A swimming pool is 5.0 m long, 4.0 m wide, and 3.0 m deep. Compute the force exerted by the water against (a) the bottom and (b) either end. (Hint: Calculate the force on a thin, horizontal strip at a depth h, and integrate this over the end of the pool.) Do not include the force due to air pressure.

The dimensions of the swimming pool are 5.0 m, 4.0 m, 3.0 m.

Part (a)

The force acting on the bottom of the swimming pool will be equal to the total weight of the water. The weight of the water is given by,

$F=\rho gV$

Here, $\rho$ is the density of the fluid, g is the acceleration due to gravity whose value is 9.8 m/s², V is the volume of the swimming pool.

The volume of the swimming pool will be,

$\begin{array}{l}V=\left(\text{5\hspace{0.17em}m}\times 4\text{\hspace{0.17em}m}\times 3\text{\hspace{0.17em}m}\right)\\ V=60{\text{\hspace{0.17em}m}}^3\end{array}$

Substitute 60 m³ for V, 9.8 m/s² for g, 1000 kg/m³ for ρ in the above equation,

$\begin{array}{c}F=\left(1000\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}\right)\times \left(9.8\text{\hspace{0.17em}m}/s^{\text{2}}\right)\times \left(60{\text{\hspace{0.17em}m}}^3\right)\left(\frac{1N}{\text{1\hspace{0.17em}kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =588000\text{\hspace{0.17em}N}\end{array}$

Thus, the force exerted by the water at the bottom is 588000 N.

Part (b)

The pressure exerted by the water may vary vertically so in order to find the force at either end of the pool. Consider a thin horizontal strip of length x equal placed between depth h and dh.

The force acting on the strip will be,

$\begin{array}{c}F=PA\\ =\left(\rho gh\right)\left(xdh\right)\end{array}$

Here,$\rho gh$ is the gauge pressure.

For net force, integrate the force on each horizontal strip.

$\begin{array}{c}F=\underset{0}{\overset{d}{\int }}\left(\rho gh\right)\left(xdh\right)\\ F=\rho gx{\left[\frac{h^2}{2}\right]}_0^d\\ F=\frac{1}{2}\rho gx{d}^2\end{array}$

Here, xis the length of the strip equals to the width of the pool, d is total depth of the pool.

Substitute 4m for x, 9.8 m/s² for g, 1000 kg/m³ for ρ, 3 m for d in the above equation,

$\begin{array}{c}F=\frac{1}{2}\left(1000\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}\right)\left(9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\right)\left(4\text{\hspace{0.17em}m}\right){\left(3\text{\hspace{0.17em}m}\right)}^2\left(\frac{1\text{\hspace{0.17em}N}}{1\text{\hspace{0.17em}kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =176400\text{\hspace{0.17em}N}\end{array}$

Thus, the force acting on the either ends of the swimming pool will be 176400 N.