Question

Find the hydrostatic force exerted on one of the long sides of a rectangular swimming pool that is 20 feet long, 12 feet wide, and 6 feet deep.

Short answer

Ans: The hydrostatic force is$269,568$pounds.

Step-by-step solution

Step 1. Given information.

given,

Find the hydrostatic force exerted on one of the long sides of a rectangular swimming pool that is 20 feet long, 12 feet wide, and 6 feet deep.

Step 2. The objective is to calculate the hydrostatic force exerted on one of the long sides of a rectangular swimming pool that is 20 feet long, 12 feet wide, and 6 feet deep.

Consider that the top of the tank is at height $y=6$and the bottom of the tank is at height $y=0$.

Draw a diagram that shows a thin representative slice of the tank at some point $y_k*$ from the bottom.

Step 3. Now,

Assume that the entire thin slice of the wall is at a depth of $d_k=6-y_k^{\ast }$units.

The area of the representative wall slice is $A_k=240△y$square feet.

The water density is $\omega =62.4$pounds per cubic foot.

The hydrostatic force exerted by the water of weight-density $\omega$and depth $d$on a horizontal wall of area $A$is given by $F=\omega Ad$.

Substitute $A_k=240\mathrm{\Delta }y,d_k=6-y_k^{\ast }$and $\omega =62.4$in $F=\omega Ad$to obtained $F=62.4\left(6-y_k^{\ast }\right)240\mathrm{\Delta }y$.

Step 4. Therefore, the hydrostatic force exerted on one of the long sides of the hot tub is F=62.46−yk∗240Δy.

The hydrostatic force on the entire sidewall is approximately $F=\sum _{k=1}^n 62.4\left(6-y_k^{\ast }\right)240\mathrm{\Delta }y$

As role="math" localid="1649309762152" $n\to \mathrm{\infty },F=\sum _{k=1}^n 62.4\left(6-y_k^{\ast }\right)240\mathrm{\Delta }y$becomes a definite integral.

Accumulate the slices from $y=0toy=6$in order to obtain the hydrostatic force on the entire sidewall.

$\begin{array}{r}W={\int }_0^6 62.4\left(240\right)(6-y)dy\\ =62.4\left(240\right){\int }_0^6 (6-y)dy\\ =62.4\left(240\right)\left[6{\int }_0^6 dy-{\int }_0^6 ydy\right]\end{array}$

Step 5. The power rule for differentiation states that ∫xndx=xn+1n+1, where n is a real number.

Use the power rule to evaluate the integral $62.4\left(8\right)\left[6{\int }_0^6 dy-{\int }_0^6 ydy\right]$

role="math" localid="1649310053685" $\begin{array}{r}W=62.4\left(240\right)\left[6{\int }_0^6 dy-{\int }_0^6 ydy\right]\\ =62.4\left(240\right){\left[6y-\frac{y^2}{2}\right]}_0^6\\ =62.4\left(240\right)\left[6\left(6\right)-\frac{(6{)}^2}{2}\right]\\ =269,568\end{array}$

The hydrostatic force is $269,568$pounds.