Thomas Calculus 13th [Solutions]

Section 6.6 Moments and Centers of Mass 479
47. When the water reaches the top of the tank the force on the movable side is
0 2
2 (62.4) 2 4 y ( y)
dy
1/2 3/2
0
(62.4) 0 2 2 2 2 3/2
2
4 y ( 2 y ) dy (62.4) 3 4 y (62.4) 3
4 332.8 ft-lb. The force
2
compressing the spring is F 100x so when the tank is full we have 332.8 100 3.33 ft.
the movable end does not reach the required 5 ft to allow drainage the tank will overflow.
48. (a) Using the given coordinate system we see that the total
width L( y ) 3 and the depth of the strip is (3 y).
Thus,
3 3
F w(3 y) L( y) dy (62.4)(3 y) 3 dy
0 0
3
(62.4)(3) (3 y) dy (62.4)(3) 3y
0
y
2
2 3
0
x x Therefore
(62.4)(3) 9 9 (62.4)(3) 9 842.4 lb
2 2
(b) Find a new water level Y such that F Y (0.75)(842.4 lb) 631.8 lb. The new depth of the strip is ( Y y)
Y
Y
and Y is the new upper limit of integration. Thus, FY
w( Y y) L( y) dy 62.4 ( Y y) 3 dy
0 0
2 Y
(62.4)(3) Y 2
2 2
0
( ) (62.4)(3) y
Y y dy Yy
2 (62.4)(3) Y Y
2 (62.4)(3) Y
2
. Therefore,
0
2F Y Y 1263.6 6.75 2.598 ft. So, Y 3 Y 3 2.598 0.402 ft 4.8 in
(62.4)(3) 187.2
6.6 MOMENTS AND CENTERS OF MASS
1. Since the plate is symmetric about the y -axis and its density is
constant, the distribution of mass is symmetric about the y-axis
and the center of mass lies on the y -axis. This means that x 0.
M
It remains to find y x
. We model the distribution of mass
M
with vertical strips. The typical strip has center of mass:
( , ) , x
2 4
2
x y x , length: 4 x width: dx,
2
2
2
area: dA 4 x dx , mass: dm dA 4 x dx
2
2 4
The moment of the strip about the x -axis is y dm x 4 4 x dx 16 x dx . The moment of the
2 2
2 5 2
4
5 5
plate about the x -axis is M 16 16 x
2 2
x y dm x dx x
16 2 16 2
2 2 2 5
2
2 5 5
3 2
2 32 128
2
32 . The mass of the plate is M (4 x ) dx 4x
x 2 8 8 32 .
2 5 5
3
2
3 3
M
Therefore y x
12 The plates center of mass is the point x, y 0, 12 .
M
5
5
128
5
32
3
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