Question

Calculate the (absolute) pressure at the bottom of a neighborhood swimming pool 30.0 m by 8.0 m whose uniform depth is 2.0 m. The pressure at the top of the pool is atmospheric pressure which is 1.013?105 Pa.
Find the (total) force exerted on the bottom of this swimming pool.
What is the pressure against the side of this pool (somewhere very near the bottom)?

Answer 1

Answer:

Total pressure= 120945[Pa]

Force exerted = 29026800 [N] or 29.02*10^6 [N]

Explanation:

We know that the total pressure is the result of the sum of the atmospheric pressure plus the manometric pressure. The equation is:

$Ptotal=Patm + Pman$

In this problem we know the atmospheric pressure 101.325x10^3 [Pa], therefore we need to find the manometric pressure.

The manometric pressure in the bottom of the swimming pool depends only on the water column of water generated (depth of the swimming pool)

$Pman = density*g*h$

where:

density = density of the water 1000 [kg/m^3]

g= gravity [m/s^2]

h= column of water (meters)

replacing the values:

$Pman= 1000 *9.81* 2 = 19620 [Pa]$

The total pressure will be:

$Ptotal= 101325+19620 = 120945 [Pa]$

The force exerte on the bottom is defined by the following expression:

$Pressure=Force/area\\\\Force= Pressure*Area\\\\Area = 30m*8m= 240 m^2Force= 120945*240\\Force= 29026800N or 2958 Ton$