# Hydrostatic force

1. Nov 18, 2007

### chocolatelover

Hi everyone,

Could you please tell me if the setup is correct?

1. The problem statement, all variables and given/known data
A circularswimming pool has a diameter of 24 feet, the sides are 5ft hight, and the depth of the water is 4 ft. How much work is required to pump all of the water out over the side? (Use the fact that water weighs 62. lb/ft^3)

2. Relevant equations
w=(force)(distance)

3. The attempt at a solution

force=(62.5 lb/ft^3)(9.8m/s^2)=
6.12.598

distance=24ft

work=int. 0 to 24(612.598)(24)ydy

Thank you very much

2. Nov 18, 2007

### dynamicsolo

You're going to have to explain your reasoning on how you came up with that work integral, because it's not clear that you have the right picture of how you are removing the water.

Consider that the cross section of the pool at every depth is a circle 24 feet in diameter. The water at the surface has to be lifted one foot to get it over the side of the pool, while the water at the bottom will need to be lifted five feet. How much water at any level y below the surface of the water or the edge of the pool (your choice) has to be lifted? (How do you figure that out?) How much work will it take to lift that layer of water to the edge of the pool? What does the integral to find the total amount of work need to look like?

3. Nov 18, 2007

### chocolatelover

Thank you very much

Could you show me how to set up the problem, please?

Thank you very much

4. Nov 18, 2007

### dynamicsolo

Think of an infinitesimally thin layer of water that is a distance y down from the top edge of the pool. It has the area of a circle of radius 12 feet, so A = (pi)·(12^2) ft^2. It has a thickness dy, so this layer has an infinitesimal volume dV = A dy ft^3. The weight of this layer is dw = (density of water) · dV = 62.5 · dV lbs.

To get this layer up out of the pool will take an infinitesimal amount of work

dW = dw · y ft-lbs ,

where y is the vertical distance we need to lift it in order to raise it to the edge of the pool (we assume it takes negligible work to shift it horizontally out from the pool).

So dW = y · (62.5) · (pi) · (12^2) · dy ft-lbs , putting all these pieces together.

We now need to add all these infinitesimal bits of work to find the total work it takes to empty the pool over the side. So that's where we need to integrate. What does that integral for W look like?

5. Nov 19, 2007

### chocolatelover

Thank you very much

Would this work?

weight=force=(volume)(weight of water)
=(3.1412^2deltayft^3)(62.5 lb/ft^3)
=(0900)(3.14lb)

distance=yft

work=int. 0 to 4 (9000)(3.14)ydy
=(226194.67 ft)(pound)

Thank you very much