Consider a rectangular swimming pool of depth 1 m with a width and length of 11 m. What is the total force on the bottom of the pool due to the water pressure?

Where the weight density of water

h=the depth of the pool and A is the area of the surface.

I didn't know that term; weight density... I'd have done that;



P is pressure
h is depth
is density of water
g acceleration due to gravity

Then ,



where P is pressure
F is force
A is area of base of pool.

Shape, width and length of the pool are of no importance. It should be approximately 1000 kg/cu.m

No, pressure is independent of shape, width and length, but force sure does.

If the pool is flat, which I can assume since the OP only states shape, then the pressure/weight of the water is evenly distubed. The answert I gave is the weight of a c/m of water. Do you see this differently?

Oh, now I see.. you're using units of force per area, and using that unit to find the total force! Ok, then I guess it's right.

That's why I said the shape, width and length have no bearing. You basically have a whole series of 1 c/m boxes of water sitting on a concrete slab. Each box weights the same and the gravitational pull is the same so I got to stick with 1000 kg/cu.m I have to leave out any consideration for temperature and purity of the water which all can change the weight.

Force on the bottom of the pool is given by Newton's per square meter.

1000 kg/m^3 is the weight density of water. That is, it weighs 1000 kilograms per cubic meter. Multiplied by gravity it is 9810 N/m^3 as the force.
There is more than a cubic meter of water in the pool. There are 121 m^3. Then, we have to multiply that by the weight per unit volume and by gravity.

Pressure is

Force is

The FORCE on the bottom of the pool is

Pressure is force per unit area. So, we have


Pascal's principle say's that fluid pressure is the same is all directions.

Force is a different matter. That is why we use integration a lot of times to find force when we have a vertically submerged (or at an angle) surface.

Now, find the force on the sides.