James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 8.3 Applications to Physics and Engineering 559
Îx
_4 15
0
x i
* 15
16
x
(a)
15
a
10
interval f0, 16g into sub intervals of equal length with endpoints x i and we choose
x i * [ fx i21 , x i g. The ith horizontal strip of the dam is approximated by a rectangle with
height Dx and width w i , where, from similar triangles in Figure 3(b),
a
16 2 x i * − 10
20
or a − 16 2 x i*
2
− 8 2 x i*
2
and so w i − 2s15 1 ad − 2s15 1 8 2 1 2 x i*d − 46 2 x i *
10
If A i is the area of the ith strip, then
a
20
16-x i
*
(b)
FIGURE 3
A i < w i Dx − s46 2 x i *d Dx
If Dx is small, then the pressure P i on the ith strip is almost constant and we can use
Equation 1 to write
P i < 1000tx i *
The hydrostatic force F i acting on the ith strip is the product of the pressure and the
area:
F i − P i A i < 1000tx i *s46 2 x i *d Dx
Adding these forces and taking the limit as n l `, we obtain the total hydrostatic force
on the dam:
F − lim
nl `
o n
1000tx i *s46 2 x i *d Dx − y 16
1000txs46 2 xd dx
i−1
0
− 1000s9.8d y 16
s46x 2 x 2 d dx − 9800F23x 2 2 x 3
0
16
3G0
< 4.43 3 10 7 N n
Example 2 Find the hydrostatic force on one end of a cylindrical drum with radius
3 ft if the drum is submerged in water 10 ft deep.
10
7
d i
y
y i *
œ9-(y œ„„„„„„„ i
*)@
Îy
SOLUTION In this example it is convenient to choose the axes as in Figure 4 so that
the origin is placed at the center of the drum. Then the circle has a simple equation,
x 2 1 y 2 − 9. As in Example 1 we divide the circular region into horizontal strips of
equal width. From the equation of the circle, we see that the length of the ith strip is
2s9 2 sy i *d 2 and so its area is
A i − 2s9 2 sy i *d 2 Dy
0
≈+¥=9
x
Because the weight density of water is − 62.5 lbyft 3 , the pressure on this strip is
approximately
d i − 62.5s7 2 y i *d
FIGURE 4
and so the force on the strip is approximately
d i A i − 62.5s7 2 y i *d 2s9 2 sy i *d 2 Dy
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