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danyzavala · May 18, 2009, 01:47 PM

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?

galactus · May 18, 2009, 01:55 PM

$F={\rho}hA$

Where the weight density of water ${\rho}=9810 \;\ N/m^{3}$

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

Unknown008 · May 19, 2009, 12:35 PM

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

$P=h{\rho}g$

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

Then ,

$P=\frac{F}{A}$

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

ballengerb1 · May 19, 2009, 01:23 PM

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

Unknown008 · May 19, 2009, 01:25 PM

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

ballengerb1 · May 19, 2009, 01:27 PM

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?

Unknown008 · May 19, 2009, 01:32 PM

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.

ballengerb1 · May 19, 2009, 01:34 PM

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.

galactus · May 19, 2009, 02:04 PM

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 $p=\frac{F}{A}={\rho}h$

Force is $F={\rho}hA$

The FORCE on the bottom of the pool is $(9810 \;\ \frac{N}{\sout{m^{3}}})(121 \;\ \sout{m^{3}})=1,187,010 \;\ N$

Pressure is force per unit area. So, we have
${\rho}h=(9810)(1)=9810 \;\ \frac{N}{m^{2}}=9810 \;\ Pa$

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.