# How do you find the total force exerted on the five exposed faces of a cube that is on a side lying on the bottom of a swimming pool that is 20ft by 15ft by 10ft deep filled with water?

Feb 28, 2015

I guess you can use the hydrostatic pressure at a depth h:

$p = {p}_{0} + \rho g h$

Where:

$p$ is your required pressure at depth $h$ ($= 10 f t = 3.05 m$), $g = 9.81 \frac{m}{{s}^{2}}$ is acceleration of gravity and ${p}_{0}$ is the pressure at the surface of water ($1 a t m = 1.013 \cdot {10}^{5} P a$).
The density of water should be: $\rho = 1000 \frac{k g}{m} ^ 3$
So you should get:
$p = 1.013 \cdot {10}^{5} + \left(1000 \cdot 9.81 \cdot 3.05\right) = 1.312 \cdot {10}^{5} P a \cong 1.3 a t m$
and:
$p r e s s u r e = \frac{f \mathmr{and} c e}{a r e a}$