=
Solution:
Guesses N 1lb= P 1lb=
Given
NW− P
c
c
2
b
2
+
⎛
⎜
⎝

⎝
⎞
⎟
⎠
Find NP,()= N 1515lb= P 25.3 lb=
Problem 8-131
The cylinder is subjected to a load that has a weight W. If the coefficients of rolling resistance for
the cylinder's top and bottom surfaces are a
A
and a
B
respectively, show that a force having a
magnitude of P = [W(a
A
+ a
B
)]/2r is required to move the load and thereby roll the cylinder
forward. Neglect the weight of the cylinder.
884
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Engineering Mechanics - Statics Chapter 8
Solution:
+
→
Σ
F
x
= 0;

Wa
A
a
B
+
()
− 0=
Since
φ
Α

and
φ
B
are very small,
cos
φ
A
()
cos
φ
B
()
= 1=
Hence from Eq.(1)
P
Wa
A
a
B

s
0.2 mm=
Solution:
P
Mg a
g
a
s
+
()
2
D
2
⎛
⎜
⎝
⎞
⎟
⎠
= P 235N=
Problem 8-133
A machine of mass M is to be moved over a level surface using a series of rollers for which
the coefficient of rolling resistance is a
g
at the ground and a
m
at the bottom surface of the
machine. Determine the appropriate diameter of the rollers so that the machine can be pushed
forward with a horizontal force
P

⎜
⎝
⎞
⎟
⎠
= r 19.2 mm= d 2 r= d 38.5mm=
886
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Engineering Mechanics - Statics Chapter 8
Problem 8-134
A single force
P
is applied to the handle of the drawer. If
friction is neglected at the bottom and the coefficient of
static friction along the sides is
μ
s
determine the largest
spacing s between the symmetrically placed handles so
that the drawer does not bind at the corners A and B
when the force
P
is applied to one of the handles.
Given:
μ
s
0.4=
a 0.3 m=

N
A
N
B
= N=
+
↑
Σ
F
y
= 0;
μ
s
N
A
μ
s
N
B
+ P− 0= P 2
μ
s
N=
Σ
M
B
= 0;
Na
μ
s

⎦
N 0=
a
μ
s
b+
μ
s
sb+()− 0= s
a
μ
s
= s 0.750 m=
Problem 8-135
The truck has mass M and a center of mass at G. Determine the greatest load it can pull if (a) the
truck has rear-wheel drive while the front wheels are free to roll, and
(
b
)
the truck has
887
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 8
()
four-wheel drive. The coefficient of static friction between the wheels and the ground is
μ
st
and

T 1N= N
C
1N= W 1N=
(a) Rear wheel drive
Given T−
μ
st
N
A
+ 0=
N
A
N
B
+ Mg− 0=
M− gb N
B
bc+()+ Ta+ 0=
T
μ
sc
N
C
− 0=
N
C
W− 0=
N
A
N

(b) Four wheel drive
Given T−
μ
st
N
A
+
μ
st
N
B
+ 0=
888
© 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 8
N
A
N
B
+ Mg− 0=
M− gb N
B
bc+()+ Ta+ 0=
T
μ
sc
N
C

N
B
, T, N
C
, W,
()
= W 15.33 kN=
Problem 8-136
The truck has M and a center of mass at G. The truck is traveling up an incline of angle
θ
.
Determine the greatest load it can pull if (a) the truck has rear-wheel drive while the front wheels
are free to roll, and (b) the truck has four-wheel drive. The coefficient of static friction between
the wheels and the ground is
μ
st
and between the crate and the ground, it is
μ
sc
.
Units Used:
kN 10
3
N=
Mg 1000 kg=
Given:
θ
10 deg=
889
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

N
A
+ Mgsin
θ
()
− 0=
N
A
N
B
+ Mgcos
θ
()
− 0=
M− gbcos
θ
()
Mgdsin
θ
()
+ N
B
bc+()+ Ta+ 0=
T
μ
sc
N
C
− W sin
θ

⎟
⎠
Find N
A
N
B
, T, N
C
, W,
()
= W 1.25 kN=
(b) Four wheel drive
Given
T−
μ
st
N
A
+
μ
st
N
B
+ Mgsin
θ
()
− 0=
N
A
N

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Engineering Mechanics - Statics Chapter 8
N
A
N
B
T
N
C
W
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find N
A

y'
= 0;
N
m
Mgcos
θ
()
− 0=
Σ
F
x'
= 0;
Mgsin
θ
()
μ
s
N
m
− 0=
μ
s
tan
θ
()
=
θ
atan
μ
s

a 2ft=
b 14 ft=
c 6ft=
W 40 lb=
Solution:
θ
asin
c
b
⎛
⎜
⎝
⎞
⎟
⎠
=
Initial guesses
P 10 lb= N
A
100 lb=
μ
A
100=
Given
Σ
F
x
= 0;
μ
A

θ
()
Pb+ 0=
P
N
A
μ
A
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
Find PN
A
,
μ
A
,
()
=
P
N
A
⎛

© 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 8
Units Used:
kip 10
3
lb=
Given:
W 8000 lb=
θ
10 deg=
μ
s
0.4=
Solution:
wedge A:
Σ
F
y
= 0;
Ncos
θ
()
μ
s
Nsin
θ
()
− W− 0=

θ
()
N sin
θ
()
+=
P' 4.96 kip=
Wedge B:
Σ
F
y
= 0;
N
C
μ
s
Nsin
θ
()
+ Ncos
θ
()
− 0=
N
C
μ
s
− N sin
θ
()

C
N sin
θ
()
+
μ
s
N cos
θ
()
+=
P 8.16 kip=
Problem 8-140
Column D is subjected to a vertical load W. It is supported on two identical wedges A and B
for which the coefficient of static friction at the contacting surfaces between A and B and
between B and C is
μ
s
. If the forces
P
and
P'
are removed, are the wedges self-locking? The
contacting surface between A and D is smooth.
893
© 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 8
Given:

θ
()
=
F 1389.2 lb=
Friction
F
max
μ
s
N=
Since
F 1389lb=
<
F
max
3151lb=
then the wedges do not slip at the contact surface AB.
Wedge B:
Σ
F
y
= 0;
N
C
F sin
θ
()
− Ncos
θ
()

()
N sin
θ
()
+=
F
C
0lb=
Friction
F
Cmax
μ
s
N
C
=
Since
F
C
0lb=
<
F
Cmax
3200lb=
then the wedges do not slip at the contact surface BC.
Therefore the wedges are self-locking.
894
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be reproduced, in any form or by any means, without permission in writing from the publisher.

a
xb
x
a
⎛
⎜
⎝
⎞
⎟
⎠
2
1
2b
a
2
x
⎛
⎜
⎝
⎞
⎟
⎠
2
+
⌠
⎮
⎮
⎮
⌡
d

rod. If the rod has a weight per unit length
γ
,
determine the vertical reaction at A and the x and y
components of reaction at the pin B.
Given:

γ
0.5
lb
ft
=
a 1ft=
b 2ft=
895
© 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 9
Solution:
yb
x
a
⎛
⎜
⎝
⎞
⎟
⎠
2

γ
L= W 1.162 lb=
x
c
1
L
0
a
xx 1
2 b
a
2
x
⎛
⎜
⎝
⎞
⎟
⎠
2
+
⌠
⎮
⎮
⎮
⌡
d=
x
c
0.620 ft=

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

π
−
2
θ
−
π
2
θ
+
α
r cos
α
()
r
⌠
⎮
⎮
⌡
d
π
−
2
θ
−
π
2
θ
+
α
r

2
θ
rcos
θ
()
r
⌠
⎮
⎮
⌡
d
π
−
2
π
2
θ
r
⌠
⎮
⎮
⌡
d
=
x
c
1.273 ft=
Σ
M
A

A
x
− B
x
+ 0= A
x
B
x
= A
x
1lb=
+
↑
Σ
F
y
= 0;
A
y
π
r
γ
− 0= A
y
π
r
γ
= A
y
3.14 lb=

a
2
=
L
0
a
x1
2 bx
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
+
⌠
⎮
⎮
⎮
⌡
d=
L 1.148ft=
x
c
1
L
0

0= O
x
0lb= O
x
0lb=
898
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 9
+
↑
Σ
F
y
= 0;
O
y
γ
L− 0= O
y
γ
L= O
y
0.574 lb=
Σ
M
O
= 0;
M

2
=
d y
d x
2 bx
a
2
=
L
0
a
x1
2 bx
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
+
⌠
⎮
⎮
⎮
⌡
d=
L 1.148ft=

⎮
⎮
⌡
d
⎡
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎦
= y
c
0.183 ft=
Problem 9-7
Locate the centroid of the parabolic area.
Solution:
a
h
b
2
=
899
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Engineering Mechanics - Statics Chapter 9

=
x
c
3
2hb
0
h
y
1
2
b
y
h
⎛
⎜
⎝
⎞
⎟
⎠
2
⌠
⎮
⎮
⌡
d=
3
8 h
2
bh
2

⎠
5
2
=
y
c
3
5
h=
Problem 9-8
Locate the centroid y
c
of the shaded area.
Given:
a 100 mm=
b 100 mm=
Solution:
yb
x
a
⎛
⎜
⎝
⎞
⎟
⎠
2
=
900
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

c
60mm=
Problem 9-9
Locate the centroid x
c
of the shaded area.
Solution:
dAydx=
x
c
x=
y
c
y
2
=
x
c
0
b
xx
h
b
2
x
2
⌠
⎮
⎮
⌡

b
x
1
2
h
b
2
x
2
⎛
⎜
⎝
⎞
⎟
⎠
2
⌠
⎮
⎮
⎮
⌡
d
0
b
x
h
b
2
x
2

) of the
centroid of the triangular area.
Solution:
A
0
a
xmx
⌠
⎮
⌡
d=
1
2
a
2
m=
x
c
2
ma
2
0
a
xxmx
⌠
⎮
⌡
d=
2
3

Problem 9-11
Determine the location (x
c
, y
c
) of the center of
gravity of the quartercircular plate. Also
determine the force in each of the supporting
wires.The plate has a weight per unit area of
γ
.
Given:
γ
5
lb
ft
2
=
a 4ft=
902
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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 9
Solution:
x
2
y
2
+ a

c
1.698 ft=
y
c
1
A
0
a
x
1
2
a
2
x
2
−
()
⌠
⎮
⎮
⌡
d=
y
c
1.698 ft=
Guesses T
A
1lb= T
B
1lb=

⎛
⎜
⎝
⎞
⎟
⎠
36.2
26.7
⎛
⎜
⎝
⎞
⎟
⎠
lb=
Problem *9-12
Locate the centroid of the shaded area.
903
© 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 9
Solution:
dAydx= x
c
x= y
c
y
2
=

L
⎛
⎜
⎝
⎞
⎟
⎠
⌠
⎮
⎮
⌡
d=
1
2
L= x
c
1
2
L=
y
c
π
2La
0
L
x
1
2
asin
π

Locate the center of gravity of the homogeneous
cantilever beam and determine the reactions at the
fixed support.The material has a density of
ρ
.
Units Used:
Mg 10
3
kg= kN 10
3
N=
Given:
ρ
8
Mg
m
3
= a 1m=
b 4m=
g 9.81
m
s
2
=
c 0.5 m=
Solution:
V
b−
0
xca

⎟
⎠
2
⌠
⎮
⎮
⌡
d=
y
c
1
V
b−
0
x
c
2
ca
x
b
⎛
⎜
⎝
⎞
⎟
⎠
2
⌠
⎮
⎮

⎥
⎦
2
⌠
⎮
⎮
⎮
⌡
d=
x
c
y
c
z
c
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
3.00−
0.25
0.30−
⎛
⎜

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

Problem 9-14
Locate the centroid (x
c
, y
c
) of the
exparabolic segment of area.
Solution:
A
a−
0
x
b
a
2
x
2
⌠
⎮
⎮
⌡
d=
1
3
ab=
x
c
3
ab
a−

a
2
x
2
⎛
⎜
⎝
⎞
⎟
⎠
2
⌠
⎮
⎮
⎮
⌡
d=
3−
10
b= y
c
3−
10
b=
905
© 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.