Finding the force of water in a pool on a vertical wall

[chenying]

Homework Statement

A swimming pool has the dimensions 24 m multiplied by 9.0 m multiplied by 2.5 m.

(a) When it is filled with water, what is the force (resulting from the water alone) on the bottom, on the short sides, and on the long sides?

Homework Equations

F = P*A

P = rho * g * h

The Attempt at a Solution

Ok, I have no idea what I am doing wrong, but I am trying to find the force of water on the short vertical side of the pool.

So, since pressure varies by height, I have to intergrate the pressure by height (h).

So P = $$\int\rho g dh$$

and A = h*w where w = width of the pool

So F = $$\int\rho g dh h*w$$

then you end up with $$\rho*w*g*(h^2)/2$$ from 0 to 9

and get 992250 Pa

I'm pretty sure what I did was correct, but I am not getting the right answer.

Can anyone help?

 

[ideasrule]

Seems good to me. Are you sure you didn't get your dimensions mixed up? That is, are you sure you used the right values for w and h?

 

[cepheid]

Well, for one thing, 2.5 m is the only sensible choice for the depth of the pool! If either of the other two is the depth, then that is one strange pool.

 

[chenying]

Why would 2.5 be the sensible height? i thought it would be the width of the pool.

 

[denverdoc]

Who has swum in a pool of 30' depth?

Edit: one other thing bothers me about the solution, the pressure at 0' depth is not 0 pressure but atmospheric pressure which I think should be accounted for if we are interested in the total pressure exerted on any of the surfaces, or am I confused?

 

[chenying]

Who has swum in a pool of 30' depth?

Edit: one other thing bothers me about the solution, the pressure at 0' depth is not 0 pressure but atmospheric pressure which I think should be accounted for if we are interested in the total pressure exerted on any of the surfaces, or am I confused?

Oh, I guess that is rather large...

But yes, you are right, atmospheric pressure is the pressure at the surface, but the problem states to not take that into account.

 

[denverdoc]

very well, I would try using 2.5 as the depth and 9 as the width as Cepheid suggests. See if you get the right answer

 

[chenying]

I did and came out with the right answer. Thanks a lot!

 

[denverdoc]

Thank Cepheid, he caught the error, but you're welcome.