91影视

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,

$F=蚁gV$

Here, $蚁$ 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,

$\begin{array}{l}V=\left(\text{5鈥尘}脳4\text{鈥尘}脳3\text{鈥尘}\right)\\ V=60{\text{鈥尘}}^3\end{array}$

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

$\begin{array}{c}F=\left(1000\text{鈥塳驳}/{\text{m}}^{\text{3}}\right)脳\left(9.8\text{鈥尘}/s^{\text{2}}\right)脳\left(60{\text{鈥尘}}^3\right)\left(\frac{1\text{鈥�}N}{\text{1鈥塳驳}鈰�\text{m}/{\text{s}}^{\text{2}}}\right)\\ =588000\text{鈥塏}\end{array}$

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,

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

Here,$蚁gh$ is the gauge pressure.

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

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

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 蚁, 3 m for d in the above equation,

$\begin{array}{c}F=\frac{1}{2}\left(1000\text{鈥塳驳}/{\text{m}}^{\text{3}}\right)\left(9.8\text{鈥尘}/{\text{s}}^{\text{2}}\right)\left(4\text{鈥尘}\right){\left(3\text{鈥尘}\right)}^2\left(\frac{1\text{鈥塏}}{1\text{鈥塳驳}鈰�\text{m}/{\text{s}}^{\text{2}}}\right)\\ =176400\text{鈥塏}\end{array}$

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