PROBLEM 5.1
Locate the centroid of the plane area shown.
SOLUTION
A, in 2
x , in.
y , in.
xA, in 3
yA, in 3
1
8 × 6 = 48
−4
9
−192
432
2
16 × 12 = 192
8
6
1536
1152
Σ
240
1344
1584
Then
X =
Σ xA 1344 in 3
=
ΣA
240 in 2
or X = 5.60 in. W
and
Y =
Σ yA 1584 in 3
=
ΣA
240 in 2
or Y = 6.60 in. W
PROBLEM 5.2
Locate the centroid of the plane area shown.
SOLUTION
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
1
× 60 × 75 = 2250
2
40
25
90 000
56 250
2
105 × 75 = 7875
112.5
37.5
885 900
295 300
Σ
10 125
975 900
351 600
Then
X =
ΣxA 975 900 mm3
=
ΣA
10 125 mm 2
or X = 96.4 mm W
and
Y =
Σ yA 351 600 mm3
=
ΣA
10 125 mm 2
or Y = 34.7 mm W
PROBLEM 5.3
Locate the centroid of the plane area shown.
SOLUTION
For the area as a whole, it can be concluded by observation that
Y =
Then
2
( 24 in.)
3
or Y = 16.00 in. W
A, in 2
x , in.
xA, in 3
1
1
× 24 × 10 = 120
2
2
(10 ) = 6.667
3
800
2
1
× 24 × 16 = 192
2
Σ
312
X =
10 +
1
(16 ) = 15.333
3
2944
3744
Σ xA 3744 in 3
=
ΣA
312 in 2
or X = 12.00 in. W
PROBLEM 5.4
Locate the centroid of the plane area shown.
SOLUTION
1
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
21 × 22 = 462
1.5
11
693
5082
2
−
1
( 6 )( 9 ) = −27
2
−6
2
162
−54
3
−
1
( 6 )(12 ) = −36
2
8
2
−288
−72
567
4956
Σ
399
Then
X =
Σ xA 567 mm3
=
ΣA
399 mm 2
or X = 1.421 mm W
and
Y =
Σ yA 4956 mm3
=
ΣA
399 mm 2
or Y = 12.42 mm W
PROBLEM 5.5
Locate the centroid of the plane area shown.
SOLUTION
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
120 × 200 = 24 000
60
120
1 440 000
2 880 000
2
−
π ( 60 )
94.5
120
−534 600
−678 600
905 400
2 201 400
Σ
2
2
= −5654.9
18 345
Then
X =
Σ xA 905 400 mm3
=
ΣA
18 345 mm 2
or X = 49.4 mm W
and
Y =
Σ yA 2 201 400 mm3
=
ΣA
18 345 mm 2
or Y = 93.8 mm W
PROBLEM 5.6
Locate the centroid of the plane area shown.
SOLUTION
A, in 2
1
π (9)
4
2
= 63.617
2
1
(15)( 9 ) = 67.5
2
Σ
131.1
x , in.
y , in.
x A, in 3
y A, in 3
−4 ( 9 )
= −3.8917
( 3π )
3.8917
−243
243
5
3
337.5
202.5
94.5
445.5
Then
X =
Σ xA
94.5 in 3
=
ΣA
131.1 in 2
or X = 0.721 in. W
and
Y =
Σ yA 445.5 in 3
=
ΣA
131.1 in 2
or Y = 3.40 in. W
PROBLEM 5.7
Locate the centroid of the plane area shown.
SOLUTION
First note that symmetry implies X = Y
A, mm 2
x , mm
xA, mm3
1
40 × 40 = 1600
20
32 000
2
−
16.98
−21 330
Σ
Then
π (40) 2
4
= −1257
343
X =
Σ xA 10 667 mm3
=
ΣA
343 mm 2
10 667
or X = 31.1 mm W
and Y = X = 31.1 mm W
PROBLEM 5.8
Locate the centroid of the plane area shown.
SOLUTION
X =0 W
First note that symmetry implies
y , in.
yA, in 3
= −25.13
1.6977
−42.67
= 56.55
2.546
144
A, in 2
1
2
−
π ( 4)
2
π ( 6)
2
Σ
Then
2
2
31.42
Y =
Σ yA 101.33 in 3
=
ΣA
31.42 in 2
101.33
or Y = 3.23 in. W
PROBLEM 5.9
For the area of Problem 5.8, determine the ratio r2 /r1 so that y = 3r1/4.
SOLUTION
1
2
Σ
A
y
yA
− r12
2
π
4r1
3π
2
− r13
3
π
4r2
3π
2 3
r2
3
2
π
r22
(r − r )
2
2
2
(
Then
Y ΣA = Σy A
or
3
π 2
2 3
r1 ×
r2 − r12 =
r2 − r13
4
2
3
(
)
(
)
2
  r 3
9π  r2 
  − 1 =  2  − 1
16  r1 
  r1 


Let
p=
r2
r1
9π
( p + 1)( p − 1)] = ( p − 1)( p 2 + p + 1)
[
16
or
)
2 3
r2 − r13
3
2
1
16 p 2 + (16 − 9π) p + (16 − 9π) = 0
PROBLEM 5.9 CONTINUED
Then
or
Taking the positive root
p=
−(16 − 9π) ± (16 − 9π) 2 − 4(16)(16 − 9π)
2(16)
p = −0.5726
p = 1.3397
r2
= 1.340 W
r1
PROBLEM 5.10
Show that as r1 approaches r2 , the location of the centroid approaches that
of a circular arc of radius ( r1 + r2 ) / 2.
SOLUTION
First, determine the location of the centroid.
y2 =
From Fig. 5.8A:
=
y1 =
Similarly
Then
(
)
2
cos α
r2 π
3
−α
2
(
2
cos α
r1
3 π2 − α
(
)
A2 =
)
A1 =
)
(
)
(
)
)
( π2 − α ) r12
(
(
Now
( π2 − α ) r22
2
cosα  π
2
cosα  π
− α r22  − r1 π
− α r12 
r2 π
2


 2

3
3
α
α
−
−
2
2
2 3
=
r2 − r13 cosα
3
π

π

Σ A =  − α  r22 −  − α  r12
2

2

π
 2
2
=  − α  r2 − r1
2

Y Σ A = Σ yA
 π
 2 3

Y  − α  r22 − r12  =
r2 − r13 cos α

 2
 3
2 r23 − r13 cos α
Y =
3 r22 − r12 π2 − α
Σ yA =
(
and
(
π
2 sin 2 − α
r2 π
3
−α
2
(
)
)
)
(
)
(
)
PROBLEM 5.10 CONTINUED
Using Figure 5.8B, Y of an arc of radius
1
( r1 + r2 ) is
2
Y =
=
Now
Let
Then
(
sin π − α
1
( r1 + r2 ) π 2
2
−α
2
(
)
)
1
cos α
(r1 + r2 ) π
2
−α
2
(
(
(1)
)
( r2 − r1 ) r2 + r1 r2 + r1
r23 − r13
=
2
2
r2 − r1
( r2 − r1 )( r2 + r1 )
2
r + r1 r2 + r12
= 2
r2 + r1
2
2
)
r2 = r + ∆
r1 = r − ∆
1
r = ( r1 + r2 )
2
( r + ∆ ) + ( r + ∆ )( r − ∆ ) + ( r − ∆ )
r23 − r13
and
=
2
2
r2 − r1
(r + ∆) + (r − ∆)
3r 2 + ∆ 2
=
2r
In the limit as ∆ → 0 (i.e., r1 = r2 ), then
2
so that
Which agrees with Eq. (1).
r23 − r13
3
= r
2
2
2
r2 − r1
3 1
= × (r1 + r2 )
2 2
2 3
cos α
Y = × ( r1 + r2 ) π
3 4
−α
2
2
or Y =
1
cos α
W
( r1 + r2 ) π
2
−α
2
PROBLEM 5.11
Locate the centroid of the plane area shown.
SOLUTION
X =0 W
First note that symmetry implies
r2 = 2 2 in., α = 45°
( ) ()
π
y2′ =
3
Σ
( )
A, in 2
y , in.
y A, in 3
1
( 4)(3) = 6
2
1
6
( 2 2 ) = 6.283
4
2 − y ′ = 0.3024
1.8997
1
( 4)( 2) = −4
2
0.6667
−2.667
1
2
2r sin α 2 2 2 sin 4
=
= 1.6977 in.
3α
3 π4
π
−
2
8.283
5.2330
Y Σ A = Σ yA
Then
(
)
Y 8.283 in 2 = 5.2330 in 3
or Y = 0.632 in. W
PROBLEM 5.12
Locate the centroid of the plane area shown.
SOLUTION
1
2
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
(40)(90) = 3600
−15
20
−54 000
72 000
π ( 40)( 60)
10
−15
6750
−10 125
−25.47
−19.099
−54 000
−40 500
−101 250
21 375
4
= 2121
3
1
(30)( 45) = 675
2
Σ
6396
XA = Σ xA
Then
(
)
X 6396 mm 2 = −101 250 mm3
or X = −15.83 mm W
YA = Σ yA
and
(
)
Y 6396 mm 2 = 21 375 mm3
or Y = 3.34 mm W
PROBLEM 5.13
Locate the centroid of the plane area shown.
SOLUTION
1
2
Σ
−
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
2
( 40)(80) = 2133
3
48
15
102 400
32 000
1
( 40)(80) = −1600
2
53.33
13.333
−85 330
−21 330
17 067
10 667
533.3
X Σ A = Σ XA
Then
(
)
X 533.3 mm 2 = 17 067 mm3
or X = 32.0 mm W
Y Σ A = Σ yA
and
(
)
Y 533.3 mm 2 = 10 667 mm3
or Y = 20.0 mm W
PROBLEM 5.14
Locate the centroid of the plane area shown.
SOLUTION
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
2
(150 )( 240 ) = 24 000
3
56.25
96
1 350 000
2 304 000
2
−
1
(150)(120) = −9000
2
50
40
−450 000
−360 000
900 000
1 944 000
Σ
15 000
X Σ A = Σ xA
Then
(
)
X 15 000 mm 2 = 900 000 mm3
or X = 60.0 mm W
Y Σ A = Σ yA
and
(
)
Y 15 000 mm 2 = 1 944 000
or Y = 129.6 mm W
PROBLEM 5.15
Locate the centroid of the plane area shown.
SOLUTION
1
2
Σ
A, in 2
x , in.
y , in.
xA, in 3
yA, in 3
1
(10)(15) = 50
3
4.5
7.5
225
375
6.366
16.366
1125
2892
1350
3267
π
4
(15)2 = 176.71
226.71
X Σ A = Σx A
Then
(
)
X 226.71 in 2 = 1350 in 3
or X = 5.95 in. W
Y ΣA = Σy A
and
(
)
Y 226.71 in 2 = 3267 in 3
or Y = 14.41 in. W
PROBLEM 5.16
Locate the centroid of the plane area shown.
SOLUTION
1
2
Σ
−
A, in 2
x , in.
y , in.
xA, in 3
yA, in 3
2
(8)(8) = 42.67
3
3
2.8
128
119.47
2
( 4)( 2) = −5.333
3
1.5
−0.8
−8
4.267
120
123.73
37.33
X Σ A = Σx A
Then
(
)
X 37.33 in 2 = 120 in 3
or X = 3.21 in. W
Y ΣA = Σy A
and
(
)
Y 37.33 in 2 = 123.73 in 3
or Y = 3.31 in. W
PROBLEM 5.17
The horizontal x axis is drawn through the centroid C of the area shown
and divides the area into two component areas A1 and A2. Determine
the first moment of each component area with respect to the x axis, and
explain the results obtained.
SOLUTION
Qx = Σ yA
Note that
Then
and
(Qx )1 =  53 m  12 × 6 × 5 m 2
or ( Qx )1 = 25.0 × 103 mm3 W
( Qx )2 =  −
2
 1

 1
 1

× 2.5 m  × 9 × 2.5  m 2 +  − × 2.5 m  × 6 × 2.5  m 2
 3
 2

 3
 2

or ( Qx ) 2 = −25.0 × 103 mm3 W
Now
Qx = ( Qx )1 + ( Qx ) 2 = 0
This result is expected since x is a centroidal axis ( thus y = 0 )
and
Qx = Σ y A = Y Σ A
( y = 0 ⇒ Qx = 0 )
PROBLEM 5.18
The horizontal x axis is drawn through the centroid C of the area shown
and divides the area into two component areas A1 and A2. Determine
the first moment of each component area with respect to the x axis, and
explain the results obtained.
SOLUTION
First, locate the position y of the figure.
A, mm 2
y , mm
yA, mm3
1
160 × 300 = 48 000
150
7 200 000
2
−150 × 80 = −16 000
160
−2 560 000
Σ
32 000
Y ΣA = Σy A
Then
(
)
Y 32 000 mm 2 = 4 640 000 mm3
or
Y = 145.0 mm
4 640 000
PROBLEM 5.18 CONTINUED
A I: Q I = Σ yA
155
115
 − ( 80 × 115) 
(160 × 155) +
=
2
2 
6
3
= 1.393 × 10 mm
W
A II : Q II = Σ yA
145
85
=−
(160 × 145) − − (80 × 85)
2
 2

6
3
= −1.393 × 10 mm
W
∴
( Qarea ) x = QI + QII = 0
Which is expected since Qx = Σ yA = yA and y = 0 ,
since x is a centroidal axis.
PROBLEM 5.19
The first moment of the shaded area with respect to the x axis is denoted
by Qx . (a) Express Qx in terms of r and θ . (b) For what value of θ is
Qx maximum, and what is the maximum value?
SOLUTION
(a) With Qx = Σ yA and using Fig. 5.8 A,
(
)
 2 r sin π − θ 
  r 2 π2 − θ  −
Qx =  3 π 2

−
θ


2


2
= r 3 cos θ − cos θ sin 2 θ
3
(
) ( 32 r sin θ )  12 × 2r cos θ × r sin θ 
(
)
or Qx =
(b) By observation, Qx is maximum when
and then
2 3
r cos3 θ W
3
θ =0W
Qx =
2 3
r W
3
PROBLEM 5.20
A composite beam is constructed by bolting four plates to four
2 × 2 × 3/8-in. angles as shown. The bolts are equally spaced along the
beam, and the beam supports a vertical load. As proved in mechanics of
materials, the shearing forces exerted on the bolts at A and B are
proportional to the first moments with respect to the centroidal x axis of
the red shaded areas shown, respectively, in parts a and b of the figure.
Knowing that the force exerted on the bolt at A is 70 lb, determine the
force exerted on the bolt at B.
SOLUTION
From the problem statement: F ∝ Qx
FA
FB
so that
=
(Qx ) A (Qx ) B
(Qx ) B
F
(Qx ) A A
and
FB =
Now
Qx = ∑ yA
So
( Qx ) A =  7.5 in. +
and
( Qx )B = ( Qx ) A + 2  7.5 in. −

0.375

in.  10 in. × ( 0.375 in.)  = 28.82 in 3
2

0.375

in.  (1.625 in.)( 0.375 in.) 
2


+ 2 ( 7.5 in. − 1 in.) ( 2 in.)( 0.375 in.) 
= 28.82 in 3 + 8.921 in 3 + 9.75 in 3
= 47.49 in 3
Then
FB =
47.49 in 3
( 70 lb) = 115.3 lb W
28.82 in 3
PROBLEM 5.21
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
First note that because the wire is homogeneous, its center of gravity will
coincide with the centroid of the corresponding line.
Then
L, in.
x , in.
y , in.
xL, in 2
yL, in 2
1
16
8
0
128
0
2
12
16
6
102
72
3
24
4
12
96
288
4
6
−8
9
−48
54
5
8
−4
6
−32
48
6
6
0
3
0
18
Σ
72
336
480
X ΣL = Σ x L
X ( 72 in.) = 336 in 2
and
or X = 4.67 in.
Y ΣL = Σ y L
Y (72 in.) = 480 in 2
or Y = 6.67 in.
PROBLEM 5.22
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the
corresponding line.
Then
L, mm
x , mm
y , mm
xL, mm 2
yL, mm 2
1
165
82.5
0
13 612
0
2
75
165
37.5
12 375
2812
3
105
112.5
75
11 812
7875
4
602 + 752 = 96.05
30
37.5
2881
3602
Σ
441.05
40 680
14 289
X ΣL = Σx L
X (441.05 mm) = 40 680 mm 2
and
or X = 92.2 mm
Y ΣL = Σ y L
Y (441.05 mm) = 14 289 mm 2
Y = 32.4 mm
PROBLEM 5.23
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the
corresponding line.
Then
L, mm
x , mm
y , mm
xL, mm 2
yL, mm 2
1
122 + 62 = 13.416
6
3
80.50
40.25
2
16
12
14
192
224
3
21
1.5
22
31.50
462
4
16
−9
14
−144
224
5
62 + 92 = 10.817
−4.5
3
−48.67
32.45
Σ
77.233
111.32
982.7
X ΣL = Σx L
X (77.233 mm) = 111.32 mm 2
and
or X = 1.441 mm
Y ΣL = Σ y L
Y (77.233 mm) = 982.7 mm 2
or Y = 12.72 mm
PROBLEM 5.24
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
First note that because the wire is homogeneous, its center of gravity will
coincide with the centroid of the corresponding line.
X =0
By symmetry
Then
L, in.
y , in.
yL, in 2
1
2
0
0
2
π ( 6)
= 3.820
72
3
2
0
0
4
π ( 4)
= 2.546
32
Σ
35.416
2 ( 6)
π
2 ( 4)
π
104
Y ΣL = Σ y L
Y (35.416 in.) = 104 in 2
or Y = 2.94 in.
PROBLEM 5.25
A 750 = g uniform steel rod is bent into a circular arc of radius 500 mm
as shown. The rod is supported by a pin at A and the cord BC. Determine
(a) the tension in the cord, (b) the reaction at A.
SOLUTION
First note, from Figure 5.8B: X =
( 0.5 m ) sin 30°
π/6
=
1.5
m
π
W = mg
= ( 0.75 kg ) 9.81 m/s 2
= 7.358 N
Then
(
)
Also note that ∆ ABD is an equilateral triangle.
Equilibrium then requires
(a) ΣM A = 0:


 1.5 
m  cos 30° ( 7.358 N ) − ( 0.5 m ) sin 60° TBC = 0
0.5 m − 
 π



or
TBC = 1.4698 N
or TBC = 1.470 N
(b) ΣFx = 0: Ax + (1.4698 N ) cos 60° = 0
or
Ax = −0.7349 N
ΣFy = 0: Ay − 7.358 N + (1.4698 N ) sin 60° = 0
or
Ay = 6.085 N
thus A = 6.13 N
83.1°
PROBLEM 5.26
The homogeneous wire ABCD is bent as shown and is supported by a pin
at B. Knowing that l = 8 in., determine the angle θ for which portion
BC of the wire is horizontal.
SOLUTION
First note that for equilibrium, the center of gravity of the wire must lie
on a vertical line through B. Further, because the wire is homogeneous,
its center of gravity will coincide with the centroid of the corresponding
line.
Thus ΣM B = 0, which implies that x = 0 or ΣxL = 0
Hence
−
2(6 in.)
π
(π × 6 in.) + 
8 in. 
 ( 8 in.)
 2 
6 in.


+  8 in. −
cosθ  ( 6 in.) = 0
2


Then
cosθ =
4
9
or θ = 63.6°
PROBLEM 5.27
The homogeneous wire ABCD is bent as shown and is supported by a pin
at B. Knowing that θ = 30°, determine the length l for which portion CD
of the wire is horizontal.
SOLUTION
First note that for equilibrium, the center of gravity of the wire must lie on
a vertical line through B. Further, because the wire is homogeneous, its
center of gravity will coincide with the centroid of the corresponding line.
Thus ΣM B = 0, which implies that x = 0
Hence
or
Σxi Li = 0
 2 ( 6 in.)

−
cos 30° + ( 6 in.) sin 30°  (π × 6 in.)
 π

 ( l in.)

+
cos 30° ( l in.)
 2

6 in. 

+ ( l in.) cos 30° −
( 6 in.) = 0
2 

or
l 2 + 12.0l − 316.16 = 0
with roots l1 = 12.77 and −24.77.
Taking the positive root
l = 12.77 in.
PROBLEM 5.28
The homogeneous wire ABCD is bent as shown and is attached to a hinge
at C. Determine the length L for which the portion BCD of the wire is
horizontal.
SOLUTION
First note that for equilibrium, the center of gravity of the wire must lie on
a vertical line through C. Further, because the wire is homogeneous, its
center of gravity will coincide with the centroid of the corresponding line.
Thus ΣM C = 0, which implies that x = 0
or
Hence
or
Σ xi Li = 0
L
( L ) + ( −4 in.)(8 in.) + ( −4 in.)(10 in.) = 0
2
L2 = 144 in 2
or L = 12.00 in.
PROBLEM 5.29
Determine the distance h so that the centroid of the shaded area is as close
to line BB′ as possible when (a) k = 0.2, (b) k = 0.6.
SOLUTION
y =
Then
y =
or
=
 (a + h) 
a
( ab ) − 
  kb ( a − h ) 
2
 2 
ba − kb ( a − h )
1 a (1 − k ) + kh
2 a(1 − k ) + kh
2
c =1− k
Let
y =
Then
ΣyA
ΣA
2
ζ =
and
h
a
a c + kζ 2
2 c + kζ
(1)
Now find a value of ζ (or h) for which y is minimum:
(
)
2
dy
a 2kζ ( c + kζ ) − k c + kζ
=
=0
dζ
2
( c + kζ ) 2
or
(
)
2ζ ( c + kζ ) − c + kζ 2 = 0
(2)
PROBLEM 5.29 CONTINUED
2cζ + 2ζ 2 − c − kζ 2 = 0
Expanding (2)
ζ=
Then
or
kζ 2 + 2cζ − c = 0
( 2c) 2 − 4 ( k ) ( c )
−2c ±
2k
Taking the positive root, since h > 0 (hence ζ > 0 )
−2 (1 − k ) + 4 (1 − k ) + 4k (1 − k )
2
h=a
(a) k = 0.2:
(b) k = 0.6:
h=a
h=a
2k
−2 (1 − 0.2 ) +
4 (1 − 0.2 ) + 4 ( 0.2 )(1 − 0.2 )
−2 (1 − 0.6 ) +
4 (1 − 0.6 ) + 4 ( 0.6 )(1 − 0.6 )
2
2 ( 0.2 )
or h = 0.472a
2
2 ( 0.6 )
or h = 0.387a
PROBLEM 5.30
Show when the distance h is selected to minimize the distance y from
line BB′ to the centroid of the shaded area that y = h.
SOLUTION
From Problem 5.29, note that Eq. (2) yields the value of ζ that minimizes h.
Then from Eq. (2)
We see
c + kζ 2
c + kζ
2ζ =
(3)
Then, replacing the right-hand side of (1) by 2ζ , from Eq. (3)
We obtain
y =
a
( 2ζ)
2
h
a
But
ζ=
So
y =h
Q.E.D.
PROBLEM 5.31
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and h.
SOLUTION
y =
For the element of area (EL) shown
h
x
a
dA = ( h − y ) dx
and
x

= h 1 −  dx
a

xEL = x
Then
1
(h + y)
2
h
x
= 1 + 
2
a
yEL =
Then area
and

x
x2 
a 
−
=
−
1
h
dx
h
x


0 

2a 
a
A = ∫ dA = ∫

x
x 
a  
x h 1 −  dx  = h 
0  
∫ xEL dA = ∫
 
a
a
=
0
1
ah
2
1 2
a h
6
x  
x   h a
x2 
1
1
1
+
h
−
dx
=
−
∫  a 2 dx

 
 
2
2 0 
a  
a 

h2 
x3 
=
x
−


2 
3a 2 
Hence
2
=
2
ah
∫ yEL dA = ∫0
0
x3 
−

3a 
 2

a
a
=
0
1 2
ah
3
xA = ∫ xEL dA
1  1
x  ah  = a 2h
2  6
x =
1
a
3
yA = ∫ yEL dA
1  1
y  ah  = ah 2
2  3
y =
2
h
3
PROBLEM 5.32
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and h.
SOLUTION
For the element (EL) shown
x = a, y = h : h = ka3
or
Then
x=
a 1/3
y
h1/3
Now
dA = xdy
At
=
xEL =
h
a3
a 1/3
y dy
h1/3
1
1 a 1/3
x=
y , yEL = y
2
2 h1/ 3
h
A = ∫ dA =∫0
Then
k =
( )
h
a 1/3  a 1/3  1 a  3 5/3 
3 2
y  1/3 y dy  =
y  =
a h
2/3 
2 h1/3
2
5
10
h 
h

0
h1
and ∫ xEL dA = ∫0
h
a 1/3
3 a
3
y dy =
y 4/3 = ah
1/3
1/3
4h
4
h
0
h
a  3 7/3 
3 2
h  a

1/3
∫ yEL dA = ∫0 y  h1/3 y dy  = h1/3  7 y  = 7 ah



0
Hence
3 2
3 
xA = ∫ xEL dA : x  ah  =
a h
 4  10
3  3
yA = ∫ yEL dA: y  ah  = ah 2
4  7
x =
2
a
5
y =
4
h
7
PROBLEM 5.33
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and h.
SOLUTION
For the element (EL) shown
x = a, y = h: h = k1a3
At
a = k 2 h3
k2 =
or
k1 =
or
h
a3
a
h3
Hence, on line 1
y =
h 3
x
a3
y =
h 1/3
x
a1/3
and on line 2
Then
h 
 h
dA =  1/3 x1/3 − 3 x3  dx
a

a
and
yEL =
1  h 1/3
h 3
 1/3 x + 3 x 
2 a
a
a
h 
1
1
a h
 3

∴ A = ∫ dA = ∫0  1/3 x1/3 − 3 x3  dx = h  1/3 x 4/3 − 3 x 4  = ah
2
a
a
a
a
4
4



0
a
a 
x
0 
∫ xELdA = ∫
h 1/3
h 
1
8 2
 3

x − 3 x3  dx = h  1/3 x7/3 − 3 x5  =
a h
1/3
5a
a
a

 7a
 0 35
a 
1/3
3 
1/3
3
∫ yEL dA = ∫0 2  a1/3 x + a3 x  a1/3 x − a3 x  dx



1
h
h
h
h
h 2 a  x 2/3 x 6 
h 2  3 x5/3 1 x 6 
=
−
dx
=
−




∫
2 0  a 2/3 a 6 
2  5 a5/3 7 a 6 
a
=
8 2
ah
35
0
From
8 2
 ah 
xA = ∫ xEL dA: x   =
a h
2
35
 
or x =
16
a
35
and
8 2
 ah 
yA = ∫ yEL dA: y   =
ah
 2  35
or y =
16
h
35
PROBLEM 5.34
Determine by direct integration the centroid of the area shown.
SOLUTION
x =0
First note that symmetry implies
For the element (EL) shown
yEL =
2r
(Figure 5.8B)
π
dA = π rd r
r
r2
π rd r = π 
r1
Then
A = ∫ dA = ∫
and
2
∫ yEL dA = ∫r
So
r
2 2
π
1
2r

( π rd r ) = 2  r 3 
1 π
3
r
(
2
=
2 3
r2 − r13
3
2
r2 − r1
 =
2
 2  r1
r2
r1
)
(
)
π
 2 3
yA = ∫ yEL dA: y  r22 − r12  =
r2 − r13
2
 3
4 r23 − r13
or y =
3π r22 − r12
(
)
(
)
PROBLEM 5.35
Determine by direct integration the centroid of the area shown.
SOLUTION
x =0
First note that symmetry implies
For the element (EL) shown
y = R cos θ, x = R sin θ
dx = R cos θ d θ
dA = ydx = R 2 cos 2θ dθ
Hence
α
1 2
α
 θ sin 2θ 
A = ∫ dA = 2∫0 R 2 cos 2 θ dθ = 2 R 2  +
 = R ( 2α sin 2α )
2
4
2

0
(
=
)
Alternatively,
2
R3
cos 2 α sin α + 2sin α
3
(
But yA = ∫ yEL dA so
or
2
31
α
2

∫ yEL dA = 2∫0 2 cosθ R cos θ dθ = R  3 cos θ sin θ + 3 sin θ 

0
α R
2
)
R3
cos 2 α sin α + 2sin α
3
y =
R2
( 2α + sin 2α )
2
(
(
)
)
cos 2 α + 2
2
y = R sin α
3
( 2α + sin 2α )
y =
2
3 − sin 2 α
R sin α
3
2α + sin 2α
PROBLEM 5.36
Determine by direct integration the centroid of the area shown.
SOLUTION
For the element (EL) shown
y =
dA = ( b − y ) dx
and
)
1
b
x = x; y = ( y + b ) =
a+ a −x )
2
2a (
b
A = ∫ dA = ∫ ( a − a − x ) dx
a
=
EL
Then
b 2
a − x2
a
(
b
a − a 2 − x 2 dx
a
2
EL
a
0
2
2
2
To integrate, let x = a sin θ : a 2 − x 2 = a cosθ , dx = a cosθ dθ
Then
π /2 b
A = ∫0
a
( a − a cosθ )( a cosθ dθ )
π /2
b
2θ  
θ
=  a 2 sin θ − a 2  + sin

a
4  0
2
π

= ab  1 − 
4

)
(
π /2
3/2  
1
a b
 b  a
and ∫ xEL dA = ∫0 x  a − a 2 − x 2 dx  =  x 2 + a 2 − x 2  
3
a
 a  2
 0
=
1 3
ab
6
(
)b (
(
)
) 
a

2
2 
2
2
∫ yEL dA = ∫0 2a a + a − x  a a − a − x dx 
b
b2 a 2
b 2  x3 
x dx =
=
 
2 ∫0
2a
2a 2  3 
( )
a
=
0
1 2
ab
6
xA = ∫ xEL dA:
 
π  1
x  ab 1 −   = a 2b
4  6
 
or x =
2a
3 ( 4 − π)
yA = ∫ yEL dA:
 
π  1
y  ab 1 −   = ab 2
4  6
 
or y =
2b
3 ( 4 − π)
PROBLEM 5.37
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.
SOLUTION
For the element (EL) shown on line 1 at
x = a, b = k2a 2
∴ y =
∴ y =
b
a2
b 2
x
a2
x = a, −2b = k1a3
On line 2 at
k2 =
or
or
k2 =
−2b
a3
−2b 3
x
a3
2b 
 b
dA =  2 x 2 + 3 x3  dx
a

a
Then
2x
a b  2
x +
0 2
x
A = ∫ dA = ∫
a 
3
b  x3 2 x 4 
dx
=
+

4a 
a 2  3

a
0
 1 1 5
= ab  +  = ab
 3 2 6
2b 
a  b 2
x
x + 3 x3  dx =
0  2
and ∫ xEL dA = ∫
a
a

b  x 4 2 x5 
+


5a 
a 2  4
a
1 2
= a 2b  + 
4 5
0
13 2
ab
20
2b 3   b 2 2b 3  
a1 b 2
∫ yEL dA = ∫0 2  a 2 x − a3 x   a 2 x + a3 x  dx 

 
 
=
a 1  b 2 
 2 x 
0
=∫
2  a
2

b 2  x5
2
 2b  
−  3 x3   dx =
− 2 x7 

4
2a  5
7a

a
 

2
a
0
2
13
 1
= b 2a5 
−  = − ab 2
70
 10 7 
Then
xA = ∫ xEL dA:
yA = ∫ yEL dA:
 5  13 2
x  ab  =
ab
 6  20
 5  13 2
y  ab  −
ab
 6  70
or x =
39
a
50
or y = −
39
b
175
PROBLEM 5.38
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.
SOLUTION
x = 0, y = b
At
b = k (0 − a)
y =
Then
2
xEL = x, yEL =
and
dA = ydx =
and
a b
2
A = ∫ dA = ∫0
a
b
a2
b
( x − a )2
a2
Now
Then
k =
or
y
b
2
=
x − a)
2(
2
2a
b
( x − a )2 dx
a2
( x − a )2 dx =
a
b 
1
3
x − a )  = ab
(
2 
0 3
3a 
∫ xEL dA = ∫0 x  a 2 ( x − a ) dx  = a 2 ∫0 ( x − 2ax + a x )dx
a
=
b

2


b
a
3
2
2
b  x4 2 3 a2 2 
1 2
− ax +
x  =
ab
2

3
2
a  4
 12
b
2 b


a
2

=
Hence
a
b2  1
5

0
∫ yEL dA = ∫0 2a 2 ( x − a )  a 2 ( x − a ) dx  = 2a 4  5 ( x − a ) 
1 2
ab
10
1 2
1 
xA = ∫ xEL dA: x  ab  =
ab
3
12


1 2
1 
yA = ∫ yEL dA: y  ab  =
ab
 3  10
x =
1
a
4
y =
3
b
10
PROBLEM 5.39
Determine by direct integration the centroid of the area shown.
SOLUTION
xEL = x
Have
yEL =
1
a
x x2 
y = 1 − + 2 
2
2
L L 

x x2 
dA = ydx = a 1 − + 2  dx
L L 

Then

x x2 
x2
2L 
a
−
+
dx
=
a
x
−
+
1


2
0
L
2L
A = ∫ dA = ∫

L 

2L
x3 
8
 = aL
3
3L2 0
2L
and
 x2
x x2  
x3
x4 
2L  
∫ xEL dA = ∫0 x  a 1 − L + L2  dx  = a  2 − 3L + 4L2 
 
 

0
10 2
=
aL
3
x x2   
x x2  
−
+
−
+
a
1
1


 
 dx 
L L2   
L L2  
2 
2L a 
∫ yEL dA = ∫0
=
a 2 EL 
x
x2
x3 x 4 
−
+
−
+
1
2
3
2

 dx
∫
L
2 0 
L2
L3 L4 
=
a2 
x 2 x3
x4
x5 
11
+ 2 − 3 + 4  = a2L
x −
L
2 
5
L
2L
5L  0
2L
Hence
 8  10 2
xA = ∫ xEL dA: x  aL  =
aL
3
3 
 1  11
yA = ∫ yEL dA: y  a  = a 2
5
8 
x =
y =
5
L
4
33
a
40
PROBLEM 5.40
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.
SOLUTION
For
y1 at x = a, y = 2b
Then
y1 =
2b = ka 2
or k =
2b
a2
2b 2
x
a2
b
x
( x + 2b) = b  2 − 
a
a
By observation
y2 = −
Now
xEL = x
and for 0 ≤ x ≤ a :
1
b
y1 = 2 x 2
2
a
and
dA = y1dx =
1
b
x
y2 =  2 − 
2
2
a
and
x

dA = y2dx = b  2 −  dx
a

yEL =
2b 2
x dx
a2
For a ≤ x ≤ 2a :
yEL =
Then
x
a 2b 2
2a 
x dx + a b  2 − dx
2
A = ∫ dA = ∫0
=
and
∫
a

a
2
 a
2b  x3 
x 
b
2
+
−
−

 

 
a  
a2  3 0
 2 
a
 2b

2


2a
a
2a
=
7
ab
6
 
x

 


0
∫ xEL dA = ∫0 x  a 2 x dx  + ∫a x b  2 − a  dx 
2a
a
=
 2 x3 
2b  x 4 
b
+


x −

3a  0
a2  4 0

=
1 2
1  2
2
2
3 
2a − ( a )  
a b + b ( 2a ) − ( a )  +

 3a 

2
=
7 2
ab
6
{
( )
PROBLEM 5.40 CONTINUED
a
b
2  2b
2


2a b 
x 
x

 


∫ yEL dA = ∫0 a 2 x  a 2 x dx  + ∫0 2  2 − a  b  2 − a  dx 

=
=
Hence

a
3
2b 2  x5 
b2  a 
x 
+
−
−
2

 

 
2  3 
a  
a4  5 0
2a
a
17 2
ab
30
7  7
xA = ∫ xEL dA: x  ab  = a 2b
6  6
 7  17 2
yA = ∫ yEL dA: y  ab  =
ab
 6  30
x =a
y =
17
b
35
PROBLEM 5.41
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.
SOLUTION
For y2
x = a, y = b : a = kb 2
at
or
k =
a
b2
b 1/2
x
a
Then
y2 =
Now
xEL = x
a
y
b x1/2
x1/2
: yEL = 2 =
, dA = y2dx = b
dx
2
2
2 a
a
and for
0≤ x≤
For
a
1
b  x 1 x1/2 
≤ x ≤ a : yEL = ( y1 + y2 ) =  − +

2
2
2a 2
a 
 x1/2 x 1 
dA = ( y2 − y1 ) dx = b 
− +  dx
 a a 2
Then
a/2
A = ∫ dA = ∫0 b
 x1/2 x 1 
x1/2
a
− +  dx
dx + ∫a/2 b 
a
 a a 2
a
a/2
 2 x3/2
b  2 3/2 
x2 1 
x
b
=
+
−
+ x


2a 2  a/2
a  3
0
3 a
=
3/2
3/2
2 b  a 
3/2
a 
  + ( a ) −   
3 a  2 
 2  
2

 1 
a  1
 a  
2
+ b  −  a −    + ( a ) −    
 2   2 
 2  
 2a 

( )
=
13
ab
24
PROBLEM 5.41 CONTINUED
and
1/2
  x1/2 x 1  

a/2  x
a
=
+
− +  dx
x
dA
x
b
dx
x


∫ EL
∫0 
∫ b 
a  a/2   a a 2  

a
a/2
 2 x5/2 x3 x 4 
b  2 5/2 
=
+
−
+
x
b



4  a/2
a  5
0
 5 a 3a
=
2 b  a 
 
5 a  2 
5/2
+ (a)
5/2
5/2
a 
−  
 2  
 1  3  a 3  1  2  a 2  
+ b  − ( a ) −    +  ( a ) −    
 2   4 
 2   
 3a 

=
71 2
ab
240
1/2 
a/2 b x
x1/2

dx 
b
∫ yEL dA = ∫0 2
a a 
1 x1/2    x1/2 x 1  
a b x
+ ∫a/2  − +
− +  dx 
 b 
2 a 2
a    a a 2  
a
a/2
3
b2  1 2 
b 2  x 2
1  x 1  


x
=
+
−
 −  
2a  2  0
2  2a 3a  a 2   

 a/2
Hence
=
2
2
3
b  a 
2
 a   b2  a 1 
−
  + ( a ) −    −


4a  2 
 2   6a  2 2 
=
11 2
ab
48
71 2
 13 
xA = ∫ xEL dA: x  ab  =
ab
 24  240
x =
 13  11 2
yA = ∫ yEL dA: y  ab  =
ab
 24  48
y =
17
a = 0.546a
130
11
b = 0.423b
26
PROBLEM 5.42
A homogeneous wire is bent into the shape shown. Determine by direct
integration the x coordinate of its centroid. Express your answer in terms
of a.
SOLUTION
First note that because the wire is homogeneous, its center of gravity
coincides with the centroid of the corresponding line
Have at
x = a, y = a : a = ka 2
or
k =
1 2
x
a
dy =
2
xdx
a
y =
Thus
and
2
1
a
2
 dy 
2 
dL = 1 +   dx = 1 +  x  dx
 dx 
a 
Then
a
a
0
∴ L = ∫ dL = ∫
=
x
a 2
4
4x2
4x2 
1 + 2 x 2 dx =  1 + 2 + ln  x + 1 + 2  
4 a
2
a
a
a 

0

(
)
a
a
5 + ln 2 + 5 = 1.4789a
2
4
a

3/2
4x2   2  a2  
4
 
1 + 2 dx  =    1 + 2 x 2  
  3  8 
a
a
 

 
0
a
x
0 
∫ xELdL = ∫
a 2 3/2
5 − 1 = 0.8484a 2
12
xL = ∫ xEL dL: x (1.4789a ) = 0.8484a 2
=
Then
(
)
x = 0.574a
PROBLEM 5.43
A homogeneous wire is bent into the shape shown. Determine by direct
integration the x coordinate of its centroid.
SOLUTION
First note that because the wire is homogeneous, its center of gravity
coincides with the centroid of the corresponding line
dL = rd θ
Now
xEL = r cos θ
Then
L = ∫ dL = ∫π /4 rdθ = r [θ ]π /4 =
and
and
7π /4
7π /4
3
πr
2
7π /4
∫ xELdL = ∫π /4 r cosθ ( rdθ )
1 
 1
7π /4
2
= r 2 [sin θ ]π /4 = r 2  −
−
 = −r 2
2
2

Thus
3 
xL = ∫ xdL : x  π r  = −r 2 2
2 
x =−
2 2
r
3π
PROBLEM 5.44
A homogeneous wire is bent into the shape shown. Determine by direct
integration the x coordinate of its centroid.
SOLUTION
First note that because the wire is homogeneous, its center of gravity
coincides with the centroid of the corresponding line
xEL = a cos3 θ
Now
dL =
and
dx 2 + dy 2
x = a cos3 θ : dx = −3a cos 2 θ sin θ dθ
Where
y = a sin 3 θ : dy = 3a sin 2 θ cosθ dθ
Then
(
) (
)
1/2
2
2

dL =  −3a cos 2 θ sin θ dθ + 3a sin 2 θ cosθ dθ 


(
= 3a cosθ sin θ cos 2 θ + sin 2 θ
) dθ
1/2
= 3a cosθ sin θ dθ
π /2
∴ L = ∫ dL = ∫0
=
and
π /2
1

3a cosθ sin θ dθ = 3a  sin 2 θ 
2

0
3
a
2
π /2
3
∫ xEL dL = ∫0 a cos θ ( 3a cosθ sin θ dθ )
π /2
 1

= 3a 2  − cos5 θ 
 5
0
Hence
=
3 2
a
5
3  3
xL = ∫ xEL dL : x  a  = a 2
2  5
x =
2
a
5
PROBLEM 5.44 CONTINUED
Alternative solution
 x
x = a cos3 θ ⇒ cos 2 θ =  
 a
 y
y = a sin θ ⇒ sin θ =  
 a
3
 x
∴  
 a
2/3
2/3
 y
+ 
 a
=1
and
Hence
(
y = a 2/3 − x 2/3
or
(
) (
1/2
dy
= a 2/3 − x 2/3
− x −1/3
dx
)
3/2
)
xEL = x
Now
Then
2/3
2
Then
and
2/3
 dy 
dL = 1 +  
 dx 
1/2
2

1/2

 
dx = 1 +  a 2/3 − x 2/3
− x −1/3   dx

 

2
(
) (
)
a
1/3
3
a a
1/3  3 2/3 
dx
=
a
x
= a


1/ 3
2
2
x

0
L = ∫ dL = ∫0
a
1/3

3
a a
1/3  3 5/3 
x
dx = a  x  = a 2
0  1/3
x

5
0 5
∫ xELdL = ∫
3  3
xL = ∫ xEL dL : x  a  = a 2
2  5
x =
2
a
5
PROBLEM 5.45
Determine by direct integration the centroid of the area shown.
SOLUTION
Have
and
2
2
r cosθ = aeθ cosθ
3
3
2
2 θ
yEL = r sin θ = ae sin θ
3
3
xEL =
dA =
1
1
( r )( rdθ ) = a 2e2θ dθ
2
2
Then
π
(
)
1 1
1
π 1

A = ∫ dA = ∫0 a 2e 2θ dθ = a 2  e 2θ  = a 2 e2π − 1 = 133.623a 2
2
2 2
4
0
and
π
 2 2θ 
θ
3 π 3θ
∫ xELdA = ∫0 3ae cosθ  2 a e dθ  = 3 a ∫0 e cosθ dθ
2
1
1


To proceed, use integration by parts, with
u = e3θ
du = 3e3θ dθ
and
dv = cosθ dθ
Then
Now let
and
3θ
3θ
3θ
∫ e cosθ dθ = e sin θ − ∫ sin θ ( 3e dθ )
u = e3θ
So that
du = 3e3θ dθ
then
dv = sin θ dθ ,
Then
v = sin θ
then
v = − cosθ
3θ
3θ
−3θ
3θ
∫ e sin θ dθ = e sin θ − 3  −e cosθ − ∫ ( − cosθ ) ( 3e dθ ) 
e3θ
3θ
∫ e cosθ dθ = 10 ( sin θ + 3cosθ )
π

1  e3θ
a3
∴ ∫ xEL dA = a3 
( sin θ + 3cos θ)  = −3e3π − 3 = −1239.26a3
3  10
0 30
Also
(
)
π
 2 2θ 
θ
3 π 3θ
∫ yEL dA = ∫0 3 ae sin θ  2 a e dθ  = 3 a ∫0 e sin θ dθ
2
1

1

PROBLEM 5.45 CONTINUED
Using integration by parts, as above, with
u = e3θ
dv = ∫ sin θ dθ
Then
du = 3e3θ dθ
and
and
v = − cosθ
3θ
3θ
3θ
∫ e sin θd θ = −e cos θ − ∫ ( − cos θ) ( 3e d θ)
So that ∫ e3θ sin θd θ =
e3θ
( − cos θ + 3sin θ)
10
π

1  e3θ
a3
∴ ∫ yEL dA = a3 
( − cos θ + 3sin θ)  = e3π + 1 = 413.09a3
3  10
0 30
(
)
yA = ∫ y dA: y (133.623a ) = 413.09a
(
Hence xA = ∫ xEL dA: x 133.623a 2 = −1239.26a3
EL
2
3
)
or x = −9.27a
or y = 3.09a
PROBLEM 5.46
Determine by direct integration the centroid of the area shown.
SOLUTION
xEL = x,
Have
yEL =
dA = ydx
and
L/2
A = ∫ dA = ∫0 x sin
and
1
πx
x sin
2
L
L/2
 L2
πx L
πx
− x cos 
dx =  2 sin
π
L
L
L 0
π
πx
L2
π2
πx 
L/2 
x = ∫ xEL dA = ∫0 x  x sin
dx 
L


L/2
 2 L2
 π x  2 L3
 π x  L 2  π x 
=  2 x sin 
 + 3 cos 
 − x sin 

 L  π
 L  π
 L 0
π
Also
=
L/2 1
y = ∫ yEL dA = ∫0
2
x sin
πx
πx 
dx 
 x sin
L 
L

1  2 L2
πx  L
2 L3 
π x
=  2 x sin
−  x − 3  cos 
2  π
L π
L 
π 
=
=
L/2
0

L2  L 
L3
1  1  L3 
1
6 + π2
−
=
(
)
   −



2
2  6  8  4π 2  2 
 96π
(
)
L3
L3
π
π3
−2
2
PROBLEM 5.46 CONTINUED
Hence
 L2 
z 
 1
xA = ∫ xEL dA: x  2  = L3  2 − 3 
π 
π
π 
or x = 0.363L
 L2 
L3  1
2 
− 3
yA = ∫ yEL dA: y  2  =
2 2
π 
 π  96π  π
or y = 0.1653L
PROBLEM 5.47
Determine the volume and the surface area of the solid obtained by
rotating the area of Problem. 5.2 about (a) the x axis, (b) the line
x = 165 mm.
SOLUTION
From the solution to Problem 5.2:
A = 10 125 mm 2 , X area = 96.4 mm, Yarea = 34.7 mm
( Area )
From the solution to Problem 5.22:
L = 441.05 mm X line = 92.2 mm, Yline = 32.4 mm
( Line )
Applying the theorems of Pappus-Guldinus, we have
(a) Rotation about the x axis:
Area = 2π Yline L = 2π ( 32.4 mm )( 441.05 mm ) = 89.786 × 103 mm 2
A = 89.8 × 103 mm 2
Volume = 2π Yarea A = 2π ( 34.7 mm )(10 125 mm ) = 2.2075 × 106 mm3
V = 2.21 × 106 mm3
(b) Rotation about x = 165 mm:
(
)
Area = 2π 165 − X line L = 2π (165 − 92.2 ) mm  ( 441.05 mm ) = 2.01774 × 105 mm 2
A = 0.202 × 106 mm 2
(
)
Volume = 2π 165 − X area A = 2π (165 − 96.4 ) mm  (10 125 mm ) = 4.3641 × 106 mm3
V = 4.36 × 106 mm3
PROBLEM 5.48
Determine the volume and the surface area of the solid obtained by
rotating the area of Problem 5.4 about (a) the line y = 22 mm, (b) the
line x = 12 mm.
SOLUTION
From the solution to Problem 5.4:
A = 399 mm 2 , X area = 1.421 mm, Yarea = 12.42 mm
(Area)
From the solution to Problem 5.23:
L = 77.233 mm, X line = 1.441 mm, Yline = 12.72 mm
(Line)
Applying the theorems of Pappus-Guldinus, we have
(a) Rotation about the line y = 22 mm:
(
)
Area = 2π 22 − Yline L = 2π ( 22 − 12.72 ) mm  ( 77.233 mm ) = 4503 mm 2
A = 4.50 × 103 mm 2
(
)
(
)
Volume = 2π 22 − Yarea A = 2π ( 22 − 12.42 ) mm  399 mm 2 = 24 016.97 mm3
V = 24.0 × 103 mm3
(b) Rotation about line x = 12 mm:
(
)
Area = 2π 12 − X line L = 2π (12 − 1.441) mm  ( 77.233 mm ) = 5124.45 mm 2
A = 5.12 × 103 mm 2
(
)
Volume = 2π (12 − 1.421) A = 2π (12 − 1.421) mm  399 mm 2 = 26 521.46 mm3
V = 26.5 × 103 mm3
PROBLEM 5.49
Determine the volume and the surface area of the solid obtained by
rotating the area of Problem 5.1 about (a) the x axis, (b) the line
x = 16 in.
SOLUTION
From the solution to Problem 5.1:
A = 240 in 2 , X area = 5.60 in., Yarea = 6.60 in.
(Area)
From the solution to Problem 5.21:
L = 72 in., X line = 4.67 in., Yline = 6.67 in.
Applying the theorems of Pappus-Guldinus, we have
(a) Rotation about the x axis:
Ax = 2π Yline L = 2π ( 6.67 in.)( 72 in.) = 3017.4 in 2
A = 3020 in 2
(
)
Vx = 2π Yarea A = 2π ( 6.60 in.) 240 in 2 = 9952.6 in 3
V = 9950 in 3
(b) Rotation about x = 16 in.:
(
)
Ax =16 = 2π 16 − X line L = 2π (16 − 4.67 ) in. ( 72 in.) = 5125.6 in 2
Ax =16 = 5130 in 2
(
)
(
)
Vx =16 = 2π 16 − X area A = 2π (16 − 5.60 ) in. 240 in 2 = 15 682.8 in 3
Vx =16 = 15.68 × 103 in 3
PROBLEM 5.50
Determine the volume of the solid generated by rotating the semielliptical
area shown about (a) the axis AA′, (b) the axis BB′, (c) the y axis.
SOLUTION
Applying the second theorem of Pappus-Guldinus, we have
(a) Rotation about axis AA′:
 π ab 
2 2
Volume = 2π yA = 2π ( a ) 
=π ab
 2 
V = π 2a 2b
 π ab 
2 2
Volume = 2π yA = 2π ( 2a ) 
 = 2π a b
 2 
V = 2π 2a 2b
 4a  π ab  2 2
Volume = 2π yA = 2π  
 = πa b
 3π  2  3
V =
(b) Rotation about axis BB′:
(c) Rotation about y-axis:
2 2
πa b
3
PROBLEM 5.51
Determine the volume and the surface area of the chain link shown,
which is made from a 2-in.-diameter bar, if R = 3 in. and L = 10 in.
SOLUTION
First note that the area A and the circumference C of the cross section of the bar are
A=
π
4
d2
and
C = πd
Observe that the semicircular ends of the link can be obtained by rotating the cross section through a horizontal
semicircular arc of radius R. Then, applying the theorems of Pappus-Guldinus, we have
Volume = 2 (Vside ) + 2 (Vend ) = 2 ( AL ) + 2 (π RA) = 2 ( L + π R ) A
2
π
= 2 10 in. + π ( 3 in.)   ( 2 in.) 
4

= 122.049 in 3
V = 122.0 in 3
Area = 2 ( Aside ) + 2 ( Aend ) = 2 ( CL ) + 2 (π RC ) = 2 ( L + π R ) C
= 2 10 in. + π ( 3 in.)  π ( 4 in.) 
= 488.198 in 2
A = 488 in 2
PROBLEM 5.52
Verify that the expressions for the volumes of the first four shapes in
Figure 5.21 on page 261 are correct.
SOLUTION
Following the second theorem of Pappus-Guldinus, in each case a
specific generating area A will be rotated about the x axis to produce the
given shape. Values of y are from Fig. 5.8A.
(1) Hemisphere: the generating area is a quarter circle
Have
 4a  π 
V = 2π yA = 2π   a 2 
 3π  4 
or V =
2 3
πa
3
(2) Semiellipsoid of revolution: the generating area is a quarter ellipse
Have
 4a  π 
V = 2π yA = 2π   ha 
 3π  4 
or V =
2 2
πa h
3
(3) Paraboloid of revolution: the generating area is a quarter parabola
Have
 3  2 
V = 2π yA = 2π  a  ah 
 8  3 
or V =
1 2
πa h
2
or V =
1 2
πa h
3
(4) Cone: the generating area is a triangle
Have
 a  1 
V = 2π yA = 2π   ha 
 3  2 
PROBLEM 5.53
A 15-mm-diameter hole is drilled in a piece of 20-mm-thick steel; the
hole is then countersunk as shown. Determine the volume of steel
removed during the countersinking process.
SOLUTION
The required volume can be generated by rotating the area shown about the y axis. Applying the second
theorem of Pappus-Guldinus, we have
 5
 1


V = 2π xA = 2π  + 7.5  mm  ×  × 5 mm × 5 mm 
3
2



 
or V = 720 mm3
PROBLEM 5.54
Three different drive belt profiles are to be studied. If at any given time
each belt makes contact with one-half of the circumference of its pulley,
determine the contact area between the belt and the pulley for each
design.
SOLUTION
Applying the first theorem of Pappus-Guldinus, the contact area AC of a
belt is given by
AC = π yL = πΣ yL
Where the individual lengths are the “Lengths” of the belt cross section
that are in contact with the pulley
Have
AC = π  2 ( y1L1 ) + y2 L2 
  2.5 mm 
2.5 
 
= π 2  60 −
 mm  

2 
 
  cos 20° 

+ ( 60 − 2.5 ) mm  (12.5 mm ) 

or AC = 3.24 × 103 mm 2
Have
AC = π  2 ( y1L1 ) 

  7.5 mm 
7.5 
= 2π  60 − 1.6 −
 mm  × 
D 
2 

  cos 20 
or AC = 2.74 × 103 mm 2
Have


2× 5
AC = π ( y1L1 ) = π  60 −
 mm  (π × 5 mm )
π 


or AC = 2.80 × 103 mm 2
PROBLEM 5.55
Determine the capacity, in gallons, of the punch bowl shown if R = 12 in.
SOLUTION
The volume can be generated by rotating the triangle and circular sector shown about the y axis. Applying the
second theorem of Pappus-Guldinus and using Fig. 5.8A, we have
V = 2π xA = 2πΣxA = 2π ( x1 A1 + x2 A2 )




 1 1   1 1


3   2 R sin 30D
π


= 2π  × R   × R ×
cos30D   R 2  
R  + 
2   3× π
 3 2   2 2
  6  

6




 R3
R3  3 3
π R3
= 2π 
+
 =
8
 16 3 2 3 
=
Since
3 3
3
π (12 in.) = 3526.03 in 3
8
1 gal = 231 in 3
V =
3526.03 in 3
= 15.26 gal
231 in 3/gal
V = 15.26 gal
PROBLEM 5.56
The aluminum shade for a small high-intensity lamp has a uniform
thickness of 3/32 in. Knowing that the specific weight of aluminum is
0.101 lb/in 3 , determine the weight of the shade.
SOLUTION
The weight of the lamp shade is given by
W = γ V = γ At
where A is the surface area of the shade. This area can be generated by rotating the line shown about the
x axis. Applying the first theorem of Pappus-Guldinus, we have
A = 2π yL = 2πΣyL = 2π ( y1L1 + y2 L2 + y3 L3 + y4 L4 )
 0.6 mm
0.60 + 0.75 
= 2π 
( 0.6 mm ) + 
 mm ×
2


 2
 0.75 + 1.25 
+
 mm ×
2


( 0.50 mm )2 + ( 0.40 mm )2
 1.25 + 1.5 
+
 mm ×
2


( 0.25 mm )2 + (1.25 mm )2 
= 22.5607 in
Then
( 0.15 mm )2 + (1.5 mm )2


2
W = 0.101 lb/in 3 × 22.5607 in 2 ×
3
in. = 0.21362 lb
32
W = 0.214 lb W
PROBLEM 5.57
The top of a round wooden table has the edge profile shown. Knowing
that the diameter of the top is 1100 mm before shaping and that the
density of the wood is 690 kg/m3 , determine the weight of the waste
wood resulting from the production of 5000 tops.
SOLUTION
All dimensions are in mm
Have
Vwaste = Vblank − Vtop
Vblank = π ( 550 mm ) × ( 30 mm ) = 9.075π × 106 mm3
2
Vtop = V1 + V2 + V3 + V4
Applying the second theorem of Pappus-Guldinus to parts 3 and 4
2
2
Vtop = π ( 529 mm ) × (18 mm )  + π ( 535 mm ) × (12 mm ) 

 

 π
4 × 12 
2
 

+ 2π   535 +
 mm  × (12 mm ) 
3
π
4



 

 
 π
4 × 18 
2

+ 2π   529 +
 mm  × (18 mm ) 
3
π
4



 

= π ( 5.0371 + 3.347 + 0.1222 + 0.2731) × 106 mm3
= 8.8671π × 106 mm3
∴ Vwaste = ( 9.0750 − 8.8671) π × 106 mm3
= 0.2079π × 10−3 m3
Finally
Wwaste = ρ wood Vwaste g N tops
(
)
= 690 kg/m3 × 0.2079π × 10−3 m3 × 9.81 m/s 2 × 5000 ( tops )
or Wwaste = 2.21 kN W
PROBLEM 5.58
The top of a round wooden table has the shape shown. Determine how
many liters of lacquer are required to finish 5000 tops knowing that each
top is given three coats of lacquer and that 1 liter of lacquer covers 12 m2.
SOLUTION
Referring to the figure in solution of Problem 5.57 and using the first theorem of Pappus-Guldinus, we have
Asurface = Atop circle + Abottom circle + Aedge
2
2
= π ( 535 mm )  + π ( 529 mm ) 

 


 π
2 × 12 
 
+ 2π  535 +
 mm  × (12 mm ) 
π 
 
 2


 π
2 × 18 
 
+ 2π  529 +
 mm  × (18 mm ) 
π 
 
 2

= 617.115π × 103 mm 2
Then
# liters = Asurface × Coverage × N tops × N coats
= 617.115π × 10−3 m 2 ×
1 liter
× 5000 × 3
12 m 2
or # liters = 2424 L W
PROBLEM 5.59
The escutcheon (a decorative plate placed on a pipe where the pipe exits
from a wall) shown is cast from yellow brass. Knowing that the specific
3
weight of yellow brass is 0.306 lb/in . determine the weight of the
escutcheon.
SOLUTION
The weight of the escutcheon is given by
W = (specific weight)V
where V is the volume of the plate. V can be generated by rotating the area A about the x axis.
Have
and
Then
a = 3.0755 in. − 2.958 in. = 0.1175 in.
sin φ =
0.5
⇒ φ = 0.16745 R = 9.5941°
3
2α = 26D − 9.5941D = 16.4059D
or
α = 8.20295D = 0.143169 rad
The area A can be obtained by combining the following four areas, as indicated.
Applying the second theorem of Pappus-Guldinus and then using Figure 5.8A, we have
V = 2π yA = 2πΣ yA
PROBLEM 5.59 CONTINUED
A, in 2
y , in.
yA, in 3
1
1
( 3.0755)(1.5) = 2.3066
2
1
(1.5) = 0.5
3
1.1533
2
−α ( 3) = −1.28851
2
1
( 2.958)( 0.5) = −0.7395
2
3
−
4
− ( 0.1755 )( 0.5 ) = −0.05875
2 ( 3) sin α
× sin (α + φ ) = 0.60921
3α
1
( 0.5 ) = 0.16667
3
1
( 0.5) = 0.25
2
–0.78497
–0.12325
–0.14688
Σ yA = 0.44296 in 3
Then
so that
(
)
V = 2π 0.44296 in 3 = 1.4476 in 3
(
)
W = 1.4476 in 3 0.306 lb/in 3 = 0.44296 lb
W = 0.443 lb W
PROBLEM 5.60
The reflector of a small flashlight has the parabolic shape shown.
Determine the surface area of the inside of the reflector.
SOLUTION
First note that the required surface area A can be generated by rotating
the parabolic cross section through 2π radians about the x axis. Applying
the first theorem of Pappus-Guldinus, we have
A = 2π yL
Now, since
2
x = ky ,
x = a : a = k ( 7.5 )
at
or
a = 56.25 k
At
x = ( a + 15 ) mm: a + 15 = k (12.5 )
or
a + 15 = 156.25k
Then
Eq. (2) a + 15 156.25k
:
=
Eq. (1)
56.25k
a
and
(1)
2
(2)
or a = 8.4375 mm
Eq. (1) ⇒ k = 0.15
∴ x = 0.15 y 2
2
1
mm
dx
= 0.3 y
dy
2
Now
So
 dx 
dL = 1 +   dy = 1 + 0.09 y 2 dy
 dy 
A = 2π yL
yL = ∫ ydL
and
12.5
∴ A = 2π ∫7.5 y 1 + 0.09 y 2 dy
12.5
3/2 
2  1 
2
= 2π  
 1 + 0.09 y

 3  0.18 
 7.5
(
= 1013 mm 2
)
or A = 1013 mm 2 W
PROBLEM 5.61
For the beam and loading shown, determine (a) the magnitude and
location of the resultant of the distributed load, (b) the reactions at the
beam supports.
SOLUTION
Resultant
R = R1 + R2
(a) Have
R1 = ( 40 lb/ft )(18 ft ) = 720 lb
R2 =
1
(120 lb/ft )(18 ft ) = 1080 lb
2
R = 1800 lb
or
The resultant is located at the centroid C of the distributed load x
Have
or
ΣM A:
1
2
(1800 lb ) x = ( 40 lb/ft )(18 ft )( 9 ft ) + (120 lb/ft )(18 ft )(12 ft )
x = 10.80 ft
R = 1800 lb W
x = 10.80 ft
(b)
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 1800 lb = 0, Ay = 1800 lb
∴ A = 1800 lb
W
or M A = 19.44 kip ⋅ ft
W
ΣM A = 0: M A − (1800 lb )(10.8 ft ) = 0
M A = 19.444 lb ⋅ ft
PROBLEM 5.62
For the beam and loading shown, determine (a) the magnitude and
location of the resultant of the distributed load, (b) the reactions at the
beam supports.
SOLUTION
(a) Have
RI = ( 300 N/m )( 6 m ) = 1800 N
RII =
ΣFy : − R = − RI − RII
Then
or
1
( 6 m )( 900 N/m ) = 1800 N
3
R = 1800 N + 1800 N = 3600 N
ΣM A : − x ( 3600 N ) = − ( 3 m )(1800 N ) − ( 4.5 m )(1800 N )
x = 3.75 m
or
R = 3600 N W
x = 3.75 m
(b) Reactions
ΣFx = 0: Ax = 0
ΣM A = 0:
or
( 6 m ) By − ( 3600 N )( 3.75 m ) = 0
By = 2250 N
B = 2250 N
W
A = 1350 N
W
ΣFy = 0: Ay + 2250 N = 3600 N
or
Ay = 1350 N
PROBLEM 5.63
Determine the reactions at the beam supports for the given loading.
SOLUTION
RI = (100 lb/ft )( 4 ft ) = 400 lb
Have
1
( 200 lb/ft )( 6 ft ) = 600 lb
2
RIII = ( 200 lb/ft )( 4 ft ) = 800 lb
RII =
ΣFx = 0: Ax = 0
Then
ΣM A = 0:
or
( 2 ft )( 400 lb ) − ( 4 ft )( 600 lb ) − (12 ft )(800 lb ) + (10 ft ) By = 0
By = 800 lb
B = 800 lb
W
A = 1000 lb
W
ΣFy = 0: Ay + 800 lb − 400 lb − 600 lb − 800 = 0
or
Ay = 1000 lb
PROBLEM 5.64
Determine the reactions at the beam supports for the given loading.
SOLUTION
Have
RI = ( 9 ft )( 200 lb/ft ) = 1800 lb
RII =
Then
1
( 3 ft )( 200 lb/ft ) = 300 lb
2
ΣFx = 0: Ax = 0
ΣM A = 0: − ( 4.5 ft )(1800 lb ) − (10 ft )( 300 lb ) + ( 9 ft ) By = 0
or
By = 1233.3 lb
B = 1233 lb
W
A = 867 lb
W
ΣFy = 0: Ay − 1800 lb − 300 lb + 1233.3 lb = 0
or
Ay = 866.7 lb
PROBLEM 5.65
Determine the reactions at the beam supports for the given loading.
SOLUTION
Have
RI =
1
( 200 N/m )( 0.12 m ) = 12 N
2
RII = ( 200 N/m )( 0.2 m ) = 40 N
Then
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + 18 N − 12 N − 40 N = 0
or
Ay = 34 N
A = 34.0 N
W
M A = 2.92 N ⋅ m
W
ΣM A = 0: M A − ( 0.8 m )(12 N ) − ( 0.22 m )( 40 N ) + ( 0.38 m )(18 N )
or
M A = 2.92 N ⋅ m
PROBLEM 5.66
Determine the reactions at the beam supports for the given loading.
SOLUTION
First, replace the given loading with the loading shown below. The two loadings are equivalent because both
are defined by a linear relation between load and distance, and the values at the end points are the same.
Have
RI = (1.8 m )( 2000 N/m ) = 3600 N
RII =
Then
1
(1.8 m )( 4500 N/m ) = 4050 N
2
ΣFx = 0: Ax = 0
ΣM B = 0: − ( 3 m ) Ay − ( 2.1 m )( 3600 N ) + ( 2.4 m )( 4050 N )
or
Ay = 270 N
A = 270 N
W
B = 720 N
W
ΣFy = 0: 270 N − 3600 N + 4050 N − By = 0
or
By = 720 N
PROBLEM 5.67
Determine the reactions at the beam supports for the given loading.
SOLUTION
Have
Then
RI =
1
( 4 m )( 2000 kN/m ) = 2667 N
3
RII =
1
( 2 m )(1000 kN/m ) = 666.7 N
3
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 2667 N − 666.7 N = 0
or
Ay = 3334 N
A = 3.33 kN
W
M A = 6.33 kN ⋅ m
W
ΣM A = 0: M A − (1 m )( 2667 N ) − ( 5.5 m )( 666.7 N )
or
M A = 6334 N ⋅ m
PROBLEM 5.68
Determine the reactions at the beam supports for the given loading.
SOLUTION
First, replace the given loading with the loading shown below. The two loadings are equivalent because both
are defined by a parabolic relation between load and distance, and the values at end points are the same.
Have
RI = ( 8 ft )(100 lb/ft ) = 800 lb
RII =
Then
2
(8 ft )( 600 lb/ft ) = 3200 lb
3
ΣFx = 0: Ax = 0
ΣM A = 0: 11B + ( 5 ft )( 800 lb ) − ( 4 ft )( 3200 ) lb = 0
or B = 800 lb
W
or A = 1600 lb
W
ΣFy = 0: Ay − 3200 lb + 800 lb + 800 lb = 0
PROBLEM 5.69
Determine (a) the distance a so that the vertical reactions at supports A
and B are equal, (b) the corresponding reactions at the supports.
SOLUTION
(a) Have
RI =
1
( a ft )(120 lb/ft ) = ( 60a ) lb
2
RII =
1
(12 − a )( 40 lb/ft ) = ( 240 − 20a ) lb
2
Then
ΣFy = 0: Ay − 60a − ( 240 − 2a ) + By = 0
or
Ay + By = 240 + 40a
Now
Ay = By ⇒ Ay = By = 120 + 20a
Also
(1)

a    1

ΣM B = 0: − (12 m ) Ay + ( 60a ) lb  12 −  ft  +  (12 − a ) ft   ( 240 − 20a ) lb  = 0
3    3


or
Ay = 80 −
140
10 2
a−
a
3
9
(2)
Equating Eqs. (1) and (2)
120 + 20a = 80 −
or
Then
Now
(b) Have
Eq. (1)
140
10
a − a2
3
9
40 2
a − 320a + 480 = 0
3
a = 1.6077 ft,
a = 22.392
a ≤ 12 ft
a = 1.608 ft W
ΣFx = 0: Ax = 0
Ay = By = 120 + 20 (1.61) = 152.2 lb
A = B = 152.2 lb
W
PROBLEM 5.70
Determine (a) the distance a so that the vertical reaction at support B is
minimum, (b) the corresponding reactions at the supports.
SOLUTION
RI =
1
( a ft )(120 lb/ft ) = 60a lb
2
RII =
1
(12 − a ) ft  ( 40 lb/ft ) = ( 240 − 20a ) lb
2
(a) Have
Then
or
Then
(b) Eq. (1)

a 
a 
ΣM A = 0: −  ft  ( 60a lb ) − ( 240 − 20a ) lb   8 +  ft  + (12 ft ) By = 0
3 
3 

By =
dBy
da
By =
10 2 20
a −
a + 160
9
3
(1)
20
20
a−
=0
9
3
or a = 3.00 ft W
=
10
20
( 3.00 )2 − ( 3.00 ) + 160
9
3
= 150 lb
and
B = 150.0 lb
W
A = 210 lb
W
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 60 ( 3.00 )  lb −  240 − 20 ( 3.00 )  lb + 150 lb = 0
or
Ay = 210 lb
PROBLEM 5.71
Determine the reactions at the beam supports for the given loading when
w0 = 1.5 kN/m.
SOLUTION
Have
RI =
1
( 9 m )( 2 kN/m ) = 9 kN
2
RII = ( 9 m )(1.5 kN/m ) = 13.5 kN
Then
ΣFx = 0: Cx = 0
ΣM B = 0: − 50 kN ⋅ m − (1 m )( 9 kN ) − ( 2.5 m )(13.5 kN ) + ( 6 m ) C y = 0
or
C y = 15.4583 kN
C = 15.46 kN
W
B = 7.04 kN
W
ΣFy = 0: By − 9 kN − 13.5 kN + 15.4583 = 0
or
By = 7.0417 kN
PROBLEM 5.72
Determine (a) the distributed load w0 at the end D of the beam ABCD
for which the reaction at B is zero, (b) the corresponding reactions at C.
SOLUTION
Have
RI =
1
( 9 m ) ( 3.5 − w0 ) kN/m  = 4.5 ( 3.5 − w0 ) kN
2
RII = ( 9 m ) ( w0 kN/m ) = 9w0 kN
(a) Then
or
so
ΣM C = 0: − 50 kN ⋅ m + ( 5 m )  4.5 ( 3.5 − w0 ) kN  + ( 3.5 m ) ( 9w0 kN ) = 0
9w0 + 28.75 = 0
w0 = −3.1944 kN/m
w0 = 3.19 kN/m
W
C = 1.375 kN
W
Note: the negative sign means that the distributed force w0 is upward.
ΣFx = 0: Cx = 0
(b)
ΣFy = 0: − 4.5 ( 3.5 + 3.19 ) kN + 9 ( 3.19 ) kN + C y = 0
or
C y = 1.375 kN
PROBLEM 5.73
A grade beam AB supports three concentrated loads and rests on soil and
the top of a large rock. The soil exerts an upward distributed load, and
the rock exerts a concentrated load RR as shown. Knowing that P = 4 kN
and wB = 12 wA , determine the values of wA and RR corresponding to
equilibrium.
SOLUTION
Have
RI = (1.2 m )( wA kN/m ) = 1.2 wA kN
RII =
1
1
(1.8 m )  wA kN/m  = 0.45 wA kN
2
2

1

RIII = (1.8 m )  wA kN/m  = 0.9 wA kN
2


Then
ΣM C = 0: − ( 0.6 m ) (1.2 wA ) kN  + ( 0.6 m ) ( 0.45 wA ) kN/m 
+ ( 0.9 m ) ( 0.9 wA ) kN/m  − (1.2 m )( 4 kN/m )
− ( 0.8 m )(18 kN/m ) + ( 0.7 m )( 24 kN/m ) = 0
or
and
wA = 6.667 kN/m
wA = 6.67 kN/m W
ΣFy = 0: RR + (1.2 m )( 6.67 kN/m ) + ( 0.45 m )( 6.67 kN/m )
+ ( 0.9 m )( 6.67 kN/m ) − 24 kN − 18 kN − 4 kN
or
RR = 29.0 kN
RR = 29.0 kN W
PROBLEM 5.74
A grade beam AB supports three concentrated loads and rests on soil and
the top of a large rock. The soil exerts an upward distributed load, and
the rock exerts a concentrated load RR as shown. Knowing that wB =
0.4wA, determine (a) the largest value of P for which the beam is in
equilibrium, (b) the corresponding value of wA.
In the following problems, use γ = 62.4 lb/ft3 for the specific weight of
fresh water and γc = 150 lb/ft3 for the specific weight of concrete if U.S.
customary units are used. With SI units, use ρ = 103 kg/m3 for the
density of fresh water and ρc = 2.40 × 103 kg/m3 for the density of
concrete. (See the footnote on page 222 for how to determine the specific
weight of a material given its density.)
j
SOLUTION
Have
RI = (1.2 m )( wA kN/m ) = 1.2 wA kN
RII =
1
(1.8 m )( 0.6 wA kN/m ) = 0.54 wA kN
2
RIII = (1.8 m )( 0.4 wA kN/m ) = 0.72 wA kN
(a) Then
ΣM A = 0:
( 0.6 m ) (1.2 wA ) kN  + (1.2 m ) RR + (1.8 m ) ( 0.54 wA ) kN 
+ ( 2.1 m ) ( 0.72 wA ) kN  − ( 0.5 m )( 24 kN )
− ( 2.0 m )(18 kN ) + ( 2.4 m ) P = 0
or
and
or
3.204 wA + 1.2 RR − 2.4 P = 48
(1)
ΣFy = 0: RR + 1.2 WA + 0.54 WA + 0.72 WA − 24 − 18 − P = 0
RR + 2.46 WA − P = 42
(2)
Now combine Eqs. (1) and (2) to eliminate wA :
( 3.204 ) Eq. 2 − ( 2.46 ) Eq. 1 ⇒ 0.252 RR = 16.488 − 2.7 P
Since RR must be ≥ 0, the maximum acceptable value of P is that for which R = 0,
or
(b) Then, from Eq. (2):
P = 6.1067 kN
2.46 WA − 6.1067 = 42
P = 6.11 kN W
or WA = 19.56 kN/m W
PROBLEM 5.75
The cross section of a concrete dam is as shown. For a dam section of
unit width, determine (a) the reaction forces exerted by the ground on the
base AB of the dam, (b) the point of application of the resultant of the
reaction forces of part a, (c) the resultant of the pressure forces exerted
by the water on the face BC of the dam.
In the following problems, use γ = 62.4 lb/ft3 for the specific weight of
fresh water and γc = 150 lb/ft3 for the specific weight of concrete if U.S.
customary units are used. With SI units, use ρ = 103 kg/m3 for the
density of fresh water and ρc = 2.40 × 103 kg/m3 for the density of
concrete. (See the footnote on page 222 for how to determine the specific
weight of a material given its density.)
SOLUTION
The free body shown consists of a 1-m thick section of the dam and the
triangular section BCD of the water behind the dam.
X1 = 6 m
Note:
X 2 = ( 9 + 3) m = 12 m
X 3 = (15 + 2 ) m = 17 m
X 4 = (15 + 4 ) m = 19 m
W = ρ gV
(a) Now
(
)(
)
(
)(
)
(
)(
)
(
)(
)
so that
1

W1 = 2400 kg/m3 9.81 m/s 2  ( 9 m )(15 m )(1 m )  = 1589 kN
2


W2 = 2400 kg/m3 9.81 m/s 2 ( 6 m )(18 m )(1 m )  = 2543 kN
1

W3 = 2400 kg/m3 9.81 m/s 2  ( 6 m )(18 m )(1 m )  = 1271 kN
2

1

W4 = 2400 kg/m3 9.81 m/s 2  ( 6 m )(18 m )(1 m )  = 529.7 kN
2

Also
P=
(
)(
)
1
1
Ap = (18 m )(1 m )   103 kg/m3 9.81 m/s 2 (18 m ) 


2
2
= 1589 kN
Then
or
ΣFx = 0: H − 1589 kN = 0
H = 1589 kN
H = 1589 kN
W
ΣFy = 0: V − 1589 kN − 2543 kN − 1271 kN − 529.7 kN
or
V = 5933 kN
V = 5.93 MN
W
PROBLEM 5.75 CONTINUED
(b) Have
ΣM A = 0: X ( 5933 kN ) + ( 6 m )(1589 kN )
− ( 6 m )(1589 kN ) − (12 m )( 2543 kN )
− (17 m )(1271 kN ) − (19 m )( 529.7 ) = 0
X = 10.48 m
or
X = 10.48 m W
to the right of A
(c) Consider water section BCD as the free body.
Have
ΣF = 0
Then
−R = 1675 kN
18.43°
or R = 1675 kN
18.43° W
Alternative solution to part (c)
Consider the face BC of the dam.
Have
and
BC =
62 + 182 = 18.9737 m
tan θ =
6
18
θ = 18.43°
(
)(
)
p = ( ρ g ) h = 1000 kg/m3 9.81 m/s 2 (18 m )
= 176.6 kN/m 2
Then
R=
(
1
1
Ap = (18.97 m )(1 m )  176.6 kN/m 2
2
2
= 1675 kN
∴ R = 1675 kN
18.43°
)
PROBLEM 5.76
The cross section of a concrete dam is as shown. For a dam section of
unit width, determine (a) the reaction forces exerted by the ground on the
base AB of the dam, (b) the point of application of the resultant of the
reaction forces of part a, (c) the resultant of the pressure forces exerted
by the water on the face BC of the dam.
SOLUTION
The free body shown consists of a 1-ft thick section of the dam and the parabolic section of water above (and
behind) the dam.
Note
x1 =
5
(16 ft ) = 10 ft
8
1


x2 = 16 + ( 6 )  ft = 19 ft
2


1


x3 =  22 + (12 )  ft = 25 ft
4


5


x4 =  22 + (12 )  ft = 29.5 ft
8


PROBLEM 5.76 CONTINUED
W = γV
Now
(
)
(
)
(
)
2

W1 = 150 lb/ft 3  (16 ft )( 24 ft ) × (1 ft )  = 38, 400 lb
3


W2 = 150 lb/ft 3 ( 6 ft )( 24 ft ) × (1 ft )  = 21, 600 lb
1

W3 = 150 lb/ft 3  (12 ft )(18 ft ) × (1 ft )  = 10,800 lb
3

(
)
2

W4 = 62.4 lb/ft 3  (12 ft )(18 ft ) × (1 ft )  = 8985.6 lb
3

Also
P=
(
)
1
1
Ap = (18 × 1) ft 2  × 62.4 lb/ft 3 × 18 ft = 10,108.8 lb
2
2
ΣFx = 0: H − 10,108.8 lb = 0
(a) Then
or H = 10.11 kips
W
ΣFy = 0: V − 38, 400 lb − 21, 600 lb − 10,800 lb − 8995.6 lb = 0
V = 79,785.6
or
V = 79.8 kips
W
ΣM A = 0: X ( 79, 785.6 lb ) − ( 6 ft )( 38, 400 lb ) − (19 ft )( 21,600 lb ) − ( 25 ft )(10,800 lb )
(b)
− ( 29.5 ft )( 8985.6 lb ) + ( 6 ft )(10,108.8 lb ) = 0
or
X = 15.90 ft
The point of application of the resultant is 15.90 ft to the right of A W
(c) Consider the water section BCD as the free body.
Have
ΣF = 0
∴ R = 13.53 kips
θ = 41.6°
On the face BD of the dam
R = 13.53 kips
41.6° W
PROBLEM 5.77
The 9 × 12-ft side AB of a tank is hinged at its bottom A and is held in
place by a thin rod BC. The maximum tensile force the rod can withstand
without breaking is 40 kips, and the design specifications require the
force in the rod not exceed 20 percent of this value. If the tank is slowly
filled with water, determine the maximum allowable depth of water d in
the tank.
SOLUTION
Consider the free-body diagram of the side.
1
1
Ap = A (γ d )
2
2
Have
P=
Now
ΣM A = 0:
( 9 ft ) T −
d
P=0
3
Then, for d max:
( 9 ft ) ( 0.2 ) ( 40 × 103 lb ) −
(
)
d max  1

3
 (12 ft ) ( d max )  62.4 lb/ft d max  = 0
3 2

or
3
216 × 103 ft 3 = 374.4 d max
or
3
d max
= 576.92 ft 3
d max = 8.32 ft W
PROBLEM 5.78
The 9 × 12-ft side of an open tank is hinged at its bottom A and is held
in place by a thin rod. The tank is filled with glycerine, whose specific
weight is 80 lb/ft 3. Determine the force T in the rod and the reactions at
the hinge after the tank is filled to a depth of 8 ft.
SOLUTION
Consider the free-body diagram of the side.
Have
P=
=
(
)
1
( 8 ft )(12 ft )  80 lb/ft 3 ( 8 ft ) = 30, 720 lb
2
ΣFy = 0: Ay = 0
Then
ΣM A = 0:
or
1
1
Ap = A (γ d )
2
2
( 9 ft ) T − 
8 
ft  ( 30, 720 lb ) = 0
3 
T = 9102.22 lb
T = 9.10 kips
W
A = 21.6 kips
W
ΣFx = 0: Ax + 30,720 lb − 9102.22 lb = 0
or
A = −21, 618 lb
PROBLEM 5.79
The friction force between a 2 × 2-m square sluice gate AB and its
guides is equal to 10 percent of the resultant of the pressure forces
exerted by the water on the face of the gate. Determine the initial force
needed to lift the gate that its mass is 500 kg.
SOLUTION
Consider the free-body diagram of the gate.
PI =
Now
(
)(
)
1
1
ApI = ( 2 × 2 ) m 2   103 kg/m3 9.81 m/s 2 ( 3 m ) 


2
2
= 58.86 kN
PII =
(
)(
)
1
1
ApII = ( 2 × 2 ) m 2   103 kg/m3 9.81 m/s 2 ( 5 m ) 


2
2
= 98.10 kN
Then
F = 0.1P = 0.1( PI + PII )
= 0.1( 58.86 + 98.10 ) kN
= 15.696 kN
Finally
(
)
ΣFy = 0: T − 15.696 kN − ( 500 kg ) 9.81 m/s 2 = 0
or T = 20.6 kN W
PROBLEM 5.80
The dam for a lake is designed to withstand the additional force caused
by silt which has settled on the lake bottom. Assuming that silt is
equivalent to a liquid of density ρ s = 1.76 × 103 kg/m3 and considering
a 1-m-wide section of dam, determine the percentage increase in the
force acting on the dam face for a silt accumulation of depth 1.5 m.
SOLUTION
First, determine the force on the dam face without the silt.
Pw =
Have
=
1
1
Apw = A ( ρ gh )
2
2
(
)(
)
1
( 6 m )(1 m )   103 kg/m3 9.81 m/s 2 ( 6 m ) 


2
= 176.58 kN
Next, determine the force on the dam face with silt.
Have
Pw′ =
(
)(
)
1
( 4.5 m )(1m )   103 kg/m3 9.81 m/s 2 ( 4.5 m ) 


2
= 99.326 kN
( Ps )I = (1.5 m )(1 m ) (103 kg/m3 )( 9.81 m/s2 ) ( 4.5 m )
= 66.218 kN
( Ps )II =
(
)(
)
1
(1.5 m )(1 m )   1.76 × 103 kg/m3 9.81 m/s 2 (1.5 m ) 


2
= 19.424 kN
Then
P′ = Pw′ + ( Ps )I + ( Ps )II = 184.97 kN
The percentage increase, % inc., is then given by
% inc. =
(184.97 − 176.58) × 100% = 4.7503%
P′ − Pw
× 100% =
176.58
Pw
% inc. = 4.75% W
PROBLEM 5.81
The base of a dam for a lake is designed to resist up to 150 percent of the
horizontal force of the water. After construction, it is found that silt
(which is equivalent to a liquid of density ρ s = 1.76 × 103 kg/m3 ) is
settling on the lake bottom at a rate of 20 mm/y. Considering a 1-m-wide
section of dam, determine the number of years until the dam becomes
unsafe.
SOLUTION
From Problem 5.80, the force on the dam face before the silt is deposited, is Pw = 176.58 kN. The maximum
allowable force Pallow on the dam is then:
Pallow = 1.5Pw = (1.5 )(176.58 kN ) = 264.87 kN
Next determine the force P′ on the dam face after a depth d of silt has settled.
Have
Pw′ =
(
)(
)
1
( 6 − d ) m × (1 m )   103 kg/m3 9.81 m/s 2 ( 6 − d ) m 


2
= 4.905 ( 6 − d ) kN
2
( Ps )I =  d (1 m ) (103 kg/m3 )( 9.81 m/s2 ) ( 6 − d ) m 
(
)
= 9.81 6d − d 2 kN
( Ps )II =
(
)(
)
1
 d (1 m )   1.76 × 103 kg/m3 9.81 m/s 2 ( d ) m 


2
= 8.6328d 2 kN
(
)
(
)
P′ = Pw′ + ( Ps )I + ( Ps )II =  4.905 36 − 12d + d 2 + 9.81 6d − d 2 + 8.6328d 2  kN


= 3.7278d 2 + 176.58 kN
PROBLEM 5.81 CONTINUED
Now required that P′ = Pallow to determine the maximum value of d.
∴
or
Finally
( 3.7278d + 176.58) kN = 264.87 kN
2
d = 4.8667 m
4.8667 m = 20 × 10−3
m
×N
year
or N = 243 years W
PROBLEM 5.82
The square gate AB is held in the position shown by hinges along its top
edge A and by a shear pin at B. For a depth of water d = 3.5 m,
determine the force exerted on the gate by the shear pin.
SOLUTION
First consider the force of the water on the gate.
Have
Then
PI =
P=
1
1
Ap = A ( ρ gh )
2
2
(
)(
)
(
)(
)
1
(18 m )2 103 kg/m3 9.81 m/s2 (1.7 m )
2
= 26.99 kN
PII =
1
(18 m )2 103 kg/m3 9.81 m/s2 (1.7 × 1.8cos 30° ) m
2
= 51.74 kN
Now
or
or
ΣM A = 0:
1
2
( LAB ) PI + ( LAB ) PII − LAB FB = 0
3
3
1
2
( 26.99 kN ) + ( 51.74 kN ) − FB = 0
3
3
FB = 43.49 kN
FB = 4.35 kN
30.0° W
PROBLEMS 5.83 AND 5.84
Problem 5.83: A temporary dam is constructed in a 5-ft-wide fresh
water channel by nailing two boards to pilings located at the sides of the
channel and propping a third board AB against the pilings and the floor of
the channel. Neglecting friction, determine the reactions at A and B when
rope BC is slack.
Problem 5.84: A temporary dam is constructed in a 5-ft-wide fresh
water channel by nailing two boards to pilings located at the sides of the
channel and propping a third board AB against the pilings and the floor of
the channel. Neglecting friction, determine the magnitude and direction
of the minimum tension required in rope BC to move board AB.
SOLUTION
First, consider the force of the water on the gate.
Have
P=
1
1
Ap = A (γ h )
2
2
So that
PI =
1
(1.5 ft )( 5 ft )   62.4 lb/ft 3 (1.8 ft ) 


2
(
)
(
)
= 421.2 lb
PII =
1
(1.5 ft )( 5 ft )   62.4 lb/ft 3 ( 3 ft ) 


2
= 702 lb
5.83
Find the reactions at A and B when rope is slack.
ΣM A = 0: − ( 0.9 ft ) B + ( 0.5 ft )( 421.2 lb ) + (1.0 ft )( 702 lb ) = 0
B = 1014 lb
or
B = 1014 lb
ΣFx = 0: 2 Ax +
W
4
4
( 421.2 lb ) + ( 702 lb ) = 0
5
5
Ax = −449.28 lb
or
Note that the factor 2 (2 Ax ) is included since Ax is the horizontal force
exerted by the board on each piling.
ΣFy = 0: 1014 lb −
or
3
3
( 421.2 lb ) − ( 702 lb ) + Ay = 0
5
5
Ay = −340.08 lb
∴ A = 563 lb
37.1° W
PROBLEMS 5.83 AND 5.84 CONTINUED
5.84
Note that there are two ways to move the board:
1. Pull upward on the rope fastened at B so that the board
rotates about A. For this case B → 0 and TBC is
perpendicular to AB for minimum tension.
2. Pull horizontally at B so that the edge B of the board moves
to the left. For this case Ay → 0 and the board remains
against the pilings because of the force of the water.
Case (1)
ΣM A = 0: −1.5TBC + ( 0.5 ft )( 421.2 lb )
+ (1.0 ft )( 702 lb ) = 0
or
Case (2)
TBC = 608.4 lb
ΣM B = 0: − (1.2 ft ) ( 2 Ax ) − ( 0.5 ft )( 702 lb )
− (1.0 ft )( 421.2 lb ) = 0
or
2 Ax = − 643.5 lb
ΣFx = 0: − TBC − 643.5 lb +
+
or
4
( 421.2 lb )
5
4
( 702 lb ) = 0
5
TBC = 255.06 lb
∴ ( TBC )min = 255 lb
W
PROBLEMS 5.85 AND 5.86
Problem 5.85: A 2 × 3-m gate is hinged at A and is held in position by
rod CD. End D rests against a spring whose constant is 12 kN/m. The
spring is undeformed when the gate is vertical. Assuming that the force
exerted by rod CD on the gate remains horizontal, determine the
minimum depth of water d for which the bottom B of the gate will move
to the end of the cylindrical portion of the floor.
Problem 5.86: Solve Problem 5.85 if the mass of the gate is 500 kg.
SOLUTION
First, determine the forces exerted on the gate by the spring and the
water when B is at the end of the cylindrical portion of the floor.
sin θ =
Have
1
2
∴ θ = 30°
Then
xsp = (1.5 m ) tan 30°
and
Fsp = kxsp
= (12 kN/m )(1.5 m ) tan 30°
= 10.39 kN
d ≥ 2m
Assume
P=
Have
PI =
Then
1
1
Ap = A ( ρ g ) h
2
2
(
)(
)
1
( 2 m )( 3 m )   103 kg/m3 9.81 m/s 2 ( d − 2 ) m 


2
= 29.43 ( d − 2 ) kN
PII =
(
)(
)
1
( 2 m )( 3 m )   103 kg/m3 9.81 m/s 2 ( d − 2 + 2cos 30° ) m 


2
= 29.43 ( d − 0.2679 ) kN
PROBLEMS 5.85 AND 5.86 CONTINUED
5.85
Find d min so that gate opens, W = 0.
Using the above free-body diagrams of the gate, we have
2 
ΣM A = 0:  m   29.43 ( d − 2 ) kN 
3 
4 
+  m   29.43 ( d − 0.2679 ) kN 
3 
− (1.5 m )(10.39 kN ) = 0
or
19.62 ( d − 2 ) + 39.24 ( d − 0.2679 ) = 15.585
58.86d = 65.3374
5.86
d = 1.110 m W
d = 1.1105 m
or
Find d min so that the gate opens.
(
)
W = 9.81 m/s 2 ( 500 kg ) = 4.905 kN
Using the above free-body diagrams of the gate, we have
2 
ΣM A = 0:  m   29.43 ( d − 2 ) kN 
3 
4 
+  m   29.43 ( d − 0.2679 ) kN 
3 
− (1.5 m )(10.39 kN ) +
− ( 0.5 m )( 4.905 kN ) = 0
or
or
19.62 ( d − 2 ) + 39.24 ( d − 0.2679 ) = 18.0375
d = 1.15171 m
d = 1.152 m W
PROBLEMS 5.87 AND 5.88
Problem 5.87: The gate at the end of a 3-ft-wide fresh water channel is
fabricated from three 240-lb, rectangular steel plates. The gate is hinged
at A and rests against a frictionless support at D. Knowing that d = 2.5 ft,
determine the reactions at A and D.
Problem 5.88: The gate at the end of a 3-ft-wide fresh water channel is
fabricated from three 240-lb, rectangular steel plates. The gate is hinged
at A and rests against a frictionless support at D. Determine the depth of
water d for which the gate will open.
SOLUTION
5.87
Note that in addition to the weights of the gate segments, the water exerts pressure on all submerged
surfaces ( p = γ h ) .
Thus, at
h = 0.5 ft, p0.5 = 62.4 lb/ft 3 ( 0.5 ) ft = 31.2 lb/ft 2
(
)
(
)
h = 2.5 ft, p2.5 = 6.24 lb/ft 3 ( 2.5 ) ft = 156.0 lb/ft 2
Then
(
)
1
( 0.5 ft )( 3 ft )  31.2 lb/ft 2 = 23.4 lb
2
P1 =
(
)
P2 = ( 2 ft )( 3 ft )  31.2 lb/ft 2 = 187.2 lb
and
or
(
)
(
)
P3 =
1
( 2 ft )( 3 ft )  31.2 lb/ft 2 = 93.6 lb
2
P4 =
1
( 2 ft )( 3 ft )  156 lb/ft 2 = 468 lb
2


1

ΣM A = 0: − 4 D + ( 2 ft )( 240 lb ) + (1 ft )( 240 lb ) −  2 + × 0.5  ft ( 23.4 lb )  − (1 ft )(187.2 lb )
3



2
1
− ( 2 ft )( 93.6 lb ) − ( 2 ft )( 468 lb ) = 0
3
3
D = 11.325 lb
∴ D = 11.33 lb
W
PROBLEMS 5.87 AND 5.88 CONTINUED
ΣFx = 0: Ax + 11.32 + 23.4 + 93.6 + 468 = 0
Ax = −596.32 lb
or
ΣFy = 0: Ay − 240 − 240 − 240 + 187.2 = 0
Ay = 532.8 lb
or
5.88
∴ A = 800 lb
h = ( d − 2 ) ft, pd − 2 = γ ( d − 2 ) lb/ft 2
At
41.8° W
γ = 62.4 lb/ft 3
where
h = d ft, pd = (γ d ) lb/ft 2
P1 =
Then
1
1
3
2
A1 pd − 2 = ( d − 2 ) ft × ( 3 ft )  γ lb/ft 3 ( d − 2 ) ft  = γ ( d − 2 ) lb
2
2
2
(Note: For simplicity, the numerical value of the density γ will be substituted into the equilibrium equations
below, rather than at this level of the calculations.)
P2 = A2 pd − 2 = ( 2 ft )( 3 ft )  γ ( d − 2 ) ft  = 6γ ( d − 2 ) lb
P3 =
P4 =
1
1
A3 pd − 2 = ( 2 ft )( 3 ft )  γ ( d − 2 ) ft  = 3γ ( d − 2 ) lb
2
2
1
1
A4 pd = ( 2 ft )( 3 ft )  γ ( d ft ) = ( 3γ d ) lb = 3γ ( d − 2 ) + 6γ  lb
2
2
As the gate begins to open, D → 0
∴
ΣM A = 0:
( 2 ft )( 240 lb ) + (1 ft )( 240 lb ) − 2 ft + ( d − 2 ) ft   γ ( d − 2 )2 lb  +

1
3
3
2

2

− (1 ft ) 6γ ( d − 2 ) lb  −  ( 2 ft )  3γ ( d − 2 ) lb 
3

1

−  ( 2 ft )  3γ ( d − 2 ) lb + 6γ lb  = 0
3

or
1
720
−4
( d − 2 )3 + 3 ( d − 2 )2 + 12 ( d − 2 ) =
γ
2
=
720
−4
62.4
= 7.53846
Solving numerically yields
d = 2.55 ft W
PROBLEM 5.89
A rain gutter is supported from the roof of a house by hangers that are
spaced 0.6 m apart. After leaves clog the gutter’s drain, the gutter slowly
fills with rainwater. When the gutter is completely filled with water,
determine (a) the resultant of the pressure force exerted by the water on
the 0.6-m section of the curved surface of the gutter, (b) the force-couple
system exerted on a hanger where it is attached to the gutter.
SOLUTION
(a) Consider a 0.6 m long parabolic section of water.
Then
P=
=
1
1
Ap = A ( ρ gh )
2
2
(
)(
)
1
( 0.08 m )( 0.6 m )  103 kg/m3 9.81 m/s2 ( 0.08 m )
2
= 18.84 N
Ww = ρ gV
(
)(
)
2

= 103 kg/m3 9.81 m/s 2  ( 0.12 m )( 0.08 m )( 0.6 m ) 
3

= 37.67 N
ΣF = 0:
Now
So that R =
( −R ) + P + Ww = 0
tan θ =
P 2 + Ww2 ,
Ww
P
= 42.12 N, θ = 63.4°
R = 42.1 N
63.4° W
(b) Consider the free-body diagram of a 0.6 m long section of water and
gutter.
Then
ΣFx = 0: Bx = 0
ΣFy = 0: By − 37.67 N = 0
or
By = 37.67 N
ΣM B = 0: M B + ( 0.06 − 0.048 ) m  ( 37.67 N ) = 0
or
M B = −0.4520 N ⋅ m
The force-couple system exerted on the hanger is then
37.7 N , 0.452 N ⋅ m
W
PROBLEM 5.90
The composite body shown is formed by removing a hemisphere of
radius r from a cylinder of radius R and height 2R. Determine (a) the
y coordinate of the centroid when r = 3R/4, (b) the ratio r/R for which
y = −1.2 R.
SOLUTION
Note, for the axes shown
V
1
(π R ) ( 2R ) = 2π R
2
2
− π r3
3
Σ

r3 
2π  R3 − 
3

2
y
yV
−R
−2π R 4
3
− r
8
1 4
πr
4
3

r4 
−2π  R 4 − 
8 

1
R4 − r 4
Σ yV
8
Y =
=−
1
ΣV
3
R − r3
3
Then
1−
=
r =
(a )
1 r 
 
8 R 
1 r 
1−  
3 R 
3
R: y = −
4
4
3
1 3
 
3 4
4
1 3
1−  
3 4
3
1−
R
or y = −1.118R W
1 r 
 
8 R 
4
1 r 
1−  
3 R 
3
1−
y = −1.2 R : − 1.2 R = −
(b)
4
or
Solving numerically
R
3
r
r
  − 3.2   + 1.6 = 0
R
R
r
= 0.884 W
R
PROBLEM 5.91
Determine the y coordinate of the centroid of the body shown.
SOLUTION
First note that the values of Y will be the same for the given body and the body shown below. Then
V
y
yV
Cone
1 2
πa h
3
1
− h
4
1
− π a 2h 2
12
2
Cylinder
1
a
−π   b = − π a 2b
2
4
 
1
− b
2
1 2 2
πa b
8
Σ
π
12
a 2 ( 4h − 3b )
−
π
24
Have
Y ΣV = ΣyV
Then
π
π

Y  a 2 ( 4h − 3b )  = − a 2 2h 2 − 3b 2
24
12

(
(
a 2 2h 2 − 3b 2
)
)
or Y = −
2h 2 − 3b 2
W
2 ( 4h − 3b )
PROBLEM 5.92
Determine the z coordinate of the centroid of the body shown. (Hint:
Use the result of Sample Problem 5.13.)
SOLUTION
First note that the body can be formed by removing a “half-cylinder” from a “half-cone,” as shown.
V
1 2
πa h
6
Half-Cone
Half-Cylinder
Σ
−
π a
2
−
π
2
  b=− ab
22
8
π
24
−
z
a
π
zV
1 3
− ah
6
*
4 a
2a
 =−
3π  2 
3π
a 2 ( 4h − 3b )
1 3
ab
12
−
1 3
a ( 2h − b )
12
From Sample Problem 5.13
Have
Then
Z ΣV = ΣzV
1
π

Z  a 2 ( 4h − 3b )  = − a3 ( 2h − b )
24
12


or Z = −
2a 2h − b
W
π 4h − 3b
PROBLEM 5.93
Consider the composite body shown. Determine (a) the value of x when
h = L/2 , (b) the ratio h/L for which x = L .
SOLUTION
x
1
L
2
1
L+ h
4
V
Rectangular prism
Lab
Pyramid
1 b
a h
3 2
Then
1 

ΣV = ab  L + h 
6 

ΣxV =
X ΣV = ΣxV
Now
xV
1 2
L ab
2
1
1 

abh  L + h 
6
4 

1  2
1 

ab 3L + h  L + h  
6 
4 

so that
 
1  1 
1 
X  ab  L + h   = ab  3L2 + hL + h 2 
6  6 
4 
 
h 1 h2 
1 h 1 

X 1 +

 = L  3 + +
L 4 L2 
6 L 6 

or
(1)
1
L
2
h
1
Substituting
=
into Eq. (1)
L
2
(a) X = ? when h =
2

1  1  1 
1 11 
X 1 +    = L 3 +   +   
6  2   6 
 2  4  2  

or
(b)
X =
57
L
104
X = 0.548L W
h
= ? when X = L
L
Substituting into Eq. (1)
or
or
h 1 h2 
1 h 1 

=
+
+
L 1 +
L
3



L 4 L2 
6 L  6 

1+
1h
1 1h
1 h2
= +
+
6L
2 6 L 24 L2
h2
= 12
L2
∴
h
= 2 3W
L
PROBLEMS 5.94 AND 5.95
Problem 5.94: For the machine element shown, determine the
x coordinate of the center of gravity.
Problem 5.95: For the machine element shown, determine the
y coordinate of the center of gravity.
SOLUTIONS
First, assume that the machine element is homogeneous so that its center of gravity coincides with the
centroid of the corresponding volume.
x , in.
y , in.
xV , in 4
yV , in 4
I
II
V , in 3
(4)(3.6)(0.75) = 10.8
(2.4)(2.0)(0.6) = 2.88
2.0
3.7
0.375
1.95
21.6
10.656
4.05
5.616
III
π(0.45)2 (0.4) = 0.2545
4.2
2.15
1.0688
0.54711
IV
−π(0.5) 2 (0.75) = − 0.5890
1.2
0.375
−0.7068
−0.22089
Σ
13.3454
32.618
9.9922
5.94
X ΣV = Σ x V
Have
(
)
X 13.3454 in 3 = 32.618 in 4
or X = 2.44 in. W
5.95
Y ΣV = Σ yV
Have
(
)
Y 13.3454 in 3 = 9.9922 in 4
or Y = 0.749 in. W
PROBLEMS 5.96 AND 5.97
Problem 5.96: For the machine element shown, locate the x coordinate
of the center of gravity.
Problem 5.97: For the machine element shown, locate the y coordinate
of the center of gravity.
SOLUTIONS
First, assume that the machine element is homogeneous so that its center of gravity coincides with the
centroid of the corresponding volume.
V , mm3
x , mm
y , mm
xV , mm 4
yV , mm 4
1
(160 )( 54 )(18) = 155 520
80
9
12 441 600
1 399 680
2
1
(120 )( 42 )( 54 ) = 136 080
2
40
32
5 443 200
4 354 560
9
3 534 114
185 508
9
−2 316 233
−130 288
19 102 681
5 809 460
π
(27) 2 (18) = 6561π
160 +
4
−π(16)2 (18) = −4608π
160
Σ
297 736
3
2
36
π
5.96
X ΣV = Σ x V
Have
(
)
X 297 736 mm3 = 19 102 681 mm 4
or X = 64.2 mm W
5.97
Y ΣV = Σ yV
Have
(
)
Y 297 736 mm3 = 5 809 460 mm 4
or Y = 19.51 mm W
PROBLEMS 5.98 AND 5.99
Problem 5.98: For the stop bracket shown, locate the x coordinate of the
center of gravity.
Problem 5.99: For the stop bracket shown, locate the z coordinate of the
center of gravity.
SOLUTIONS
First, assume that the bracket is homogeneous so that its center of gravity coincides with the centroid of the
corresponding volume.
Z II = 24 mm +
1
(90 + 86) mm = 112 mm
2
Z III = 24 mm +
1
(102) mm = 58 mm
3
X III = 68 mm +
1
( 20) mm = 78 mm
2
Have..
Z IV = 110 mm +
2
( 90) mm = 170 mm
3
X IV = 60 mm +
2
(132) mm = 156 mm
3
V , mm3
x , mm
z , mm
xV , mm 4
zV , mm 4
I
( 200 )(176 )( 24 ) = 844 800
100
12
84 480 000
1 013 760
II
( 200 )( 24 )(176 ) = 844 800
100
112
84 480 000
94 617 600
78
58
9 865 440
733 840
156
170
−22 239 360
−24 235 200
156 586 080
8 785 584
III
IV
Σ
1
( 20 )(124 )(102 ) = 126 480
2
1
− ( 90 )(132 )( 24 ) = −142 560
2
1 673 520
5.98
X ΣV = ΣxV
Have
(
)
X 1 673 520 mm3 = 156 586 080 mm 4
or X = 93.6 mm W
5.99
Z ΣV = ΣzV
Have
(
)
Z 1 673 520 mm3 = 8 785 584 mm 4
or Z = 52.5 mm W
PROBLEM 5.100
Locate the center of gravity of the sheet-metal form shown.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity coincides with the centroid of
the corresponding area.
A, mm 2
1
2
3
1
( 90 )( 60 )
2
= 2700
( 90 )( 200 )
= 18 000
− ( 45 )(100 )
= −4500
π
( 45)2
4
2
= 1012.5π
Σ
Have
19 380.9
x , mm
30
y , mm
z , mm
xA, mm3
yA, mm3
zA, mm3
0
81 000
378 000
0
120 + 20
= 140
45
60
80
810 000
1 080 000
1 440 000
22.5
30
120
−101 250
−135 000
−540 000
143 139
0
569 688
932 889
1 323 000
1 469 688
45
0
160 +
( 4 )( 45)
3π
= 179.1
X ΣA = ΣxA:
(
)
X 19 380.9 mm 2 = 932 889 mm3
= 48.1 mm
or
X = 48.1 mm W
Y ΣA = Σ yA
(
)
Y 19 380.9 mm 2 = 1 323 000 mm3
Y = 68.3 mm
or
Y = 68.3 mm W
Z ΣA = Σ zA
(
)
Z 19 380.9 mm 2 = 1 469 688 mm3
or
Z = 75.8 mm
Z = 75.8 mm W
PROBLEM 5.101
A mounting bracket for electronic components is formed from sheet
metal of uniform thickness. Locate the center of gravity of the bracket.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the bracket coincides with
the centroid of the corresponding area. Then (see diagram)
zV = 22.5 −
4 ( 6.25 )
3π
= 19.85 mm
π
AV = −
2
( 6.25)2
= −61.36 mm 2
A, mm 2
IV
( 25)( 60) = 1500
(12.5)( 60 ) = 750
( 7.5)( 60 ) = 450
− (12.5 )( 30 ) = −375
V
Σ
−61.36
2263.64
I
II
III
x , mm
y , mm
z , mm
xA, mm3
yA, mm3
zA, mm3
12.5
0
30
18 750
0
45 000
25
−6.25
30
18 750
−4687.5
22 500
28.75
−12.5
30
12 937.5
−5625
13 500
10
0
37.5
−3750
0
−14 062.5
10
0
19.85
−613.6
46 074
0
−10 313
−1218.0
65 720
X ΣA = ΣxA
Have
(
)
X 2263.64 mm 2 = 46 074 mm3
or X = 20.4 mm W
Y ΣA = Σ yA
(
)
Y 2263.64 mm 2 = −10 313 mm3
or Y = −4.55 mm W
Z ΣA = Σ zA
(
)
Z 2263.64 mm 2 = 65 720 mm3
or Z = 29.0 mm W
PROBLEM 5.102
Locate the center of gravity of the sheet-metal form shown.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the form coincides with the
centroid of the corresponding area.
yI = 6 +
zI =
4
= 7.333 in.
3
1
( 6) = 2 in.
3
1
xII = yII =
xIV = 11 −
I
II
III
IV
Σ
A, in 2
12
56.55
30
−3.534
95.01
x , in.
y , in.
z , in.
0
3.8197
8.5
10.363
7.333
3.8197
0
0
2
3
3
3
π
( 2)( 6) = 3.8197 in.
1
( 4)(1.5) = 10.363 in.
3π
x A, in 3
0
216
255
−36.62
434.4
y A, in 3
88
216
0
0
304
z A, in 3
24
169.65
90
−10.603
273.0
X ΣA = Σ x A
Have
(
)
X 95.01 in 2 = 434 in 3
or X = 4.57 in. W
Y ΣA = Σ y A
(
)
Y 95.01 in 2 = 304.0 in 3
or Y = 3.20 in. W
Z ΣA = Σ z A
(
)
Z 95.01 in 2 = 273.0 in 3
or Z = 2.87 in. W
PROBLEM 5.103
An enclosure for an electronic device is formed from sheet metal of
uniform thickness. Locate the center of gravity of the enclosure.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the form coincides with the
centroid of the corresponding area.
Consider the division of the back, sides, and top into eight segments according to the sketch.
Z = 6.00 in. W
Note that symmetry implies
and
A8 = A2
A7 = A 3
A6 = A5
Thus
1
2
3
4
5
6
7
8
Σ
Have
A, in 2
x , in.
y , in.
xA, in 3
yA, in 3
(12 )( 9 ) = 108
(11.3)( 9 ) = 101.7
0
4.5
0
486
5.6
4.5
569.5
457.6
7.467
9.8
101.25
132.89
5.6
10.2
769.72
1402.0
7.467
10.6
64.15
91.06
7.467
7.467
5.6
10.6
9.8
4.5
64.15
101.25
569.5
2239.5
91.06
132.89
457.6
3251.1
1
(11.3)( 2.4 ) = 13.56
2
(12 )(11.454 ) = 137.45
1
(1.5)(11.454 ) = 8.591
2
8.591
13.56
101.7
493.2
(
)
Y ΣA = Σ yA: Y ( 493.2 in ) = 3251.1 in
X ΣA = ΣxA: X 493.2 in 2 = 2239.5 in 3
2
3
or X = 4.54 in. W
or Y = 6.59 in. W
PROBLEM 5.104
A 200-mm-diameter cylindrical duct and a 100 × 200-mm rectangular
duct are to be joined as indicated. Knowing that the ducts are fabricated
from the same sheet metal, which is of uniform thickness, locate the center
of gravity of the assembly.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides with the
centroid of the corresponding area. Also note that symmetry implies
Z = 0W
A, m 2
x, m
y, m
xA, m3
yA, m3
π ( 0.2 )( 0.3) = 0.1885
0
0.15
0
0.028274
= 0.06366
0.25
−0.02000
−0.007854
−4 ( 0.1)
= −0.04244
3π
0.30
−0.000667
0.004712
0.15
0.30
0.00900
0.001800
5
( 0.3)( 0.2 ) = 0.060
( 0.3)( 0.2 ) = 0.060
0.15
0.20
0.00900
0.001200
6
−
π
0.20
−0.000667
−0.003142
7
0.15
0.25
0.004500
0.007500
8
( 0.3)( 0.1) = 0.030
( 0.3)( 0.1) = 0.030
4 ( 0.1)
3π
0.15
0.25
Σ
0.337080
0.004500
0.023667
0.007500
0.066991
1
2
3
4
Have
−
π
2
( 0.2 )( 0.1) = −0.0314
2 ( 0.1)
π
π
2
( 0.1) = 0.01571
2
2
( 0.1)2 = −0.1571
X ΣA = Σ xA: X (0.337080 mm 2 ) = 0.023667 mm3
X = 0.0702 m
or
(
X = 70.2 mm W
)
Y ΣA = Σ yA: Y 0.337080 mm 2 = 0.066991 mm3
or
Y = 0.19874 m
Y = 198.7 mm W
PROBLEM 5.105
An elbow for the duct of a ventilating system is made of sheet metal of
uniform thickness. Locate the center of gravity of the elbow.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides with the
centroid of the corresponding area. Also, note that the shape of the duct implies
Y = 1.5 in. W
xI = zI = 16 in. −
Note that
xII = 16 in. −
zII = 12 in. −
2
π
2
π
2
π
(16 in.) = 5.81408 in.
(8 in.) = 10.9070 in.
(8 in.) = 6.9070 in.
xIV = zIV = 16 in. −
4
(16 in.) = 9.2094 in.
3π
4
(8 in.) = 12.6047 in.
3π
4
zV = 12 in. −
(8 in.) = 8.6047 in.
3π
xV = 16 in. −
Also note that the corresponding top and bottom areas will contribute equally when determining x and z .
Thus
A, in 2
x , in.
z , in.
xA, in 3
zA, in 3
(16 )( 3) = 75.3982
5.81408
5.81408
438.37
438.37
(8)( 3) = 37.6991
10.9070
6.9070
411.18
260.39
III
4 ( 3) = 12
8
14
96.0
168.0
IV
 π
2
2   (16) = 402.1239
 4
9.2094
9.2094
3703.32
3703.32
V
π 2
−2   ( 8 ) = −100.5309
4
12.6047
8.6047
−1267.16
−865.04
VI
−2 ( 4 )( 8 ) = −64
12
14
−768.0
−896.0
Σ
362.69
2613.71
2809.04
I
II
Have
π
2
π
2
(
)
X ΣA = Σ xA: X 362.69 in 2 = 2613.71 in 3
or X = 7.21 in. W
(
)
Z ΣA = Σ zA: Z 362.69 in 2 = 2809.04 in 3
or Z = 7.74 in. W
PROBLEM 5.106
A window awning is fabricated from sheet metal of uniform thickness.
Locate the center of gravity of the awning.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the awning coincides with
the centroid of the corresponding area.
yII = yVI = 80 +
( 4 )( 500 ) = 292.2 mm
3π
4
500
( )( ) = 212.2 mm
zII = zVI =
3π
( 2 )( 500 ) = 398.3 mm
yIV = 80 +
π
zIV =
( 2 )( 500 ) = 318.3 mm
π
AII = AVI =
AIV =
2
4
( 500 )2 = 196 350 mm 2
( 500 )( 680 ) = 534 071 mm2
A, mm 2
(80)(500) = 40 000
y , mm
z , mm
yA, mm3
zA, mm3
40
250
1.6 × 106
10 × 106
292.2
212.2
57.4 × 106
41.67 × 106
III
196 350
(80)(680) = 54 400
40
500
0.2176 × 106
27.2 × 106
IV
534 071
398.3
318.3
212.7 × 106
170 × 106
V
(80)(500) = 40 000
40
250
1.6 × 106
10 × 106
VI
196 350
292.2
212.2
57.4 × 106
41.67 × 106
Σ
1.061 × 106
332.9 × 106
300.5 × 106
I
II
Now, symmetry implies
and
π
π
(
6
Y ΣA = Σ yA: Y 1.061 × 10 mm
2
) = 332.9 × 10 mm
6
X = 340 mm W
3
or Y = 314 mm W
(
)
Z ΣA = Σ zA: Z 1.061 × 106 mm 2 = 300.5 × 106 mm3
or Z = 283 mm W
PROBLEM 5.107
The thin, plastic front cover of a wall clock is of uniform thickness.
Locate the center of gravity of the cover.
SOLUTION
First, assume that the plastic is homogeneous so that the center of gravity of the cover coincides with the
centroid of the corresponding area.
X = 150.0 mm
Next, note that symmetry implies
A, mm 2
1
2
3
4
5
6
7
Σ
Have
( 300 )( 280 )
= 84 000
( 280 )( 50 )
= 14 000
( 300 )( 50 )
= 15 000
( 280 )( 50 )
= 14 000
−π (100 )
y , mm
z , mm
y A, mm3
0
140
0
11 760 000
25
140
350 000
1 960 000
25
0
375 000
0
25
140
350 000
1 960 000
0
130
0
−4 084 070
z , mm3
2
= −31 416
−π
( 30 )2
4
= −706.86
−π
( 30 )2
4
= −706.86
0
260 −
( 4 )( 30 ) = 247.29
0
−174 783
0
260 −
( 4 )( 30 ) = 247.29
0
−174 783
1 075 000
11 246 363
3π
3π
94 170
(
)
Y Σ A = Σ yA: Y 94 170 mm 2 = 1 075 000 mm3
or Y = 11.42 mm W
(
)
Z Σ A = Σ zA: Z 94 170 mm 2 = 11 246 363 mm3
or Z = 119.4 mm W
PROBLEM 5.108
A thin steel wire of uniform cross section is bent into the shape shown,
where arc BC is a quarter circle of radius R. Locate its center of gravity.
SOLUTION
First, assume that the wire is homogeneous so that its center of gravity coincides with the centroid of the
corresponding line.
L, in 2
x , in.
y , in.
z ,in.
x L, in 2
y L, in 2
z L, in 2
1
15
0
7.5
0
0
112.5
0
2
14
7
0
0
98
0
0
13
 5
9 
 12 
0
6
149.5
0
78
135.0
225.0
180.0
382.5
337.5
258.0
3
π
4
Σ
Have
(15)
2
= 23.56
= 11.5
3  2 × 15 


5 π 
30
24
π
π
= 5.73
= 9.549
= 7.639
65.56
X ΣL = Σ x L: X ( 65.56 in.) = 382.5 in 2
or X = 5.83 in. W
Y ΣL = Σ y L: Y ( 65.56 in.) = 337.5 in 2
or Y = 5.15 in. W
Z ΣL = Σ z L: Z ( 65.56 in.) = 258.0 in 2
or Z = 3.94 in. W
PROBLEM 5.109
A thin steel wire of uniform cross section is bent into the shape shown,
where arc BC is a quarter circle of radius R. Locate its center of gravity.
SOLUTION
First, assume that the wire is homogeneous so that its center of gravity coincides with the centroid of the
corresponding line
x2 = z2 =
Have
L1 =
( 2 )(8) = 16 ft
π
π
82 + 32 = 8.5440 ft
8π
L 2 = 2 = 4π ft
4
Have
L, ft
x , ft
y , ft
z , ft
1
8.5440
4
1.5
2
4π
16π
0
0
16
3
4
8
3
32.110
0
0
0
1.5
x L, ft 2
34.176
y L, ft 2
12.816
z L, ft 2
0
64.0
0
64.0
0
0
98.176
0
4.5
17.316
32
0
96.0
π
4
0
X ΣL = Σ x L: X ( 32.110 ft ) = 98.176 ft 2
or X = 3.06 ft W
Y ΣL = Σ y L: Y ( 32.110 ft ) = 17.316 ft 2
or Y = 0.539 ft W
Z ΣL = Σ z L: Z ( 32.110 ft ) = 96.0 ft 2
or Z = 2.99 ft W
PROBLEM 5.110
The frame of a greenhouse is constructed from uniform aluminum
channels. Locate the center of gravity of the portion of the frame shown.
SOLUTION
First, assume that the channels are homogeneous so that the center of gravity of the frame coincides with the
centroid of the corresponding line.
Note
x8 = x9 =
( 2 )( 0.9 ) = 0.57296 m
π
y8 = y9 = 1.5 +
L 7 = L8 =
1
2
3
4
5
6
7
8
9
10
Σ
Have
L, m
x, m
y, m
z, m
0.6
0.9
1.5
1.5
2.4
0.6
0.9
1.4137
1.4137
0.6
11.827
0.9
0.45
0.9
0.9
0
0.9
0.45
0.573
0.573
0
0
0
0.75
0.75
1.2
1.5
1.5
2.073
2.073
2.4
0.3
0.6
0
0.6
0.6
0.3
0.6
0
0.6
0.3
xL, m 2
0.540
0.4050
1.350
1.350
0
0.540
0.4050
0.8100
0.8100
0
6.210
π
2
( 2 )( 0.9 ) = 2.073 m
π
( 0.9 ) = 1.4137 m
yL, m 2
0
0
1.125
1.125
2.880
0.9
1.350
2.9306
2.9306
1.440
14.681
zL, m 2
0.18
0.54
0
0.9
1.44
0.18
0.54
0
0.8482
0.18
4.8082
X ΣL = Σ xL: X (11.827 m ) = 6.210 m 2
or X = 0.525 m W
Y ΣL = Σ yL: Y (11.827 m ) = 14.681 m 2
or Y = 1.241 m W
Z ΣL = Σ zL: Z (11.827 m ) = 4.8082 m 2
or Z = 0.406 m W
*
PROBLEM 5.111
The decorative metalwork at the entrance of a store is fabricated from
uniform steel structural tubing. Knowing that R = 1.2 m, locate the center
of gravity of the metalwork.
SOLUTION
First, assume that the tubes are homogeneous so that the center of gravity of the metalwork coincides with the
centroid of the corresponding line.
Z = 0W
Note that symmetry implies
x, m
L, m
Have
1
3
2
3
(1.2) cos 45° = 0.8485
(1.2) cos 45° = 0.8485
3
1.2π
0
4
1.2π
( 2 )(1.2 ) = 0.7639
5
0.6π
( 2 )(1.2 ) = 0.7639
Σ
15.425
π
π
y, m
xL, m 2
yL, m 2
1.5
2.5456
4.5
1.5
2.5456
4.5
3.7639
0
14.1897
3
2.88
11.3097
3.7639
1.44
7.0949
9.4112
41.594
X ΣL = Σ xL: X (15.425 m ) = 9.4112 m 2
or X = 0.610 m W
Y ΣL = Σ yL: Y (15.425 m ) = 41.594 m 2
or Y = 2.70 m W
PROBLEM 5.112
A scratch awl has a plastic handle and a steel blade and shank. Knowing
that the specific weight of plastic is 0.0374 lb/in 3 and of steel is
0.284 lb/in 3, locate the center of gravity of the awl.
SOLUTION
Y = Z = 0W
First, note that symmetry implies
xI =
5
2π
( 0.5 in.) = 0.3125 in., WI = 0.0374 lb/in 3   ( 0.5 in.)3 = 0.009791 lb
8
 3 
(
)
(
)
xII = 1.6 in. + 0.5 in. = 2.1 in. WII = 0.0374 lb/in 3 (π )( 0.5 in.) ( 3.2 in.) = 0.093996 lb
2
2
π 
xIII = 3.7 in. − 1 in. = 2.7 in., WIII = − 0.0374 lb/in 3   ( 0.12 in.) ( 2 in.) = −0.000846 lb
4
(
)
2
2
π 
xIV = 7.3 in. − 2.8 in. = 4.5 in., WIV = 0.284 lb/in 3   ( 0.12 in.) ( 5.6 in.) = 0.017987 lb
4
(
xV = 7.3 in. +
1
π
( 0.4 in.) = 7.4 in., WV = 0.284 lb/in 3   ( 0.06 in.)2 ( 0.4 in.) = 0.000428 lb
4
3
(
I
II
III
IV
V
Σ
Have
)
W , lb
0.009791
0.093996
−0.000846
0.017987
0.000428
0.12136
)
x , in.
0.3125
2.1
2.7
4.5
7.4
xW , in ⋅ lb
0.003060
0.197393
−0.002284
0.080942
0.003169
0.28228
X ΣW = ΣxW : X ( 0.12136 lb ) = 0.28228 in.⋅ lb
or X = 2.33 in. W
PROBLEM 5.113
A bronze bushing is mounted inside a steel sleeve. Knowing that the
density of bronze is 8800 kg/m3 and of steel is 7860 kg/m3 , determine
the center of gravity of the assembly.
SOLUTION
X = Z = 0W
First, note that symmetry implies
W = ( ρ g )V
Now
 π 

yI = 4 mm , WI = 7860 kg/m3 9.81 m/s2    0.0362 − 0.0152 m 2  ( 0.008 m ) 


 4 

(
)(
)
(
)
= 0.51887 N
 π 

yII = 18 mm, WII = 7860 kg/m3 9.81 m/s 2    0.02252 − 0.052 m 2  ( 0.02 m ) 

 4  

(
)(
)
(
)
= 0.34065 N
 π 

yIII = 14 mm, WIII = 8800 kg/m3 9.81 m/s2    0.152 − 0.102 m 2  ( 0.028 m ) 


 4 

(
)(
)
(
)
= 0.23731 N
Y ΣW = Σ yW
Have
Y =
( 4 mm)( 0.5189 N ) + (18 mm)( 0.3406 N ) + (14 mm )( 0.2373 N )
0.5189 N + 0.3406 N + 0.2373 N
or Y = 10.51 mm W
( above base )
PROBLEM 5.114
A marker for a garden path consists of a truncated regular pyramid carved
from stone of specific weight 160 lb/ft3. The pyramid is mounted on a
steel base of thickness h. Knowing that the specific weight of steel is
490 lb/ft3 and that steel plate is available in 14 in. increments, specify the
minimum thickness h for which the center of gravity of the marker is
approximately 12 in. above the top of the base.
SOLUTION
First, locate the center of gravity of the stone. Assume that the stone is
homogeneous so that the center of gravity coincides with the centroid of
the corresponding volume.
Have
y1 =
3
( 56 in.) = 42 in.,
4
V1 =
1
(12 in.)(12 in.)( 56 in.)
3
= 2688 in 3
y2 =
3
( 28 in.) = 21 in.,
4
V2 = −
1
( 6 in.)( 6 in.)( 28 in.)
3
= −366 in 3
Vstone = 2688 in 3 − 366 in 3
Then
= 2352 in 3
and
Y =
=
ΣyV
ΣV
( 42 in.) ( 2688 in 3 ) + ( 21 in.) ( −366 in 3 )
2352 in 3
= 45 in.
Therefore, the center of gravity of the stone is ( 45 − 28 ) in. = 17 in.
above the base.
Now
(
Wstone = γ stoneVstone = 160 lb/ft
3
)(
 1 ft 
2352 in 

 12 in. 
3
)
= 217.78 lb
Wsteel = γ steelVsteel
 1 ft 
= 490 lb/ft (12 in.)(12 in.) h  

 12 in. 
(
3
)
= ( 40.833h ) 1b
3
3
PROBLEM 5.114 CONTINUED
Then
Ymarker =
=
or
ΣyW
= 12 in.
ΣW
(17 in.)( 217.78 lb ) +  −
h 
in.  ( 40.833 h ) lb
 2 
( 217.78 + 40.833h ) lb
h 2 + 24h − 53.334 = 0
With positive solution h = 2.0476 in.
∴ specify h = 2 in. W
PROBLEM 5.115
The ends of the park bench shown are made of concrete, while the seat
and back are wooden boards. Each piece of wood is 36 × 120 ×
1180 mm. Knowing that the density of concrete is 2320 kg/m3 and of
wood is 470 kg/m3 , determine the x and y coordinates of the center of
gravity of the bench.
SOLUTION
First, note that we will account for the two concrete ends by counting twice the weights of components 1, 2,
and 3.
(
)(
)
W1 = ( ρc g )V1 = 2320 kg/m3 9.81 m/s 2 ( 0.480 m )( 0.408 m )( 0.072 m ) 
= 320.9 N
(
)(
)
W2 = − ( ρc g )V2 = − 2320 kg/m3 9.81 m/s 2 ( 0.096 m )( 0.048 m )( 0.072 m ) 
= −7.551 N
(
)(
)
W3 = ( ρc g )V3 = 2320 kg/m3 9.81 m/s 2 ( 0.096 m )( 0.384 m )( 0.072 m ) 
= 60.41 N
W4 = W5 = W6 = W7 = ρ wVboard
(
)(
)
= 470 kg/m3 9.81 m/s 2 ( 0.120 m )( 0.036 m )(1.180 m ) 
= 23.504 N
PROBLEM 5.115 CONTINUED
Have
W, N
x , mm
y , mm
x W, mm ⋅ N
y W, mm ⋅ N
1
2 ( 320.4 ) = 641.83
312
−204
200 251.4
−130 933.6
2
2 ( −7.551) = −15.10
312
−384
−4711.8
5799.1
3
4
5
6
7
8
Σ
2 ( 60.41) = 120.82
84
228
360
442
124.7
160.1
192
18
18
18
328.3
139.6
10 148.5
5358.8
8461.3
10 388.5
2930.9
3762.9
236 590
23 196.5
423.1
423.1
423.1
7716.2
3281.1
−89 671
23.504
23.504
23.504
23.504
23.504
865.06
X ΣW = Σ xW : X ( 865.06 N ) = 236 590 mm ⋅ N
or X = 274 mm W
Y ΣW = Σ yW : Y ( 865.06 N ) = −89 671 mm ⋅ N
or Y = −103.6 mm W
PROBLEM 5.116
Determine by direct integration the values of x for the two volumes
obtained by passing a vertical cutting plane through the given shape of
Fig. 5.21. The cutting plane is parallel to the base of the given shape and
divides the shape into two volumes of equal height.
A hemisphere.
SOLUTION
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = πr 2dx, xEL = x
The equation of the generating curve is x 2 + y 2 = a 2 so that
r 2 = a 2 − x 2 and then
(
)
)

x3 
dx = π a 2 x − 
3 0

dV = π a 2 − x 2 dx
Component 1
(
a/2
π a2 − x2
0
V1 = ∫
11 3
πa
24
=
and
a/2
∫1 xEL dV = ∫0 x π ( a − x ) dx 
a/2
2
2
a/2
 x2 x4 
= π a 2
−

4 0
 2
=
Now
7
π a4
64
7
 11

x1V1 = ∫1 xEL dV : x1  π a3  =
π a4
24
64


or x1 =
Component 2
a
π
a /2
V2 = ∫
(
a

x3 
a − x dx = π  a 2 x − 
3  a/2

2
2
)


a3 

a
= π a 2 ( a ) −  − a2   −

3

   2 

5
=
πa 3
24
( a2 )  
3

3 
 
21
aW
88
PROBLEM 5.116 CONTINUED
and
a
 2 x2 x4 
a
2
2
∫2 xELdV = ∫a/2 x π a − x dx  = π  a 2 − 4 

 a/2
(
)
2

2
4

a )   2 a2
(
 2 (a)


=π a
−
− a
−
2
4  
2
 




9
=
π a4
64
( )
Now
( a2 )  
4

4 
 
9
 5

x2V2 = ∫2 xELdV : x2  π a3  =
π a4
24
64


or x2 =
27
aW
40
PROBLEM 5.117
Determine by direct integration the values of x for the two volumes
obtained by passing a vertical cutting plane through the given shape of
Fig. 5.21. The cutting plane is parallel to the base of the given shape and
divides the shape into two volumes of equal height.
A semiellipsoid of revolution.
SOLUTION
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = πr 2dx, xEL = x
The equation of the generating curve is
r2 =
x2
y2
+
= 1 so that
h2 a 2
a2 2
h − x 2 and then
h2
(
)
dV = π
a2 2
h − x 2 dx
h2
(
)
Component 1
2
h/2 a
π
2
0
V1 = ∫
=
and
h
(
h/2
a2 
x3 
h − x dx = π 2  h 2 x − 
3 0
h 
2
2
)
11 2
πa h
24
 a2

∫1 xEL dV = ∫0 x π h 2 ( h − x ) dx 
h/2
2
2


h/2
=π
a2  2 x2 x4 
− 
h
4 0
h2  2
7
π a 2h 2
64
7
 11

x1V1 = ∫1 xEL dV : x1  π a 2h  =
π a 2h 2
 24
 64
=
Now
or x1 =
21
hW
88
PROBLEM 5.117 CONTINUED
Component 2
h
V2 = ∫h/2 π
h
a2 2
a2 
x3 
h − x 2 dx = π 2 h 2 x − 
2
3 h/2
h
h 
(
)
()
3

3

h 
h)   2  h 
a 2  2
(
2 
− h   −
= π 2 h ( h ) −

3    2
3 
h 






5
=
πa 2 h
24
and
 a2

∫2 xELdV = ∫h/2 x π h2 ( h − x ) dx 
h
2
2


h
=π
a 2  2 x2 x4 
−
h

4  h/2
h2  2
2

2
4

h )   2 h2
(
a2  2 ( h )
− h
= π 2 h
−
−
2
4  
2
h 




9
=
π a 2h 2
64
( )
Now
( h2 )  
4

4 
 
9
 5

x2V2 = ∫2 xEL dV : x2  π a 2h  =
π a 2h 2
24
64


or x2 =
27
hW
40
PROBLEM 5.118
Determine by direct integration the values of x for the two volumes
obtained by passing a vertical cutting plane through the given shape of
Fig. 5.21. The cutting plane is parallel to the base of the given shape and
divides the shape into two volumes of equal height.
A paraboloid of revolution.
SOLUTION
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = πr 2dx, xEL = x
The equation of the generating curve is x = h −
r2 =
h 2
y so that
a2
a2
( h − x ) and then
h
dV = π
a2
( h − x ) dx
h
Component 1
h/2
V1 = ∫0 π
a2
( h − x ) dx
h
h/2
=π
=
and
a2 
x2 
 hx −

h 
2 0
3 2
πa h
8
2

h/2  a
=
x
dV
x
π
∫1 EL
∫0  h ( h − x ) dx 


h/2
a 2  x 2 x3 
=π
− 
h
3 0
h  2
=
Now
1
π a 2h 2
12
1
3

x1V1 = ∫1 xEL dV : x1  π a 2h  = π a 2h 2
8
 12
or x1 =
2
hW
9
PROBLEM 5.118 CONTINUED
Component 2
h
h
V2 = ∫h/2 π
a2
a2 
x2 
( h − x ) dx = π  hx − 
h
h 
2  h/2

2

h)    h 
(
a2 


= π  h (h) −
− h  −
h 
2   2




1
= πa 2 h
8
( h2 )  
2

2 
 
h
and
 a2

a 2  x 2 x3 
h
∫2 xELdV = ∫h/2 x π h ( h − x ) dx  = π h  h 2 − 3 



 h/2
2

2
3

h )   h2
(
a2  ( h )


=π
−
− h
−
 h
3   2
h  2





1
= π a 2h 2
12
( )
Now
( h2 )  
3

3 
 
1
1

x2V2 = ∫2 xEL dV : x2  π a 2h  =
π a 2h2
8
12


or x2 =
2
hW
3
PROBLEM 5.119
Locate the centroid of the volume obtained by rotating the shaded area
about the x axis.
SOLUTION
y = 0W
First note that symmetry implies
z = 0W
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = πr 2dx, xEL = x

x2 
Now r = b1 − 2  so that
a 

2
2
x2 
dV = πb 1 − 2  dx
a 

2
Then


x2 
2x2 x4 
a
a
V = ∫0 π b 2 1 − 2  dx = ∫0 π b 2  1 − 2 + 4  dx
a 
a
a 


2
x5 
2 x3
= π b  x − 2 + 4 
3a
5a 

a
0
2 1

= π ab 2 1 − + 
3 5

8
= π ab 2
15
and
2x2 x4 
a
2 
=
−
+ 4  dx
x
dV
π
b
x
1

∫ EL
∫0

a2
a 

2 x
2x4
x6 
= π b 
− 2 + 4 
4a
6a 
 2
2
1 1 1
= π a 2b 2  − + 
2 2 6
=
1 2 2
πa b
6
a
0
PROBLEM 5.119 CONTINUED
Then
1
 8

xV = ∫ xEL dV : x  π ab 2  = π a 2b 2
 15
 16
or x =
15
aW
6
PROBLEM 5.120
Locate the centroid of the volume obtained by rotating the shaded area
about the x axis.
SOLUTION
y = 0W
First, note that symmetry implies
z = 0W
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = πr 2dx, xEL = x
Now r = 1 −
1
so that
x
2
1

dV = π 1 −  dx
x

2
1 

= π 1 − + 2  dx
x x 

3
Then
2
1 
1
3 

V = ∫1 π 1 − + 2  dx = π  x − 2 ln x − 
x x 
x 1



1 
1 
= π  3 − 2 ln3 −  − 1 − 2 ln 1 −  
3 
1 

= ( 0.46944π ) m3
and
2
3  
x π 1− +
1  
∫ x ELdV = ∫
 
x
3
 x2

1  
=
dx
π
 − 2 x + ln x 

2
x  
2
1
  32
 13
 
= π   − 2 ( 3) + ln 3 −  − 2 (1) + ln1 
 2
 
  2
= (1.09861π ) m
Now
(
)
xV = ∫ x EL dV : X 0.46944π m3 = 1.09861π m 4
or x = 2.34 m W
PROBLEM 5.121
Locate the centroid of the volume obtained by rotating the shaded area
about the line x = h.
SOLUTION
x = hW
First, note that symmetry implies
z = 0W
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = πr 2dy, yEL = y
Now x 2 =
Then
h2 2
h 2
a − y2
a − y 2 so that r = h −
2
a
a
2
h2
dV = π 2 a − a 2 − y 2 dy
a
(
)
)
(
a
)
(
2
h2
2
2
a
a
y
dy
−
−
a2
and
V = ∫0 π
Let
y = a sin θ ⇒ dy = a cos θ d θ
Then
V =π
=π
)
(
2
h 2 π /2
2
2
2
−
−
sin
θ
a
a
a
a cosθ dθ
∫
a2 0
h 2 π /2  2
2
2
2
∫ a − 2a ( a cosθ ) + a − a sin θ  a cosθ dθ
a2 0 
(
π /2
= π ah 2 ∫0
)
( 2 cosθ − 2 cos θ − sin θ cosθ ) dθ
2
2
π /2

 θ sin 2θ  1 3 
= π ah 2  2sin θ − 2  +
 − sin θ 
4  3
2

0

 π  1
= π ah 2  2 − 2  2  − 
2 3


 
= 0.095870π ah 2
PROBLEM 5.121 CONTINUED
and
 h2

(
2
)
(
=π
h2 a
2a 2 y − 2ay a 2 − y 2 − y 3 dy
2 ∫0
a
=π
3/2
1 
h2  2 2 2
− y4 
a y + a a2 − y2
2 
3
4 0
a 
=π
3/2  
1  2
h2  2
2
a ( a ) − a4  −  a a2  
2 
4  3
a 

=
Now
) 
a
2
2
∫ y ELdV = ∫0 y π a 2 a − a − y dy 
(
a
)
( )
1
π a 2h 2
12
(
)
yV = ∫ y EL dV : y 0.095870π ah 2 =
1
π a 2h 2
12
or y = 0.869a W
PROBLEM 5.122
Locate the centroid of the volume generated by revolving the portion of
the sine curve shown about the x axis.
SOLUTION
y = 0W
z = 0W
Choose as the element of volume a disk of radius r and thickness dx.
Then
First, note that symmetry implies
dV = πr 2dx, xEL = x
πx
Now
r = b sin
so that
dV = πb 2 sin 2
Then
2a
πx
2a
dx
2a
V = ∫a π b 2 sin 2
πx
2a
dx
2a
 x sin π2ax 
= πb  −

2 πa 
 2
a
2
( ) ( a2 )
= π b 2  22a −

1
= π ab 2
2
and
2a


2πx


∫ x ELdV = ∫a x  π b sin 2a dx 
2
Use integration by parts with
u = x
du = dx
dV = sin 2
V =
x
−
2
πx
2a
sin π x
2π
a
a
PROBLEM 5.122 CONTINUED
Then
2a
 

sin πax  
x sin πax  
2a  x
π
=
−
−
−
x
dV
b
x
dx







∫ EL
∫a  2
2π
2π


 
   2
a
a


 

a

2
2a

π x  
a2

 2a 
 a   1
= π b 2   2a   − a    −  x 2 +
cos
 
a a 
2π 2
 2   4
   2 

 3  1
1
a2
a 2  
2
2

= π b 2  a 2  −  ( 2a ) +
−
+
a
(
)

2π 2 4
2π 2  
 2   4
1 
3
= π a 2b 2  − 2 
4
π 

Now
= 0.64868π a 2b 2
1

xV = ∫ x EL dV : x  π ab 2  = 0.64868π a 2b 2
2

or x = 1.297a W
PROBLEM 5.123
Locate the centroid of the volume generated by revolving the portion of
the sine curve shown about the y axis. (Hint: Use a thin cylindrical shell
of radius r and thickness dr as the element of volume.)
SOLUTION
x = 0W
First note that symmetry implies
z = 0W
Choose as the element of volume a cylindrical shell of radius r and
thickness dr.
1
dV = ( 2πr )( y )( dr ) , yEL = y
Then
2
Now
so that
Then
y = b sin
πr
2a
dV = 2πbr sin
2a
V = ∫a 2πbr sin
πr
2a
πr
2a
dr
dr
Use integration by parts with
u =r
dv = sin
du = dr
Then
v=−
2a
π
πr
2a
cos
dr
πr
2a
2a
  2a
πr  
πr  
2a  2a
V = 2πb ( r )  −
cos   − ∫a  cos  dr 
 π
 π
2a  a
2a  


2a
 2a
 4a 2
πr  

( 2a ) ( −1)  +  2 sin  
= 2πb −
2a a 
π
 π

2
2
 4a
4a 
− 2 
V = 2πb 
π 
 π
1

= 8a 2b 1 − 

π
= 5.4535 a 2b
Also
πr  
2a 
πr

∫ y ELdV = ∫a  12 b sin 2a   2πbr sin 2a dr 
2a
= πb 2 ∫a r sin 2
πr
dr
2a
PROBLEM 5.123 CONTINUED
Use integration by parts with
Then
u =r
dv = sin 2
du = dr
v=
r
−
2
πr
dr
2a
sin πar
2π
a
2a
 

sin πar  
r sin πar  
2a  r
∫ y EL dV = π b  ( r )  2 − 2π   − ∫a  2 − 2π  dr 
  
a
a
  a

 

2
2a

a2
π r  

 2a 
 a   r 2
cos
= π b 2   ( 2a )   − ( a )    −  +
 
a a 
2π 2
 2 
 2   4
 

3
 ( 2a ) 2
( a )2 + a 2  
a2

= π b2  a 2 − 
+
−

4
2π 2
2π 2  
 4
 2

1 
3
= π a 2b 2  − 2 
4 π 
= 2.0379a 2b 2
Now
(
)
yV = ∫ y EL dV : y 5.4535a 2b = 2.0379a 2b 2
or y = 0.374b W
PROBLEM 5.124
Show that for a regular pyramid of height h and n sides ( n = 3, 4, … ) the
centroid of the volume of the pyramid is located at a distance h / 4 above
the base.
SOLUTION
Choose as the element of a horizontal slice of thickness dy. For any
number N of sides, the area of the base of the pyramid is given by
Abase = kb 2
where k = k ( N ) ; see note below. Using similar triangles, have
or
Then
and
s
h−y
=
b
h
b
s = (h − y)
h
b2
2
dV = Aslicedy = ks 2dy = k 2 ( h − y ) dy
h
h
V = ∫0 k
1 2
kb h
3
=
Also
h
b2
b2  1
2
3
h
y
dy
k
−
=
− (h − y) 
(
)
2
2 
h
h  3
0
yEL = y
2

b2 h
h  b
2
so then ∫ y EL dV = ∫0 y k 2 ( h − y ) dy  = k 2 ∫0 h 2 y − 2hy 2 + y 3 dy
h
 h

(
)
h
=k
Now
1 2 2
b2  1 2 2 2 3 1 4 
h y − hy + y  =
kb h
2 
3
4 0 12
h 2
1 2 2
1

yV = ∫ y EL dV : y  kb 2h  =
kb h
3
 12
or y =
Note
b 
1
Abase = N  × b × 2 π 
tan N
2

=
N
b2
4 tan Nπ
= k ( N ) b2
1
h Q.E.D. W
4
PROBLEM 5.125
Determine by direct integration the location of the centroid of one-half of a
thin, uniform hemispherical shell of radius R.
SOLUTION
First note that symmetry implies
x = 0W
The element of area dA of the shell shown is obtained by cutting the shell
with two planes parallel to the xy plane. Now
dA = (π r )( Rdθ )
yEL = −
2r
π
r = R sin θ
dA = π R 2 sin θ dθ
where
so that
yEL = −
2R
π
sin θ
π
π
A = ∫02 π R 2 sin θ dθ = π R 2 [ − cosθ ]02
Then
= π R2
and
∫ y ELdA = ∫02  − π sin θ  (π R sin θ dθ )
π 


2R

2
π
3 θ
sin 2θ  2
= −2 R  −
4  0
2
=−
Now
π
2
R3
(
)
yA = ∫ y EL dA: y π R 2 = −
π
2
R3
1
or y = − R W
2
Symmetry implies
z = y∴ z =−
1
RW
2
PROBLEM 5.126
The sides and the base of a punch bowl are of uniform thickness t. If
t << R and R = 350 mm, determine the location of the center of gravity
of (a) the bowl, (b) the punch.
SOLUTION
(a) Bowl
First note that symmetry implies
x = 0W
z = 0W
for the coordinate axes shown below. Now assume that the bowl may be
treated as a shell; the center of gravity of the bowl will coincide with the
centroid of the shell. For the walls of the bowl, an element of area is
obtained by rotating the arc ds about the y axis. Then
dAwall = ( 2π R sin θ )( Rdθ )
and ( yEL ) wall = − R cos θ
Then
and
π /2
π /2
Awall = ∫π /6 2π R 2 sin θ dθ = 2π R 2 [ − cosθ ]π /6
= π 3R 2
ywall Awall = ∫ ( yEL )wall dA
π /2
(
= ∫π /6 ( − R cosθ ) 2π R 2 sin θ dθ
= π R3 cos 2 θ 
)
π /2
π /6
3
= − π R3
4
By observation
Now
Abase =
π
4
R2,
ybase = −
3
R
2
y ΣA = ΣyA
or
π
3
π  3 


y  π 3R 2 + R 2  = − π R3 + R 2  −
R
4 
4
4  2 

or
y = −0.48763R
R = 350 mm
∴ y = −170.7 mm W
(b) Punch
First note that symmetry implies
x = 0W
z = 0W
and that because the punch is homogeneous, its center of gravity will
coincide with the centroid of the corresponding volume. Choose as the
element of volume a disk of radius x and thickness dy. Then
dV = π x 2dy, yEL = y
PROBLEM 5.126 CONTINUED
Now
Then
x2 + y 2 = R2
0
π
− 3/2 R
V =∫
(
(
)
dV = π R 2 − y 2 dy
so that
)
0
1 

R 2 − y 2 dy = π  R 2 y − y 3 
3 − 3/2 R

3
 
 1
  3
3
3
2
= −π  R  −
−
R −
R  = π 3R3
  2  3  2   8


and
(
0
)
0
1 2 2 1 4
2
2
∫ yEL dV = ∫− 3/2 R ( y ) π R − y dy  = π  2 R y − 4 y 

 − 3/2 R
2
4

1 2
3 
1
3  
15

R −
R − −
R  = − π R4
= −π




2  2 
4 2  
64


Now
or
15
3

yV = ∫ yEL dV : y  π 3 R3  = − π R 4
8
64


y =−
5
8 3
R
R = 350 mm
∴ y = −126.3 mm W
PROBLEM 5.127
After grading a lot, a builder places four stakes to designate the corners of
the slab for a house. To provide a firm, level base for the slab, the builder
places a minimum of 60 mm of gravel beneath the slab. Determine the
volume of gravel needed and the x coordinate of the centroid of the
volume of the gravel. (Hint: The bottom of the gravel is an oblique plane,
which can be represented by the equation y = a + bx + cz.)
SOLUTION
First, determine the constants a, b, and c.
At
x = 0, z = 0 : y = −60 mm
∴ −60 mm = a; a = −60 mm
x = 7200 mm, z = 0 : y = −100 mm
At
∴ − 100 mm = −60 + b ( 7200)
1
180
x = 0 , z = 12 000 mm: y = −120 mm
b=−
At
∴ − 120 mm = −60 mm + c (12000 )
c=−
1
200
It follows that y = −60 −
1
1
x−
z where all dimensions
180
120
are in mm.
Choose as the element of volume a filament of base dx × dz and height
| y |. Then
dV = y dxdz, xEL = x
dV = −60 −
1
1
x−
z dxdz
180
200
12000 7200 
 60 +
0
1
1 
x+
z  dxdz
180
200 
or
Then
V = ∫0
∫

7200
12000 
= ∫0
1 2
1

60 x + 360 x + 200 xz 

0

2
7200 )
(
( 7200 ) z dz
+
( )( 7200 ) +
12000
 60
0
=∫

dz
360
200

12000
36 2 

z
= 576000 z +
2  0

= 9.504 × 109 mm3 = 9.50 m3
V = 9.50 m3 W
PROBLEM 5.127 CONTINUED
and
12000 7200 
x  60 +
0
∫ x ELdV = ∫0
∫

12000  60
= ∫0
12000
= ∫0
2

1
1 
x+
z  dxdz
180
200 
7200
x2 +
1 3
1 2 
x +
x z
540
400
0
dz
( 2246.4 × 10 + 129 600z ) dz
6
12000
129 600 2 

z 
=  2246.4 × 106 z −
2

0
= 2.695 × 1013 + 0.933 × 1013
= 3.63 × 1013 mm 4 = 36.3 m 4
Now
(
)
xV = ∫ x EL dV : x 9.50 m3 = 36.3 m 4
or x = 3.82 m W
PROBLEM 5.128
Determine by direct integration the location of the centroid of the volume
between the xz plane and the portion shown of the surface
16h ax − x 2 bz − z 2
y =
.
a 2b 2
(
)(
)
SOLUTION
a
2
b
z =
2
x =
First note that symmetry implies
Choose as the element of volume a filament of base dx × dz and height y. Then
dV = ydxdz, yEL =
(
dV =
Then
V = ∫0 ∫0
)
( ax − x )(bz − z ) dxdz
ab
b a 16h
2 2
=
and
)(
16h
ax − x 2 bz − z 2 dxdz
a 2b 2
or
V =
1
y
2
(
2
2
a
)
16h b
1 
a
bz − z 2  x 2 − x3  dz
2 2 ∫0
3 0
ab
z
b
( )
16h  a 2
1 3 b
1 
8ah  b 2 1 3  4
a − ( a )   z 2 − z 3  = 2  ( b ) − ( b )  = abh
2 2 
3
3 0 3b  2
3
a b 2
 2
 9
∫ yELdV = ∫0 ∫0 2  a 2b2 ( ax − x )( bz − z )  a 2b2 ( ax − x )( bz − z ) dxdz 
b a 1  16h
2
2

2
(
  16h

)(
2


2
)
=
128h b a 2 2
a x − 2ax3 + x 4 b 2 z 2 − 2bz 3 + z 4 dxdz
4 4 ∫0 ∫0
ab
=
2
128h 2 b 2 2
a 4 1 5
3
4 a
3
−
+
−
b
z
2
bz
z
x
x + x  dz

∫
2
5 0
a 2b 4 0
3
=
128h 2  a 2
a
1 5   b2 3 b 4 1 5 
3
4
−
+
a
a
(
)
(
)
(a)   Z − Z + Z 

5 0
Z
2
5
a 4b 4  3
 3
=

64ah 2  b3
1
32
b
abh 2
( b )3 − ( b )4 + ( b )5  =
4 
2
5
225
15b  3

(
)
a
b
PROBLEM 5.128 CONTINUED
Now
32
4

yV = ∫ y EL dV : y  abh  =
abh 2
9
225


or y =
8
h
25
PROBLEM 5.129
Locate the centroid of the section shown, which was cut from a circular
cylinder by an inclined plane.
SOLUTION
x =0
First note that symmetry implies
Choose as the element of volume a vertical slice of width 2x, thickness dz, and height y. Then
1
dV = 2 xy dz, yEL = y, z EL = z
2
h
h
h
z
Now
y = −
z = 1 − 
x = a2 − z 2
and
a
2 2a
2
z

dV = h a 2 − z 2 1 −  dz

a
So
Then
a
h
0
V =∫
=
=
3/2 

z
 1 
 z 1 2
a − z 1 −  dz = h   z a 2 − z 2 + a 2 sin −1    +
a − z2 

 a   3a
a

 2 
2
2
1 2  −1
a h sin (1) − sin −1 ( −1) 
2
π
2
a 2h
(
)
a
−a
PROBLEM 5.129 CONTINUED
a 1
h
z  


 
2
2
z




∫ yEL dV = ∫−a  2 × 2 1 − a    h a − z 1 − a  dz 
Then
=

h2 a
z
z2 
a 2 − z 2 1 − 2 + 2  dz
∫
−a
a a 
4

=
3
2 2

h 2  1 
−1  z  
2
2
2
a − z2 2 
  z a − z + a sin    + 
4  2 
 a    3a

(
)
3
1  z
a2z 2
a4
 z   
+ 2 − a2 − z 2 2 +
a − z2 +
sin −1    
8
8
a  4
 a   
(
=
)
a
−a
5h 2a 2  −1
sin (1) − sin −1 ( −1) 
32 
 π a 2  5h 2a 2
yV = ∫ yEL dV : y 
h  =
(π )
32
 2 
Then
or y =
and
a

2
2
z




5
h
16
∫ zELdV = ∫−a z  h a − z 1 − a  dz 

a
3
3
1 z
a2 z 2
a4
 1
 z   
= h − a 2 − z 2 2 −  − a 2 − z 2 2 +
a − z2 +
sin −1    
a 4
8
8
 a   
 3
−a
(
=−
)
(
)
a3h  −1
sin (1) − sin −1 ( −1) 

8
 π a 2h 
π a3h
zV = ∫ z EL dV : z 
 = −
8
 2 
or z = −
a
4
PROBLEM 5.130
Locate the centroid of the plane area shown.
SOLUTION
A, in 2
x , in.
y , in.
xA, in 3
yA, in 3
1
14 × 20 = 280
7
10
1960
2800
2
−π ( 4 ) = −16π
6
12
−301.59
−603.19
Σ
229.73
1658.41
2196.8
2
X Σ A = ΣxA
Then
(
)
X 229.73 in 2 = 1658.41 in 3
or X = 7.22 in.
Y Σ A = ΣyA
and
(
)
Y 229.73 in 2 = 2196.8 in 3
or Y = 9.56 in.
PROBLEM 5.131
For the area shown, determine the ratio a/b for which x = y.
SOLUTION
A
1
2
Σ
Then
2
ab
3
1
− ab
2
1
ab
6
x
3
a
8
1
a
3
y
xA
yA
3
b
5
2
b
3
2
ab
4
a 2b
−
6
2ab 2
5
ab 2
−
3
a 2b
12
ab 2
15
X Σ A = ΣxA
 1  a 2b
X  ab =
6 
12
or
or
Now
1
a
2
Y Σ A = ΣyA
X =
 1  ab 2
Y  ab =
6 
15
2
Y = b
5
1
2
X =Y ⇒
a= b
2
5
or
4
a
=
b
5
PROBLEM 5.132
Locate the centroid of the plane area shown.
SOLUTION
Dimensions in mm
1
2
3
Σ
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
126 × 54 = 6804
1
× 126 × 30 = 1890
2
1
× 72 × = 1728
2
10 422
9
27
61 236
183 708
30
64
56 700
120 960
48
−16
82 944
−27 648
200 880
277 020
X ΣA = Σ xA
Then
(
)
X 10 422 m 2 = 200 880 mm 2
or X = 19.27 mm W
Y Σ A = Σ yA
and
(
)
Y 10 422 m 2 = 270 020 mm3
or Y = 26.6 mm W
PROBLEM 5.133
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and h.
SOLUTION
By observation
h
x+h
a
x

= h 1 − 

a
y1 = −
For y2:
At
At
x = 0, y = h: h = k (1 − 0)
(
x = a, y = 0: 0 = h 1 − ca 2
xEL = x
 x x2 
= h  − 2  dx
a a 
1
h 
x 
x2  
yEL = ( y1 − y2 ) = 1 −  + 1 − 2  
2
2 
a 
a  
h
x x2 
−
−
2
a a 2 
2 
a
 x2
x2 
a  x
h
−
dx
=
h



2
0 
A = ∫ dA = ∫
=
and
1
a2

x2  
x 
dA = ( y2 − y1 ) dx = h 1 − 2  − 1 −   dx
a  
a  

=
Then
C =
or

x2 
y2 = h 1 − 2 
a 

Then
Now
)
k =h
or
a
a 
x3 
− 2
 2a 3a  0
1
ah
6
 x 3
x2  
a   x
−
x h  − 2  dx  = h 
0  
∫ x ELdA = ∫
  a
1 2
=
a h
12
a 

 3a
x4 

4a 2  
a
0
PROBLEM 5.133 CONTINUED
x x2    x x2  
−
−
−
dx 
2
h
a a 2    a a 2  
2 
ah
∫ y EL dA = ∫0
=
h2 a  x
x2
x4 
2 − 3 2 + 4  dx
∫

0
2  a
a
a 
a
h2  x2
x3
x5 
1 2
=
−
+
=
ah

2
4
2 a
a
5a 0 10
1 2
1 
xA = ∫ x EL dA: x  ah =
a h
 6  12
x =
1
aW
2
1 2
1 
yA = ∫ y EL dA: y  ah =
a h
 6  10
y =
3
hW
5
PROBLEM 5.134
Member ABCDE is a component of a mobile and is formed from a single
piece of aluminum tubing. Knowing that the member is supported at C
and that l = 2 m, determine the distance d so that portion BCD of the
member is horizontal.
SOLUTION
First note that for equilibrium, the center of gravity of the component
must lie on a vertical line through C. Further, because the tubing is
uniform, the center of gravity of the component will coincide with the
centroid of the corresponding line. Thus, X = 0
So that
Then
Σ xL = 0
0.75


cos 55°  m × ( 0.75 m )
− d −
2


+ ( 0.75 − d ) m × (1.5 m )

1

+ (1.5 − d ) m −  × 2 m × cos 55°   × ( 2 m ) = 0
2


or
( 0.75 + 1.5 + 2 ) d =  ( 0.75)2 − 2 cos 55° + ( 0.75)(1.5) + 3
1
2

or d = 0.739 m W
PROBLEM 5.135
A cylindrical hole is drilled through the center of a steel ball bearing
shown here in cross section. Denoting the length of the hole by L, show
that the volume of the steel remaining is equal to the volume of a sphere
of diameter L.
SOLUTION
Calculate volumes by rotating cross sections about a line and using Theorem II of Pappus-Guldinus
y AA =
For the sector:
2R sin α
3α
A = α R2
2
For the triangle:
h=
1
L
R2 −   =
4R 2 − L2
2
2
A=
1
( L )( h )
2
=
y AA =
2
1
h=
4R 2 − L2 ,
3
3
1
L 4R 2 − L2
4
Using Theorem II of Pappus-Guldinus
Vball = 2π ( y AA )1 A1 − 2π ( y AA )2 A2
 2 R sin α
1
 1

α R 2 −  4R 2 − L2  L 4R 2 − L2  
= 2π 
3
α
3
4




(
)
(
)
L
2

4 R 2 − L2 
= 2π  R3 sin α −
12
3

Now
Then
R sin α =
L
2
2  L 
1
1 3
V = 2π    R 2 − LR 2 +
L
3
12 
3  2 
=
Note Vsphere =
L
If r = , then
2
Therefore,
π
6
L3
4
πr where r is the radius
3
3
4 L
π
Vsphere = π   = L3
3 2
6
Vball = Vsphere =
π
6
L3 Q.E.D.
PROBLEM 5.136
For the beam and loading shown, determine (a) the magnitude and
location of the resultant of the distributed load, (b) the reactions at the
beam supports.
SOLUTION
(a) Have
RI = ( 4 m )( 200 N/m ) = 800 N
Then
2
( 4 m )( 600 N/m ) = 1600 N
3
ΣFy : −R = − RI − RII
or
R = 800 + 1600 = 2400 N
RII =
and
ΣM A: − X ( 2400 ) = −2 ( 800 ) − 2.5 (1600 )
or
X =
7
m
3
∴ R = 2400 N , X = 2.33 m
(b) Reactions
ΣFx = 0: Ax = 0
7 
ΣM A = 0: ( 4 m ) By −  m ( 2400 N ) = 0
3 
or
By = 1400 N
ΣFy = 0: Ay + 1400 N − 2400 N = 0
or
Ay = 1000 N
∴ A = 1000 N , B = 1400 N
PROBLEM 5.137
Determine the reactions at the beam supports for the given loading.
SOLUTION
Have
RI =
1
( 3 ft )( 480 lb/ft ) = 720 lb
2
RII =
1
( 6 ft )( 600 lb/ft ) = 1800 lb
2
RIII = ( 2 ft )( 600 lb/ft ) = 1200 lb
ΣFx = 0: Bx = 0
Then
ΣM B = 0:
( 2 ft )( 720 lb ) − ( 4 ft )(1800 lb )
+ ( 6 ft ) Cy − ( 7 ft )(1200 lb ) = 0
or
C y = 2360 1b
C = 2360 lb
ΣFy = 0: − 720 lb + By − 1800 lb + 2360 lb − 1200 lb = 0
or
By = 1360 lb
B = 1360 lb
PROBLEM 5.138
Locate the center of gravity of the sheet-metal form shown.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with
the centroid of the corresponding area.
yI = −
1
(1.2) = −0.4 m
3
1
(3.6) = 1.2 m
3
4 (1.8)
2.4
m
xIII = −
=−
3π
π
zI =
A, m 2
I
II
III
Σ
Have
1
( 3.6) (1.2) = 2.16
2
(3.6)(1.7) = 6.12
π
2
(1.8) 2 = 5.0894
x, m
y, m
z, m
xA, m3
yA, m3
zA, m3
1.5
−0.4
1.2
3.24
−0.864
2.592
0.75
0.4
1.8
4.59
2.448
11.016
0.8
1.8
−3.888
4.0715
9.1609
3.942
5.6555
22.769
−
2.4
π
13.3694
(
)
X ΣV = ΣxV : X 13.3694 m 2 = 3.942 m3
(
)
or X = 0.295 m
Y ΣV = ΣyV : Y 13.3694 m 2 = 5.6555 m3
or Y = 0.423 m
(
)
Z ΣV = Σ zV : Z 13.3694 m 2 = 22.769 m3
or Z = 1.703 m
PROBLEM 5.139
The composite body shown is formed by removing a semiellipsoid of
revolution of semimajor axis h and semiminor axis a2 from a hemisphere of
radius a. Determine (a) the y coordinate of the centroid when h = a/2, (b)
the ratio h/a for which y = −0.4a.
SOLUTION
V
y
yV
Hemisphere
2 3
πa
3
3
− a
8
1
− πa 4
4
2
Semiellipsoid
2 a
1
− π   h = − πa 2 h
3 2
6
3
− h
8
ΣV =
Then
π
6
a 2 ( 4a − h )
ΣyV = −
π
16
(
Y ΣV = Σ yV
so that
π
π

Y  a 2 ( 4a − h )  = − a 2 4a 2 − h 2
16
6

or
2
h
3 

 h 
Y  4 −  = − a 4 −   

 a 
a
8 

Substituting
1
πa 2 h 2
16
a 2 4a 2 − h 2
Now
(
(a) Y = ? when h =
+
)
)
(1)
a
2
h 1
= into Eq. (1)
a
2
2
1
3 

1 
Y  4 −  = − a 4 −   
2
8 

 2  
or
Y =−
45
a
112
Y = −0.402a
PROBLEM 5.139 CONTINUED
(b)
h
= ? when Y = −0.4a
a
Substituting into Eq. (1)
( −0.4a )  4 −

2
h
3 
h 
a
4
=
−
−


  
a
8 
 a  
2
or
Then
 h
 h
3   − 3.2   + 0.8 = 0
 a
 a
2
h 3.2 ± ( −3.2 ) − 4 ( 3)( 0.8 )
=
a
2 ( 3)
=
3.2 ± 0.8
6
or
h
2
h
2
=
and
=
a
5
a
3
PROBLEM 5.140
A thin steel wire of uniform cross section is bent into the shape shown.
Locate its center of gravity.
SOLUTION
First assume that the wire is homogeneous so that its center of gravity will coincide with the centroid of the
corresponding line.
x1 = 0.3sin 60° = 0.15 3 m
z1 = 0.3cos 60° = 0.15 m
 0.6sin 30° 
x2 = 
 sin 30°
π

6


0.9
=
m
π
 0.6sin 30° 
z2 = 
 cos 30°
π

6


0.9
=
3m
π
π 
L2 =   ( 0.6 ) = ( 0.2π ) m
3
Have
L, m
x, m
y, m
z, m
xL, m 2
yL, m 2
zL, m 2
1
1.0
0.15 3
0.4
0.15
0.25981
0.4
0.15
2
0.2π
0.18
0
0.31177
3
4
∑
0.8
0.6
3.0283
0
0
0.43981
0.32
0.48
1.20
0.48
0.18
1.12177
0.9
π
0
0
0
0.4
0.8
0.9 3
π
0.6
0.3
X Σ L = Σ x L: X ( 3.0283 m ) = 0.43981 m 2
or X = 0.1452 m
Y Σ L = Σ yL: Y ( 3.0283 m ) = 1.20 m 2
or Y = 0.396 m
Z Σ L = Σ z L: Z ( 3.0283 m ) = 1.12177 m 2
or Z = 0.370 m
PROBLEM 5.141
Locate the centroid of the volume obtained by rotating the shaded area
about the x axis.
SOLUTION
y =0
First note that symmetry implies
and z = 0
y = k ( X − h)
Have
x = 0, y = a: a = k ( −h )
At
k =
or
2
2
a
h2
Choose as the element of volume a disk of radius r and thickness dx.
Then
dV = π r 2dx, X EL = x
a
2
r = 2 ( x − h)
h
Now
dV = π
so that
Then
h
V = ∫0 π
=
and
a2
( x − h)4 dx
h4
h
a2
π a2 
4
5
−
=
x
h
dx
x − h ) 
(
)
(
4
4 
0
5h 
h
1 2
πa h
5
 a2

∫ x EL dV = ∫0 x π h 4 ( x − h ) dx 
h

2
4

(
)
=π
a h 5
x − 4hx 4 + 6h 2 x3 − 4h3 x 2 + h 4 x dx
4 ∫0
h
=π
a2  1 6 4 5 3 2 4 4 3 3 1 4 2 
x − hx + h x − h x + h x 
5
2
3
2
h 4  6
0
h
=
Now
1
π a 2h 2
30
π
 π 2 2
xV = ∫ x EL dV : x  a 2h  =
a h
5
 30
or x =
1
h
6