Calculating Force on the Bottom and Sides of a Swimming Pool

In summary, the force on the horizontal sliver of area dA at a depth h is dF = (2.1m)(width of the slice)+(0m)(depth of the slice) = 2.1 N
  • #1
abpandanguyen
33
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Homework Statement


A swimming pool has dimensions 28.0 m 12.0 m and a flat bottom. When the pool is filled to a depth of 2.10 m with fresh water, what is the force caused by the water on the bottom?

On each end? (The ends are 12.0 m.)

On each side? (The sides are 28.0 m.)

Homework Equations


d = density
P = dgV
P = F/A

The Attempt at a Solution



I got my first answer and it was solved by using the above equation
(28.0)(12.0)(2.10)m3 x (1000) kg/m3 x 9.80 m/s2

= 6914880 N

for each end I am using
(12.0)(2.10)m2 x (9.80)m/s2 x (1000) kg/m3 =
which comes around to 246960
It wants the answers in kN and it says my answer is within 10% of the correct answer. I did the same for the third part of the problem and got the same response. I even looked at the solutions manual and they seemed to follow the same procedure so I am not sure what I am doing wrong.
 
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  • #2
Thats the correc tmethod for the bottom.

But think about the pressure on the sides and the ends.
Whats the water pressure at the bottom of the sides?
What about the top (at the surface of the water) ?
 
  • #3
Sorry... I am not really following you. I think I'm having a hard time picturing the setting in question. I am thinking of this question as it was asking for the pressure the water had on the side as in a rectangle. Am I thinking of this right?
 
  • #4
Yes, but on the bottom you have the full depth (and mass) of water pressing down.

On the sides you have a varying height of water, how much water is pressing down (and out) at the bottom compared to the surface?
 
  • #5
The units of your answer for the sides are wrong. It doesn't come out to N/m^2.

What mgb is getting at is that the pressure varies with depth, so you generally can't just take the pressure at an arbitrary depth and multiply it by the area to get the total force.
 
  • #6
The question is force not pressure
so the answer should be width (m)* depth(m) * height (m) * g (m/s^2) * density (kg/m^3) = kg m/s^2 = N

(remember, F = ma = kg m/s^2 = N )
 
  • #7
mgb_phys said:
The question is force not pressure
so the answer should be width (m)* depth(m) * height (m) * g (m/s^2) * density (kg/m^3) = kg m/s^2 = N

(remember, F = ma = kg m/s^2 = N )
D'oh! That's what I meant. The second calculation didn't come out to Newtons, but rather to N/m.
 
  • #8
Okay so now I am doing this

1000 x 9.80 x 2.10 to get Pascals (N/m2 = kg/(m)(s)2)

I am then multiplying this number by the respective areas of the sides and ends to get Newtons but if my original answers were within 10% of the correct answer, then these are definitely way out of the ballpark.
 
  • #9
I am still slightly confused ;_; I think of my idea of how pressure is distributed is skewed.
 
  • #10
You calculated the pressure at the bottom of the pool, where the pressure is the highest, and the pressure at the top is 0. So your calculation is going to give you an overestimate of the force on the wall.

Think of a horizontal sliver of area dA at a depth h. Calculate the force dF on it, and then integrate from the top to the bottom of the pool.
 
  • #11
Another way to think of it (if you haven't done calculus)
At the top of the side (just at the water surface) the pressure is zero
At the bottom the pressure is 2.1m as you worked out

Now, what would the average pressure be.
 
  • #12
Okay, so I'm basically adding the pressure at the top and the bottom, then dividing it by two (averaging it)? That makes sense if its right >.>.
 
  • #13
Yes. It works because the pressure varies linearly with depth.
 
  • #14
Okay, that makes sense now. Thank you very much!
 

FAQ: Calculating Force on the Bottom and Sides of a Swimming Pool

What is pressure over an area?

Pressure over an area, also known as pressure per unit area or surface pressure, is a measure of the force applied over a certain area. It is commonly used in physics and engineering to describe the distribution of force or weight over a surface.

How is pressure over an area calculated?

Pressure over an area is calculated by dividing the force (in newtons) applied over a surface by the area (in square meters) that the force is being applied to. The formula for pressure is P = F/A, where P is pressure, F is force, and A is area.

What is the SI unit for pressure over an area?

The SI unit for pressure over an area is pascal (Pa). 1 pascal is equal to 1 newton per square meter (N/m²). Other common units for pressure include pounds per square inch (psi) and atmospheres (atm).

What are some real-world examples of pressure over an area?

There are many examples of pressure over an area in everyday life. Some common examples include the pressure exerted by a person's feet while standing, the pressure of air in a balloon, the pressure of water in a swimming pool, and the pressure of a car's tires on the road.

How does pressure over an area affect objects?

Pressure over an area can affect objects in different ways. For example, increased pressure over an area can cause objects to compress or deform. It can also create buoyant forces, which can make objects float or sink. Pressure over an area also plays a crucial role in determining the stability and integrity of structures and materials.

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