Engineering Mechanics Statics - Examples Part 16 ppsx

Engineering Mechanics - Statics Chapter 10
I
max
I
x
I
y
+
2
R+= I
max
4.92 10
6
× mm
4
=
I
min
I
x
I
y
+
2
R−= I
min
1364444.44 mm
4
=
Problem 10-88
Determine the directions of the principal axes with

origin located at point O, and the principal moments
of inertia for the area about these axes. Solve using
Mohr's circle
Given:
a 4in=
b 2in=
c 2in=
d 2in=
r 1in=
Solution:
I
x
1
3
cd+()ab+()
3
π
r
4
4
π
r
2
a
2
+
⎛
⎜
⎝
⎞

⎟
⎠
−= I
x
236.95 in
4
=
I
y
1
3
ab+()cd+()
3
π
r
4
4
π
r
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
−= I

y
114.65 in
4
=
I
xy
ab+
2
⎛
⎜
⎝
⎞
⎟
⎠
dc+
2
⎛
⎜
⎝
⎞
⎟
⎠
ab+()dc+()da
π
r
2
−= I
xy
118.87 in
4

=
RI
x
I
x
I
y
+
2
⎛
⎜
⎝
⎞
⎟
⎠
−
⎡
⎢
⎣
⎤
⎥
⎦
2
I
xy
2
+= R 133.67 in
4
=
I

max
I
x
I
y
+
2
R+= I
max
309in
4
=
I
min
I
x
I
y
+
2
R−= I
min
42.1 in
4
=
1051
© 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 10

θ
p1
1−
2
asin
I
xy
R
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
p1
31.39− deg=
θ
p2
θ
p1
π
2
+=
θ
p2
58.61 deg=
Problem 10-89
The area of the cross section of an airplane wing has the listed properties about the x and y axes

passing through the centroid C. Determine the orientation of the principal axes and the principal
moments of inertia. Solve using Mohr's circle.
Given:
I
x
450 in
4
=
I
y
1730 in
4
=
I
xy
138 in
4
=
Solution:
RI
x
I
x
I
y
+
2
⎛
⎜
⎝

⎞
⎟
⎠
−
⎡
⎢
⎣
⎤
⎥
⎦
2
I
xy
2
+= R 654.71 in
4
=
I
max
I
x
I
y
+
2
R+
⎛
⎜
⎝
⎞

⎟
⎠
= I
max
1.74 10
3
× in
4
=
I
min
I
x
I
y
+
2
R−
⎛
⎜
⎝
⎞
⎟
⎠
= I
min
435in
4
=
1052

© 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 10
θ
p1
1
2
asin
I
xy
R
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
p1
6.08 deg=
θ
p2
θ
p1
90 deg+=
θ
p2
96.08 deg=

Problem 10-90
The right circular cone is formed by revolving the shaded area
around the x axis. Determine the moment of inertia l
x
and
express the result in terms of the total mass m of the cone. The
cone has a constant density
ρ
.
Solution:
m
0
h
x
ρπ
rx
h
⎛
⎜
⎝
⎞
⎟
⎠
2
⌠
⎮
⎮
⌡
d=
1

3
h
ρπ
r
2
=
l
x
3m
π
hr
2
0
h
x
1
2
π
rx
h
⎛
⎜
⎝
⎞
⎟
⎠
4
⌠
⎮
⎮

⌡
d=
3
10
mr
2
=
l
x
3
10
mr
2
=
Problem 10-91
Determine the moment of inertia of the thin ring about the z axis. The ring has a mass m.
1053
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Engineering Mechanics - Statics Chapter 10
Solution:
m
ρ
2
π
R=
ρ
m
2

π
R
=
I
0
2
π
θ
m
2
π
R
⎛
⎜
⎝
⎞
⎟
⎠
R
2
R
⌠
⎮
⎮
⌡
d= mR
2
= ImR
2
=

Problem 10-92
The solid is formed by revolving the shaded area
around the y axis. Determine the radius of gyration
k
y
. The specific weight of the material is
γ
.
Given:
a 3in=
b 3in=
γ
380
lb
ft
3
=
Solution:
m
0
b
y
γπ
a
y
b
⎛
⎜
⎝
⎞

⎟
⎠
3
⎡
⎢
⎣
⎤
⎥
⎦
2
⌠
⎮
⎮
⎮
⌡
d=
m 2.66lb=
I
y
0
b
y
γπ
a
y
b
⎛
⎜
⎝
⎞

⎟
⎠
3
⎡
⎢
⎣
⎤
⎥
⎦
2
1
2
a
y
b
⎛
⎜
⎝
⎞
⎟
⎠
3
⎡
⎢
⎣
⎤
⎥
⎦
2
⌠

⎮
⎮
⎮
⌡
d=
I
y
6.46 lb in
2
⋅=
k
y
I
y
m
= k
y
1.56 in=
1054
© 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 10
Problem 10-93
Determine the moment of inertia I
x
for the sphere and express the result in terms of
the total mass m of the sphere. The sphere has a constant density
ρ
.

Solution:
m
ρ
4
π
r
3
3
=
ρ
3m
4
π
r
3
=
I
x
r−
r
x
1
2
3m
4
π
r
3
⎛
⎜

⎝
⎞
⎟
⎠
π
r
2
x
2
−
()
r
2
x
2
−
()
⌠
⎮
⎮
⌡
d=
2
5
mr
2
= I
x
2
5

mr
2
=
Problem 10-94
Determine the radius of gyration k
x
of the paraboloid. The density of the material is
ρ
.
Units Used:
Mg 1000 kg=
Given:
ρ
5
Mg
m
3
= a 200 mm= b 100 mm=
1055
© 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 10
Solution:
m
p
0
a
x
ρπ

b
2
x
a
⎛
⎜
⎝
⎞
⎟
⎠
⌠
⎮
⎮
⌡
d=
m
p
15.71 kg=
I
x
0
a
x
1
2
ρπ
b
2
x
a

⎛
⎜
⎝
⎞
⎟
⎠
b
2
x
a
⎛
⎜
⎝
⎞
⎟
⎠
⌠
⎮
⎮
⌡
d=
I
x
52.36 10
3−
× kg m
2
⋅=
k
x

I
x
m
p
= k
x
57.7 mm=
Problem 10-95
Determine the moment of inertia of the semi-ellipsoid with respect to the x axis and
express the result in terms of the mass m of the semiellipsoid. The material has a
constant density
ρ
.
Solution:
m
0
a
x
ρπ
b
2
1
x
2
a
2
−
⎛
⎜
⎜

⎝
⎞
⎟
⎟
⎠
⌠
⎮
⎮
⎮
⌡
d=
2
3
a
ρπ
b
2
=
ρ
3m
2
π
ab
2
=
I
x
0
a
x

1
2
3m
2
π
ab
2
⎛
⎜
⎝
⎞
⎟
⎠
π
b
2
1
x
2
a
2
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠

b
2
1
x
2
a
2
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⌠
⎮
⎮
⎮
⌡
d=
2
5
mb
2
= I
x
2
5

mb
2
=
1056
© 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 10
Problem 10-96
Determine the radius of gyration k
x
of the body. The specific weight of the material is
γ
.
Given:
γ
380
lb
ft
3
=
a 8in=
b 2in=
Solution:
m
b
0
a
x
γπ

b
2
x
a
⎛
⎜
⎝
⎞
⎟
⎠
2
3
⌠
⎮
⎮
⎮
⎮
⌡
d= m
b
13.26 lb=
I
x
0
a
x
1
2
γπ
b

2
x
a
⎛
⎜
⎝
⎞
⎟
⎠
2
3
b
2
x
a
⎛
⎜
⎝
⎞
⎟
⎠
2
3
⌠
⎮
⎮
⎮
⎮
⌡
d= I

x
0.59 slug in
2
⋅=
k
x
I
x
m
b
= k
x
1.20 in=
Problem 10-97
Determine the moment of inertia for the ellipsoid with respect to the x axis and express the result in
terms of the mass m of the ellipsoid. The material has a constant density
ρ
.
1057
© 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 10
Solution:
m
a−
a
x
ρπ
b

2
1
x
2
a
2
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⌠
⎮
⎮
⎮
⌡
d=
4
3
a
ρπ
b
2
=
ρ
3m

4
π
ab
2
=
I
x
a−
a
x
1
2
3m
4
π
ab
2
π
b
2
1
x
2
a
2
−
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
b
2
1
x
2
a
2
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⌠
⎮
⎮
⎮
⌡
d=
2
5
mb
2

= I
x
2
5
mb
2
=
Problem 10-98
Determine the moment of inertia of
the homogeneous pyramid of mass m
with respect to the z axis. The density
of the material is
ρ
.
Suggestion: Use a
rectangular plate element having a
volume of dV = (2x)(2y) dz.
Solution:
V
0
h
za 1
z
h
−
⎛
⎜
⎝
⎞
⎟

⎠
⎡
⎢
⎣
⎤
⎥
⎦
2
⌠
⎮
⎮
⌡
d=
1
3
ha
2
=
ρ
m
V
=
3m
a
2
h
=
I
z
3m

a
2
h
0
h
z
1
6
a 1
z
h
−
⎛
⎜
⎝
⎞
⎟
⎠
⎡
⎢
⎣
⎤
⎥
⎦
4
⌠
⎮
⎮
⌡
d=

1
10
ma
2
= I
z
1
10
ma
2
=
1058
© 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 10
Problem 10-99
The concrete shape is formed by rotating the shaded area about the y axis. Determine the moment of
inertia I
y
. The specific weight of concrete is
γ
.
Given:
γ
150
lb
ft
3
=

a 6in=
b 4in=
c 8in=
Solution:
I
y
1
2
γπ
ab+()
2
ca b+()
2
0
c
y
1
2
γπ
a
2
y
c
⎛
⎜
⎝
⎞
⎟
⎠
a

2
y
c
⌠
⎮
⎮
⌡
d−=
I
y
2.25 slug ft
2
⋅=
Problem 10-100
Determine the moment of inertia of the thin plate about an axis perpendicular to the page and passing
through the pin at O. The plate has a hole in its center. Its thickness is c, and the material has a
density of
ρ
Given:
a 1.40 m= c 50 mm=
b 150 mm=
ρ
50
kg
m
3
=
Solution:
I
G

1
12
ρ
a
2
ca
2
a
2
+
()
1
2
ρπ
b
2
cb
2
−=
I
G
1.60 kg m
2
⋅=
I
0
I
G
md
2

+=
1059
© 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 10
m
ρ
a
2
c
ρπ
b
2
c−=
m 4.7233 kg=
I
0
I
G
masin 45 deg()()
2
+=
I
0
6.23 kg m
2
⋅=
Problem 10-101
Determine the moment of inertia I

z
of the frustum of the cone which has a conical depression. The
material has a density
ρ
.
Given:
ρ
200
kg
m
3
=
a 0.4 m=
b 0.2 m=
c 0.6 m=
d 0.8 m=
Solution:
h
da
ab−
=
I
z
3
10
ρ
1
3
π
a

2
h
⎛
⎜
⎝
⎞
⎟
⎠
⎡
⎢
⎣
⎤
⎥
⎦
a
2
3
10
ρ
1
3
π
a
2
c
⎛
⎜
⎝
⎞
⎟

⎠
⎡
⎢
⎣
⎤
⎥
⎦
a
2
−
3
10
ρ
1
3
π
b
2
hd−()
⎡
⎢
⎣
⎤
⎥
⎦
⎡
⎢
⎣
⎤
⎥

⎦
b
2
−=
I
z
1.53 kg m
2
⋅=
Problem 10-102
Determine the moment of inertia for the assembly about an axis which is perpendicular to the
page and passes through the center of mass G. The material has a specific weight
γ
.
Given:
a 0.5 ft= d 0.25 ft=
1060
© 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 10
b 2ft= e 1ft=
c 1ft=
γ
90
lb
ft
3
=
Solution:

I
G
1
2
γπ
ab+()
2
ea b+()
2
1
2
γπ
b
2
ed−()b
2
−
1
2
γπ
c
2
dc
2
−=
I
G
118slug ft
2
⋅=

Problem 10-103
Determine the moment of inertia for the assembly about an axis which is perpendicular to the page and
passes through point O. The material has a specific weight
γ
.
Given:
a 0.5 ft= d 0.25 ft=
b 2ft= e 1ft=
c 1ft=
γ
90
lb
ft
3
=
Solution:
I
G
1
2
γπ
ab+()
2
ea b+()
2
1
2
γπ
b
2

ed−()b
2
−
1
2
γπ
c
2
dc
2
−=
I
G
118slug ft
2
⋅=
M
γπ
ab+()
2
e
γπ
b
2
ed−()−
γπ
c
2
d−= M 848.23 lb=
I

O
I
G
Ma b+()
2
+= I
O
283slug ft
2
⋅=
1061
© 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 10
Problem 10-104
The wheel consists of a thin ring having a mass M
1
and four spokes made from slender rods, each
having a mass M
2
. Determine the wheel’s moment of inertia about an axis perpendicular to the page
and passing through point A.
Given:
M
1
10 kg=
M
2
2kg=

a 500 mm=
Solution:
I
G
M
1
a
2
4
1
3
M
2
a
2
+=
I
A
I
G
M
1
4M
2
+
()
a
2
+=
I

A
7.67 kg m
2
⋅=
Problem 10-105
The slender rods have a weight density
γ
.
Determine the moment of inertia for the assembly
about an axis perpendicular to the page and passing through point A.
Given:
γ
3
lb
ft
=
a 1.5 ft=
b 1ft=
c 2ft=
Solution:
I
1
3
γ
bc+()bc+()
2
1
12
γ
2a 2a()

2
+
γ
2ac
2
+=
I 2.17 slug ft
2
⋅=
1062
© 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 10
Problem 10-106
Each of the three rods has a mass m. Determine the
moment of inertia for the assembly about an axis which is
perpendicular to the page and passes through the center
point O.
Solution:
I
O
3
1
12
ma
2
m
asin 60 deg()
3

⎛
⎜
⎝
⎞
⎟
⎠
2
+
⎡
⎢
⎣
⎤
⎥
⎦
=
I
O
1
2
ma
2
=
Problem 10-107
The slender rods have weight density
γ
.
Determine the moment of inertia for the assembly
about an axis perpendicular to the page and passing through point A
Given:
γ

3
lb
ft
=
a 1.5 ft=
b 2ft=
Solution:
I
A
1
3
γ
bb
2
1
12
γ
2a 2a()
2
+
γ
2a()b
2
+= I
A
1.58 slug ft
2
⋅=
Problem 10-108
The pendulum consists of a plate having weight W

p
and a slender rod having weight W
r .
Determine the radius of gyration of the pendulum about an axis perpendicular to the page and
passing through point O.
1063
© 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 10
Given:
W
p
12 lb= a 1ft=
W
r
4lb= b 1ft=
c 3ft=
d 2ft=
Solution:
I
0
1
12
W
r
cd+()
2
W
r

cd+
2
c−
⎛
⎜
⎝
⎞
⎟
⎠
2
+
1
12
W
p
a
2
b
2
+
()
+ W
p
c
b
2
+
⎛
⎜
⎝

⎞
⎟
⎠
2
+=
k
0
I
0
W
p
W
r
+
= k
0
3.15 ft=
Problem 10-109
Determine the moment of inertia for the overhung crank about the x axis. The material is steel
having density
ρ
.
Units Used:
Mg 1000 kg=
Given:
ρ
7.85
Mg
m
3

=
a 20 mm=
b 20 mm=
c 50 mm=
d 90 mm=
e 30 mm=
Solution:
m
ρπ
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
c= m 0.12 kg=
M
ρ
2dbe= M 0.85 kg=
1064
© 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 10
I
x
2

1
2
m
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
md e−()
2
+
⎡
⎢
⎣
⎤
⎥
⎦
1
12
M 2d()
2
e
2
+
⎡
⎣

⎤
⎦
+=
I
x
3.25 10
3−
× kg m
2
⋅=
Problem 10-110
Determine the moment of inertia for the overhung crank about the x' axis. The material is steel
having density
ρ
.
Units used:
Mg 1000 kg=
Given:
ρ
7.85
Mg
m
3
=
a 20 mm=
b 20 mm=
c 50 mm=
d 90 mm=
e 30 mm=
Solution:

m
ρπ
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
c= m 0.12 kg=
M
ρ
2dbe= M 0.85 kg=
I
x
2
1
2
m
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2

md e−()
2
+
⎡
⎢
⎣
⎤
⎥
⎦
1
12
M 2d()
2
e
2
+
⎡
⎣
⎤
⎦
+=
I
x'
I
x
M 2m+()de−()
2
+= I
x'
7.19 10

3−
× kg m
2
⋅=
1065
© 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 10
Problem 10-111
Determine the moment of inertia for the solid steel assembly about the x axis. Steel has a specific
weight
γ
st
.
Given:
a 2ft=
b 3ft=
c 0.5 ft=
d 0.25 ft=
γ
st
490
lb
ft
3
=
Solution:
h
ca

cd−
=
I
x
γ
st
π
c
2
b
c
2
2
⎛
⎜
⎝
⎞
⎟
⎠
π
3
c
2
h
3c
2
10
⎛
⎜
⎝

⎞
⎟
⎠
+
π
3
d
2
ha−()
3d
2
10
⎛
⎜
⎝
⎞
⎟
⎠
−
⎡
⎢
⎣
⎤
⎥
⎦
=
I
x
5.64 slug ft
2

⋅=
Problem 10-112
The pendulum consists of two slender rods AB and OC which have a
mass density
ρ
r
.
The thin plate has a mass density
ρ
p.
Determine the
location y
c
of the center of mass G of the pendulum, then calculate
the moment of inertia of the pendulum about an axis perpendicular to
the page and passing through G.
Given:
ρ
r
3
kg
m
=
ρ
s
12
kg
m
2
=

a 0.4 m=
b 1.5 m=
1066
© 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 10
c 0.1 m=
d 0.3 m=
Solution:
y
c
b
ρ
r
b
2
π
d
2
ρ
s
bd+()+
π
c
2
ρ
s
bd+()−
b

ρ
r
π
d
2
ρ
s
+
π
c
2
ρ
s
−
ρ
r
2a+
= y
c
0.888 m=
I
G
1
12
2a
ρ
r
2a()
2
2a

ρ
r
y
c
2
+
1
12
b
ρ
r
b
2
+
b
ρ
r
b
2
y
c
−
⎛
⎜
⎝
⎞
⎟
⎠
2
1

2
π
d
2
ρ
s
d
2
+
π
d
2
ρ
s
bd+ y
c
−
()
2
++

1
2
π
c
2
ρ
s
c
2

π
c
2
ρ
s
bd+ y
c
−
()
2
−+

=
I
G
5.61 kg m
2
⋅=
Problem 10-113
Determine the moment of inertia for the shaded area about the x axis.
Given:
a 2in=
b 8in=
Solution:
I
x
0
b
yy
2

a
y
b
⎛
⎜
⎝
⎞
⎟
⎠
1
3
⌠
⎮
⎮
⎮
⎮
⌡
d=
I
x
307in
4
=
1067
© 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 10
Problem 10-114
Determine the moment of inertia for the shaded area about the y axis.

Given:
a 2in=
b 8in=
Solution:
I
y
0
a
xx
2
bb
x
a
⎛
⎜
⎝
⎞
⎟
⎠
3
−
⎡
⎢
⎣
⎤
⎥
⎦
⌠
⎮
⎮

⌡
d=
I
y
10.67 in
4
=
Problem 10-115
Determine the mass moment of inertia I
x
of the body and express the result in terms of the total
mass m of the body. The density is constant.
Solution:
m
0
a
x
ρπ
bx
a
b+
⎛
⎜
⎝
⎞
⎟
⎠
2
⌠
⎮

⎮
⌡
d=
7
3
a
ρπ
b
2
=
ρ
3m
7
π
ab
2
=
1068
© 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 10
I
x
0
a
x
1
2
3m

7
π
ab
2
⎛
⎜
⎝
⎞
⎟
⎠
π
bx
a
b+
⎛
⎜
⎝
⎞
⎟
⎠
2
bx
a
b+
⎛
⎜
⎝
⎞
⎟
⎠

2
⌠
⎮
⎮
⎮
⌡
d=
93
70
mb
2
=
I
x
93
70
mb
2
=
Problem 10-116
Determine the product of inertia for the shaded area with respect to the x and y axes.
Given:
a 1m=
b 1m=
Solution:
I
xy
0
b
y

1
2
ya
y
b
⎛
⎜
⎝
⎞
⎟
⎠
1
3
a
y
b
⎛
⎜
⎝
⎞
⎟
⎠
1
3
⌠
⎮
⎮
⎮
⎮
⌡

d=
I
xy
0.1875 m
4
=
Problem 10-117
Determine the area moments of inertia I
u
and I
v
and
the product of inertia I
uv
for the semicircular area.
Given:
r 60 mm=
θ
30 deg=
1069
© 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 10
Solution:
I
x
π
r
4

8
= I
y
I
x
=
I
xy
0mm
4
=
I
u
I
x
I
y
+
2
I
x
I
y
−
2
cos 2
θ
()
+ I
xy

sin 2
θ
()
−= I
u
5.09 10
6
× mm
4
=
I
v
I
x
I
y
+
2
I
x
I
y
−
2
cos 2
θ
()
− I
xy
sin 2

θ
()
−= I
v
5.09 10
6
× mm
4
=
I
uv
I
x
I
y
−
2
sin 2
θ
()
I
xy
cos 2
θ
()
+= I
uv
0m
4
=

Problem 10-118
Determine the moment of inertia for the shaded area about the x axis.
Given:
a 3in=
b 9in=
Solution:
I
x
0
b
yy
2
a 1
y
b
−
⌠
⎮
⎮
⌡
d=
I
x
333in
4
=
Problem 10-119
Determine the moment of inertia for the shaded area about the y axis.
1070
© 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 10
Given:
a 3in=
b 9in=
Solution:
I
y
0
a
xx
2
b 1
x
2
a
2
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⌠
⎮
⎮

⎮
⌡
d=
I
y
32.4 in
4
=
Problem 10-120
Determine the area moment of inertia of the area about the x axis. Then, using the parallel-axis
theorem, find the area moment of inertia about the x' axis that passes through the centroid C of the
area.
Given:
a 200 mm=
b 200 mm=
Solution:
I
x
0
b
yy
2
2a
y
b
⌠
⎮
⎮
⌡
d=

I
x
914 10
6
× mm
4
=
Find the area and the distance to the centroid
A
0
b
y2a
y
b
⌠
⎮
⎮
⌡
d=
A 53.3 10
3
× mm
2
=
1071
© 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 10
y

c
1
A
0
b
yy2a
y
b
⌠
⎮
⎮
⌡
d=
y
c
120.0 mm=
I
x'
I
x
Ay
c
2
−= I
x'
146 10
6
× mm
4
=

Problem 10-121
Determine the area moment of inertia for the triangular area about (a) the x axis, and
(b) the centroidal x' axis.
Solution:
I
x
0
h
yy
2
b
h
hy−()
⌠
⎮
⎮
⌡
d=
1
12
h
3
b⋅⋅= I
x
1
12
bh
3
=
I

x'
bh
3
12
1
2
bh
h
3
⎛
⎜
⎝
⎞
⎟
⎠
2
−=
1
36
h
3
b⋅⋅= I
x'
1
36
bh
3
=
Problem 10-122
Determine the product of inertia of the shaded area with respect to the x and y axes.

Given:
a 2in=
b 1in=
1072
© 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 10
Solution:
I
xy
0
b
y
a
2
y
b
ya
y
b
⌠
⎮
⎮
⌡
d=
I
xy
0.667 in
4

=
1073
© 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 11
Problem 11-1
The thin rod of weight W rests against the smooth wall and
floor. Determine the magnitude of force
P
needed to hold it
in equilibrium.
Solution:
x
p
L cos
θ
()
=
δ
x
p
L− sin
θ
()
δθ
=
y
w
L

2
⎛
⎜
⎝
⎞
⎟
⎠
sin
θ
()
=
δ
y
w
L
2
⎛
⎜
⎝
⎞
⎟
⎠
cos
θ
()
δθ
=
δ
U = 0;
P−

δ
x
p
W
δ
y
w
− 0=
P− L− sin
θ
()
δθ
()
W
L
2
cos
θ
()
δθ
⎛
⎜
⎝
⎞
⎟
⎠
− 0=
δθ
PLsin
θ

()
WL
2
⎛
⎜
⎝
⎞
⎟
⎠
cos
θ
()
−
⎡
⎢
⎣
⎤
⎥
⎦
0=
Since

δθ
0≠

PLsin
θ
()
WL
2

⎛
⎜
⎝
⎞
⎟
⎠
cos
θ
()
− 0=
P
W
2
cot
θ
()
=
Problem 11-2
The disk has a weight W and is subjected to a vertical force
P
and a
couple moment M. Determine the disk’s rotation
θ


if the end of the
spring wraps around the periphery of the disk as the disk turns. The
spring is originally unstretched.
Given:
W 10 lb=

P 8lb=
M 8lbft⋅=
a 1.5 ft=
1074
© 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 11
k 12
lb
ft
=
Solution:
δ
UPa
δ
θ M
δθ
+ ka
θ
a
δθ
−= Pa M+ ka
2
θ
−
()
δθ
= 0=
Pa M+ ka

2
θ
− 0=
θ
Pa M+
ka
2
=
θ
42.4 deg=
Problem 11-3
The platform supports a load W. Determine the
horizontal force
P
that must be supplied by the
screw in order to support the platform when the
links are at the arbitrary angle
θ
.
Solution:
xlcos
θ
()
=
δ
xl− sin
θ
()
δθ
=

y 2 lsin
θ
()
=
δ
y 2 lcos
θ
()
δθ
=
δ
UW−
δ
yP
δ
x−= 0=
W− 2 lcos
θ
()
δθ
()
Pl− sin
θ
()
δθ
()
− 0=
2− W cos
θ
()

P sin
θ
()
+ 0=
P 2 W cot
θ
()
=
Problem 11-4
Each member of the pin-connected mechanism has mass m
1
. If the spring is unstretched
when
θ
= 0
ο
,
determine the angle

θ
for equilibrium.
1075
© 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.