Question

A trough is filled with water and its vertical ends have the shape of the parabolic region in the figure. Find the hydrostatic force on one end of the trough.

Short answer

The hydrostatic force on one end of the trough is \({\rm{2,133}}{\rm{.33}}\;{\rm{lb}}\).

Step-by-step solution

Explanation of the Solution

Given:

The parabolic shape of the trough has the width of\({\rm{8ft}}\)and the height of\({\rm{4ft}}\).

Calculation:

Draw the parabolic shape of the trough with the dimensions as shown in Figure.

Refer Figure.

The parabolic curve makes an equation of\({\rm{y = }}\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{4}}}\)…(1)

Rearrange Equation (1) as shown below:

\(\begin{aligned}{\rm{4y}} = {{\rm{x}}^{\rm{2}}}\\{\rm{x}} = {\rm{2}}\sqrt {\rm{y}} \end{aligned}\)

Let's divide the parabolic region into horizontal strips of equal width.

Let the horizontal strip is approximated by a rectangle of dimensions height\({\rm{\Delta y}}\)and width\({{\rm{w}}_{\rm{i}}}\)

From the parabolic curve equation,

The width of the\({i^{th}}\)strip\(\left( {{{\rm{w}}_{\rm{i}}}} \right)\)is\({\rm{2}}\left( {{\rm{2}}\sqrt {\rm{y}} } \right)\)and depth\({{\rm{d}}_{\rm{i}}} = {\rm{4 - }}{{\rm{y}}_{\rm{i}}}^{\rm{*}}\)

Calculate the area of the\({i^{th}}\)rectangular strip as follows:

\({{\rm{A}}_{\rm{i}}} = {{\rm{w}}_{\rm{i}}}{\rm{\Delta y}}\)…(2)

Substitute\({\rm{2}}\left( {{\rm{2}}\sqrt {\rm{y}} } \right)\)for\({{\rm{w}}_{\rm{i}}}\)in Equation (2).

\({{\rm{A}}_{\rm{i}}}{\rm{ = 2}}\left( {{\rm{2}}\sqrt {\rm{y}} } \right){\rm{\Delta y}}\)

Calculating the Force

Calculate the pressure of the \({i^{th}}\) rectangular strip \(\left( {{{\rm{P}}_{\rm{i}}}} \right)\) as follows:

\({{\rm{P}}_{\rm{i}}} = {\rm{\gamma }}{{\rm{d}}_i}\)… (3)

Here, the unit weight of water is\({\rm{\gamma }}\)and depth of strip is\({{\rm{d}}_{\rm{i}}}\).

Let the unit weight of water is\({\rm{\gamma }} = {\rm{62}}{\rm{.5lb/f}}{{\rm{t}}^{\rm{3}}}\)

Substitute\({\rm{62}}{\rm{.5 lb/f}}{{\rm{t}}^{\rm{3}}}\)for \({\rm{\gamma }}\)and \({\rm{4 - }}{{\rm{y}}_{\rm{i}}}^{\rm{*}}\)for \({\rm{i}}\)Equation (3).

The expression to find the hydrostatic force on one end of the trough is shown below:

Here, the lower limit is\({\rm{a}}\)and the upper limit is\({\rm{b}}\)

Substitute\({\rm{0}}\)for \({\rm{a}}\), \({\rm{4ft}}\) for \({\rm{b}}\), \({\rm{62}}{\rm{.5(4 - y)}}\) for \({{\rm{P}}_{\rm{i}}}\), and \({\rm{2(2}}\sqrt {\rm{y}} {\rm{)dy}}\)for \({\rm{A}}{}_{\rm{i}}\)in equation (4).

\(\begin{aligned}{\rm{F}} = \int\limits_{\rm{0}}^{\rm{4}} {{\rm{62}}{\rm{.5}}\left( {{\rm{4 - y}}} \right){\rm{2}}\left( {{\rm{2}}\sqrt {\rm{y}} } \right){\rm{dy}}} \\ = {\rm{4}}\left( {{\rm{62}}{\rm{.5}}} \right)\int\limits_{\rm{0}}^{\rm{4}} {\left( {{\rm{4 - y}}} \right)} \sqrt {\rm{y}} {\rm{dy}}\\ = {\rm{250}}\int\limits_{\rm{0}}^{\rm{4}} {\left( {{\rm{4}}{{\rm{y}}^{\frac{{\rm{1}}}{{\rm{2}}}}}{\rm{ - }}{{\rm{y}}^{\frac{{\rm{3}}}{{\rm{2}}}}}} \right)} {\rm{dy}}\;\;\;\;\;\;\;\;\;\;\;\;...\left( 5 \right)\end{aligned}\)

Integrate Equation (5) and apply the limits.

\(\begin{aligned}{\rm{F}} = {\rm{250}}\left( {{\rm{4}}\frac{{{{\rm{y}}^{\frac{{\rm{3}}}{{\rm{2}}}}}}}{{\frac{{\rm{3}}}{{\rm{2}}}}}{\rm{ - }}\frac{{{{\rm{y}}^{\frac{{\rm{5}}}{{\rm{2}}}}}}}{{\frac{{\rm{5}}}{{\rm{2}}}}}} \right)_{\rm{0}}^{\rm{4}}\\ = {\rm{250}}\left( {\frac{{\rm{8}}}{{\rm{3}}}{{\rm{y}}^{\frac{{\rm{3}}}{{\rm{2}}}}}{\rm{ - }}\frac{{\rm{2}}}{{\rm{5}}}{{\rm{y}}^{\frac{{\rm{5}}}{{\rm{2}}}}}} \right)_{\rm{0}}^{\rm{4}}\\ = {\rm{250}}\left( {\left( {\frac{{\rm{8}}}{{\rm{3}}}{{\left( {\rm{4}} \right)}^{\frac{{\rm{3}}}{{\rm{2}}}}}{\rm{ - }}\frac{{\rm{2}}}{{\rm{5}}}{{\left( {\rm{4}} \right)}^{\frac{{\rm{5}}}{{\rm{2}}}}}} \right){\rm{ - }}\left( {\rm{0}} \right)} \right)\\ = {\rm{250}}\left( {\frac{{{\rm{64}}}}{{\rm{3}}}{\rm{ - }}\frac{{{\rm{64}}}}{{\rm{5}}}} \right)\\ = {\rm{2,133}}{\rm{.33 lb}}\end{aligned}\)

Hence, the hydrostatic force on one end of the trough is \({\rm{2,133}}{\rm{.33 lb}}\)