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?

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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?

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Answer 1 · 2015-02-28T14:25:44

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

$p = p_{0} + ρ g h$ $p=p_0+rhogh$

Where:

$p$ $p$ is your required pressure at depth $h$ $h$ ( $= 10 f t = 3.05 m$ $=10ft=3.05m$ ), $g = 9.81 \frac{m}{s^{2}}$ $g=9.81m/(s^2)$ is acceleration of gravity and $p_{0}$ $p_0$ is the pressure at the surface of water ( $1 a t m = 1.013 \cdot 10^{5} P a$ $1 atm=1.013*10^5Pa$ ).
The density of water should be: $ρ = 1000 \frac{k g}{m^{3}}$ $rho=1000 (kg)/m^3$
So you should get:
$p = 1.013 \cdot 10^{5} + (1000 \cdot 9.81 \cdot 3.05) = 1.312 \cdot 10^{5} P a ≅ 1.3 a t m$ $p=1.013*10^5+(1000*9.81*3.05)=1.312*10^5Pa~=1.3atm$
and:
$p r e s s u r e = \frac{f or c e}{a r e a}$ $pressure=(force)/(area)$

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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?

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