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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–1.
Locate the center of mass of the homogeneous rod bent
into the shape of a circular arc.
y
30
300 mm
x
SOLUTION
dL = 300 d u
30
'
x = 300 cos u
'
y = 300 sin u
x =
L
2p
3
'
x dL
=
L
L-2p3
300 cos u (300du)
2p
3
dL
L-2p3
300d u
(300)2 C sin u D -3 2p3
2p
=
4
300 a p b
3
Ans.
= 124 mm
y = 0
Ans.
(By symmetry)
Ans:
x = 124 mm
y = 0
879
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–2.
y
Determine the location (x, y) of the centroid of the wire.
2 ft
4 ft
SOLUTION
Length and Moment Arm: The length of the differential element is dL =
2dx2 + dy2 = ¢
B
1 + a
y = x2
2
dy
dy
b ≤ dx and its centroid is ∼
= 2x.
y = y = x2. Here,
dx
dx
x
Centroid: Due to symmetry
∼
Ans.
x = 0
Applying Eq. 9–7 and performing the integration, we have
2 ft
∼
∼
y =
LL
ydL
=
dL
LL
L-2 ft
x 21 + 4x2 dx
2 ft
L-2 ft
=
2
21 + 4x2 dx
16.9423
= 1.82 ft
9.2936
Ans.
Ans:
x = 0
y = 1.82 ft
880
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–3.
y
Locate the center of gravity x of the homogeneous rod. If
the rod has a weight per unit length of 100 N>m, determine
the vertical reaction at A and the x and y components of
reaction at the pin B.
1m
B
y x2
A
1m
x
Solution
Length
And
Arm. The length of the differential element is
dy
dy 2
dL = 2dx2 + dy2 = c 1 + a b d dx and its centroid is ~
x = x. Here
= 2x.
A
dx
dx
Perform the integration
L =
Moment
LL
dL =
L0
= 2
1m
L0
= cx
21 + 4x2 dx
1m
A
A
x2 +
x2 +
= 1.4789 m
LL
x~ dL =
L0
= 2
1
dx
4
1
1
1 1m
+ ln ax +
x2 + b d
4
4
A
4 0
1m
x21 + 4x2 dx
L0
1m
x
A
x2 +
1
dx
4
2
1 3>2 1 m
= c ax2 + b d
3
4
0
= 0.8484 m2
Centroid.
x =
~
0.8484 m2
1L x dL
= 0.5736 m = 0.574 m
=
1.4789 m
1L dL
Ans.
Equations of Equilibrium. Refering to the FBD of the rod shown in Fig. a
+ ΣFx = 0;
S
a+ΣMB = 0;
Ans.
Bx = 0
100(1.4789) (0.4264) - Ay(1) = 0
Ans.
Ay = 63.06 N = 63.1 N
a+ΣMA = 0;
By(1) - 100(1.4789) (0.5736) = 0
Ans.
By = 84.84 N = 84.8 N
Ans:
x = 0.574 m
Bx = 0
Ay = 63.1 N
By = 84.8 N
881
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–4.
y
Locate the center of gravity y of the homogeneous rod.
1m
B
y x2
A
1m
x
Solution
Length
And
Arm. The length of the differential element is
dy
dy 2
dL = 2dx2 + dy2 = c 1 + a b d dx and its centroid is ~
y = y. Here
= 2x.
A
dx
dx
Perform the integration,
L =
Moment
LL
L0
dL =
= 2
1m
21 + 4x2 dx
L0
= cx
1m
A
A
x2 +
= 1.4789 m
LL
y~ dL =
L0
= 2
1
dx
4
1
1
1 1m
+ ln ax +
x2 + b d
4
4
A
4 0
1m
x2 21 + 4x2 dx
L0
= 2c
x2 +
1m
x
2
A
x2 +
1
dx
4
x
1
1
1
1 1m
1 3
2
x x2 + ln ax + x2 + b d
ax + b 4A
32 A
4 128
A
4 0
4
= 0.6063 m2
Centroid.
0.6063 m2
1L y dL
y =
= 0.40998 m = 0.410 m
=
1.4789 m
1L dL
~
Ans.
Ans:
y = 0.410 m
882
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–5.
y
Determine the distance y to the center of gravity of the
homogeneous rod.
1m
2m
Solution
Length
And
y 2x3
x
Moment
Arm. The length of the differential element is
dy 2
y = y. Here
dL = 2dx2 + dy2 = a 1 + a b b dx and its centroid is at ~
A
dx
dy
= 6x2. Evaluate the integral numerically,
dx
L =
LL
~
LL
dL =
y dL =
L0
L0
1m
1m
21 + 36x4 dx = 2.4214 m
3
2x 21 + 36x4 dx = 2.0747 m2
Centroid. Applying Eq. 9–7,
y =
~
2.0747 m2
1L y dL
= 0.8568 = 0.857 m
=
2.4214 m
1L dL
Ans.
Ans:
y = 0.857 m
883
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–6.
Locate the centroid y of the area.
y
y
1
1 – – x2
4
1m
x
2m
SOLUTION
Area and Moment Arm: The area of the differential element is
y
1
1
1
'
= a1 - x 2 b .
dA = ydx = a1 - x2 b dx and its centroid is y =
4
2
2
4
Centroid: Due to symmetry
Ans.
x = 0
Applying Eq. 9–4 and performing the integration, we have
2m
'
ydA
y =
LA
=
1 2
1
1
¢ 1 - x ≤ ¢ 1 - x2 ≤ dx
2
4
4
L- 2m
2m
dA
LA
L- 2m
=
¢
¢1 -
1 2
x ≤ dx
4
x
x3
x 5 2m
+
≤`
2
12
160 - 2m
x
¢x ≤`
12 - 2m
3
2m
=
2
m
5
Ans.
Ans:
y =
884
2
m
5
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–7.
Determine the area and the centroid x of the parabolic area.
y
h
SOLUTION
y
Differential Element:The area element parallel to the x axis shown shaded in Fig. a
will be considered. The area of the element is
dA = x dy =
a
h1>2
h x2
––
a2
x
a
y 1>2 dy
x
a
'
'
Centroid: The centroid of the element is located at x =
=
y 1>2 and y = y.
2
2h1>2
Area: Integrating,
h
A =
LA
LA
=
dA
LA
h
'
xdA
x =
dA =
L0
¢
a
1>2
L0 h
y 1>2 dy =
2a
3h
a
y1>2 ≤ ¢ 1>2 y 1>2 dy ≤
1>2
2h
h
a
2
ah
3
1>2
A y3>2 B 2 =
h
0
2
ah
3
Ans.
a2 y2 h
a
¢ ≤`
y dy
2h 2 0
3
L0 2h
=
=
= a Ans.
2
2
8
ah
ah
3
3
h 2
Ans:
x =
885
3
a
8
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–8.
Locate the centroid of the shaded area.
y
y a cospx
L
a
L
2
x
L
2
Solution
Area And Moment Arm. The area of the differential element shown shaded in
y
p
a
p
Fig. a is dA = ydx = a cos x dx and its centroid is at y~ = = cos x.
2
2
2
L
Centroid. Perform the integration
y =
~
1A y dA
1A dA
L>2
=
p
p
a
a cos xbaa cos x dxb
L
L
L-L>2 2
L>2
L-L>2
L>2
=
=
L-L>2
p
x dx
L
a2
2p
x + 1b dx
acos
4
L
L>2
L-L>2
a cos
p
x dx
L
L>2
a2 L
2p
a
sin
x + xb `
4 2p
L
-L>2
a
=
a cos
L>2
aL
p
sin xb `
p
L
-L>2
a2 L>4
2aL>p
=
p
a
8
Ans.
x = 0
Ans.
Due to Symmetry,
Ans:
p
a
8
x = 0
y =
886
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–9.
y
Locate the centroid x of the shaded area.
4m
y
1 2
x
4
x
4m
Solution
Area And Moment Arm. The area of the differential element shown shaded in Fig. a
1
is dA = x dy and its centroid is at ~
x = x. Here, x = 2y1>2
2
Centroid. Perform the integration
x =
~
1A x dA
1A dA
=
=
L0
4m
1
a2y1>2 ba2y1>2 dyb
2
L0
4m
2y1>2 dy
3
m
2
Ans.
Ans:
x =
887
3
m
2
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–10.
y
Locate the centroid y of the shaded area.
4m
y
1 2
x
4
x
4m
Solution
Area And Moment Arm. The area of the differential element shown shaded in Fig. a
y = y. Here, x = 2y1>2.
is dA = x dy and its centroid is at ∼
Centroid. Perform the integration
y =
~
1A y dA
1A dA
=
=
=
L0
4m
L0
y a2y1>2 dyb
4m
2y1>2 dy
4m
4
a y 5>2 b `
5
0
4m
4
a y3>2 b `
3
0
12
m
5
Ans.
Ans:
y =
888
12
m
5
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–11.
y
Locate the centroid x of the area.
h
y —2 x2
b
SOLUTION
h
dA = y dx
'
x = x
x =
LA
b
'
x dA
LA
=
dA
h 3
x dx
2
L0 b
b
h 2
x dx
2
L0 b
=
B
h 4
x R
4b2
0
h
B 2 x3 R
3b
0
b
x
b
b
=
3
b
4
Ans.
Ans:
x =
889
3
b
4
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–12.
y
Locate the centroid y of the shaded area.
h
y —2 x2
b
SOLUTION
h
dA = y dx
y
'
y =
2
y =
LA
b
'
y dA
LA
=
dA
2
h 4
x dx
4
L0 2b
b
h 2
x dx
2
b
L0
=
h2 5 b
x R
B
10b4
0
B
h 3
x R
3b2
0
b
x
b
=
3
h
10
Ans.
Ans:
y =
890
3
h
10
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–13.
y
Locate the centroid x of the shaded area.
SOLUTION
4m
1
dA = 14 - y2dx = a x 2 b dx
16
'
x = x
x =
LA
8
'
xdA
LA
x = 6m
=
dA
L0
L0
8
xa
a
y
4
1 2
––
x
16
8m
x2
b dx
16
1 2
x b dx
16
Ans.
Ans:
x = 6m
891
x
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–14.
y
Locate the centroid y of the shaded area.
SOLUTION
4m
dA = 14 - y2dx = a
y =
y =
1 2
x b dx
16
y
4
1 2
––
x
16
8m
4 + y
2
LA
8
'
ydA
LA
=
dA
x2
x2
1
b a b dx
¢8 2 L0
16
16
8
L0
a
1 2
x b dx
16
Ans.
y = 2.8 m
Ans:
y = 2.8 m
892
x
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–15.
Locate the centroid x of the shaded area. Solve the problem by
evaluating the integrals using Simpson’s rule.
y
y = 0.5ex2
SOLUTION
At x = 1 m
2
y = 0.5e1 = 1.359 m
1
LA
dA =
L0
1
(1.359 - y) dx =
L0
x
2
a 1.359 = 0.5 ex b dx = 0.6278 m2
1m
x = x
1
x dA =
LA
L0
2
x a 1.359 - 0.5 ex b dx
= 0.25 m3
x =
x dA
LA
LA
dA
=
0.25
= 0.398 m
0.6278
Ans.
Ans:
x = 0.398 m
893
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–16.
Locate the centroid y of the shaded area. Solve the problem by
evaluating the integrals using Simpson’s rule.
y
y = 0.5ex2
SOLUTION
1
dA =
L0
LA
y =
1
(1.359 - y) dx =
L0
a 1.359 - 0.5ex b dx = 0.6278 m
2
2
1.359 + y
2
1
y dA =
LA
L0
x
1m
a
x2
1.359 + 0.5 e
2
b A 1.359 - 0.5 ex B dx
2
1
=
y =
1
2
a 1.847 - 0.25 e2x b dx = 0.6278 m3
2 L0
y dA
LA
dA
=
0.6278
= 1.00 m
0.6278
Ans.
LA
Ans:
y = 1.00 m
894
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–17.
y
Locate the centroid y of the area.
SOLUTION
y
4 in.
Area: Integrating the area of the differential element gives
A =
LA
dA =
8 in.
8 in.
x
L0
2
––
x3
2>3
3
dx = c x 5>3 d 2
0
5
x
= 19.2 in.2
8 in.
'
1
Centroid: The centroid of the element is located at y = y>2 = x2>3. Applying
2
Eq. 9–4, we have
8 in.
'
y =
y dA
LA
LA
c
=
L0
=
dA
8 in.
1 2>3 2>3
1 4>3
x A x B dx
x dx
L
2
2
0
=
19.2
19.2
8 in.
3 7>3 2
x d
14
0
19.2
Ans.
= 1.43 in.
Ans:
y = 1.43 in.
895
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–18.
y
Locate the centroid x of the area.
y
SOLUTION
h
h n
—
nx
a
h
dA = y dx
'
x = x
x =
LA
x =
=
L0
dA
h
2
B hx -
h n+1
x ≤ dx
an
a
h n
x ≤ dx
an
¢h -
h1xn + 22
a 1n + 22
n
h1x
2
n+1
a n1n + 12
h
h
b a2
2
n + 2
ah -
¢ hx -
L0
B x2 -
a
a
'
x dA
LA
=
x
a
h
ba
n + 1
=
R
R
a
0
a
0
a(1 + n)
2(2 + n)
Ans.
Ans:
x =
896
a(1 + n)
2(2 + n)
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–19.
y
Locate the centroid y of the area.
y
SOLUTION
h
h n
—
x
an
h
dA = y dx
y =
LA
y =
a
'
y dA
LA
=
x
a
y
'
y =
2
=
h2
1
h2
¢ h2 - 2 n xn + 2n x2n ≤ dx
2 L0
a
a
a
dA
L0
¢h -
h n
x ≤ dx
an
2h21xn + 12
h21x 2n + 12 a
1 2
+ 2n
Bh x - n
R
2
a 1n + 12
a 12n + 12 0
B hx -
h1x n + 12
n
a 1n + 12
2n2
h
21n + 1212n + 12
n
n + 1
=
R
a
0
hn
2n + 1
Ans.
Ans:
y =
897
hn
2n + 1
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–20.
Locate the centroid y of the shaded area.
y
y
h xn
––
an
h
SOLUTION
dA = y dx
x
a
y
y =
2
'
ydA
y =
LA
1
2
=
dA
LA
a
L0
L0
h2
a 2n
x 2n dx
=
a
h
an
xn dx
h2(a 2n + 1)
2a 2n(2n + 1)
h(a n + 1)
a n(n + 1)
=
hn + 1
2(2n + 1)
Ans.
Ans:
y =
898
hn + 1
2(2n + 1)
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–21.
y
Locate the centroid x of the shaded area.
1
y (4 x2 )2
16 ft
4 ft
x
4 ft
Solution
Area And Moment Arm. The area of the differential element shown shaded
in Fig. a is dA = y dx = ( 4 - x1>2 ) 2 dx = (x - 8x1>2 + 16)dx and its centroid is
at ~
x = x.
Centroid. Perform the integration
x =
~
1A x dA
1A dA
=
=
L0
4 ft
L0
a
a
= 1
x(x - 8x1>2 + 16)dx
4 ft
( x - 8x1>2 + 16) dx
4 ft
x3
16 5>2
x + 8x2 b `
3
5
0
4 ft
16 3>2
x2
x + 16xb `
2
3
0
3
ft
5
Ans.
Ans:
3
x = 1 ft
5
899
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–22.
y
Locate the centroid y of the shaded area.
1
y (4 x2 )2
16 ft
4 ft
x
4 ft
Solution
Area And Moment Arm. The area of the differential element shown shaded in
1
Fig. a is dA = y dx = ( 4 - x 2 ) 2 dx = (x - 8x1>2 + 16)dx and its centroid is at
y
1
= ( 4 - x1>2 ) 2.
y~ =
2
2
Centroid. Perform the integration,
y =
~
1A y dA
1A dA
=
=
=
L0
4 ft
1
( 4 - x1>2 ) 2 ( x - 8x1>2 + 16 ) dx
2
L0
L0
4 ft
4 ft
(x - 8x1>2 + 16)dx
1
a x2 - 8x3>2 + 48x - 128x1>2 + 128bdx
2
L0
a
= 4
4 ft
( x - 8x1>2 + 16 ) dx
4 ft
x3
16 5>2
256 3>2
x + 24x2 x + 128xb `
6
5
3
0
8
ft
55
a
4 ft
16 3>2
x2
x + 16xb `
2
3
0
Ans.
Ans:
y = 4
900
8
ft
55
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–23.
y
Locate the centroid x of the shaded area.
y h2 x2h
a
h
x
a
Solution
Area And Moment Arm. The area of the differential element shown shaded in Fig. a
h
is dA = y dx = a - 2 x2 + hbdx and its centroid is at ~
x = x.
a
Centroid. Perform the integration,
x =
~
1A x dA
1A dA
=
L0
a
L0
x aa
a-
h 2
x + hbdx
a2
h
a
2
2
x + hbdx
h 4
h 2 a
x
+
x b`
2
4a2
0
=
a
h 3
a - 2 x + hxb `
3a
0
a-
=
3
a
8
Ans.
Ans:
x =
901
3
a
8
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–24.
y
Locate the centroid y of the shaded area.
y h2 x2h
a
h
x
a
Solution
Area And Moment Arm. The area of the differential element shown shaded in Fig. a
y
1
h2
h
is dA = y dx = a - 2 x2 + hbdx and its centroid is at ~
y = = a - x2 + hb.
2
2
a
a
Centroid. Perform the integration,
y =
a
~
1A y dA
1
h
h
a - 2 x2 + hba - 2 x2 + hbdx
a
a
L0 2
=
a
h
dA
2
1A
a - 2 x + hbdx
L0
a
a
1 h2 5
2h2 3
2
a 4x x
+
h
xb
`
2 5a
3a2
0
=
a
h 3
a - 2 x + hxb `
3a
0
=
2
h
5
Ans.
Ans:
y =
902
2
h
5
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–25.
z
The plate has a thickness of 0.25 ft and a specific weight of
g = 180 lb>ft3. Determine the location of its center of
gravity. Also, find the tension in each of the cords used to
support it.
B
SOLUTION
1
2
Area and Moment Arm: Here, y = x - 8x + 16. The area of the differential
1
'
element is dA = ydx = 1x - 8x2 + 162dx and its centroid is x = x and
1
1
'
y = 1x - 8x2 + 162. Evaluating the integrals, we have
2
16 ft
A =
16 ft
16 ft
LA
dA =
L0
A
1
y2
1
x2
C
y
4
x
1x - 8x2 + 162dx
1
16 ft
1
16 3
= a x2 x2 + 16x b `
= 42.67 ft2
2
3
0
LA
'
xdA =
16 ft
1
x31x - 8x 2 + 162dx4
L0
16 ft
1
16 5
= a x3 x2 + 8x2 b `
= 136.53 ft3
3
5
0
LA
'
ydA =
=
16 ft
L0
1
1
1
1x - 8x2 + 16231x - 8x2 + 162dx4
2
16 ft
1 1 3
32 5
512 3
a x x2 + 48x2 x2 + 256xb `
2 3
5
3
0
= 136.53 ft3
Centroid: Applying Eq. 9–6, we have
x =
LA
'
xdA
LA
y =
LA
=
136.53
= 3.20 ft
42.67
Ans.
=
136.53
= 3.20 ft
42.67
Ans.
dA
'
ydA
LA
dA
Equations of Equilibrium: The weight of the plate is W = 42.6710.25211802 = 1920 lb.
©Mx = 0;
192013.202 - TA1162 = 0 TA = 384 lb
Ans.
©My = 0;
TC1162 - 192013.202 = 0 TC = 384 lb
Ans.
©Fz = 0;
TB + 384 + 384 - 1920 = 0
Ans.
TB = 1152 lb = 1.15 kip
903
Ans:
x = 3.20 ft
y = 3.20 ft
TA = 384 lb
TC = 384 lb
TB = 1.15 kip
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–26.
y
Locate the centroid x of the shaded area.
4 ft
y
1 2
x
4
Solution
Area And Moment Arm. The area of the differential element shown shaded in Fig. a
1
is dA = y dx = x2 dx and its centroid is at ~
x = x.
4
Centroid. Perform the integration
x =
~
1A x dA
1A dA
=
=
L0
4 ft
L0
a
a
x
4 ft
1
x a x2 dxb
4
4 ft
1 2
x dx
4
1 4 4 ft
x b`
16
0
1 3 4 ft
x b`
12
0
= 3 ft
Ans.
Ans:
x- = 3 ft
904
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–27.
y
Locate the centroid y of the shaded area.
4 ft
y
1 2
x
4
Solution
Area And Moment arm. The area of the differential element shown shaded in Fig. a
y
1 1
1
1
is dA = y dx = x2 dx and its centroid is located at ~
y = = a x2 b = x2.
2
2 4
8
4
Centroid. Perform the integration,
y =
~
1A y dA
1A dA
=
=
L0
4 ft
4 ft
1 2 1 2
x a x dxb
8
4
L0
6
ft
5
x
4 ft
1 2
x dx
4
Ans.
Ans:
y =
905
6
ft
5
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–28.
y
Locate the centroid x of the shaded area.
yx
100 mm
y 1 x2
100
100 mm
x
Solution
1 2
x . Thus the area of the
100
1 2
x ) dx
differential element shown shaded in Fig. a is dA = ( y2 - y1 ) dx = ( x 100
~
and its centroid is at x = x.
Area And Moment arm. Here, y2 = x and y1 =
Centroid. Perform the integration
x =
~
1A x dA
1A dA
=
=
L0
100 mm
L0
a
a
x ax -
100 mm
ax -
1 2
x bdx
100
1 2
x bdx
100
x3
1 4 100 mm
x b`
3
400
0
x2
1 3 100 mm
x b`
2
300
0
= 50.0 mm
Ans.
Ans:
x = 50.0 mm
906
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–29.
Locate the centroid y of the shaded area.
y
yx
100 mm
y 1 x2
100
100 mm
x
Solution
Area And Moment arm. Here, x2 = 10y 1>2 and x1 = y. Thus, the area of the
differential element shown shaded in Fig. a is dA = ( x2 - x1 ) dy = ( 10y1>2 - y ) dy
and its centroid is at ~
y = y.
Centroid. Perform the integration,
y =
~
1A y dA
1A dA
=
=
L0
100 mm
L0
y a10y1>2 - ybdy
100 mm
a4y 5>2 -
a
1>2
a10y
- ybdy
y3 100 mm
b`
3 0
y2 100 mm
20 3>2
y - b`
3
2 0
= 40.0 mm
Ans.
Ans:
y = 40.0 mm
907
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–30.
y
Locate the centroid x of the shaded area.
a
h
h
y ––
ax
y(
h
)(xb)
ab
x
b
Solution
a
a - b
b y + b. Thus the area
y and x2 = a
h
h
a
a - b
by + b - y d dy =
of the differential element is dA = ( x2 - x1 ) dy = c a
h
h
x -x
b
a
b
b
1
( b - y ) dy and its centroid is at ~x = x1 + 2 1 = ( x2 + x1 ) = y - y + .
h
h
2h
2
2
2
Area And Moment arm. Here x1 =
Centroid. Perform the integration,
x =
h
~
1A x dA
b
b
b
a
a y y + b c ab - ybdy d
h
2h
2
h
L0
=
h
b
dA
1A
ab - ybdy
h
L0
=
c
h
b
b
b2
(a - b)y2 +
(b - 2a)y3 +
yd `
2
2h
2
6h
0
b 2 h
aby y b`
2h
0
bh
(a + b)
6
=
bh
2
1
= (a + b)
3
Ans.
Ans:
x =
908
1
(a + b)
3
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–31.
y
Locate the centroid y of the shaded area.
a
h
h
y ––
ax
y(
h
)(xb)
ab
x
b
Solution
a - b
a
by + b. Thus the area
y and x2 = a
h
h
a
a - b
by + b - y d dy =
of the differential element is dA = ( x2 - x1 ) dy = c a
h
h
b
ab - ybdy and its centroid is at ~
y = y.
h
Centroid. Perform the integration,
Area And Moment arm. Here, x1 =
y =
~
1A y dA
1A dA
=
=
L0
h
L0
y ab -
h
ab -
b
ybdy
h
b
ybdy
h
b
b 3 h
a y2 y b`
2
3h
0
aby -
b 2 h
y b`
2h
0
1 2
bh
h
6
=
=
1
3
bh
2
Ans.
Ans:
y =
909
h
3
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–32.
y
Locate the centroid x of the shaded area.
y a sin
SOLUTION
a
Area and Moment Arm: The area of the differential element is
x
dA = ydx = a sin dx and its centroid are x = x
a
x =
LA
x
a
pa
'
xdA
LA
=
dA
L0
x a a sin
pa
L0
=
=
c a3 sin
a sin
x
ap
x
dx b
a
x
dx
a
pa
x
x
- x a a2 cos b d `
a
a
0
a -a2 cos
x pa
b`
a 0
p
a
2
Ans.
Ans:
x =
910
p
a
2
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–33.
y
Locate the centroid y of the shaded area.
y
SOLUTION
a sin
x
a
a
Area and Moment Arm: The area of the differential element is
y
a
x
x
= sin .
dA = ydx = a sin dx and its centroid are y =
a
a
2
2
x
ap
pa
1
2x
1 2
x
x
a
bd`
c a a x - a sin
sin aa sin dxb
a
4
2
a
a
0
pa
L0 2
y = LA
=
=
=
pa
pa
8
x
x
2
dA
a sin dx
a -a cos b `
a
LA
L0
a 0
pa
ydA
Ans.
Ans:
y =
911
pa
8
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–34.
y
The steel plate is 0.3 m thick and has a density of
7850 kg>m3. Determine the location of its center of mass.
Also compute the reactions at the pin and roller support.
y2 2x
2m
SOLUTION
x
y1 = - x1
A
y 22 = 2x2
2m
dA = 1y2 - y12 dx =
'
x = x
A 22x + x B dx
y x
B
2m
y2 + y1
22x - x
'
=
y =
2
2
x=
LA
=
LA
y =
LA
2
'
x dA
dA
LA
2
L0
A 22x + x B dx
=
c
2
222 5>2
1
x + x3 d
5
3
0
2
2 22 3>2
1
c
x + x2 d
3
2
0
2
'
y dA
=
dA
L0
x A 22x + x B dx
22x - x
A 22x + x B dx
2
L0
2
L0
A 22x + x B dx
=
c
= 1.2571 = 1.26 m Ans.
2
x2
1
- x3 d
2
6
0
2
2 22 3>2
1
c
x + x2 d
3
2
0
= 0.143 m
Ans.
A = 4.667 m2
W = 785019.81214.667210.32 = 107.81 kN
a + ©MA = 0;
-1.25711107.812 + NB A 2 22 B = 0
Ans.
NB = 47.92 = 47.9 kN
+
: ©Fx = 0;
- A x + 47.92 sin 45° = 0
Ans.
A x = 33.9 kN
+ c ©Fy = 0;
A y + 47.92 cos 45° - 107.81 = 0
Ans.
A y = 73.9 kN
Ans:
x = 1.26 m
y = 0.143 m
NB = 47.9 kN
Ax = 33.9 kN
Ay = 73.9 kN
912
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–35.
y
Locate the centroid x of the shaded area.
h
y h —n xn
a
h
h
yh—
a x
x
a
Solution
h n
h
x and y1 = h - x. Thus, the
an
a
area of the differential element shown shaded in Fig. a is dA = ( y2 - y1 ) dx
Area And Moment Arm. Here, y2 = h -
= ch -
h n
h
h
h
x - ah - xb d dx = a x - n xn bdx and its centroid is ~
x = x.
an
a
a
a
Centroid. Perform the integration
x =
a
~
1A x dA
h
h
x a x - n xn bdx
a
a
L0
=
a
h
h n
dA
1A
a x - n x bdx
a
L0 a
=
c
c
a
h 3
h
x - n
xn + 2 d `
3a
a (n + 2)
0
a
h 2
h
x - n
xn + 1 b `
2a
a (n + 1)
0
ha2 (n - 1)
=
3(n + 2)
ha(n - 1)
2(n + 1)
= c
2(n + 1)
3(n + 2)
da
Ans.
Ans:
x = c
913
2(n + 1)
3(n + 2)
da
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–36.
y
Locate the centroid y of the shaded area.
h
y h —n xn
a
h
h
yh—
a x
x
a
Solution
h n
h
x and y1 = h - x. Thus, the
an
a
area of the differential element shown shaded in Fig. a is dA = ( y2 - y1 ) dx
h
h
h
h
= c h - n xn - ah - xb d dx = a x - n xn bdx and its centroid is at
a
a
a
a
Area And Moment Arm. Here, y2 = h -
∼
y = y1 + a
y2 - y1
1
1
h
h
1
h
h
b = ( y2 + y1 ) = ah - n xn + h - xb = a2h - n xn - xb.
2
2
2
a
a
2
a
a
Centroid. Perform the integration
y =
a
~
1A y dA
h
1
h
h
h
a2h - n xn - xba x - n xn bdx
2
a
a
a
a
L0
=
a
h n
h
dA
1A
a x - n x bdx
a
a
L0
=
a
1 h2 2
h2
2h2
h2
x2n + 1 d `
c x - 2 x3 - n
xn + 1 + 2n
2 a
a (n + 1)
3a
a (2n + 1)
0
c
=
h2a c
= c
a
h 2
h
x - n
xn + 1 b `
2a
a (n + 1)
0
(4n + 1) (n - 1)
6(n + 1)(2n + 1)
hac
n - 1
d
2(n + 1)
(4n + 1)
3(2n + 1)
d
Ans.
d h
Ans:
y = c
914
(4n + 1)
3(2n + 1)
d h
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–37.
y
Locate the centroid x of the circular sector.
r
a
C
x
a
x
Solution
Area And Moment Arm. The area of the differential element shown in Fig. a is
1
2
dA = r 2 du and its centroid is at x~ = r cos u.
2
3
Centroid. Perform the integration
x =
=
~
1A x dA
1A dA
L-a
a
=
2
1
a r cos u ba r 2 du b
3
2
1 2
r du
2
L-a
a
a
1
a r 3 sin u b `
3
-a
a
1
a r2 ub `
2
-a
2 3
r sin a
3
=
r2 a
=
2 r sin a
a
b
3
a
Ans.
Ans:
x =
915
2 r sin a
a
b
3
a
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–38.
Determine the location r of the centroid C for the loop of the
lemniscate, r2 = 2a2 cos 2u, ( -45° … u … 45°).
r2 = 2a2 cos 2θ
r
O
θ
C
_
r
SOLUTION
1
1
(r) r du = r2 du
2
2
dA =
45°
A = 2
L0
1
(2a2 cos 2u)du = a2 C sin 2u D 45°
= a2
0
2
45°
2
x dA
x =
LA
L0
=
A 23 r cos u B A 12r2du B
dA
2
3
=
a2
45°
L0
r3 cos u du
2
a
LA
x dA =
LA
x =
2
3 L0
45°
3
r cos u du =
2
3 L0
45°
A 2a2 B 3/2 cos u (cos 2 u)3/2du = 0.7854 a3
0.7854 a 3
= 0.785 a
a2
Ans.
Ans:
x = 0.785 a
916
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–39.
Locate the center of gravity of the volume. The material is
homogeneous.
z
2m
SOLUTION
y 2 = 2z
2m
Volume and Moment Arm: The volume of the thin disk differential element is
'
dV = py2dz = p12z2dz = 2pzdz and its centroid z = z.
y
Centroid: Due to symmetry about z axis
- = y
-= 0
x
Ans.
Applying Eq. 9–3 and performing the integration, we have
Lv
z =
Lv
=
z12pzdz2
L0
dV
2m
L0
z3
3
2m
z2
2p
2
2m
2p
=
2m
'
z dV
0
=
2pzdz
4
m
3
Ans.
0
Ans:
x = y = 0
4
z = m
3
917
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–40.
Locate the centroid y of the paraboloid.
z
z2 = 4y
4m
y
SOLUTION
Volume and Moment Arm: The volume of the thin disk differential element is
'
dV = pz2dy = p14y2dy and its centroid y = y.
4m
Centroid: Applying Eq. 9–3 and performing the integration, we have
y =
LV
4m
'
ydV
LV
=
dV
L0
y3p14y2dy4
4m
p14y2dy
L0
4p
=
y3
3
4m
2
4m
y
4p
2
0
Ans.
= 2.67 m
0
Ans:
y = 2.67 m
918
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–41.
Locate the centroid z of the frustum of the right-circular
cone.
z
r
h
R
SOLUTION
Volume
y =
and
Moment
1r - R2z + Rh
h
Arm:
From
the
geometry,
y - r
h - z
=
,
R - r
h
y
x
. The volume of the thin disk differential element is
dV = py2dz = pc a
=
1r - R2z + Rh
h
2
b d dz
p
c1r - R22z2 + 2Rh1r - R2z + R2h2 ddz
h2
and its centroid z = z.
Centroid: Applying Eq. 9–5 and performing the integration, we have
z =
LV
h
'
z dV
LV
=
dV
=
=
L0
zb
p
31r - R22z2 + 2Rh1r - R2z + R2h24dz r
h2
h
p
3r - R22z2 + 2Rh1r - R2z + R2h24dz
2
L0 h
4
2
h
z3
p
2 z
2 2 z
1r
R2
+
2Rh1r
R2
+
R
h
B
¢
≤
¢
≤
¢
≤
R
`
4
3
2
h2
0
3
p
2 z
1r
R2
3
h2
+ 2Rh1r - R2
z2
2
+ R2h21z2 `
R2 + 3r2 + 2rR
h
4 R2 + r2 + rR
h
0
Ans.
Ans:
z =
919
R2 + 3r 2 + 2rR
h
4(R2 + r 2 + rR)
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–42.
z
Determine the centroid y of the solid.
y
z –– (y 1)
6
SOLUTION
1 ft
y
Differential Element: The thin disk element shown shaded in Fig. a will be considered. The x
volume of the element is
2
y
p 4
dV = pz2 dy = pc (y - 1) d dy =
(y - 2y3 + y2)dy
6
36
3 ft
Centroid: The centroid of the element is located at yc = y. We have
y =
LV
3 ft
y~ dV
LV
=
dV
L0
L0
yc
3 ft
p 4
A y - 2y3 + y2 B dy d
36
p 4
A y - 2y3 + y2 B dy
36
3 ft
=
L0
3 ft
L0
p 5
A y - 2y4 + y3 B dy
36
p 4
A y - 2y3 + y2 B dy
36
=
y4 3 ft
p y6
2
c
- y5 +
d2
36 6
5
4 0
y4
y 3 3 ft
p y5
c
+
d2
36 5
2
3 0
Ans.
= 2.61 ft
Ans:
y = 2.61 ft
920
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–43.
Locate the centroid of the quarter-cone.
z
h
SOLUTION
a
'
z = z
r =
y
a
(h - z)
h
x
dV =
p 2
p a2
r dz =
(h - z)2 dz
4
4 h2
dV =
p a2 2
z3 h
p a2
d
(h2 - 2hz + z2) dz =
c h z - hz2 +
2
2
3 0
4 h L0
4h
h
L
=
pa2 h
p a2 h3
a b =
2
3
12
4h
2
h
z3
z4 h
p a 2 2 z2
pa
'
2
2
2h
+
d
(h
2hz
+
z
)
z
dz
=
ch
z dV =
2
2
3
4 0
4 h2
4 h L0
L
=
pa2h2
p a2 h4
b
=
a
48
4 h2 12
p a2 h2
h
48
L
z =
=
=
4
p a2h
dV
12
L
'
z dV
h
Ans.
h
pa2
pa 2
4r
4a
'
(h - z)2 dz =
(h3 - 3h2 z + 3hz2 - z3) dz
xdV =
2
2
3p
3
4
h
4
h
L0
L0 p h
L
=
3h4
h4
p a2 4a
4
4
+
h
b
a
h
2
4
4 h2 3ph
=
a3 h
p a2 a h3
b
=
a
12
4 h2 3p
a3 h
12
a
L
=
x = y =
=
p
p a2h
dV
12
L
'
xdV
Ans.
Ans:
h
4
a
x = y =
p
z =
921
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–44.
z
The hemisphere of radius r is made from a stack of very thin
plates such that the density varies with height r = kz,
where k is a constant. Determine its mass and the distance
to the center of mass G.
G
_
z
r
SOLUTION
y
Mass and Moment Arm: The density of the material is r = kz. The mass of the thin
disk differential element is dm = rdV = rpy2dz = kz3p(r2 - z2) dz4 and its
'
centroid z = z. Evaluating the integrals, we have
x
r
m =
Lm
dm =
L0
kz3p(r2 - z2) dz4
= pk ¢
Lm
'
z dm =
r2z2
z4 r
pkr4
- ≤` =
2
4 0
4
Ans.
r
L0
z5kz3p(r2 - z2) dz46
= pk ¢
2pkr5
r2z3
z5 r
- ≤` =
3
5 0
15
Centroid: Applying Eq. 9–3, we have
z =
Lm
'
z dm
Lm
=
dm
2pkr5>15
pkr4>4
=
8
r
15
Ans.
Ans:
pkr 4
4
8
z =
r
15
m =
922
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–45.
z
Locate the centroid z of the volume.
1m
y2 0.5z
2m
Solution
y
Volume And Moment arm. The volume of the thin disk differential element shown
shaded in Fig. a is dV = py2 dz = p(0.5z)dz and its centroid is at ~
z = z.
Centroid. Perform the integration
z =
~
1V z dV
1V dV
=
=
=
L0
2m
L0
x
z[p(0.5z)dz]
2m
p(0.5z)dz
0.57 3 2 m
z `
3
0
0.5p 2 2 m
z `
2
0
4
m
3
Ans.
Ans:
z =
923
4
m
3
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–46.
Locate the centroid of the ellipsoid of revolution.
z
y2 z2
1
b2 a 2
a
x
SOLUTION
b
dV = p z2 dy
b
L
L
dV =
L0
'
ydV =
p a2 a 1 b
L0
y
y2
b
p a2y a1 -
b dy = p a2 c y 2
y
b
2
b dy = p a2 c
2
y3
b
3b
0
d =
2
2pa2b
3
2
y4 b
y
p a2b2
d =
2
2
4
4b 0
pa2b2
3
4
LV
= b
=
y =
8
2pa2b
dV
3
LV
'
ydV
x = z = 0
Ans.
Ans.
(By symmetry)
Ans:
3
b
8
x = z = 0
y =
924
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–47.
z
Locate the center of gravity z of the solid.
2
––
z 4y 3
SOLUTION
16 in.
Differential Element: The thin disk element shown shaded in Fig. a will be
considered. The volume of the element is
y
2
1
p 3
dV = py2 dz = pc z3>2 d dz =
z dz
8
64
8 in.
x
Centroid: The centroid of the element is located at zc = z. We have
z =
LV
16 in.
~
z dV
LV
=
dV
L0
L0
zc
16 in.
p 3
z dz d
64
p 3
z dz
64
16 in.
=
L0
L0
16 in.
p 4
z dz
64
p 3
z dz
64
=
p z5 16 in.
a b2
64 5 0
p z4 2 16 in.
a b
64 4 0
= 12.8 in. Ans.
Ans:
z = 12.8 in.
925
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–48.
z
Locate the center of gravity y of the volume. The material is
homogeneous.
z 1 y2
100
4 in.
1 in.
10 in.
y
10 in.
Solution
Volume And Moment Arm. The volume of the thin disk differential element shown
p
1 2 2
shaded in Fig. a is dV = pz2 dy = pa
y b dy =
y4 dy and its centroid is
100
10000
y = y.
at ~
Centroid. Perform the integration
y =
~
1V y dV
1V dV
20 in.
=
=
L10 in.
ya
p
y4 dyb
10000
20 in.
p
y4 dy
L10 in. 10000
a
a
20 in.
p
y6 b `
60000
10 in.
20 in.
p
y5 b `
50000
10 in.
Ans.
= 16.94 in. = 16.9 in.
Ans:
y = 16.9 in.
926
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–49.
Locate the centroid z of the spherical segment.
z
z2
a2
y2
1a
—
2
C
a
SOLUTION
z
2
2
2
dV = py dz = p(a - z ) dz
y
z = z
z =
LV
a
'
z dV
LV
p
=
dV
L
p
=
a
2
La2
x
z (a2 - z2)dz
a
(a 2 - z2)dz
p B a2 a
z4 a
z2
b - a bR
a
2
4
2
z3 a
p B a (z) - a b R
a
3
2
2
=
pB
a4
a4
a4
a4
+
R
2
4
8
64
a3
a3
a3
+
pBa R
3
2
24
3
=
pB
9a 4
R
64
pB
5a 3
R
24
Ans.
z = 0.675 a
Ans:
z = 0.675a
927
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–50.
Determine the location z of the centroid for the
tetrahedron. Hint: Use a triangular “plate” element parallel
to the x–y plane and of thickness dz.
z
b
a
SOLUTION
z = ca1 -
1
1
yb = ca1 - xb
a
b
c
L
dV =
c
y
c
1
z
abc
z
1
(x)(y)dz =
a a 1 - b ba 1 - b dz =
c
c
2 L0
6
L0 2
x
c
z
a b c2
z
1
'
z a a 1 - b ba 1 - b dz =
z dV =
c
c
2 L0
24
L
a b c2
c
24
L
=
z =
=
abc
4
dV
6
L
'
z dV
Ans.
Ans:
z =
928
c
4
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–51.
The truss is made from five members, each having a length
of 4 m and a mass of 7 kg>m. If the mass of the gusset plates
at the joints and the thickness of the members can be
neglected, determine the distance d to where the hoisting
cable must be attached, so that the truss does not tip
(rotate) when it is lifted.
y
d
B
4m
C
4m
4m
SOLUTION
60
'
©xM = 4(7)(1+ 4 + 2 + 3 + 5) = 420 kg # m
A
©M = 4(7)(5) = 140 kg
d = x =
4m
'
420
©xM
=
= 3m
©M
140
4m
D
Ans.
Ans:
d = 3m
929
x
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–52.
z
Determine the location (x, y, z) of the centroid of the
homogeneous rod.
200 mm
x
30
600 mm
100 mm
Solution
y
Centroid. Referring to Fig. a, the length of the segments and the locations of their
respective centroids are tabulated below
Segment L(mm)
x~(mm)
y~(mm)
z~(mm)
x~ L(mm2)
0
0
100
0
1
200
2
600
300 cos 30° 300 sin 30°
0
3
100
600 cos 30° 600 sin 30°
–50
a
900
y~ L(mm2) z~ L(mm2)
0
155.88 ( 103 ) 90.0 ( 103 )
20.0 ( 103 )
0
51.96 ( 103 ) 30.0 ( 103 ) - 5.0 ( 103 )
207.85 ( 103 ) 120.0 ( 103 ) 15.0 ( 103 )
Thus,
x =
207.85 ( 103 ) mm2
Σ~
xL
=
= 230.94 mm = 231 mm
ΣL
900 mm
Ans.
y =
120.0 ( 103 ) mm2
Σ~
yL
=
= 133.33 mm = 133 mm
ΣL
900 mm
Ans.
z =
15.0 ( 103 ) mm2
Σ~
zL
=
= 16.67 mm = 16.7 mm
ΣL
900 mm2
Ans.
Ans:
x = 231 mm
y = 133 mm
z = 16.7 mm
930
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–53.
A rack is made from roll-formed sheet steel and has the
cross section shown. Determine the location 1x, y2 of the
centroid of the cross section. The dimensions are indicated
at the center thickness of each segment.
y
30 mm
80 mm
SOLUTION
50 mm
©L = 15 + 50 + 15 + 30 + 30 + 80 + 15 = 235 mm
'
©xL = 7.5(15) + 0(50) + 7.5(15) + 15(30) + 30(30) + 45(80) + 37.5(15) = 5737.50 mm2
'
©yL = 0(15) + 25(50) + 50(15) + 65(30) + 80(30) + 40(80) + 0(15) = 9550 mm2
x =
'
©xL
5737.50
=
= 24.4 mm
©L
235
Ans.
y =
'
©yL
9550
=
= 40.6 mm
©L
235
Ans.
x
15 mm
15 mm
Ans:
x = 24.4 mm
y = 40.6 mm
931
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–54.
Locate the centroid (x, y) of the metal cross section. Neglect
the thickness of the material and slight bends at the corners.
y
50 mm
150 mm
SOLUTION
x
Centroid: The length of each segment and its respective centroid are tabulated below.
Segment
L (mm)
'
y (mm)
'
y L (mm2)
1
50p
168.17
26415.93
2
180.28
75
13520.82
3
400
0
0
4
180.28
75
13520.82
©
917.63
53457.56
- = 0
Due to symmetry about y axis, x
y =
50 mm 100 mm 100 mm 50 mm
Ans.
'
©yL
53457.56
=
= 58.26 mm = 58.3 mm
©L
917.63
Ans.
Ans:
x = 0
y = 58.3 mm
932
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–55.
Locate the center of gravity 1x, y, z2 of the homogeneous
wire.
z
400 mm
SOLUTION
2 (300) p
'
a b (300) = 165 000 mm2
©xL = 150(500) + 0(500) +
p
2
y
300 mm
p
©L = 500 + 500 + a b (300) = 1471.24 mm
2
x =
x
'
©xL
165 000
=
= 112 mm
©L
1471.24
Ans.
Due to symmetry,
Ans.
y = 112 mm
p
'
© z L = 200(500) + 200(500) + 0 a b (300) = 200 000 mm2
2
z =
'
© zL
200 000
=
= 136 mm
©L
1471.24
Ans.
Ans:
x = 112 mm
y = 112 mm
z = 136 mm
933
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–56.
The steel and aluminum plate assembly is bolted together
and fastened to the wall. Each plate has a constant width in
the z direction of 200 mm and thickness of 20 mm. If the
density of A and B is rs = 7.85 Mg>m3, and for C,
ral = 2.71 Mg>m3, determine the location x of the center of
mass. Neglect the size of the bolts.
y
100 mm
200 mm
A
SOLUTION
C
B
300 mm
©m = 2 C 7.85(10)3(0.3)(0.2)(0.02) D + 2.71(10)3(0.3)(0.2)(0.02) = 22.092 kg
'
©xm = 150{2 C 7.85(10)3(0.3)(0.2)(0.02) D }+350 C 2.71(10)3(0.3)(0.2)(0.02) D
= 3964.2 kg.mm
x =
'
3964.2
©xm
=
= 179 mm
©m
22.092
Ans.
Ans:
x = 179 mm
934
x
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–57.
Locate the center of gravity G1x, y2 of the streetlight. Neglect
the thickness of each segment. The mass per unit length of
each segment is as follows: rAB = 12 kg>m, rBC = 8 kg>m,
rCD = 5 kg>m, and rDE = 2 kg>m.
1m
y
90 1 m
1m
D
E
1m
C
SOLUTION
G (x, y)
3m
2(1) p
'
b a b(5)
©xm = 0(4)(12) + 0(3)(8) + 0(1)(5) + a 1 p
2
B
+ 1.5 (1) (5) + 2.75 (1.5) (2) = 18.604 kg # m
©m = 4 (12)+3 (8)+ 1(5)+
x =
1.5 m
p
(5) + 1(5) + 1.5 (2) = 92.854 kg
2
4m
A
'
©xm
18.604
=
= 0.200 m
©m
92.854
x
Ans.
2(1) p
'
b a b(5)
©ym = 2 (4) (12) + 5.5 (3)(8) + 7.5(1) (5) + a 8 +
p
2
+ 9 (1) (5) + 9(1.5) (2) = 405.332 kg # m
'
©ym
405.332
=
= 4.37 m
y =
©m
92.854
Ans.
Ans:
x = 0.200 m
y = 4.37 m
935
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–58.
Determine the location y of the centroidal axis x - x of the
beam’s cross-sectional area. Neglect the size of the corner
welds at A and B for the calculation.
150 mm
15 mm
B
y
15 mm
150 mm
C
x
SOLUTION
A
'
©yA = 7.5(15) (150) + 90(150) (15) + 215(p) (50)2
50 mm
= 1 907 981.05 mm2
©A = 15(150) + 150(15) + p(50)2
= 12 353.98 mm2
y =
'
©yA
1 907 981.05
=
= 154 mm
©A
12 353.98
Ans.
Ans:
y = 154 mm
936
x
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–59.
y
Locate the centroid (x, y) of the shaded area.
6 in.
6 in.
x
6 in.
6 in.
Solution
Centroid. Referring to Fiq. a, the areas of the segments and the locations of their
respective centroids are tabulated below
y (in.)
3
x∼A(in. )
∼A(in.3)
0
0
0
0
-4
4
72.0
-72.0
72.0
-72.0
Segment
A(in.2)
x∼(in.)
1
12(12)
2
1
- (6)(6)
2
Σ
126
∼
y
Thus,
Σx~A
72.0 in.3
=
= 0.5714 in. = 0.571 in.
ΣA
126 in.2
Σy~A
- 72.0 in.3
=
= - 0.5714 in. = - 0.571 in.
y =
ΣA
126 in.2
Ans.
x =
Ans.
Ans:
x = 0.571 in.
y = - 0.571 in.
937
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–60.
Locate the centroid y for the beam’s cross-sectional area.
120 mm
240 mm
x
y
Solution
240 mm
Centroid. The locations of the centroids measuring from the x axis for segments
1 and 2 are indicated in Fig. a. Thus
300(120)(600) + 120(240)(120)
Σy~A
y =
=
ΣA
120(600) + 240(120)
120 mm
240 mm
= 248.57 mm = 249 mm
Ans:
y = 249 mm
938
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–61.
Determine the location y of the centroid C of the beam
having the cross-sectional area shown.
150 mm
15 mm
B
150 mm
y
C
x
15 mm
15 mm
A
100 mm
Solution
Centroid. The locations of the centroids measuring from the x axis for segments 1,
2 and 3 are indicated in Fig. a. Thus
y =
7.5(15)(150) + 90(150)(15) + 172.5(15)(100)
Σ~
yA
=
ΣA
15(150) + 150(15) + 15(100)
Ans.
= 79.6875 mm = 79.7 mm
Ans:
y = 79.7 mm
939
x
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–62.
y
Locate the centroid (x, y) of the shaded area.
6 in.
3 in.
6 in.
x
Solution
6 in.
Centroid. Referring to Fig. a, the areas of the segments and the locations of their respective
centroids are tabulated below
Segment
A(in.2)
x~(in.)
y~(in.)
1
1
(6)(9)
2
2
6
54.0
162.0
2
1
(6)(3)
2
-2
7
-18.0
63.0
3
6(6)
-3
3
-108.0
108.00
Σ
72.0
-72.0
333.0
x~A(in.3)
y~A(in.3)
Thus,
Σx~A
- 72.0 in.3
=
= - 1.00 in.
ΣA
72.0 in.2
Σy~A
333.0 in.3
y =
=
= 4.625 in.
ΣA
72.0 in.2
Ans.
x =
Ans.
Ans:
x = - 1.00 in.
y = 4.625 in.
940
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–63.
Determine the location y of the centroid of the beam’s crosssectional area. Neglect the size of the corner welds at A and B
for the calculation.
35 mm
A
SOLUTION
110 mm
2
35
35
'
b = 393 112 mm3
©yA = p(25)2(25) + 15(110)(50 + 55) + pa b a50 + 110 +
2
2
©A = p(25)2 + 15(110) + p a
35 2
b = 4575.6 mm2
2
C
15 mm
B
y
50 mm
'
©yA
393 112
=
= 85.9 mm
y =
©A
4575.6
Ans.
Ans:
y = 85.9 mm
941
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–64.
y
Locate the centroid (x, y) of the shaded area.
1 in.
3 in.
x
3 in.
3 in.
Solution
Centroid. Referring to Fig. a, the areas of the segments and the locations of their respective
centroids are tabulated below
Segment
A(in.2)
x~(in.)
y~(in.)
1
p 2
(3 )
4
4
p
4
p
9.00
9.00
2
3(3)
- 1.5
1.5
-13.50
13.50
3
1
(3)(3)
2
-4
1
-18.00
4.50
4
-
0
4
3p
0
-0.67
Σ
18.9978
-22.50
26.33
p 2
(1 )
2
x~A(in.3)
y~A(in.3)
Thus,
Σx~A
- 22.50 in.3
=
= - 1.1843 in. = -1.18 in.
ΣA
18.9978 in.2
Σy~A
26.33 in.3
y =
=
= 1.3861 in. = 1.39 in.
ΣA
18.9978 in.2
x =
Ans.
Ans.
Ans:
x = - 1.18 in.
y = 1.39 in.
942
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–65.
y
Determine the location (x, y) of the centroid C of the area.
1.5 in.
1.5 in.
1.5 in.
1.5 in.
x
Solution
1.5 in.
Centroid. Referring to Fig. a, the areas of the segments and the locations of their respective
centroids are tabulated below.
Segment
A(in.2)
x~(in.)
y~(in.)
1
3(3)
1.5
1.5
13.5
p
(1.52)
4
2
p
2
p
-1.125
1
(1.5)(1.5)
2
2.5
2.5
-2.8125
-2.8125
9.5625
9.5625
-
2
3
Σ
-
6.1079
x~A(in.3)
y~A(in.3)
13.5
-1.125
Thus
Σx~A
9.5625 in.3
=
= 1.5656 in. = 1.57 in.
ΣA
6.1079 in.2
Σy~A
9.5625 in.3
y =
=
= 1.5656 in. = 1.57 in.
ΣA
6.1079 in.2
x =
Ans.
Ans.
Ans:
x = 1.57 in.
y = 1.57 in.
943
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–66.
Determine the location y of the centroid C for a beam having
the cross-sectional area shown. The beam is symmetric with
respect to the y axis.
y
C
y
SOLUTION
'
©yA = 6(4)(2) - 1(1)(0.5) - 3(1)(2.5) = 40 in3
3 in.
1 in.
2 in. 1 in. 2 in. 1 in.
©A = 6(4) - 1(1) - 3(1) = 20 in2
y =
'
©yA
40
=
= 2 in.
©A
20
Ans.
Ans:
y = 2 in.
944
x
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–67.
Locate the centroid y of the cross-sectional area of the
beam constructed from a channel and a plate. Assume all
corners are square and neglect the size of the weld at A.
20 mm y
350 mm
C
A
10 mm
70 mm
SOLUTION
325 mm
325 mm
Centroid: The area of each segment and its respective centroid are tabulated below.
Segment
A (mm2)
'
y (mm)
'
y A (mm3)
1
350(20)
175
1 225 000
2
630(10)
355
2 236 500
3
70(20)
385
539 000
©
14 700
4 000 500
Thus,
'
©yA
4 000 500
'
y =
=
= 272.14 mm = 272 mm
©A
14 700
Ans.
Ans:
y = 272 mm
945
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–68.
A triangular plate made of homogeneous material has a
constant thickness which is very small. If it is folded over as
shown, determine the location y of the plate’s center of
gravity G.
z
1 in.
1 in.
SOLUTION
y
1
©A = (8) (12) = 48 in2
2
G
z
6 in.
1 in.
3 in.
1
1
'
©yA = 2(1) a b (1)(3) + 1.5(6)(3) + 2(2) a b(1)(3)
2
2
3 in.
x
= 36 in3
y =
3 in.
'
©yA
36
=
= 0.75 in.
©A
48
3 in.
y
1 in.
Ans.
Ans:
y = 0.75 in.
946
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–69.
A triangular plate made of homogeneous material has a
constant thickness which is very small. If it is folded over as
shown, determine the location z of the plate’s center of
gravity G.
z
1 in.
1 in.
SOLUTION
©A =
y
1
(8)(12) = 48 in2
2
G
z
6 in.
1 in.
3 in.
1
1
'
© z A = 2(2) a b (2)(6) + 3(6)(2) + 6 a b (2)(3)
2
2
3 in.
x
= 78 in3
z =
3 in.
~
78
©zA
=
= 1.625 in.
©A
48
3 in.
y
1 in.
Ans.
Ans:
z = 1.625 in.
947
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–70.
Locate the center of mass z of the forked lever, which
is made from a homogeneous material and has the
dimensions shown.
z
0.5 in.
3 in.
2 in.
SOLUTION
G
1
1
©A = 2.5(0.5) + c p (2.5)2 - p (2)2 d + 2 [(3)(0.5)] = 7.7843 in2
2
2
4(2.5) 1
2.5
'
© zA =
(2.5) (0.5) + a 5 b a p (2.5)2 b
2
3p
2
- a5 z =
2.5 in.
4(2) 1
b a p (2)2 b + 6.5(2)(3)(0.5) = 33.651 in3
3p
2
'
33.651
© zA
=
= 4.32 in.
©A
7.7843
x
z
0.5 in.
Ans.
Ans:
z = 4.32 in.
948
y
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–71.
Determine the location x of the centroid C of the shaded
area which is part of a circle having a radius r.
y
r
a
SOLUTION
x
2r
1
1
2
'
©xA = r 2 aa
sin a b - (r sin a) (r cos a) a r cos ab
2
3a
2
3
r3
r3
sin a sin a cos2 a
3
3
=
r3 3
sin a
3
©A =
=
x
a
Using symmetry, to simplify, consider just the top half:
=
C
1
1 2
r a - (r sin a) (r cos a)
2
2
1 2
sin2a
r aa b
2
2
'
©xA
=
x =
©A
r3
3
1
2
sin3 a
r2 A a -
sin 2 a
2
B
=
2
3
r sin3a
a -
Ans.
sin 2a
2
Ans:
x =
949
2
3r
sin3 a
a -
sin 2a
2
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–72.
A toy skyrocket consists of a solid conical top,
rt = 600 kg>m3, a hollow cylinder, rc = 400 kg>m3, and a
stick having a circular cross section, rs = 300 kg>m3.
Determine the length of the stick, x, so that the center of
gravity G of the skyrocket is located along line aa.
a
3 mm
5 mm
100 mm
20 mm
10 mm
a G
x
SOLUTION
20
1
x
'
©xm = a b c a b p (5)2 (20) d (600) - 50 C p A 52 - 2.52 B (100) D (400) - C (x) p (1.5)2 D (300)
4
3
2
= - 116.24 A 106 B - x2(1060.29) kg # mm4>m3
1
©m = c p (5)2 (20) d (600) + p A 52 - 2.52 B (100)(400) + C xp (1.5)2 D (300)
3
= 2.670 A 106 B + 2120.58x kg # mm3/m3
x =
'
-116.24(106) - x2(1060.29)
©xm
= - 100
=
©m
2.670(106) + 2120.58x
- 116.24 A 106 B - x2 (1060.29) = - 267.0 A 106 B - 212.058 A 103 B x
1060.29x2 - 212.058 A 103 B x - 150.80 A 106 B = 0
Solving for the positive root gives
Ans.
x = 490 mm
Ans:
x = 490 mm
950
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–73.
Locate the centroid y for the cross-sectional area of the
angle.
a
a
–y
C
SOLUTION
t
t
Centroid : The area and the centroid for segments 1 and 2 are
A 1 = t1a - t2
t
t
22
a - t
'
+ b cos 45° +
=
1a + 2t2
y1 = a
2
2
2cos 45°
4
A 2 = at
t
t
22
a
'
y2 = a - b cos 45° +
=
1a + t2
2
2
2cos 45°
4
Listed in a tabular form, we have
Segment
A
'
y
'
yA
1
t1a - t2
22
1a + 2t2
4
22t 2
1a + at - 2t22
4
2
at
22
1a + t2
4
22t 2
1a + at2
4
©
t12a - t2
22t 2
1a + at - t22
2
Thus,
'
©yA
y =
=
©A
=
22t 2
1a + at - t22
2
t12a - t2
22 a2 + at - t2
Ans.
2 2a - t
Ans:
y =
951
22 ( a2 + at - t 2 )
2(2a - t)
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–74.
Determine the location (x, y) of the center of gravity of the
three-wheeler. The location of the center of gravity of each
component and its weight are tabulated in the figure. If the
three-wheeler is symmetrical with respect to the x–y plane,
determine the normal reaction each of its wheels exerts on
the ground.
1.
2.
3.
4.
y
Rear wheels
18 lb
Mechanical components 85 lb
Frame
120 lb
Front wheel
8 lb
3
2
SOLUTION
'
©xW = 4.51182 + 2.31852 + 3.111202
4
1 ft
= 648.5 lb # ft
1
2 ft
A
B
2.30 ft
©W = 18 + 85 + 120 + 8 = 231 lb
x =
1.50 ft
1.30 ft
x
1.40 ft
0.80 ft
'
648.5
©xW
=
= 2.81 ft
©W
231
Ans.
'
©yW = 1.301182 + 1.51852 + 211202 + 1182
= 398.9 lb # ft
y =
a + ©MA = 0;
'
©yW
398.9
=
= 1.73 ft
©W
231
Ans.
21NB214.52 - 23112.812 = 0
Ans.
NB = 72.1 lb
+ c ©Fy = 0;
NA + 2172.12 - 231 = 0
Ans.
NA = 86.9 lb
Ans:
x = 2.81 ft
y = 1.73 ft
NB = 72.1 lb
NA = 86.9 lb
952
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–75.
Locate the center of mass (x, y, z) of the homogeneous
block assembly.
z
250 mm
200 mm
x
SOLUTION
100 mm
150 mm
Centroid: Since the block is made of a homogeneous material, the center of mass of
the block coincides with the centroid of its volume. The centroid of each composite
segment is shown in Fig. a.
150 mm
150 mm
1
(75)(150)(150)(550) + (225)(150)(150)(200) + (200) a b(150)(150)(100)
'
2.165625(109)
2
©xV
= 120 mm
x =
=
=
©V
1
18(106)
(150)(150)(550) + (150)(150)(200) + (150)(150)(100)
2
Ans.
1
(275)(150)(150)(550) + (450)(150)(150)(200) + (50) a b(150)(150)(100)
'
©yV
5.484375(109)
2
= 305 mm
y =
=
=
©V
1
18(106)
(150)(150)(550) + (150)(150)(200) + (150)(150)(100)
2
Ans.
1
(75)(150)(150)(550) + (75)(150)(150)(200) + (50) a b(150)(150)(100)
'
1.321875(109)
2
© zV
z =
= 73.4 mm
=
=
©V
1
18(106)
(150)(150)(550) + (150)(150)(200) + (150)(150)(100)
2
Ans.
y
Ans:
x = 120 mm
y = 305 mm
z = 73.4 mm
953
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–76.
The sheet metal part has the dimensions shown. Determine
the location 1x, y, z2 of its centroid.
z
D
3 in.
C
SOLUTION
A
©A = 4(3) +
4 in.
B
1
(3)(6) = 21 in2
2
x
1
'
©xA = - 2(4)(3) + 0 a b (3)(6) = - 24 in3
2
y
6 in.
2
1
'
©yA = 1.5(4)(3) + (3) a b (3)(6) = 36 in3
3
2
1
1
'
© z A = 0(4)(3) - (6) a b (3)(6) = - 18 in3
3
2
x =
'
- 24
©xA
=
= - 1.14 in.
©A
21
Ans.
y =
'
©yA
36
=
= 1.71 in.
©A
21
Ans.
z =
'
- 18
© zA
=
= - 0.857 in.
©A
21
Ans.
Ans:
x = - 1.14 in.
y = 1.71 in.
z = - 0.857 in.
954
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–77.
The sheet metal part has a weight per unit area of 2 lb>ft2
and is supported by the smooth rod and at C. If the cord is
cut, the part will rotate about the y axis until it reaches
equilibrium. Determine the equilibrium angle of tilt,
measured downward from the negative x axis, that AD
makes with the -x axis.
z
D
3 in.
C
SOLUTION
A
Since the material is homogeneous, the center of gravity coincides with the centroid.
See solution to Prob. 9-74.
u = tan
4 in.
B
x
y
6 in.
-1
1.14
b = 53.1°
a
0.857
Ans.
Ans:
u = 53.1°
955
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–78.
The wooden table is made from a square board having a
weight of 15 lb. Each of the legs weighs 2 lb and is 3 ft long.
Determine how high its center of gravity is from the floor.
Also, what is the angle, measured from the horizontal,
through which its top surface can be tilted on two of its
legs before it begins to overturn? Neglect the thickness of
each leg.
4 ft
4 ft
3 ft
SOLUTION
z =
'
15(3) + 4(2) (1.5)
© zW
=
= 2.48 ft
©W
15 + 4(2)
u = tan - 1 a
Ans.
2
b = 38.9°
2.48
Ans.
Ans:
z = 2.48 ft
u = 38.9°
956
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–79.
The buoy is made from two homogeneous cones each
having a radius of 1.5 ft. If h = 1.2 ft, find the distance z to
the buoy’s center of gravity G.
h
1.5 ft
z
4 ft
G
SOLUTION
1
1
1.2
4
'
b + p(1.5)2 (4) a b
© z V = p (1.5)2 (1.2) a3
4
3
4
= 8.577 ft4
©V =
1
1
p (1.5)2 (1.2) + p(1.5)2 (4)
3
3
= 12.25 ft3
'
© zV
8.577
z- =
=
= 0.70 ft
©V
12.25
Ans.
Ans:
z = 0.70 ft
957
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–80.
The buoy is made from two homogeneous cones each
having a radius of 1.5 ft. If it is required that the buoy’s
center of gravity G be located at z = 0.5 ft, determine the
height h of the top cone.
h
1.5 ft
z
4 ft
G
SOLUTION
1
1
h
4
'
© z V = p (1.5)2 (h) a- b + p(1.5)2 (4) a b
3
4
3
4
= -0.5890 h2 + 9.4248
©V =
1
1
p (1.5)2 (h) + p(1.5)2 (4)
3
3
= 2.3562 h + 9.4248
'
- 0.5890 h2 + 9.4248
© zV
'
=
= 0.5
z =
©V
2.3562 h + 9.4248
-0.5890 h2 + 9.4248 = 1.1781 h + 4.7124
Ans.
h = 2.00 ft
Ans:
h = 2.00 ft
958
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–81.
The assembly is made from a steel hemisphere,
rst = 7.80 Mg>m3,
and
an
aluminum
cylinder,
ral = 2.70 Mg>m3. Determine the mass center of the
assembly if the height of the cylinder is h = 200 mm.
z
80 mm
G
_
z
SOLUTION
© z m = C 0.160 - 38 (0.160) D A 23 B p(0.160)3 (7.80) + A 0.160 +
0.2
2
©m =
3
160 mm
y
B p(0.2)(0.08)2(2.70)
= 9.51425(10 - 3) Mg # m
2
3
h
x
2
A B p(0.160) (7.80) + p (0.2)(0.08) (2.70)
= 77.7706(10 - 3) Mg
z =
9.51425(10 - 3)
© zm
=
= 0.122 m = 122 mm
©m
77.7706(10 - 3)
Ans.
Ans:
z = 122 mm
959
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–82.
The assembly is made from
and
an
rst = 7.80 Mg>m3,
ral = 2.70 Mg>m3. Determine the
so that the mass center of the
z = 160 mm.
a steel hemisphere,
aluminum
cylinder,
height h of the cylinder
assembly is located at
z
80 mm
G
_
z
SOLUTION
© z m = C 0.160 - 38 (0.160) D A 23 B p(0.160)3 (7.80) + A 0.160 +
h
2
©m =
3
160 mm
y
B p (h)(0.08)2(2.70)
= 6.691(10 - 3) + 8.686(10 - 3) h + 27.143(10 - 3) h2
2
3
h
x
2
A B p(0.160) (7.80) + p (h)(0.08) (2.70)
= 66.91(10 - 3) + 54.29(10 - 3) h
z =
6.691(10 - 3) + 8.686(10 - 3) h + 27.143(10 - 3) h2
© zm
= 0.160
=
©m
66.91(10 - 3) + 54.29(10 - 3) h
Solving
Ans.
h = 0.385 m = 385 mm
Ans:
h = 385 mm
960
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–83.
The car rests on four scales and in this position the scale
readings of both the front and rear tires are shown by FA
and FB. When the rear wheels are elevated to a height of 3 ft
above the front scales, the new readings of the front wheels
are also recorded. Use this data to compute the location x
and y to the center of gravity G of the car. The tires each
have a diameter of 1.98 ft.
G
_
y
B
A
_
x
FB
975 lb
SOLUTION
In horizontal position
3.0 ft
B
984 lb
9.40 ft
FA
1959 lb
1129 lb
1168 lb
G
W = 1959 + 2297 = 4256 lb
a + ©MB = 0;
A
2297(9.40) - 4256 x = 0
FA
1269 lb
1307 lb
2576 lb
Ans.
x = 5.0733 = 5.07 ft
u = sin - 1 a
2297 lb
3 - 0.990
b = 12.347°
9.40
With rear whells elevated
a + ©MB = 0;
2576(9.40 cos 12.347°) - 4256 cos 12.347°(5.0733)
- 4256 sin 12.347° y¿ = 0
y¿ = 2.86 ft
Ans.
y = 2.815 + 0.990 = 3.80 ft
Ans:
x = 5.07 ft
y = 3.80 ft
961
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–84.
z
Determine the distance h to which a 100-mm diameter hole
must be bored into the base of the cone so that the center of
mass of the resulting shape is located at z = 115 mm. The
material has a density of 8 Mg>m3.
1
0.5
2
3 p10.152 10.52 4
1
2
3 p10.152 10.52
A B - p10.05221h2 A h2 B
- p10.05221h2
= 0.115
500 mm
0.4313 - 0.2875 h = 0.4688 - 1.25 h2
C
h
50 mm
h2 - 0.230 h - 0.0300 = 0
_
z
150 mm
Choosing the positive root,
Ans.
h = 323 mm
x
Ans:
h = 323 mm
962
y
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–85.
Determine the distance z to the centroid of the shape which
consists of a cone with a hole of height h = 50 mm bored into
its base.
z
500 mm
SOLUTION
1
0.5
0.05
'
b - p (0.05)2 (0.05) a
b
© z V = p (0.15)2 A 0.5 B a
3
4
2
-3
C
h
50 mm
4
= 1.463(10 ) m
©V =
1
p (0.15)2 (0.5) - p (0.05)2 (0.05)
3
z
y
150 mm
x
3
= 0.01139 m
z =
'
1.463 (10 - 3)
©z V
=
= 0.12845 m = 128 mm
©V
0.01139
Ans.
Ans:
z = 128 mm
963
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–86.
Locate the center of mass z of the assembly. The cylinder
and the cone are made from materials having densities of
5 Mg>m3 and 9 Mg>m3, respectively.
z
0.6 m
0.4 m
SOLUTION
Center of mass: The assembly is broken into two composite segments, as shown in
Figs. a and b.
z =
=
'
© zm
=
©m
0.2 m
1
5000(0.4) C p(0.2 2)(0.8) D + 9000(0.8 + 0.15)c p(0.4 2)(0.6) d
3
x
1
5000 C p(0.2 2)(0.8) D + 9000c p(0.4 2)(0.6) d
3
1060.60
= 0.754 m = 754 mm
1407.4
0.8 m
y
Ans.
Ans:
z = 754 mm
964
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–87.
Major floor loadings in a shop are caused by the weights of
the objects shown. Each force acts through its respective
center of gravity G. Locate the center of gravity (x, y) of all
these components.
z
y
450 lb
1500 lb
G2
G1
9 ft
6 ft
SOLUTION
Centroid: The floor loadings on the floor and its respective centroid are tabulated
below.
Loading
W (lb)
x (ft)
y (ft)
xW(lb # ft)
yW(lb # ft)
1
2
3
4
450
1500
600
280
6
18
26
30
7
16
3
8
2700
27000
15600
8400
3150
24000
1800
2240
©
2830
53700
31190
600 lb
7 ft
280 lb
G3
G4
4 ft
5 ft
3 ft
12 ft
8 ft
x
Thus,
x =
©xW
53700
=
= 18.98 ft = 19.0 ft
©W
2830
Ans.
y =
©yW
31190
=
= 11.02 ft = 11.0 ft
©W
2830
Ans.
Ans:
x = 19.0 ft
y = 11.0 ft
965
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–88.
The assembly consists of a 20-in. wooden dowel rod and a
tight-fitting steel collar. Determine the distance x to its
center of gravity if the specific weights of the materials are
gw = 150 lb>ft3 and gst = 490 lb>ft3. The radii of the dowel
and collar are shown.
5 in.
5 in.
10 in.
G
x
x
2 in.
1 in.
SOLUTION
©xW =
E 10p (1)2 (20)(150) + 7.5p (5)(2 2 - 12)(490) F
1
(12)3
= 154.8 lb # in.
©W =
E p (1)2 (20)(150) + p (5)(2 2 - 12)(490) F
1
(12)3
= 18.82 lb
x =
©xW
154.8
=
= 8.22 in.
©W
18.82
Ans.
Ans:
x = 8.22 in.
966
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–89.
The composite plate is made from both steel (A) and brass
(B) segments. Determine the mass and location 1x, y, z2 of
its mass center G. Take rst = 7.85 Mg>m3 and
rbr = 8.74 Mg>m3.
z
A
225 mm
G
150 mm
B
150 mm
30 mm
SOLUTION
x
1
1
©m = ©rV = c 8.74 a (0.15)(0.225)(0.03)b d + c 7.85a (0.15)(0.225)(0.03) b d
2
2
+ [7.85(0.15)(0.225)(0.03)]
= C 4.4246 A 10 - 3 B D + C 3.9741 A 10 - 3 B D + C 7.9481 A 10 - 3 B D
= 16.347 A 10 - 3 B = 16.4 kg
2
1
(0.150) b (4.4246) A 10 - 3 B + a 0.150 + (0.150) b (3.9741) A 10 - 3 B
3
3
©xm = a0.150 +
+
Ans.
1
(0.150)(7.9481)(10 -3) = 2.4971(10 -3) kg # m
2
1
2
0.225
b(7.9481) A 10 - 3 B
©zm = a (0.225) b(4.4246) A 10 - 3 B + a (0.225)b (3.9471) A 10 - 3 B + a
3
3
2
= 1.8221 A 10 - 3 B kg # m
x =
2.4971(10 - 3)
©xm
=
= 0.153 m = 153 mm
©m
16.347(10 - 3)
Ans.
Due to symmetry:
Ans.
y = - 15 mm
z =
1.8221(10 - 3)
©zm
= 0.1115 m = 111 mm
=
©m
16.347(10 - 3)
Ans.
Ans:
Σm = 16.4 kg
x = 153 mm
y = - 15 mm
z = 111 mm
967
y
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–90.
Determine the volume of the silo which consists of a
cylinder and hemispherical cap. Neglect the thickness of
the plates.
10 ft 10 ft
10 ft
SOLUTION
V = ©u r A = 2p c
80 ft
4(10) 1
a b p (10)2 + 5(80)(10) d
4
3p
= 27.2 A 103 B ft3
Ans.
Ans:
V = 27.2 ( 103 ) ft 3
968
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–91.
Determine the outside surface area of the storage tank.
15 ft
4 ft
30 ft
SOLUTION
Surface Area: Applying the theorem of Pappus and Guldinus, Eq.9–7. with u = 2p,
L1 = 2152 + 4 2 = 2241 ft, L2 = 30 ft, r1 = 7.5 ft and r2 = 15 ft, we have
A = u©r~L = 2p C 7.5 A 2241 B + 15(30) D = 3.56 A 103 B ft2
Ans.
Ans:
A = 3.56 ( 103 ) ft 2
969
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–92.
Determine the volume of the storage tank.
15 ft
4 ft
30 ft
SOLUTION
Volume: Applying the theorem of Pappus and Guldinus, Eq. 9–8 with u = 2p,
1
r1 = 5 ft, r2 = 7.5 ft, A 1 = (15)(4) = 30.0 ft2 and A 2 = 30(15) = 450 ft2, we have
2
V = u©rA = 2p[5(30.0) + 7.5(450)] = 22.1 A 103 B ft3
Ans.
Ans:
V = 22.1 ( 103 ) ft 3
970
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–93.
Determine the surface area of the concrete sea wall,
excluding its bottom.
8 ft
SOLUTION
30 ft
Surface Area: Applying Theorem of Pappus and Guldinus, Eq. 9–9 with
5
50
bp =
p rad, L1 = 30 ft, L2 = 8 ft, L3 = 272 + 302 = 2949 ft,
u = a
180
18
N1 = 75 ft, N2 = 71 ft and N3 = 63.5 ft as indicated in Fig. a,
A 1 = u©NL =
60 ft
50
15 ft
5
p [75(30) + 71(8) + 63.5( 2949)]
18
= 4166.25 ft2
The surface area of two sides of the wall is
1
A 2 = 2 c (8 + 15)(30) d = 690 ft2
2
Thus the total surface area is
A = A 1 + A 2 = 4166.25 + 690
= 4856.25 ft2
= 4856 ft2
Ans.
Ans:
A = 4856 ft2
971
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–94.
A circular sea wall is made of concrete. Determine the total
weight of the wall if the concrete has a specific weight of
gc = 150 lb>ft3.
8 ft
30 ft
SOLUTION
V = ©ur~ A = a
60 ft
50
2
1
50°
b p[a 60 + (7)b a b (30) (7) + 71(30)(8)]
180°
3
2
15 ft
= 20 795.6 ft3
W = gV = 150(20 795.6) = 3.12(106) lb
Ans.
Ans:
W = 3.12 ( 106 ) lb
972
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–95.
A ring is generated by rotating the quartercircular area about
the x axis. Determine its volume.
a
SOLUTION
2a
Volume: Applying the theorem of Pappus and Guldinus, Eq. 9–10, with u = 2p,
4a
p
6p + 4
r = 2a +
=
a and A = a2, we have
3p
3p
4
V = urA = 2p
6p + 4
a
3p
p 2
a
4
=
p(6p + 4) 3
a
6
Ans.
x
Ans:
V =
973
p(6p + 4)
6
a3
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–96.
A ring is generated by rotating the quartercircular area
about the x axis. Determine its surface area.
a
SOLUTION
2a
Surface Area: Applying the theorem of Pappus and Guldinus, Eq. 9–11, with u = 2p ,
2(p + 1)
pa
5
, r1 = 2a, r 2 =
L1 = L3 = a, L2 =
a and r3 = a, we have
p
2
2
A = u©rL = 2p 2a(a) +
2(p + 1)
pa
5
a a
b + a (a)
p
2
2
x
= p(2p + 11)a2
Ans.
Ans:
A = p(2p + 11)a2
974
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–97.
Determine the volume of concrete needed to construct
the curb.
100 mm
150 mm
30
4m
150 mm
150 mm
Solution
p
p
1
V = Σu A r = a b[(0.15)(0.3)(4.15)] + a b c a b(0.15)(0.1)(4.25)d
6
6
2
V = 0.114 m3
Ans.
Ans:
V = 0.114 m3
975
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–98.
Determine the surface area of the curb. Do not include the
area of the ends in the calculation.
100 mm
150 mm
30
4m
150 mm
150 mm
Solution
A = ΣurL =
p
{4(0.15) + 4.075(0.15) + (4.15 + 0.075) ( 20.152 + 0.12 )
6
+ 4.3(0.25) + 4.15(0.3)}
A = 2.25 m2
Ans.
Ans:
A = 2.25 m2
976
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–99.
A ring is formed by rotating the area 360° about the x – x
axes. Determine its surface area.
80 mm
30 mm
50 mm
30 mm
100 mm
x
x
Solution
Surface Area. Referring to Fig. a, L1 = 110 mm, L2 = 2302 + 802 = 27300 mm
L3 = 50 mm, r1 = 100 mm, r2 = 140 mm and r3 = 180 mm. Applying the theorem
of pappus and guldinus, with u = 2p rad,
A = uΣrL
= 2p 3 100(110) + 2 (140) ( 17300 2 + 180 (50) 4
= 275.98 ( 103 ) mm2
= 276 ( 103 ) mm2
Ans.
Ans:
A = 276(103) mm2
977
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–100.
A ring is formed by rotating the area 360° about the x – x
axes. Determine its volume.
80 mm
30 mm
50 mm
30 mm
100 mm
x
x
Solution
1
(60)(80) = 2400 mm2, A2 = 50(80) = 4000 mm2,
2
r1 = 126.67 mm and r2 = 140 mm. Applying the theorem of pappus and guldinus,
Volume. Referring to Fig. a, A1 =
with u = 2p rad,
V = uΣrA
= 2p 3 126.67(2400) + 140(4000) 4
= 5.429 ( 106 ) mm3
= 5.43 ( 106 ) mm3
Ans.
Ans:
V = 5.43 ( 106 ) mm3
978
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–101.
The water-supply tank has a hemispherical bottom and
cylindrical sides. Determine the weight of water in the tank
when it is filled to the top at C. Take gw = 62.4 lb>ft3.
6 ft
C
8 ft
6 ft
SOLUTION
~ = 2p e 3182162 +
V = ©urA
4162
3p
1
a b 1p21622 f
4
V = 1357.17 ft3
Ans.
W = g V = 62.411357.172 = 84.7 kip
Ans:
W = 84.7 kip
979
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–102.
Determine the number of gallons of paint needed to paint
the outside surface of the water-supply tank, which consists
of a hemispherical bottom, cylindrical sides, and conical top.
Each gallon of paint can cover 250 ft2.
6 ft
C
8 ft
6 ft
SOLUTION
~ = 2p e 3 A 6 22 B + 6182 +
A = ©urL
2162
p
a
2162p
4
bf
= 687.73 ft2
Number of gal. =
687.73 ft2
= 2.75 gal.
250 ft2>gal.
Ans.
Ans:
Number of gal. = 2.75 gal
980
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–103.
Determine the surface area and the volume of the ring
formed by rotating the square about the vertical axis.
b
a
a
SOLUTION
~ = 2c 2p a b A = ©urL
= 4p cba -
45
a
a
sin 45°b(a) d + 2 c 2p a b + sin 45°b(a) d
2
2
a2
a2
sin 45° + ba +
sin 45° d
2
2
Ans.
= 8pba
Also
A = ©urL = 2p(b)(4a) = 8pba
~ = 2p(b)(a)2 = 2pba 2
V = ©urA
Ans.
Ans:
A = 8pba
V = 2pba2
981
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–104.
Determine the surface area of the ring. T he cross section is
circular as shown.
8 in.
4 in.
SOLUTION
'
A = ur L = 2p (3) 2p (1)
= 118 in.2
Ans.
Ans:
A = 118 in.2
982
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–105.
The heat exchanger radiates thermal energy at the rate of
2500 kJ h for each square meter of its surface area.
Determine how many joules (J) are radiated within a
5-hour period.
0.5 m
0.75 m
0.75 m
SOLUTION
A = ©u r L = (2p) B 2 a
0.75 + 0.5
b 2(0.75)2 + (0.25)2 + (0.75)(1.5) + (0.5)(1) R
2
0.75 m
= 16.419 m2
Q = 2500 A 103 B a
1.5 m
1m
h
#
J
m2
b A 16.416 m2 B (5 h) = 205 MJ
Ans.
0.5 m
Ans:
Q = 205 MJ
983
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–106.
Determine the interior surface area of the brake piston. It
consists of a full circular part. Its cross section is shown in
the figure.
40 mm
60 mm
80 mm
SOLUTION
20 mm
A = © u r L = 2 p [20(40) + 552(30)2 + (80)2 + 80(20)
+ 90(60) + 100(20) + 110(40)]
40 mm 30 mm 20 mm
A = 119(103) mm2
Ans.
Ans:
A = 119 ( 103 ) mm2
984
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–107.
The suspension bunker is made from plates which are
curved to the natural shape which a completely flexible
membrane would take if subjected to a full load of coal. This
curve may be approximated by a parabola, y = 0.2x2.
Determine the weight of coal which the bunker would
contain when completely filled. Coal has a specific weight of
g = 50 lb>ft3, and assume there is a 20% loss in volume due
to air voids. Solve the problem by integration to determine
the cross-sectional area of ABC; then use the second
theorem of Pappus–Guldinus to find the volume.
y
10 ft
C
B
20 ft
SOLUTION
y 0.2x2
x
x =
2
A
x
'
y = y
dA = x dy
20
y
3
2
dy =
y2 2 = 133.3 ft2
0
L0 C 0.2
3 20.2
20
LA
dA =
20
LA
x dA =
x =
L0
y
y2 20
2 = 500 ft3
dy =
0.4
0.8 0
x dA
LA
LA
dA
=
500
= 3.75 ft
133.3
V = u r A = 2p (3.75) (133.3) = 3142 ft3
W = 0.8 g V = 0.8(50)(3142) = 125 664 lb = 126 kip
Ans.
Ans:
W = 126 kip
985
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–108.
Determine the height h to which liquid should be poured
into the cup so that it contacts three-fourths the surface
area on the inside of the cup. Neglect the cup’s thickness for
the calculation.
40 mm
160 mm
h
Solution
Surface Area. From the geometry shown in Fig. a,
r
40
=
;
h
160
r =
1
h
4
1
217
1 2
2
h, Fig. b. Applying the theorem of pappus and
h and L =
a hb + h =
A 4
8
4
Guldinus, with u = 2p rad,
Thus, r =
1
217
p217 2
A = u Σr L = 2p a hba
hb =
h
8
4
16
For the whole cup, h = 160 mm. Thus
Ao = a
It is required that A =
p217
b 1 1602 2 = 1600p217 mm2
16
3
3
A = 1 1600p217 2 = 1200p217 mm2. Thus
4 o
4
1200p217 =
p217 2
h
16
Ans.
h = 138.56 mm = 139 mm
Ans:
h = 139 mm
986
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–109.
Determine the surface area of the roof of the structure if it
is formed by rotating the parabola about the y axis.
y
y
16
(x2/16)
16 m
SOLUTION
Centroid: The
= ¢
C
1 + a
x
length
of
the
differential
element
is
dL = 2dx2 + dy 2
16 m
dy
dy
x
b ≤ dx and its centroid is x = x. Here,
= - . Evaluating the
dx
dx
8
2
integrals, we nave
16 m
L =
LL
L
dL =
'
xdL =
L0
16 m
'
x¢
L0
¢
C
C
1 +
1 +
x2
≤ dx = 23.663 m
64
x2
≤ dx = 217.181 m2
64
Applying Eq. 9–5, we have
x =
LL
'
xdL
LL
=
dL
217.181
= 9.178 m
23.663
Surface Area: Applying the theorem of Pappus and Guldinus, Eq. 9–7, with u = 2p,
L = 23.663 m, r = x = 9.178, we have
A = urL = 2p(9.178) (23.663) = 1365 m2
Ans.
Ans:
A = 1365 m2
987
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–110.
A steel wheel has a diameter of 840 mm and a cross section
as shown in the figure. Determine the total mass of the
wheel if r = 5 Mg>m3.
100 mm
A
30 mm
60 mm
420 mm
250 mm
30 mm
840 mm
80 mm
SOLUTION
Volume: Applying the theorem of Pappus and Guldinus, Eq. 9–12, with u = 2p,
r1 = 0.095 m,
r2 = 0.235 m,
r3 = 0.39 m,
A 1 = 0.110.032 = 0.003 m2,
2
A 2 = 0.2510.032 = 0.0075 m and A 3 = 10.1210.062 = 0.006 m2, we have
A
Section A–A
V = u©rA = 2p30.09510.0032 + 0.23510.00752 + 0.3910.00624
= 8.775p110-32m3
The mass of the wheel is
m = rV = 51103238.775110-32p4
Ans.
= 138 kg
Ans:
m = 138 kg
988
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–111.
Half the cross section of the steel housing is shown in the
figure. There are six 10-mm-diameter bolt holes around its
rim. Determine its mass. The density of steel is 7.85 Mg m3.
The housing is a full circular part.
30 mm
20 mm
40 mm
10 mm
30 mm
10 mm
10 mm
10 mm
SOLUTION
V = 2p[ (40)(40)(10) + (55)(30)(10) + (75)(30)(10)] - 6 C p (5)2(10) D = 340.9 A 103 B mm3
m = rV = a7850
= 2.68 kg
kg
m3
b (340.9) A 103 B A 10-9 B m3
Ans.
Ans:
m = 2.68 kg
989
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–112.
y
The water tank has a paraboloid-shaped roof. If one liter of
paint can cover 3 m2 of the tank, determine the number of
liters required to coat the roof.
1 (144 x2)
y ––
96
x
2.5 m
SOLUTION
Length and Centroid: The length of the differential element shown shaded in
Fig. a is
dL = 2dx2 + dy2 =
where
dy
1
= - x. Thus,
dx
48
dL =
B
12 m
LL
dL =
1 + a
dy 2
b dx
dx
2
x2
1
1
x b dx =
1 + 2 dx =
2482 + x2 dx
48
B
48
48
1 + a-
Integrating,
L =
A
12 m
L0
1
2482 + x2 dx = 12.124 m
48
The centroid x of the line can be obtained by applying Eq. 9–5 with xc = x.
x =
LL
12 m
x~ dL
=
L0
xc
1
2482 + x2 dx d
48
73.114
=
= 6.031 m
12.124
12.124
dL
LL
Surface Area: Applying the first theorem of Pappus and Guldinus and using the
results obtained above with r = x = 6.031 m, we have
A = 2prL = 2p(6.031)(12.124) = 459.39 m2
Thus, the amount of paint required is
# of liters =
459.39
= 153 liters
3
Ans.
Ans:
153 liters
990
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–113.
Determine the volume of material needed to make the
casting.
2 in.
6 in.
SOLUTION
6 in.
Side View
4 in.
Front View
V = ©uAy
4(6)
4(2)
1
1
b + 2(6)(4) (3) - 2 a pb (2)2 a6 bd
= 2 p c2a pb(6)2 a
4
3p
2
3p
= 1402.8 in3
V = 1.40(103) in3
Ans.
Ans:
V = 1.40 ( 10 3 ) in3
991
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–114.
Determine the height h to which liquid should be poured into
the cup so that it contacts half the surface area on the inside
of the cup. Neglect the cup’s thickness for the calculation.
30 mm
50 mm
h
SOLUTION
10 mm
A = uz ~
rL = 2p{202(20)2 + (50)2 + 5(10)}
= 2p(1127.03) mm2
x =
20h
2h
=
50
5
2p b 5(10) + a 10 +
2h 2
h
1
b a b + h2 r = (2p)(1127.03)
5 B 5
2
10.77h + 0.2154h2 = 513.5
Ans.
h = 29.9 mm
Ans:
h = 29.9 mm
992
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–115.
p
The pressure loading on the plate varies uniformly along
each of its edges. Determine the magnitude of the resultant
force and the coordinates ( x , y) of the point where the line of
action of the force intersects the plate. Hint: The equation
defining the boundary of the load has the form p ax
by c, where the constants a, b, and c have to be determined.
40 lb/ft
30 lb/ft
20 lb/ft
10 lb/ft
SOLUTION
y
5 ft
10 ft
x
p = ax + by + c
At x = 0,
y = 0;
p = 40
40 = 0 + 0 + c;
At x = 5,
c = 40
y = 0,
p = 30
30 = a (5) + 0 + 40;
At x = 0;
a = -2
y = 10,
p = 20
20 = 0 + b (10) + 40;
b = -2
Thus,
p = - 2x - 2y + 40
5
FR =
p(x,y)dA =
LA
10
L0 L0
( -2x - 2y + 40) dy dx
= - 2 A 12(5)2 B (10) - 2 A 12(10)2 B 5 + 40(5)(10)
Ans.
= 1250 lb
5
xp(x,y) dA =
LA
10
2
( -2x - 2 yx + 40 x) dy dx
L0 L0
= - 2 A 13(5)2 B (10) - 2 A 12 (10)2 B A 12(5)2 B + 40 A 12(5)2 B (10)
= 2916.67 lb # ft
x =
xp(x,y) dA
LA
=
p(x,y)dA
2916.67
= 2.33 ft
1250
Ans.
LA
5
yp(x,y) dA =
LA
10
L0 L0
( -2x y - 2y2 + 40y) dy dx
= - 2 A 12(5)2 B A 12(10)2 B - 2 A 13(10)3 B (5) + 40(5) A 12(10)2 B
= 5416.67 lb # ft
y =
yp(x,y) dA
LA
p(x,y) dA
LA
=
5416.67
= 4.33 ft
1250
Ans.
Ans:
FR = 1250 lb
x = 2.33 ft
y = 4.33 ft
993
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–116.
p
The load over the plate varies linearly along the sides of the
plate such that p = (12 - 6x + 4y) kPa. Determine the
magnitude of the resultant force and the coordinates ( x, y )
of the point where the line of action of the force intersects
the plate.
12 kPa
18 kPa
x
1.5 m
6 kPa
2m
Solution
Centroid. Perform the double integration.
FR =
LA
r(x, y)dA =
=
=
L0
1.5 m
L0
1.5 m
L0
1.5 m
L0
2m
(12 -6x + 4y)dxdy
1 12x
2m
- 3x2 + 4xy 2 `
dy
0
(8y + 12)dy
= (4y2 + 12y) `
1.5 m
0
Ans.
= 27.0 kN
LA
xr(x, y)dA =
=
=
=
L0
1.5 m
L0
1.5 m
L0
1.5 m
L0
2m
1 12x
(6x2 - 2x3 + 2x2y) `
LA
=
=
2m
dy
0
(8y + 8)dy
1 4y2
+ 8y 2 `
= 21.0 kN # m
yr(x, y)dA =
- 6x2 + 4xy 2 dx dy
L0
1.5 m
L0
1.5 m
L0
1.5 m
L0
2m
1.5 m
0
1 12y
1 12xy
- 6xy + 4y2 2 dx dy
- 3x2y + 4xy2 2 `
(8y2 + 12y)dy
1.5 m
8
= a y3 + 6y2 b `
3
0
= 22.5 kN # m
994
2m
dy
0
y
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–116. Continued
Thus,
x =
y =
LA
xp(x, y)dA
LA
LA
=
21.0 kN # m
7
= m = 0.778 m 27.0 kN
9
Ans.
=
22.5 kN # m
= 0.833 m 27.0 kN
Ans.
p(x, y)dA
yp(x, y)dA
LA
p(x, y)dA
Ans:
FR = 27.0 kN
x = 0.778 m
y = 0.833 m
995
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–117.
The load over the plate varies linearly along the sides of the
plate such that p = 23[x14 - y2] kPa. Determine the resultant
force and its position 1x, y2 on the plate.
p
8 kPa
y
3m
x
4m
SOLUTION
Resultant Force and its Location: The volume of the differential element is
2
'
'
dV = d FR = pdxdy = (xdx)[(4 - y)dy] and its centroid is at x = x and y = y.
3
3m
FR =
LFk
=
dFR =
LFR
=
'
ydFR =
LFR
=
x =
L0
2 2
(x dx)
3
L0
(4 - y)dy
Ans.
4m
(4 - y)dy
y2 4 m
x3 3 m
2
b 2 R = 48.0 kN # m
a 4y Ba b 2
3
3 0
2 0
3m
L0
2
(xdx)
3
L0
4m
y(4 - y)dy
y3 4 m
2
x2 3 m
B a b 2 a 2y2 - b 2 R = 32.0 kN # m
3
2 0
3 0
'
xdFR
LFR
=
48.0
= 2.00 m
24.0
Ans.
=
32.0
= 1.33 m
24.0
Ans.
dFR
LFR
'
ydFR
y =
4m
y2 4 m
x2 3 m
2
a 4y b 2 R = 24.0 kN
Ba b 2
3
2 0
2 0
3m
xdFR =
L0
2
(xdx)
3
L0
LFR
dFR
LFR
Ans:
FR = 24.0 kN
x = 2.00 m
y = 1.33 m
996
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–118.
p
The rectangular plate is subjected to a distributed load over
its entire surface. The load is defined by the expression
p = p0 sin (px>a) sin (py>b), where p0 represents the
pressure acting at the center of the plate. Determine the
magnitude and location of the resultant force acting on
the plate.
p0
y
x
SOLUTION
a
b
Resultant Force and its Location: The volume of the differential element is
py
px
dy b .
dx b asin
dV = dFR = pdxdy = p0 a sin
a
b
a
FR =
dFR = p0
LFR
L0
a sin
= p0 B a =
b
py
px
dx b
a sin
dy b
a
b
L0
a
b
px 2 a
px 2 b
b R
cos
b a - cos
p
a
p
b
0
0
4ab
p0
p2
Ans.
Since the loading is symmetric, the location of the resultant force is at the center of
the plate. Hence,
x =
a
2
y =
b
2
Ans.
Ans:
FR =
4ab
p0
p2
a
2
b
y =
2
x =
997
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–119.
A wind loading creates a positive pressure on one side of
the chimney and a negative (suction) pressure on the other
side, as shown. If this pressure loading acts uniformly along
the chimney’s length, determine the magnitude of the
resultant force created by the wind.
p
p p0 cos u
Solution
FRx =
p
2l 2p
L- 2
u
l
(p0 cos u) cos u r du =
p
2
2rlp0
p
L- 2
p
= 2rlp0 a b
2
cos2 u du
Ans.
FRx = plrp0
FRy = 2l
p
2
p
L- 2
(p0 cos u) sin u r du = 0
Thus,
Ans.
FR = plrp0
Ans:
p
FRx = 2rlp0 a b
2
FR = plrp0
998
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–120.
When the tide water A subsides, the tide gate automatically
swings open to drain the marsh B. For the condition of high
tide shown, determine the horizontal reactions developed
at the hinge C and stop block D. The length of the gate is
6 m and its height is 4 m. rw = 1.0 Mg/m3.
C
4m
A
3m
B
2m
D
SOLUTION
Fluid Pressure: The fluid pressure at points D and E can be determined using
Eq. 9–13, p = rgz.
pD = 1.01103219.812122 = 19 620 N>m2 = 19.62 kN>m2
pE = 1.01103219.812132 = 29 430 N>m2 = 29.43 kN>m2
Thus,
wD = 19.62162 = 117.72 kN>m
wE = 29.43162 = 176.58 kN>m
Resultant Forces:
FR1 =
1
1176.582132 = 264.87 kN
2
FR2 =
1
1117.722122 = 117.72 kN
2
Equations of Equilibrium:
a + ©MC = 0;
264.87132 - 117.7213.3332 - Dx 142 = 0
Ans.
Dx = 100.55 kN = 101 kN
+ ©F = 0;
:
x
264.87 - 117.72 - 100.55 - Cx = 0
Ans.
Cx = 46.6 kN
Ans:
Dx = 101 kN
Cx = 46.6 kN
999
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–121.
The tank is filled with water to a depth of d = 4 m.
Determine the resultant force the water exerts on side A and
side B of the tank. If oil instead of water is placed in the tank,
to what depth d should it reach so that it creates the same
resultant forces? ro = 900 kg>m3 and rw = 1000 kg>m3.
2m
3m
A
B
d
SOLUTION
For water
At side A:
wA = b rw g d
= 2(1000)(9.81) (4)
= 78 480 N/m
FRA =
1
(78 480)(4) = 156 960 N = 157 kN
2
Ans.
At side B:
wB = b rw g d
= 3(1000)(9.81)(4)
= 117 720 N>m
FRB =
1
(117 720)(4) = 235 440 N = 235 kN
2
Ans.
For oil
At side A:
wA = b ro g d
= 2(900)(9.81)d
= 17 658 d
FRA =
1
(17 658 d)(d) = 156 960 N
2
Ans.
d = 4.22 m
Ans:
For water: FRA = 157 kN
FRB = 235 kN
For oil: d = 4.22 m
1000
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–122.
The concrete “gravity” dam is held in place by its own
weight. If the density of concrete is rc = 2.5 Mg>m3, and
water has a density of rw = 1.0 Mg>m3, determine the
smallest dimension d that will prevent the dam from
overturning about its end A.
1m
6m
A
d–1
d
Solution
Loadings. The computation will be based on b = 1 m width of the dam. The pressure
at the base of the dam is.
P = rgh = 1000(9.81)(6) = 58.86 ( 103 ) pa = 58.86 kPa
Thus
w = pb = 58.86(1) = 58.86 kN>m
The forces that act on the dam and their respective points of application, shown in
Fig. a, are
W1 = 2500 3 1(6)(1) 4 (9.81) = 147.15 ( 103 ) N = 147.15 kN
1
W2 = 2500c (d - 1)(6)(1) d (9.81) = 73.575 ( d - 1 )( 103 ) = 73.575(d - 1) kN
2
1
2
( FR )v = 1000c (d - 1)(6)(1) d (9.81) = 29.43 ( d - 1 )( 103 ) = 29.43(d - 1) kN
( FR ) h =
x1 = 0.5
y =
1
(58.86)(6) = 176.58 kN
2
x2 = 1 +
1
1
(d - 1) = (d + 2)
3
3
x3 = 1 +
2
1
(d - 1) = (2d + 1)
3
3
1
(6) = 2 m
3
Equation of Equilibrium. Write the moment equation of equilibrium about A by
referring to the FBD of the dam, Fig. a,
1
a+ΣMA = 0; 147.15(0.5) + [73.575(d - 1)]c (d + 2) d
3
1
+ [29.43(d - 1)] c (2d + 1) d - 176.58(2) = 0
3
44.145d 2 + 14.715d - 338.445 = 0
Solving and chose the positive root
Ans.
d = 2.607 m = 2.61 m
Ans:
d = 2.61 m
1001
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–123.
y
The factor of safety for tipping of the concrete dam is defined as the
ratio of the stabilizing moment due to the dam’s weight divided by
the overturning moment about O due to the water pressure.
Determine this factor if the concrete has a density of
rconc = 2.5 Mg>m3 and for water rw = 1 Mg>m3.
1m
6m
O
x
4m
Solution
Loadings. The computation will be based on b = 1m width of the dam. The pressure
at the base of the dam is
P = pwgh = 1000(9.81)(6) = 58.86(103)pa = 58.86 kPa
Thus,
w = Pb = 58.86(1) = 58.86 kN>m
The forces that act on the dam and their respective points of application, shown in
Fig. a, are
W1 = (2500)[(1(6)(1)](9.81) = 147.15(103)] N = 147.15 kN
1
W2 = (2500) c (3)(6)(1) d (9.81) = 220.725(103) N = 220.725 kN
2
FR =
1
(58.86)(6) = 176.58 kN
2
x1 = 3 +
1
2
1
(1) = 3.5 ft x2 = (3) = 2m y = (6) = 2 m
2
3
3
Thus, the overturning moment about O is
MOT = 176.58(2) = 353.16 kN # m
And the stabilizing moment about O is
Ms = 147.15(3.5) + 220.725(2) = 956.475 kN # m
Thus, the factor of safety is
F.S. =
Ms
956.475
=
= 2.7083 = 2.71
MOT
353.16
Ans.
Ans:
F.S. = 2.71
1002
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–124.
The concrete dam in the shape of a quarter circle.
Determine the magnitude of the resultant hydrostatic force
that acts on the dam per meter of length. The density of
water is rw = 1 Mg>m3.
3m
SOLUTION
Loading: The hydrostatic force acting on the circular surface of the dam consists of
the vertical component Fv and the horizontal component Fh as shown in Fig. a.
Resultant Force Component: The vertical component Fv consists of the weight
of water contained in the shaded area shown in Fig. a. For a 1-m length of dam,
we have
Fv = rgAABCb = (1000)(9.81) B (3)(3) -
p 2
(3 ) R (1) = 18947.20 N = 18.95 kN
4
The horizontal component Fh consists of the horizontal hydrostatic pressure.
Since the width of the dam is constant (1 m), this loading can be represented
by a triangular distributed loading with an intensity of wC = rghCb =
1000(9.81)(3)(1) = 29.43 kN>m at point C, Fig. a.
Fh =
1
(29.43)(3) = 44.145 kN
2
Thus, the magnitude of the resultant hydrostatic force acting on the dam is
FR = 3F h 2 + F v 2 = 344.1452 + 18.952 = 48.0 kN
Ans.
Ans:
FR = 48.0 kN
1003
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–125.
The tank is used to store a liquid having a density of
80 lb>ft3. If it is filled to the top, determine the magnitude of
force the liquid exerts on each of its two sides ABDC
and BDFE.
C
A
D
4 ft
6 ft
B
F
12 ft
3 ft
E
Solution
w1 = 80(4)(12) = 3840 lb>ft
w2 = 80(10)(12) = 9600 lb>ft
ABDC :
F1 =
1
(3840)(5) = 9.60 kip
2
Ans.
BDEF :
F2 =
1
(9600 - 3840)(6) + 3840(6) = 40.3 kip
2
Ans.
Ans:
F1 = 9.60 kip
F2 = 40.3 kip
1004
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–126.
y
The parabolic plate is subjected to a fluid pressure that
varies linearly from 0 at its top to 100 lb>ft at its bottom B.
Determine the magnitude of the resultant force and its
location on the plate.
2 ft
2 ft
y x2
4 ft
x
Solution
FR =
LA
p dA =
L0
= 2
4
(100 - 25y)(2x dy)
L0
4
B
1
(100 - 25y)ay 2 dyb
2 3
2 5 4
= 2c 100 a by 2 - 25a by 2 d = 426.7 lb = 427 lb
3
5
0
FR y =
LA
yp dA;
426.7 y = 2
L0
4
Ans.
1
y(100 - 25y)y 2 dy
2 5
2 7 4
426.7 y = 2c 100 a by 2 - 25a by2 d
5
7
0
426.7 y = 731.4
Ans.
y = 1.71 ft
Due to symmetry,
Ans.
x = 0
Ans:
FR = 427 lb
y = 1.71 ft
x = 0
1005
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–127.
The 2-m-wide rectangular gate is pinned at its center A and
is prevented from rotating by the block at B. Determine the
reactions at these supports due to hydrostatic pressure.
rw = 1.0 Mg>m3.
6m
1.5 m
A
B
SOLUTION
1.5 m
w1 = 1000(9.81)(3)(2) = 58 860 N/m
w2 = 1000(9.81)(3)(2) = 58 860 N/m
F1 =
1
(3)(58 860) = 88 290
2
F2 = (58 860)(3) = 176 580
a + ©MA = 0;
88 290(0.5) - FB (1.5) = 0
Ans.
FB = 29 430 N = 29.4 kN
+ ©F = 0;
:
x
88 290 + 176 580 - 29 430 - FA = 0
Ans.
FA = 235 440 N = 235 kN
Ans:
FB = 29.4 kN
FA = 235 kN
1006
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–128.
The tank is filled with a liquid which has a density of
900 kg>m3. Determine the resultant force that it exerts on
the elliptical end plate, and the location of the center of
pressure, measured from the x axis.
y
1m
1m
4 y2
x2
1
0.5 m
x
0.5 m
SOLUTION
Fluid Pressure: The fluid pressure at an arbitrary point along y axis can be
determined using Eq. 9–13, p = g(0.5 - y) = 900(9.81)(0.5 - y) = 8829(0.5 - y).
Resultant Force and its Location: Here, x = 21 - 4y2. The volume of the
differential element is dV = dFR = p(2xdy) = 8829(0.5 - y)[221 - 4y 2] dy.
Evaluating integrals using Simpson’s rule, we have
the
0.5 m
FR =
LFR
d FR = 17658
L-0.5 m
(0.5 - y)(21 - 4y2)dy
Ans.
= 6934.2 N = 6.93 kN
0.5 m
LFR
yd FR = 17658
y(0.5 - y)(21 - 4y2)dy
= - 866.7 N # m
'
ydFR
y =
L-0.5 m
LFR
=
dFR
- 866.7
= - 0.125 m
6934.2
Ans.
LFR
Ans:
FR = 6.93 kN
y = - 0.125 m
1007
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–129.
Determine the magnitude of the resultant force acting on
the gate ABC due to hydrostatic pressure. The gate has a
width of 1.5 m. rw = 1.0 Mg>m3.
1.5 m
B
1.25 m
C
SOLUTION
2m
w1 = 1000(9.81)(1.5)(1.5) = 22.072 kN>m
60°
A
w2 = 1000(9.81)(2)(1.5) = 29.43 kN>m
Fx =
1
(29.43)(2) + (22.0725)(2) = 73.58 kN
2
F1 = c (22.072) a 1.25 +
F2 =
2
b d = 53.078 kN
tan 60°
1
2
(1.5)(2)a
b (1000)(9.81) = 16.99 kN
2
tan 60°
Fy = F1 + F2 = 70.069 kN
F = 2F2x + F2y = 2(73.58)2 + (70.069)2 = 102 kN
Ans.
Ans:
F = 102 kN
1008
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–130.
The semicircular drainage pipe is filled with water.
Determine the resultant horizontal and vertical force
components that the water exerts on the side AB of the
pipe per foot of pipe length; gw = 62.4 lb>ft3.
A
2 ft
B
SOLUTION
Fluid Pressure: The fluid pressure at the bottom of the drain can be determined
using Eq. 9–13, p = gz.
p = 62.4(2) = 124.8 lb>ft2
Thus,
w
w = 124.8(1) = 124.8 lb>ft
Resultant Forces: The area of the quarter circle is A =
1 2
1
pr = p(2 2) = p ft2.
4
4
Then, the vertical component of the resultant force is
Ans.
FR v = gV = 62.4[p(1)] = 196 lb
and the horizontal component of the resultant force is
FR h =
1
(124.8)(2) = 125 lb
2
Ans.
Ans:
FRv = 196 lb
FRh = 125 lb
1009