Engineering Mechanics - Statics Episode 3 Part 3 docx

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

0.5747 in=
Equilibrium:
+
↑
Σ
F
y
= 0;
R
y
F− 0= R
y
F= R
y
20.00 lb=
+
→
Σ
F
x
= 0;
PR
x
− 0= R
x
P=
RR
x
2
R

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
0.75 in=
881
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Engineering Mechanics - Statics Chapter 8
d
2
2in=
μ
s
0.2=
Solution:
r
fA
1
2
d
2
μ
s
= r
fA
0.2 in=
r
fB
1
2
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

1
2
d
2
μ
s
= r
fA
7.50 mm=
r
fB
1
2
d
1
μ
s
= r
fB
3mm=
882
© 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 8
The lawn roller has mass M. If the arm BA is held at angle
θ
from the horizontal and the
coefficient of rolling resistance for the roller is r, determine the force
P

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

= 0;
r− Mg Psin
θ
()
r− P cos
θ
()
acos
θ
1
()
+ 0=
P
rMg
sin
θ
()
− r cos
θ
()
a cos
θ
1
()
+
=
P 96.7 N=
Problem 8-130
The handcart has wheels with a diameter D. If a crate having a weight W is placed on the cart,
determine the force

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
+
⎛
⎜
⎝

⎞
⎟
⎠
− 0=
b
b
2
c
2
+
⎛
⎜
⎝
⎞
⎟
⎠
PN
2a
D
⎛
⎜
⎝
⎞
⎟
⎠
=
N
P
⎛
⎜

⎝
⎞
⎟
⎠
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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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 8
Solution:
+
→
Σ
F
x
= 0;

R
Ax
P− 0= R
Ax
P=
+
↑
Σ
F
y
= 0;
R
Ay
W− 0= R
Ay
P=
Σ
M
B
= 0;
Prcos
φ
A
()
rcos
φ
B
()
+
()

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

+
()
2r
=
(QED)
Problem 8-132
A steel beam of mass M is moved over a level surface using a series of rollers of diameter D
for which the coefficient of rolling resistance is a
g
at the ground and a
s
at the bottom surface
of the beam. Determine the horizontal force
P
needed to push the beam forward at a constant
speed. Hint: Use the result of Prob. 8–131.
Units Used:
Mg 1000 kg=
885
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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 8
Given:
M 1.2 Mg=
D 30 mm=
a
g
0.4 mm=
a

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

. Hint: Use the result of Prob. 8-131.
Units Used:
Mg 1000 kg=
Given:
M 1.4 Mg=
a
g
0.5 mm=
a
m
0.2 mm=
P 250 N=
Solution:
P
Mga
g
a
m
+
()
2 r
=
rMg
a
g
a
m
+
2 P
⎛

⎜
⎝
⎞
⎟
⎠
= r 19.2 mm= d 2 r= d 38.5mm=
886
© 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 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=

b 1.25 m=
Solution:
Equation of Equilibrium and Friction :

If
the drawer does not bind at corners A and
B, slipping would have to occur at points
A and B. Hence, F
A
=
μ
N
A
and F
B
=
μ
N
B
+
→
Σ
F
x
= 0;
N
B
N
A
− 0=

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

Nb+ P
sb+
2
⎛
⎜
⎝
⎞
⎟
⎠
− 0=
a
μ
s
b+ 2
μ
s
sb+
2
⎛
⎜
⎝
⎞
⎟
⎠
−
⎡
⎢
⎣
⎤
⎥

⎦
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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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 8
()
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:
M 1.25 Mg=
μ
st
0.5= a 600 mm=
b 1.5 m=
μ
sc
0.4=
c 1m=
g 9.81
m
s
2
= d 800 mm=
Solution:
Guesses N
A
1N= N
B
1N=

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
T
N
C
W
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find N
A
N
B
, T, N
C
, W,
()
= W 6.97 kN=

(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.
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 8
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
T
N
C
W
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find N
A

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.

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 8
M 1.25 Mg= a 600 mm=
μ
st
0.5= b 1.5 m=
μ
sc
0.4= c 1m=
d 800 mm=
g 9.81
m
s
2
=
Solution:
Guesses N
A
1N= N
B
1N=
T 1N= N
C
1N= W 1N=
(a) Rear wheel drive
Given
T−
μ
st

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

()
− 0=
N
C
W cos
θ
()
− 0=
N
A
N
B
T
N
C
W
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟

⎟
⎠
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

B
+ Mgcos
θ
()
− 0=
M− gbcos
θ
()
Mgdsin
θ
()
+ N
B
bc+()+ Ta+ 0=
T
μ
sc
N
C
− W sin
θ
()
− 0=
N
C
W cos
θ
()
− 0=
890

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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
T
N
C
W
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find N
A

N
B
, T, N
C
, W,
()
= W 6.89 kN=
Problem 8-137
A roofer, having a mass M, walks slowly in an upright position down along the surface of a
dome that has a radius of curvature
ρ
. If the coefficient of static friction between his shoes
and the dome is
μ
s

determine the angle
θ

at which he first begins to slip.
Given:
M 70 kg=
ρ
20 m=
μ
s
0.7=
Solution:
Σ
F

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

()
=
θ
35.0 deg=
Problem 8-138
A man attempts to lift the uniform ladder of weight W to an upright position by applying a
force
P
perpendicular to the ladder at rung R. Determine the coefficient of static friction
b
etween the ladder and the
g
round at
A
if the ladder be
g
ins to sli
p
on the
g
round when his
891
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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 8
g
gpg
hands reach height c.
Given:

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

N
A
P sin
θ
()
− 0=
Σ
F
y
= 0;
N
A
W− P cos
θ
()
+ 0=
Σ
M
A
= 0;
W−
ba+
2
⎛
⎜
⎝
⎞
⎟
⎠
cos

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

⎜
⎝
⎞
⎟
⎠
20.7
21.3
⎛
⎜
⎝
⎞
⎟
⎠
lb=
μ
A
0.41=
Problem 8-139
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
.
Determine the force
P
needed to raise the column and the equilibrium force
P'
needed to hold wedge A stationary. The contacting surface between A and D is smooth.
892

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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=

N
W
cos
θ
()
μ
s
sin
θ
()
−
=
N 8739.8 lb=
Σ
F
x
= 0;
μ
s
Ncos
θ
()
Nsin
θ
()
+ P'− 0=
P'
μ
s
N cos

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
()

N cos
θ
()
+=
N
C
8000lb=
Σ
F
x
= 0;
P
μ
s
N
C
− Nsin
θ
()
−
μ
s
Ncos
θ
()
− 0=
P
μ
s
N

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.
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 8
Given:

W 8000 lb=
θ
10 deg=
μ
s
0.4=
Solution:
Wedge A:
Σ
F
y
= 0;
NWcos
θ
()
− 0=
NWcos
θ
()
=
N 7878.5 lb=
Σ
F
x
= 0;
W sin
θ
()
F− 0=
FWsin

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
()

− 0=
N
C
F sin
θ
()
N cos
θ
()
+=
N
C
8000lb=
Σ
F
x
= 0;
F
C
F cos
θ
()
+ Nsin
θ
()
− 0=
F
C
F− cos
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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Engineering Mechanics - Statics Chapter 9
Problem 9-1
Locate the center of mass of the
homogeneous rod bent in the form of a
parabola.
Given:

a 1m=
b 2m=
Solution:
yb
x
a
⎛
⎜
⎝
⎞
⎟
⎠
2
=
d y
d x
2b
a
2
x=
y
c
0

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

0
a
x1
2b
a
2
x
⎛
⎜
⎝
⎞
⎟
⎠
2
+
⌠
⎮
⎮
⎮
⌡
d
= y
c
0.912 m= x
c
0m=
Problem 9-2
Locate the center of gravity x
c
of the homogeneous

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.
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Engineering Mechanics - Statics Chapter 9
Solution:
yb
x
a
⎛
⎜
⎝
⎞
⎟
⎠
2

=
d y
d x
2 b
a
2
x=
L
0
a
x1
2 b
a
2
x
⎛
⎜
⎝
⎞
⎟
⎠
2
+
⌠
⎮
⎮
⎮
⌡
d=
W

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=

Guesses A
y
1lb= B
x
1lb= B
y
1lb=
Given B
x
0= A
y
B
y
+ W− 0= A
y
− aWax
c
−
()
+ 0=
B
x
B
y
A
y
⎛
⎜
⎜
⎜

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

⎠
0.000
0.720
0.442
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
Problem 9-3
Locate the center of mass of the homogeneous rod bent into the shape of a circular arc.
Given:
r 300 mm=
θ
30 deg=
Solution:
y
c
0= Symmetry
896
© 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
x
c

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

⌠
⎮
⎮
⌡
d
= x
c
124.049 mm=
Problem 9-4
Locate the center of gravity x
c
of the homogeneous rod bent in the form of a semicircular arc.
The rod has a weight per unit length
γ
. Also, determine the horizontal reaction at the smooth
support B and the x and y components of reaction at the pin A.
Given:
γ
0.5
lb
ft
=
r 2ft=
Solution:
x
c
π
−
2
π

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

= 0;
π
− r
γ
x
c
B
x
2 r()+ 0= B
x
π
r
γ
x
c
2r
= B
x
1lb=
897
© 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
+
→
Σ
F
x
= 0;

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=

Problem 9-5
Determine the distance x
c
to the center of gravity of the homogeneous rod bent into the parabolic
shape. If the rod has a weight per unit length
γ
determine the reactions at the fixed support O.
Given:
γ
0.5
lb
ft
=
a 1ft=
b 0.5 ft=
Solution:
yb
x
a
⎛
⎜
⎝
⎞
⎟
⎠
2
=
d y
d x
2 bx

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

a
xx 1
2 bx
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
+
⌠
⎮
⎮
⎮
⌡
d=
x
c
0.531 ft=
+
→
Σ
F
x
= 0;
O
x

0= O
x
0lb= O
x
0lb=
898
© 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
+
↑
Σ
F
y
= 0;
O
y
γ
L− 0= O
y
γ
L= O
y
0.574 lb=
Σ
M
O
= 0;
M

O
γ
Lx
c
− 0= M
O
γ
Lx
c
= M
O
0.305 lb ft⋅=
Problem 9-6
Determine the distance y
c
to the center of gravity of the homogeneous rod bent into the parabolic
shape.
Given:
a 1ft=
b 0.5 ft=
Solution:
yb
x
a
⎛
⎜
⎝
⎞
⎟
⎠

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

y
c
1
L
0
a
xb
x
a
⎛
⎜
⎝
⎞
⎟
⎠
2
1
2 bx
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
+
⌠
⎮

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

dAxdy=
x
c
x
2
=
y
c
y=
A
0
h
yb
y
h
⌠
⎮
⎮
⌡
d= hb
y
h
⎛
⎜
⎝
⎞
⎟
⎠
1
2

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

=
x
c
3
8
b=
y
c
3
2hb
0
h
yyb
y
h
⌠
⎮
⎮
⌡
d=
3
5
h
h
h
⎛
⎜
⎝
⎞
⎟

⎠
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.

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
xa
y
b
=
y
c
0
b
yy2a
y
b
⌠
⎮
⎮
⌡
d
0
b
y2a
y
b
⌠
⎮
⎮
⌡
d
= y

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
⌠
⎮
⎮
⌡

d
0
b
xh
x
2
b
2
⌠
⎮
⎮
⎮
⌡
d
=
3
4
b
4
b
3
=
x
c
3
4
b=
y
c
0

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

⌠
⎮
⎮
⌡
d
=
3
10
b
5
h
2
b
5
h
= y
c
3
10
h=
901
© 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
Problem 9-10
Determine the location (x
c
, y
c

) 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

a= x
c
2
3
a=
y
c
2
ma
2
0
a
x
1
2
mx()
2
⌠
⎮
⎮
⌡
d=
1
3
am=
y
c
m
3
a=

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
© 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:
x
2
y
2
+ a

2
=
ya
2
x
2
−=
A
π
a
2
4
= WA
γ
= W 62.832lb=
x
c
1
A
0
a
xxa
2
x
2
−
⌠
⎮
⌡
d= x

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=

Given T
A
T
B
+ W− 0= T
B
aWx
c
− 0=
T
A
T
B
⎛
⎜
⎝
⎞
⎟
⎠
Find T
A
T
B
,
()
=
T
A
T
B

⎛
⎜
⎝
⎞
⎟
⎠
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
=

A
0
L
xasin
π
x
L
⎛
⎜
⎝
⎞
⎟
⎠
⌠
⎮
⎮
⌡
d=
2
π
La=
x
c
π
2La
0
L
xxasin
π
x

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

x
L
⎛
⎜
⎝
⎞
⎟
⎠
⎛
⎜
⎝
⎞
⎟
⎠
2
⌠
⎮
⎮
⌡
d=
1
8
π
a= y
c
1
8
π
a=
Problem 9-13

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

x
b
⎛
⎜
⎝
⎞
⎟
⎠
2
⌠
⎮
⎮
⌡
d=
W
ρ
gV=
x
c
1
V
b−
0
xxca
x
b
⎛
⎜
⎝
⎞

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

⌡
d=
904
© 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
z
c
1
V
b−
0
x
c−
2
a
x
b
⎛
⎜
⎝
⎞
⎟
⎠
2
⎡
⎢
⎣
⎤

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

⎜
⎝
⎞
⎟
⎟
⎠
m=
Guesses A
x
1N= A
y
1N= A
z
1N= M
A
1Nm⋅=
Given A
x
0= A
y
0= A
z
W− 0= M
A
Wb x
c
+
()
− 0=
A

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

A
x
A
y
A
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
0.00
0.00
52.32
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN= M
A
52.32 kN m⋅=

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−

0
xx
b
a
2
x
2
⌠
⎮
⎮
⌡
d=
3−
4
a= x
c
3−
4
a=
y
c
3
ab
a−
0
x
1
2
−
b

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.