Expert: Socrates - 3/2/2010

Question
A swimming pool is 5 feet deep at one end and 10 feet deep at the other, and the bottom is an inclined plane. The length and width of the pool are 40 feet and 20 feet. If the pool is full of water, what is the fluid force on each of the vertical walls?

Answer
You will need to draw some pictures for this . The pool has 4 sides . Two sides are rectangles. One rectangle has length 20ft and height 5 ft . The other rectangle has length 20 ft and height 10ft . The other two sides are identical quadrilaterals. Each quadrilateral has a 40ft horizontal top , vertical sides with lengths 10ft and 5 ft , a sloping bottom having length
5√65.

Water at depth h feet exerts pressure equal to 62.43h lbs per square foot.

For the 20 by 5 rectangle ,imagine the rectangle divided into horizontal strips , with the midpoint of the strip at depth h ,the width of rectangle dh and the length 20.
The total pressure on this rectangle will be estimated by (62.43h)(20)(dh).
Use an integral to "sum up" these pressures as h varies from 0 to 5

S(62.43h)(20)dh from 0 to 5 =

S 1248.6 h dh from 0 to 5 =

15,607.5 lbs

The other rectangular side is done the same way ,except h varies from 0 to 10

S(62.43h)(20)dh from 0 to 10 =

S 1248.6 h dh from 0 to 10 =

62,430 lbs

The quadrilateral side is more difficult because the length of the strip varies with the depth once the depth is below 5 ft. The quadrilateral side can be divided into two pieces ,a rectangle upper portion with length 40 and height 5 . A right triangle lower portion with horizontal leg 40, vertical leg 5 and sloping bottom hypotenuse 5√65

First get the pressure on the rectangular part.

S(62.43h)(40)dh from 0 to 5 =

31,215 lbs

Add this to the pressure on the right triangle part when h is greater than 5.

To find this pressure , use similar triangles to find that the length at depth h will be

40 - 4h

so the integral for pressure will be

S (62.43h)(40-4h)dh  from 5 to 10 =

S 2497.2 h - 249.72 h^2 dh from 5 to 10 =

= 20,810

adding this to the pressure on the rectangular part

20,810 + 31,215 = 52,025 lbs.


To summarize , the pressures on the walls are

15,607.5 lbs ,  62,430 lbs , 52,025 lbs and 52,025 lbs