Question

A swimming pool has dimensions $30.0m\times 10.0m$and a flat bottom. When the pool is filled to a depth of $20.0m$with fresh water, what is the force exerted by the water on (a) the bottom? (b) On each end? (c) On each side?

Short answer

(a) The force exerted by the water on the bottom is $F_b=5.88\times {10}^{6}N\left(down\right)$.

(b) The force exerted by the water on each end is$F=196kN\left(outward\right)$ .

(c) The force exerted by the water on each side is$F=588kN\left(outward\right)$ .

Step-by-step solution

Pressure:

The pressure in a fluid is the force per unit area exerted by the fluid on a surface:

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

Where, P is the pressure, A is area, and F is the force exerted on the fluid.

(a) The force exerted by the water on the bottom:

The pressure on the bottom due to the water is

$$\begin{gathered} P_b=\rho gY \\ =1.96\times {10}^{4}Pa \end{gathered}$$

The force exerted by the water on the bottom is then

$$\begin{gathered} F_b=P_{b}A=P_b\left(L\times B\right) \\ =\left(1.96\times {10}^{4}Pa\right)\left(30.\text{0 m}\times 10.0\text{m}\right) \\ =5.88\times {10}^{6}N\left(down\right) \end{gathered}$$

Hence, the force exerted by the water on the bottom is $F_b=5.88\times {10}^{6}N\left(down\right)$.

(b) The force exerted by the water on each end:

Pressure varies with depth. On a strip of height dy and length L, the force

$$\begin{gathered} F=PdA \\ =PLdY \\ =\rho gYLdY \end{gathered}$$

Which gives the integral as below.

$$\begin{gathered} F=\underset{0}{\overset{h}{\int }}\rho ghLdY \\ =\frac{1}{2}\rho gL{h}^2 \\ =\left(\frac{1}{2}\rho gh\right)Lh \\ =P_{average}A \end{gathered}$$

On each end,

$$\begin{gathered} F=P_{average}A \\ =\left(9.8\times {10}^{3}Pa\right)\left(20.0m^2\right) \\ =196kN\left(outward\right) \end{gathered}$$

Hence, the force exerted by the water on each end isrole="math" localid="1663669149245" $196kN\left(outward\right)$ .

(c) The force exerted by the water on each side:

On the side,

$$\begin{gathered} F=P_{average}A \\ =\left(9.8\times {10}^{3}Pa\right)\left(60.0m^2\right) \\ =588kN\left(outward\right) \end{gathered}$$

Hence, the force exerted by the water on each side is$F=588kN\left(outward\right)$ .