pressure...

[SoccaCrazy24]

An above-ground backyard swimming pool is shaped like a large hockey puck, with a circular bottom and a vertical wall forming its perimeter. The diameter of the pool is 4.9 m and its depth is 1.3 m. Find the total outward force exerted on the vertical wall of the pool.

so P= F / A
and i used P as Pat = 101000
and the area i found as surface area = (pi)(d)(h) = 20.01
so i multiplied P * A to equal F
20.01*101000 = 2021 kN .... where did i go wrong???

[Fermat]

The force on the wall comes from the pressure of the fluid acting against that wall.
What fluid is pressing against the wall ?
What is the pressure at any point in that fluid ?

[SoccaCrazy24]

well in the problem it does not say any liquid is in there so the only fluid i can think of is the atmospheric pressure or pressure of air which is 101000 N/m2 ...... is this right?

[Fermat]

Hmmm, you're right. There is no mention of water. I assumed that since it was a swimming pool it would be filled with water!

And if it is only air in the pool, then I can't see where you have gone wrong either.

Do you know the right answer ?

[SoccaCrazy24]

umm no.... but there is a problem with similar numbers that vary maybe by a little bit such as a height of 1.4 m and diameter of 4.8 m im not sure... but the answer was 160 kN .... i can get you the information for the problem that comes out to be 160 kN tomorrow...

[Fermat]

OK, thanks.

[SoccaCrazy24]

well if anyone else can help please feel free... im a little bit lost on this one...

[FredGarvin]

I would think that you'll have to assume that it's filled with water. The thing about that is that the outward force is going to be dependent on depth. The force distribution should be triangular with the max pressure at the bottom going to zero at the surface. The average force can then be calculated.

Check out page 7 and 8 here:
http://www.cbu.edu/~bbbeard/313/Day_05.pdf

[Fermat]

Just noticed something. The pool is above ground.

That means that, if there were no water in it, then both the inside and outside of the pool wall be subjected to atmospheric pressure, which would cancel - leaving a net force of zero acting on the vertical wall surface.
So it looks like it must be force from the pressure of water only.

The pressure in the water will depend upon its depth and also the pressure of the atmosphere on its surface. But the outside of the pool wall is also subject to atmosphere, so the effects of atmosperic pressure will cancel out and you will only need to work out the force on the wall from the water.
Have alook at that pdf file FredGarvin attached.

Also, doing it this way, I got 145 kN/m2 for that 160 Kn/m² problem with the dimensions you gave. But at least it's a reasonable ballpark figure as opposed to 2021 Kn/m²