Force in Varying Places in a Swimming Pool

In summary, the conversation discusses the calculation of the force on the bottom, long sides, and short sides of a swimming pool when filled with water. The solution involves using integrals and the equation p = p0 + ρ*density*g*h to calculate the pressure at different depths. It is also mentioned that the force on the floor can be found by multiplying the total pressure at the bottom by the pool floor area.
  • #1
Okazaki
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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 (Fb) of the pool? On the long side (Fl)? On the short sides (Fw)? (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?
 
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  • #2
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?
 
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  • #3
Okazaki said:
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?
 
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  • #4
Ok, so I tried again:

Fbottom = ∫ (p0 + ρdensity*l*w)dh
=p0 * l * w * h + ρdensity*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?
 
  • #5
Okazaki said:
the units won't add up
That's because this term is wrong:
Okazsaki said:
ρdensity*l*w
ρdensity*l*w.dh gives a mass.
What equation do you know for pressure at a depth in a liquid?
 
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  • #6
haruspex said:
That's because this term is wrong:

ρdensity*l*w.dh gives a mass.
What equation do you know for pressure at a depth in a liquid?
haruspex said:
That's because this term is wrong:

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

only p = p0 + ρdensity*g*h. It's literally the only equation my book gives.
 
  • #7
Okazaki said:
only p = p0 + ρdensity*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:
Okazaki said:
Fbottom = ∫ (p0 + ρdensity*l*w)dh
=p0 * l * w * h + ρdensity*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?
 
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  • #8
haruspex said:
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?
 
  • #9
Okazaki said:
The vertical forces would just be the atmosphere, and the weight of water above it?
One more... what stops it descending?
Okazaki said:
But what about the sides?
Those would be horizontal forces.
 
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  • #10
haruspex said:
One more... what stops it descending?

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

haruspex said:
Those would be horizontal forces.

How do I solve for that?
 
  • #11
Okazaki said:
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.
 
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  • #12
haruspex said:
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
Fbottom = 32349779.5 N
 
  • #13
Okazaki said:
But
Fbottom = 32349779.5 N
Does that mean that
Fl = ∫ (p0 + ρ*g*h)A*dl
= p0 ∫A * dl + ρ*g*h∫A dl
...None of those units add up?
 
  • #14
Okazaki said:
But Fbottom = 32349779.5 N
Are you saying that's the given answer or what you calculated? Looks about right to me. Please post your working.
 
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  • #15
Okazaki said:
Does that mean that
Fl = ∫ (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)?
 
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  • #16
haruspex said:
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
 
  • #17
haruspex said:
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.
 
  • #18
Okazaki said:
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.
Okazaki said:
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 Fl)?
 
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  • #19
haruspex said:
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 Fl)?

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

And the precision thing makes sense. It's just, the integrals don't.
 
  • #20
Okazaki said:
Well, part of the question asks to find the Force on the long side of the pool (Fl). 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?
 
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  • #21
haruspex said:
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.

Fl = ∫(p0 + ρdensity*g*h)A
from 0 to 3 m
 
  • #22
Okazaki said:
That would give you the exact same equation I had before.

Fl = ∫(p0 + ρdensity*g*h)A
from 0 to 3 m
Where h and A are what, exactly?
 
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  • #23
haruspex said:
Where h and A are what, exactly?

A is area and h is height or depth.
 
  • #24
Okazaki said:
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?
 
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  • #25
haruspex said:
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)
 
  • #26
Okazaki said:
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.
 
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  • #27
haruspex said:
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.

Fl = ∫(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
 
  • #28
Okazaki said:
Fl = ∫(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.
 
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  • #29
haruspex said:
Much better, except l*dh is just the area of a strip of height dh, yes? Now integrate that.

Fl = ∫(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 Fw is basically the same, just with w x dh instead of l x dh
 
  • #30
Okazaki said:
Fl = ∫(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 Fw is basically the same, just with w x dh instead of l x dh
looks right.
 
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Related to Force in Varying Places in a Swimming Pool

1. What is force in a swimming pool?

Force in a swimming pool refers to the amount of pressure or resistance exerted on an object or person in the water. It is typically measured in units of Newtons (N).

2. How does force vary in different areas of a swimming pool?

The force in a swimming pool can vary depending on the depth and location of the object or person in the water. For example, the force will be greater at the bottom of the pool compared to the surface due to the weight of the water above.

3. What factors can affect force in a swimming pool?

The force in a swimming pool can be affected by various factors, such as the depth of the water, the density of the water, the speed of the object or person moving through the water, and any external forces such as wind or currents.

4. How is force related to buoyancy in a swimming pool?

Force and buoyancy are closely related in a swimming pool. Buoyancy is the upward force that water exerts on an object, and it is directly proportional to the force of gravity on the object. In other words, the greater the force of gravity, the greater the buoyant force, and the easier it is for an object to float in the water.

5. How can force in a swimming pool be measured?

Force in a swimming pool can be measured using a device called a force meter or a spring scale. These tools can be used to measure the amount of force exerted on an object or person in the water, and are typically calibrated in units of Newtons (N).

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