# Homework Help: Fluid force in a pool

1. Dec 7, 2004

### agro

Let us define a pool. Viewed from top, we see a rectangle of sides 16 ft and 10 ft. When we look from the side (so the 16 ft side is perpendicular to our view), it is a trapezoid with width 16 ft and 4 ft, 16 ft being the height. It is roughly like this:

From top:

Code (Text):

16 ft
------------------
|                |
|                |
|                |  10 ft
------------------

From one side (and the other opposite side):

Code (Text):

16 ft
------------------
4ft |                |
....            |
....        |  8 ft
....    |
....|

The two other sides are rectangular (4 ft x 10 ft and 8 ft x 10 ft). The floor (OK, it's also a "side") is a rectangle with dimension 10 ft x $\sqrt{16^2+4^2}$ ft.

Let the pool be filled completely with water with weight density $\pi$ = 62.4 lb/ft^3.

The question is: what is the fluid force on the bottom of the pool?

Let us use the second image for this discussion. I define an vertical x axis with x = 0 located at 4 ft below the water surface and the positive x is below x = 0. I define h(x) = 4 + x as the height of water surface at a particular x location and w(x) = 10 as the width of the pool.

To find the fluid force F, I use the equation

$$F = \int_0^4\pi h(x)w(x)\sec t\;dx$$

Where t is the angle between the x axis and the inclined pool floor.

Since sec t = $\sqrt{16^2+4^2}/4$,

$$F = \int_0^4 (62.4) (4 + x) (10) \frac{\sqrt{16^2+4^2}}{4} dx$$

However the answer given by my book isn't the same as mine (the book's answer is bigger). Since I ruled out the possibility of miscalculation the integral, it must be my formulation of the integral (or the book's answer is wrong).

However I can't see what's wrong with my integral... I tried doing it "low level", formulating the Riemann Sums first, but it results in the integral I use.

What's wrong?

Thanks a lot.

2. Dec 8, 2004

### Staff: Mentor

I don't understand your choice of variable. I would pick x as measured along a horizontal axis, from x = 0 to x = 16. Then the water depth is h(x) = 4 + x/4.

If you rewrite your integral using the variable x as I have defined it, you should be OK.