Engineering Mechanics - Statics Episode 3 Part 5 pps

Engineering Mechanics - Statics Chapter 9
r
3
r
2
r
1
−=
Solution:
V 2
π
2
π
4
⎛
⎜
⎝
⎞
⎟
⎠
r
2
2
4r
2
3
π
⎛
⎜
⎝
⎞

⎟
⎠
2r
2
2r
3
()
r
2
2
⎛
⎜
⎝
⎞
⎟
⎠
+ 2
π
2
⎛
⎜
⎝
⎞
⎟
⎠
r
3
2
r
2

4r
3
3
π
−
⎛
⎜
⎝
⎞
⎟
⎠
−
⎡
⎢
⎣
⎤
⎥
⎦
=
V 1.40 10
3
× in
3
=
Problem 9-94
A circular sea wall is made of concrete. Determine the total weight of the wall if the concrete has
a specific weight
γ
c
.

Given:
γ
c
150
lb
ft
3
=
a 60 ft=
b 15 ft=
c 8ft=
d 30 ft=
θ
50 deg=
Solution:
W
γ
c
θ
a
1
2
db c−()
⎡
⎢
⎣
⎤
⎥
⎦
2

3
bc−()
1
2
db c−()
⎡
⎢
⎣
⎤
⎥
⎦
+ ab+
c
2
−
⎛
⎜
⎝
⎞
⎟
⎠
dc+
⎡
⎢
⎣
⎤
⎥
⎦
=
W 3.12 10

6
× lb=
Problem 9-95
Determine the surface area of the tank, which consists of a cylinder and hemispherical cap.
961
© 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
Given:

a 4m=
b 8m=
Solution:
A 2
π
ab
2a
π
π
a
2
+
⎛
⎜
⎝
⎞
⎟
⎠
=

A 302 m
2
=
Problem 9-96
Determine the volume of the tank, which consists of a cylinder and hemispherical cap.
Given:

a 4m=
b 8m=
Solution:
V 2
π
4a
3
π
π
a
2
4
⎛
⎜
⎝
⎞
⎟
⎠
a
2
ba()+
⎡
⎢

⎣
⎤
⎥
⎦
=
V 536 m
3
=
962
© 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 9-97
Determine the surface area of the silo which consists of a cylinder and hemispherical cap. Neglect the
thickness of the plates.
Given:

a 10 ft=
b 10 ft=
c 80 ft=
Solution:
A 2
π
2a
π
π
a
2
⎛

⎜
⎝
⎞
⎟
⎠
ac+
⎡
⎢
⎣
⎤
⎥
⎦
=
A 5.65 10
3
× ft
2
=
Problem 9-98
Determine the volume of the silo which
consists of a cylinder and hemispherical
cap. Neglect the thickness of the plates.
Given:

a 10 ft=
b 10 ft=
c 80 ft=
Solution:
V 2
π

4a
3
π
π
a
2
4
⎛
⎜
⎝
⎞
⎟
⎠
ca
a
2
⎛
⎜
⎝
⎞
⎟
⎠
+
⎡
⎢
⎣
⎤
⎥
⎦
=

V 27.2 10
3
× ft
3
=
963
© 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 9-99
The process tank is used to store liquids during manufacturing. Estimate both the volume of the
tank and its surface area. The tank has a flat top and the plates from which the tank is made
have negligible thickness.
Given:
a 4m=
b 6m=
c 3m=
Solution:
V 2
π
c
3
ca
2
⎛
⎜
⎝
⎞
⎟

⎠
c
2
cb()+
⎡
⎢
⎣
⎤
⎥
⎦
=
V 207m
3
=
A 2
π
c
2
ccb+
c
2
a
2
c
2
++
⎛
⎜
⎝
⎞

⎟
⎠
=
A 188 m
2
=
Problem 9-100
Determine the height h to which liquid should be poured into the cup so that it contacts half the
surface area on the inside of the cup. Neglect the cup's thickness for the calculation.
Given:
a 30 mm=
b 50 mm=
c 10 mm=
Solution:
Total area
A
total
2
π
c
c
2
ac+
2
b
2
ac−()
2
++
⎡

⎢
⎣
⎤
⎥
⎦
=
Guess h 1mm= e 1mm=
964
© 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
Given
ac−
b
ec−
h
=
A
total
2
2
π
c
c
2
ec+
2
h
2

ec−()
2
++
⎡
⎢
⎣
⎤
⎥
⎦
=
e
h
⎛
⎜
⎝
⎞
⎟
⎠
Find eh,()= e 21.942 mm= h 29.9 mm=
Problem 9-101
Using integration, compute both the area and the centroidal distance x
c
of the shaded region. Then,
using the second theorem of Pappus–Guldinus, compute the volume of the solid generated by
revolving the shaded area about the aa axis.
Given:

a 8in=
b 8in=
Solution:

A
0
a
xb
x
a
⎛
⎜
⎝
⎞
⎟
⎠
2
⌠
⎮
⎮
⌡
d=
x
c
2a
1
A
0
a
xxb
x
a
⎛
⎜

⎝
⎞
⎟
⎠
2
⌠
⎮
⎮
⌡
d−=
A 21.333 in
2
= x
c
10in=
V 2
π
Ax
c
= V 1.34 10
3
× in
3
=
965
© 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 9-102

Using integration, determine the area and the centroidal
distance y
c
of the shaded area. Then, using the second
theorem of Pappus–Guldinus, determine the volume of a
solid formed by revolving the area about the x axis.
Given:

a 0.5 ft=
b 2ft=
c 1ft=
Solution:
A
a
b
x
c
2
x
⌠
⎮
⎮
⌡
d=
A 1.386 ft
2
=
y
c
1

A
a
b
x
1
2
c
2
x
⎛
⎜
⎝
⎞
⎟
⎠
2
⌠
⎮
⎮
⎮
⌡
d=
y
c
0.541 ft=
V 2
π
Ay
c
= V 4.71 ft

3
=
Problem 9-103
Determine the surface area of the roof
of the structure if it is formed by
rotating the parabola about the y axis.
Given:
a 16 m=
b 16 m=
966
© 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:
Centroid : The length of the differential element is
dL dx
2
dy
2
+= 1
dy
dx
⎛
⎜
⎝
⎞
⎟
⎠
2

+
⎡
⎢
⎣
⎤
⎥
⎦
dx=
and its centroid is
x
c
x=
Here,
dy
dx
2−
bx
a
2
=
Evaluating the integrals, we have
L
0
a
x1
4b
2
x
2
a

4
+
⌠
⎮
⎮
⎮
⌡
d=
L 23.663 m=
x
c
1
L
0
a
xx 1
4b
2
x
2
a
4
+
⌠
⎮
⎮
⎮
⌡
d=
x

c
9.178 m=
A 2
π
x
c
L= A 1.365 10
3
× m
2
=
Problem 9-104
The suspension bunker is made from plates which are curved to the natural shape which a completely
flexible membrane would take if subjected to a full load of coal.This curve may be approximated by a
parabola, y/b = (x/a)
2
. Determine the weight of coal which the bunker would contain when completely
filled. Coal has a specific weight of
γ
, and assume there is a fraction loss p in volume due to air voids.
Solve the problem by integration to determine the cross-sectional area of ABC; then use the second
theorem of Pappus–Guldinus to find the volume.
Units Used:
kip 10
3
lb=
Given:
a 10 ft=
b 20 ft=
967

© 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
γ
50
lb
ft
3
=
p 0.2=
Solution:
A
0
b
ya
y
b
⌠
⎮
⎮
⌡
d=
A 133.3 ft
2
=
x
c
1
A

0
b
y
1
2
a
y
b
⎛
⎜
⎝
⎞
⎟
⎠
2
⌠
⎮
⎮
⌡
d=
x
c
3.75 ft=
V 2
π
Ax
c
= V 3.142 10
3
× ft

3
=
W 1 p−()
γ
V= W 125.7 kip=
Problem 9-105
Determine the interior surface area of the
brake piston. It consists of a full circular
part. Its cross section is shown in the figure.
Given:
a 40 mm=
b 30 mm=
c 20 mm=
d 20 mm=
e 80 mm=
968
© 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
f 60 mm=
g 40 mm=
Solution:
A 2
π
a
2
aa
b
2

+
⎛
⎜
⎝
⎞
⎟
⎠
b
2
e
2
++ ca b+
c
2
+
⎛
⎜
⎝
⎞
⎟
⎠
+ ab+ c+()f+ ab+
3c
2
+
⎛
⎜
⎝
⎞
⎟

⎠
c+
ab+ 2c+()g+

⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
=
A 119 10
3
× mm
2
=
Problem 9-106
Determine the magnitude of the resultant hydrostatic force acting on the dam and its location H,
measured from the top surface of the water. The width of the dam is w; the mass density is
ρ
w
.
Units Used:
Mg 10
3
kg=
MN 10
6

N=
Given:
w 8m=
ρ
w
1
Mg
m
3
=
h 6m=
g 9.81
m
s
2
=
Solution:
ph
ρ
w
g= p 58860
N
m
2
=
F
1
2
hwp=
F 1.41 MN=

H
2
3
⎛
⎜
⎝
⎞
⎟
⎠
h= H 4m=
969
© 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 9-107
The tank is filled with water to a depth d. Determine the resultant force the water exerts on side A
and side B of the tank. If oil instead of water is placed in the tank, to what depth d should it reach so
that it creates the same resultant forces? The densities are
ρ
0
and
ρ
w
.
Given:
kN 10
3
N=
d 4m=

a 3m=
b 2m=
ρ
o
900
kg
m
3
=
ρ
w
1000
kg
m
3
=
g 9.81
m
s
2
=
Solution:
For water
At side A:
W
A
b
ρ
w
gd= W

A
78480
N
m
=
F
RA
1
2
W
A
d= F
RA
157kN=
At side B:
W
B
a
ρ
w
gd= W
B
117720
N
m
=
F
RB
1
2

W
B
d= F
RB
235kN=
For oil
At side A:
F
RA
1
2
b
ρ
o
gd
1
d
1
=
d
1
2F
RA
b
ρ
o
g
= d
1
4.216 m=

970
© 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 9-108
The factor of safety for tipping of the concrete dam is defined as the ratio of the stabilizing moment
about O due to the dam’s weight divided by the overturning moment about O due to the water pressure.
Determine this factor if the concrete has specific weight
γ
conc
and water has specific weight
γ
w
.
Given:
a 3ft=
b 15 ft=
c 9ft=
γ
w
62.4
lb
ft
3
=
γ
conc
150
lb

ft
3
=
Solution:
For a 1-ft thick section:
W
γ
w
b 1ft()=
W 936
lb
ft
=
F
1
2
Wb= F 7020lb=
W
1
γ
conc
1ft()ab= W
1
6750lb=
W
2
γ
conc
1
2

ca−()b 1ft()=
W
2
6750lb=
Moment to overturn:
M
O
F
1
3
b= M
O
35100lb ft=
Moment to stabilize:
M
S
W
1
ca−()
a
2
+
⎡
⎢
⎣
⎤
⎥
⎦
W
2

2
3
ca−()
⎡
⎢
⎣
⎤
⎥
⎦
+= M
S
77625lb ft⋅=
F
s
M
S
M
O
= F
s
2.21=
971
© 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 9-109
The concrete "gravity" dam is held in place by its own weight. If the density of concrete is
ρ
c

and
water has a density
ρ
w
, determine the smallest dimension d that will prevent the dam from
overturning about its end A.
Units Used:
Mg 10
3
kg=
Given:
ρ
c
2.5
Mg
m
3
=
ρ
w
1.0
Mg
m
3
=
h 6m=
g 9.81
m
s
2

=
Solution:
Consider a dam of width
a 1m=
.
w
ρ
w
gha= w 58860
N
m
= F
1
2
wh= F 176580N=
W
1
2
ρ
c
gdha=
Equilibrium W
2d
3
F
h
3
− 0=
1
2

ρ
c
gdha
2d
3
Fh
3
=
d
F
ρ
c
ga
= d 2.683 m=
Problem 9-110
The concrete dam is designed so that its face AB has a gradual slope into the water as shown.
Because of this, the frictional force at the base BD of the dam is increased due to the hydrostatic
force of the water acting on the dam. Calculate the hydrostatic force acting on the face AB of the
dam. The dam has width w, the water density is
γ
w
.
972
© 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
Units Used:
kip 10
3

lb=
Given:
w 60 ft=
γ
w
62.4
lb
ft
3
=
a 18 ft=
b 12 ft=
Solution:
F
AB
1
2
w
γ
w
ba
2
b
2
+= F
AB
486kip=
Problem 9-111
The symmetric concrete “gravity” dam is held in place by its own weight. If the density of concrete is
ρ

c
and water has a density
ρ
w
,

determine the smallest distance d at its base that will prevent the dam
from overturning about its end A.The dam has a width w.
Units Used:
Mg 10
3
kg= MN 10
6
N=
Given:
a 1.5 m=
ρ
c
2.5
Mg
m
3
=
b 9m=
ρ
w
1.0
Mg
m
3

=
w 8m=
Solution:
Guesses
d 3m= F
h
1MN=
F
v
1MN= W 1MN=
973
© 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
Given
F
v
da−
4
bw
ρ
w
g=
F
h
1
2
ρ
w

gbwb=
W
ρ
c
gw ab
da−
2
⎛
⎜
⎝
⎞
⎟
⎠
b+
⎡
⎢
⎣
⎤
⎥
⎦
=
W
d
2
F
v
d
da−
6
−

⎛
⎜
⎝
⎞
⎟
⎠
+ F
h
b
3
− 0=
F
v
F
h
W
d
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
Find F

v
F
h
, W, d,
()
=
F
v
F
h
W
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
0.379
3.178
4.545
⎛
⎜
⎜
⎝
⎞
⎟

⎟
⎠
MN= d 3.65 m=
Problem 9-112
The tank is used to store a liquid having a specific weight
γ
.
If it is filled to the top, determine
the magnitude of force the liquid exerts on each of its two sides ABDC and BDFE.
Units used:
kip 10
3
lb=
Given:
γ
80
lb
ft
3
=
a 6ft=
b 6ft=
c 12 ft=
d 8ft=
e 4ft=
974
© 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:
Fluid Pressure:

The fluid pressure at points B
and E can be determined using
p
B
γ
e= p
B
320
lb
ft
2
=
p
E
γ
ed+()= p
E
960
lb
ft
2
=
Thus
w
B
p
B

c= w
B
3.84
kip
ft
=
w
E
p
E
c= w
E
11.52
kip
ft
=
Resultant Forces: The resultant Force acts on surface ABCD is
F
R1
1
2
w
B
e
2
b
2
+= F
R1
13.8 kip=

and on surface BDFE is
F
R2
1
2
w
B
w
E
+
()
d= F
R2
61.4 kip=
Problem 9-113
The rectangular gate of width w is pinned
at its center A and is prevented from
rotating by the block at B. Determine the
reactions at these supports due to
hydrostatic pressure.
Units Used:
Mg 10
3
kg= kN 10
3
N=
Given:
a 1.5 m=
ρ
w

1.0
Mg
m
3
=
b 6m=
975
© 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
w 2m= g 9.81
m
s
2
=
Solution:
w
1
ρ
w
gb 2a−()w= w
1
59
kN
m
=
w
2
ρ

w
g2aw= w
2
59
kN
m
=
F
1
1
2
2aw
1
= F
1
88kN=
F
2
w
2
2a= F
2
177kN=
Σ
M
A
= 0;

F
1

a
3
F
B
a− 0= F
B
1
3
F
1
= F
B
29.4 kN=
Σ
F
x
= 0;
F
1
F
2
+ F
B
− F
A
− 0= F
A
F
1
F

2
F
B
−+= F
A
235kN=
Problem 9-114
The gate AB has width w. Determine the horizontal and vertical components of force acting on
the pin at B and the vertical reaction at the smooth support A. The density of water is
ρ
w
.
Units Used:
Mg 10
3
kg=
kN 10
3
N=
MN 10
6
N=
Given:
w 8m=
ρ
w
1.0
Mg
m
3

=
a 5m=
b 4m=
976
© 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
c 3m=
g 9.81
m
s
2
=
Solution:
Fluid Pressure: The fluid pressure at points A and B can be determined using Eq. 9-15,
p
A
ρ
w
ga b+()= p
A
88.29
kN
m
2
= w
A
p
A

w= w
A
706.32
kN
m
=
p
B
ρ
w
ga= p
B
49.05
kN
m
2
= w
B
p
B
w= w
B
392.4
kN
m
=
Equilibrium
w
B
b

2
c
2
+
2
1
2
w
A
w
B
−
()
b
2
c
2
+
3
+ A
y
c− 0=
A
y
w
B
b
2
c
2

+
2
1
2
w
A
w
B
−
()
2 b
2
c
2
+
()
3
+
c
= A
y
2.507 MN=
A
y
w
B
c−
1
2
w

A
w
B
−
()
c− B
y
− 0=
B
y
A
y
w
B
c−
1
2
w
A
w
B
−
()
c−= B
y
858.92 kN=
B
x
− w
B

b+
1
2
w
A
w
B
−
()
b+ 0=
B
x
w
B
b
1
2
w
A
w
B
−
()
b+= B
x
2.197 MN=
Problem 9-115
The storage tank contains oil having a specific weight
γ
. If the tank has width w, calculate the

resultant force acting on the inclined side BC of the tank, caused by the oil, and specify its location
along BC, measured from B. Also compute the total resultant force acting on the bottom of the tank.
977
© 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
Units Used:
kip 10
3
lb=
Given:
γ
56
lb
ft
3
=
c 8ft=
w 6ft= d 4ft=
a 10 ft= e 3ft=
b 2ft= f 4ft=
Solution:
w
B
w
γ
b= w
B
672

lb
ft
= w
C
w
γ
bc+()= w
C
3360
lb
ft
=
F
h1
w
B
c= F
h2
1
2
w
C
w
B
−
()
c= F
v1
γ
wbe= F

v2
1
2
γ
wce=
The resultant force
F
Rx
F
h1
F
h2
+= F
Ry
F
v1
F
v2
+= F
R
F
Rx
2
F
Ry
2
+= F
R
17.225 kip=
The location h measured from point B

Guess h 1ft= Given
F
v1
e
2
F
v2
2e
3
+ F
h1
c
2
+ F
h2
2c
3
+ F
Rx
ch
c
2
e
2
+
F
Ry
eh
c
2

e
2
+
+=
h Find h()= h 5.221 ft=
On the bottom of the tank
F
bot
γ
wf b c+ d+()= F
bot
18.816 kip=
Problem 9-116
The arched surface AB is shaped in the form of a quarter circle. If it has a length L, determine the
horizontal and vertical components of the resultant force caused by the water acting on the surface.
The density of water is
ρ
w
.
978
© 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
Units Used:
Mg 10
3
kg=
kN 10
3

N=
Given:
L 8m=
ρ
w
1.0
Mg
m
3
=
a 3m=
b 2m=
g 9.81
m
s
2
=
Solution:
F
3
ρ
w
gabL= F
3
470.88 kN=
F
2
ρ
w
gabL= F

2
470.88 kN=
F
1
ρ
w
g
b
2
bL= F
1
156.96 kN=
Wb
2
π
b
2
4
−
⎛
⎜
⎝
⎞
⎟
⎠
L
ρ
w
g= W 67.368 kN=
F

x
F
1
F
2
+= F
x
628kN=
F
y
F
2
W+= F
y
538kN=
Problem 9-117

The rectangular bin is filled with coal, which creates a pressure distribution along wall A that
varies as shown, i.e. p = p
0
(z/b)
3
. Determine the resultant force created by the coal and specify
its location measured from the top surface of the coal.
Units used:
kip 10
3
lb=
979
© 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
Given:
a 4ft=
b 10 ft=
p
0
4000
lb
ft
2
=
Solution:
Resultant Force and its location:
F
0
b
zp
0
z
b
⎛
⎜
⎝
⎞
⎟
⎠
3
a

⌠
⎮
⎮
⌡
d=
F 40kip=
z
c
1
F
0
b
zzp
0
z
b
⎛
⎜
⎝
⎞
⎟
⎠
3
a
⌠
⎮
⎮
⌡
d=
z

c
8ft=
Problem 9-118


The semicircular drainage pipe is filled with water. Determine the resultant horizontal and
vertical force components that the water exerts on the side AB of the pipe per foot of pipe
length; water has density
.
γ

Given:
γ
62.4
lb
ft
3
=
r 2ft=
Solution:
w
γ
r= w 124.8
lb
ft
2
=
Resultant forces (per unit foot):
F
Rh

1
2
wr= F
Rh
124.8
lb
ft
=
980
© 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
F
Rv
γ
π
r
2
4
= F
Rv
196.0
lb
ft
=
Problem 9-119
The load over the plate varies linearly along the sides of the plate such that p = k y (a-x). Determine the
magnitude of the resultant force and the coordinates (x
c

, y
c
) of the point where the line of action of the
force intersects the plate.
Given:
a 2ft=
b 6ft=
k 10
lb
ft
4
=
Solution:
pxy,()ky a x−()=
F
R
0
a
x
0
b
ypxy,()
⌠
⎮
⌡
d
⌠
⎮
⌡
d= F

R
360lb=
x
c
1
F
R
0
a
x
0
b
yxp x y,()
⌠
⎮
⌡
d
⌠
⎮
⌡
d= x
c
0.667 ft=
y
c
1
F
R
0
a

x
0
b
yyp x y,()
⌠
⎮
⌡
d
⌠
⎮
⌡
d= y
c
4ft=
Problem 9-120
The drum is filled to its top (y = a) with oil having a density
γ
.

Determine the resultant force of the oil
pressure acting on the flat end of plate A of the drum and specify its location measured from the top of
the drum.
981
© 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
Given:

a 1.5 ft=

γ
55
lb
ft
3
=
Solution:
F
R
a−
a
y
γ
2 a
2
y
2
− ay−()
⌠
⎮
⌡
d= F
R
583lb=
da
1
F
R
a−
a

yy
γ
2 a
2
y
2
− ay−()
⌠
⎮
⌡
d−= d 1.875 ft=
Problem 9-121
The gasoline tank is constructed with elliptical ends on each side of the tank. Determine the resultant
force and its location on these ends if the tank is half full.
Given:

a 3ft=
b 4ft=
γ
41
lb
ft
3
=
Solution:
F
R
a−
0
y

γ
− y2
b
a
a
2
y
2
−
⎛
⎜
⎝
⎞
⎟
⎠
⌠
⎮
⎮
⌡
d=
F
R
984lb=
982
© 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
y
c

1
F
R
a−
0
yy
γ
− y2
b
a
a
2
y
2
−
⎛
⎜
⎝
⎞
⎟
⎠
⎡
⎢
⎣
⎤
⎥
⎦
⌠
⎮
⎮

⌡
d=
y
c
1.767− ft= x
c
0ft=
Problem 9-122
The loading acting on a square plate is represented by a parabolic pressure distribution. Determine the
magnitude of the resultant force and the coordinates (x
c
, y
c
) of the point where the line of action of
the force intersects the plate. Also, what are the reactions at the rollers B and C and the
ball-and-socket joint A? Neglect the weight of the plate.
Units Used:
kPa 10
3
Pa=
kN 10
3
N=
Given:
a 4m=
p
0
4 kPa=
Solution:
Due to symmetry

x
c
0=
F
R
0
a
yp
0
y
a
a
⌠
⎮
⎮
⌡
d=
F
R
42.667 kN=
y
c
1
F
R
0
a
yyp
0
y

a
a
⌠
⎮
⎮
⌡
d=
y
c
2.4 m=
Equilibrium Guesses A
y
1kN= B
y
1kN= C
y
1kN=
Given A
y
B
y
+ C
y
+ F
R
− 0=
B
y
C
y

+
()
aF
R
y
c
− 0=
B
y
a
2
C
y
a
2
− 0=
983
© 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
A
y
B
y
C
y
⎛
⎜
⎜

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

⎟
⎠
17.067
12.8
12.8
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
Problem 9-123


The tank is filled with a liquid which has density
.
ρ
Determine the resultant force that it exerts
on the elliptical end plate, and the location of the center of pressure, measured from the x axis.

Units Used:
kN 10
3
N=
Given:
a 1m=
b 0.5 m=

ρ
900
kg
m
3
=
g 9.81
m
s
2
=
Solution:
F
R
b−
b
y
ρ
g2a 1
y
b
⎛
⎜
⎝
⎞
⎟
⎠
2
− by−()
⌠

⎮
⎮
⌡
d=
F
R
6.934 kN=
y
c
1
F
R
b−
b
yy
ρ
g2a 1
y
b
⎛
⎜
⎝
⎞
⎟
⎠
2
− by−()
⌠
⎮
⎮

⌡
d=
y
c
0.125− m=
Problem 9-124
A circular V-belt has an inner radius r and a cross-sectional area as shown. Determine the volume
of material required to make the belt.
984
© 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
Given:
r 600 mm=
a 25 mm=
b 50 mm=
c 75 mm=
Solution:
V 2
π
r
c
3
+
⎛
⎜
⎝
⎞
⎟

⎠
2
1
2
⎛
⎜
⎝
⎞
⎟
⎠
ac r
c
2
+
⎛
⎜
⎝
⎞
⎟
⎠
bc+
⎡
⎢
⎣
⎤
⎥
⎦
= V 22.4 10
3−
× m

3
=
Problem 9-125
A circular V-belt has an inner radius r and a cross-sectional area as shown. Determine the surface
area of the belt.
Given:
r 600 mm=
a 25 mm=
b 50 mm=
c 75 mm=
Solution:
A 2
π
rb 2 r
c
2
+
⎛
⎜
⎝
⎞
⎟
⎠
a
2
c
2
++ rc+()b 2a+()+
⎡
⎢

⎣
⎤
⎥
⎦
= A 1.246 m
2
=
985
© 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.