Force in Varying Places in a Swimming Pool

[Okazaki]

Homework Statement

A swimming pool has dimensions 25 m x 10 m x 3m (length x width x height.) When it is filled with water, what is the force on the bottom (F_b) of the pool? On the long side (F_l)? On the short sides (F_w)? (note that integrals are required.) If you are concerned with whether or not the concrete walls and floor will collapse, is it appropriate to take the atmospheric pressure into account?

Homework Equations

d = m/v
F = ρA

The Attempt at a Solution

I'm having a lot of trouble setting up the integral. I legitimately am kind of lost on how to do it. Would you just do

F = ∫ ρA (from the top of the pool to the bottom). And would you use the density of the entire pool for ρ?

 

[SteamKing]

Hint: Use Pascal's Law for this problem.

And how is the density of the entire pool different from the density of any random unit volume of water in the pool?

 

[haruspex]

F = ∫ ρA (from the top of the pool to the bottom). And would you use the density of the entire pool for ρ?

Are you using ρ for pressure, density, or mixing the two?

 

[Okazaki]

Ok, so I tried again:

F_bottom = ∫ (p₀ + ρ_density*l*w)dh
=p₀ * l * w * h + ρ_density*l*w*h
for the bottom of the pool, the integral doesn't matter so much.

...but...I'm confused. How do you find p₀ (which is pressure at the surface of the liquid due to the atmosphere) and even if you do find it, the units won't add up?

 

[haruspex]

the units won't add up

That's because this term is wrong:

ρ_density*l*w

ρ_density*l*w.dh gives a mass.
What equation do you know for pressure at a depth in a liquid?

 

[Okazaki]

That's because this term is wrong:

ρ_density*l*w.dh gives a mass.
What equation do you know for pressure at a depth in a liquid?

That's because this term is wrong:

ρ_density*l*w.dh gives a mass.
What equation do you know for pressure at a depth in a liquid?

only p = p₀ + ρ_density*g*h. It's literally the only equation my book gives.

 

[haruspex]

only p = p₀ + ρ_density*g*h. It's literally the only equation my book gives.

That's right, but it's not what you posted before.
p₀ also features wrongly in these two equations:

Fbottom = ∫ (p₀ + ρ_density*l*w)dh
=p₀ * l * w * h + ρ_density*l*w*h

The first one would give p₀'s contribution to the force at the bottom as p₀h, a pressure multiplied by a distance, while in the second you have a pressure multiplied by a volume.
The force at the bottom should be easy, no need for integrals. Take the whole body of water as a unit. What are the vertical forces acting on it?

 

[Okazaki]

That's right, but it's not what you posted before.
p₀ also features wrongly in these two equations:

The first one would give p₀'s contribution to the force at the bottom as p₀h, a pressure multiplied by a distance, while in the second you have a pressure multiplied by a volume.
The force at the bottom should be easy, no need for integrals. Take the whole body of water as a unit. What are the vertical forces acting on it?

The vertical forces would just be the atmosphere, and the weight of water above it?

But what about the sides?

 

[haruspex]

The vertical forces would just be the atmosphere, and the weight of water above it?

One more... what stops it descending?

But what about the sides?

Those would be horizontal forces.

 

[Okazaki]

One more... what stops it descending?

What do you mean by "what stops it descending"?

Those would be horizontal forces.

How do I solve for that?

 

[haruspex]

What do you mean by "what stops it descending"?

I asked about the forces acting on the body of water. You mentioned atmosphere and its own weight (gravity). Both of those act downwards.
However, I guess you instead tried to answer what forces act on the floor of the pool. The short answer to that is the pressure of the water. (The pool floor doesn't know anything about the atmosphere, and feels no direct force from it. The force is transferred by the water.)
But in the end, yes, the force on the floor is the sum of the force from atmospheric pressure and the weight of the water. So what do those add up to?

Note that this shortcut works because the pool has vertical sides. The more correct approach is to figure out the total pressure at the floor of the pool (using the equation you posted in post#6) and multiply by the pool floor area. That works for any pool profile.

 

[Okazaki]

I asked about the forces acting on the body of water. You mentioned atmosphere and its own weight (gravity). Both of those act downwards.
However, I guess you instead tried to answer what forces act on the floor of the pool. The short answer to that is the pressure of the water. (The pool floor doesn't know anything about the atmosphere, and feels no direct force from it. The force is transferred by the water.)
But in the end, yes, the force on the floor is the sum of the force from atmospheric pressure and the weight of the water. So what do those add up to?

Note that this shortcut works because the pool has vertical sides. The more correct approach is to figure out the total pressure at the floor of the pool (using the equation you posted in post#6) and multiply by the pool floor area. That works for any pool profile.

But
F_bottom = 32349779.5 N

 

[Okazaki]

But
F_bottom = 32349779.5 N

Does that mean that
F_l = ∫ (p₀ + ρ*g*h)A*dl
= p₀ ∫A * dl + ρ*g*h∫A dl
...None of those units add up?

 

[haruspex]

But F_bottom = 32349779.5 N

Are you saying that's the given answer or what you calculated? Looks about right to me. Please post your working.

 

[haruspex]

Does that mean that
F_l = ∫ (p₀ + ρ*g*h)A*dl

p₀ + ρ*g*h is the pressure at the bottom.
(p₀ + ρ*g*h)A is the force on the bottom.
What does it mean to integrate that dl (whatever l is)?

 

[Okazaki]

Are you saying that's the given answer or what you calculated? Looks about right to me. Please post your working.

F = (p₀ + ρ*g*h)A
= A * (1.0 x 10⁵ kg/(m*s²) + 999.97 kg/m³ * 9.8 m/s² * 3 m)
= 250 m² * 129399.118 kg/(m*s²)
= 32349779.5 N

 

[Okazaki]

p₀ + ρ*g*h is the pressure at the bottom.
(p₀ + ρ*g*h)A is the force on the bottom.
What does it mean to integrate that dl (whatever l is)?

I don't know. I figured I would integrate with respect to length here.

 

[haruspex]

F = (p₀ + ρ*g*h)A
= A * (1.0 x 10⁵ kg/(m*s²) + 999.97 kg/m³ * 9.8 m/s² * 3 m)
= 250 m² * 129399.118 kg/(m*s²)
= 32349779.5 N

Yes, except that the precision makes no sense. The atmospheric pressure is only accurate to two or three decimal places since it varies, so an answer should look more like 3.23*10⁸N.

I don't know. I figured I would integrate with respect to length here.

By what logic? And what do you mean by length here... the length of the pool or its depth? And what are you trying to calculate with this integral (i.e. what do you mean by F_l)?

 

[Okazaki]

Yes, except that the precision makes no sense. The atmospheric pressure is only accurate to two or three decimal places since it varies, so an answer should look more like 3.23*10⁸N.

By what logic? And what do you mean by length here... the length of the pool or its depth? And what are you trying to calculate with this integral (i.e. what do you mean by F_l)?

Well, part of the question asks to find the Force on the long side of the pool (F_l). I only really guessed here.

And the precision thing makes sense. It's just, the integrals don't.

 

[haruspex]

Well, part of the question asks to find the Force on the long side of the pool (F_l). I only really guessed here.

For the forces on the sides you will need integrals, but you can't use the force on the floor as the integrand. The force on the side won't depend on the pool area.
What is the pressure at depth x? What force does that exert on a wall over a horizontal strip of height dx?

 

[Okazaki]

For the forces on the sides you will need integrals, but you can't use the force on the floor as the integrand. The force on the side won't depend on the pool area.
What is the pressure at depth x? What force does that exert on a wall over a horizontal strip of height dx?

That would give you the exact same equation I had before.

F_l = ∫(p0 + ρdensity*g*h)A
from 0 to 3 m

 

[haruspex]

That would give you the exact same equation I had before.

F_l = ∫(p0 + ρdensity*g*h)A
from 0 to 3 m

Where h and A are what, exactly?

 

[Okazaki]

Where h and A are what, exactly?

A is area and h is height or depth.

 

[haruspex]

A is area and h is height or depth.

No, I said exactly, i.e. in the present context. The area of what, and what is it in terms of pool dimensions and the depth variable? h is the depth of what?

 

[Okazaki]

No, I said exactly, i.e. in the present context. The area of what, and what is it in terms of pool dimensions and the depth variable? h is the depth of what?

h is the depth of the pool (so 3 m) and the Area would be length x width (l x w or 25 m * 10 m)

 

[haruspex]

h is the depth of the pool (so 3 m) and the Area would be length x width (l x w or 25 m * 10 m)

No.
The pressure is different at different heights, so you cannot simply multiply by the entire area of a wall. Worse, you have hA in your integral, and according to what you specified above that would be the whole volume.

We need an expression for the force on a thin strip of wall, the full length of the wall, at some depth x, and of height dx. For small dx, we can take the pressure to be uniform over that area. Please post an expression for that force.

 

[Okazaki]

No.
The pressure is different at different heights, so you cannot simply multiply by the entire area of a wall. Worse, you have hA in your integral, and according to what you specified above that would be the whole volume.

We need an expression for the force on a thin strip of wall, the full length of the wall, at some depth x, and of height dx. For small dx, we can take the pressure to be uniform over that area. Please post an expression for that force.

F_l = ∫(ρ*g*h * l *dh)
from depth 0 m to depth 3 m
l*dh being the area of the long side of the pool

 

[haruspex]

F_l = ∫(ρ*g*h * l *dh)
from depth 0 m to depth 3 m
l*dh being the area of the long side of the pool

Much better, except l*dh is just the area of a strip of height dh, yes? Now integrate that.

 

[Okazaki]

Much better, except l*dh is just the area of a strip of height dh, yes? Now integrate that.

F_l = ∫(ρ*g*h * l *dh)
= ρ*g*l * ∫h dh
= ρ*g*l * h²/2
= 999.97 kg/m³ * 9.8 m/s² * 25 m * (3m)²/2

Which, when you evaluate it all:
= ~1.1 x 10⁶ N

And the formula for F_w is basically the same, just with w x dh instead of l x dh

 

[haruspex]

F_l = ∫(ρ*g*h * l *dh)
= ρ*g*l * ∫h dh
= ρ*g*l * h²/2
= 999.97 kg/m³ * 9.8 m/s² * 25 m * (3m)²/2

Which, when you evaluate it all:
= ~1.1 x 10⁶ N

And the formula for F_w is basically the same, just with w x dh instead of l x dh

looks right.