A trough, whose cross section is a trapezoid, measures 2 m across at the bottom and 4 m across at the top, and is 2 m deep. If the trough is filled with a liquid of mass density $\rho$, what is the force due to hydrostatic pressure on one end of the trough?

Solution We position the trough in a rectangular coordinate system, as shown in Figure 62.

Figure 62

The sides of the end of the trough are lines that pass through the points $(0,2),$ $(1,0)$ and the points $(3,0),$ $(4,2)$, respectively. The equations of these lines are $$\begin{equation*} y-2=\dfrac{0-2}{1-0}(x-0)=-2x\qquad\hbox{or equivalently}\qquad x=g(y)=-\dfrac{1}{2}y+1 \end{equation*}$$

and $$\begin{equation*} y-0=\dfrac{2-0}{4-3}(x-3)=2x-6\qquad\hbox{or equivalently}\qquad x=f(y)=\dfrac{1}{2}y+3 \end{equation*}$$

Then the hydrostatic force on the $i$th interval of the trough is $$\begin{equation*} F_{i}=\,\underset{\color{#0066A7}{\hbox{Weight}}}{\underbrace{\rho \cdot g\,}}\underset{\color{#0066A7}{\hbox{ Depth}}}{\underbrace{( H-v_{i}) }\,}\underset{\color{#0066A7}{\hbox{Area}}}{ \underbrace{[ f( v_{i}) -g( v_{i}) ] \Delta y}}\underset{\color{#0066A7}{\hbox{$H=2$}}}{\underset{\color{#0066A7}{\uparrow }}{=}}\rho g( 2-v_{i}) [ f( v_{i}) -g( v_{i}) ] \Delta y \end{equation*}$$

The liquid fills the trough from $y=0$ to $y=2,$ so the hydrostatic force $F$ due to the pressure of the liquid on an end of the trough is $$\begin{eqnarray*} F &=&\int_{0}^{2}\rho g(2-y)[f(y)-g(y)] ~{\it dy}\\[5pt] &=&\int_{0}^{2}\rho g(2-y)\left[ \left( \dfrac{1}{2}y+3\right) -\left( -\dfrac{1}{2}y+1\right) \right] {\it dy}=\rho g\int_{0}^{2}(2-y)(y+2)~{\it dy} \notag \\[5pt] &=&\rho g\int_{0}^{2}(-y^{2}+4)~{\it dy}=\rho g\left[ -\dfrac{y^{3}}{3}+4y\right] _{0}^{2}=\rho g\!\left( -\dfrac{8}{3}+8\right) =\dfrac{16}{3}\rho g\hbox{ } \textrm{ N} \end{eqnarray*}$$