# Force on the sides of a swimming pool, differeniation

• TFM
In summary, the conversation discusses the task of calculating the force exerted by water against either end of a swimming pool, given its dimensions. The initial solution presented involves turning the pool on its end, but it is later suggested to keep it in its original position. The formula used is force = (density)(gravity)(depth)(area), with an integral used to determine the area at the end of the pool. After some troubleshooting and corrections, the final answer is determined to be 156594.2 N, which is accepted by Mastering Physics as the correct answer.
TFM
[SOLVED] Force on the sides of a swimming pool, differeniation

## Homework Statement

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.

f = pgya

## 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:
Why doesn't it seem to work?

Edit: I think you need to integrate more carefully.

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

TFM

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:
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

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

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

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?

Mapes said:
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?

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

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

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 apparently isn't the answer)
TFM

Last edited:
x g?

x g

D'oh

With g in, gives right answer:

156594.2

Although Mastering Physics only accepts 160000

Thanks, Mapes

TFM

You're welcome, good luck.

## What causes the force on the sides of a swimming pool?

The force on the sides of a swimming pool is caused by the weight of the water and the pressure exerted by the water against the walls of the pool. This force is known as hydrostatic force and is dependent on the density of the water, the depth of the water, and the surface area of the pool.

## How does the force on the sides of a swimming pool differ from the force at the bottom?

The force on the sides of a swimming pool is different from the force at the bottom because the walls of the pool are vertical while the bottom is horizontal. This results in a difference in the direction and distribution of the force.

## What is differentiation in relation to the force on the sides of a swimming pool?

Differentiation is the mathematical process of finding the rate of change of a function. In the context of the force on the sides of a swimming pool, differentiation is used to calculate the change in force as the depth of the water changes.

## How does the depth of the water affect the force on the sides of a swimming pool?

The depth of the water directly affects the force on the sides of a swimming pool. As the depth of the water increases, so does the hydrostatic force, resulting in a greater force on the walls of the pool. This is because the weight and pressure of the water increase with depth.

## Can the force on the sides of a swimming pool be calculated?

Yes, the force on the sides of a swimming pool can be calculated using the equation F = ρgV, where F is the force, ρ is the density of the water, g is the acceleration due to gravity, and V is the volume of the water. However, this calculation may be more complex in real-life situations due to varying depth and shape of the pool.

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