# Force on the sides of a swimming pool, differeniation

1. Feb 16, 2008

### TFM

[SOLVED] Force on the sides of a swimming pool, differeniation

1. The problem statement, all variables and given/known data

A swimming pool measures a length of 5.7m , width 3.8m , and depth 2.9m. Compute the force exerted by the water against either end. Do not include the force due to air pressure.

I have already calculated the force on the bottom as 6.2*10^5N

Hint: Calculate the force on a thin, horizontal strip at a depth , and integrate this over the end of the pool.

2. Relevant equations

f = pgya

3. The attempt at a solution

pretend the swimming pool is on its end, using formula
f = intergration,/' pgdya
from 0 to 5.7

getting F = 1/2 pg(y^2)a

But this doesn't seem to work?

Any help/suggestions/ideas

TFM

Last edited: Feb 16, 2008
2. Feb 16, 2008

### Mapes

Why doesn't it seem to work?

Edit: I think you need to integrate more carefully.

Last edited: Feb 16, 2008
3. Feb 16, 2008

### TFM

I get an answer of 1754395.02, which Mastering physics says is wrong. Any idea where in my integration I have gone wrong?

TFM

4. Feb 16, 2008

### Mapes

I think I see the problem. Don't turn the pool on its end; that's changing the problem. We want to find the force on either end in its orginal configuration. This will affect a couple of your variables.

Last edited: Feb 16, 2008
5. Feb 16, 2008

### TFM

Keeping the Pool in the same way, I get F = 892586.94
However, if i ignore the 1/2 from the integration, I get 1785173.88

Is this the answer, because there are two sides?

TFM

6. Feb 16, 2008

### Mapes

I don't think so. Can you show how you integrated and what variables you multiplied together to get this result?

7. Feb 16, 2008

### TFM

I integrated:

f = intergration,/' p*g*y*a dy
and got
F = 1/2 pg(y^2)a

I then used
p = 1000 (density of water)
g = 9.8
y = 2.9
a = 5.7*3.8

Giving me 892586.94

TFM

8. Feb 16, 2008

### Mapes

Right away I can tell that the units are off: your units are N-m for a force answer, which is incorrect of course. That's why checking units is so useful is troubleshooting a problem.

I agree with setting force equal to (density)(g)(depth)(area). However, it seems to me that your area should be a rectangular differential element at the end of the pool. This differential element would have a width of w and a height of dy. Does this make sense?

9. Feb 16, 2008

### TFM

I'm not sure...? Would it be:
Area = width*length
Area = ,/' width*length dlength
giving area = 1/2 * width*length^2
putting bvack into the equation,

F = pgya

F = pgy(1/2 * width*length^2)

If not, then I'm don't understand, sorry

TFM

10. Feb 16, 2008

### Mapes

How about $dF=pgy\,da=pgy(w\,dy)$, $F=\int_0^{\mathrm{depth}}dF=\frac{1}{2}pgw(\mathrm{depth})^2$?

11. Feb 16, 2008

### TFM

That makes sense - I was still using the area as being the floor. I now get an answer of 15979, alkthough this is ten times smaller than the force on the floor? (and apparantly isn't the answer)
TFM

Last edited: Feb 16, 2008
12. Feb 16, 2008

### Mapes

x g?

13. Feb 16, 2008

### TFM

D'oh

With g in, gives right answer:

156594.2

Although Mastering Physics only accepts 160000

Thanks, Mapes

TFM

14. Feb 16, 2008

### Mapes

You're welcome, good luck.