The tank shown in $FIGURECP14.72$is completely filled with a liquid of density $\rho$. The right face is not permanently attached to the tank but, instead, is held against a rubber seal by the tension in a spring. To prevent leakage, the spring must both pull with sufficient strength and prevent a torque from pushing the bottom of the right face out.

a. What minimum spring tension is needed?

b. If the spring has the minimum tension, at what height $d$from the bottom must it be attached?

(a)

The surface area of the container will be affected by the pressure at a specific depth in a water body. This area will be subjected to some force due to the pressure. The pressure force is the force exerted by the fluid on the container's walls.

The expression for the pressure force is:

$F=\int pdA$

Where, $p$is the fluid pressure and $dA$is the elemental area. Thus, the pressure force is:

$F=w{\int }_0^h{p}_0+\frac{1}{2}pg(h-y)dy$

$=w\left[p_{0}h+\frac{1}{2}\rho g{h}^2\right]$

$F=wh\left[p_0+\rho gh\right]$

(b)

The spring must be horizontally in the middle to produce no torque. The center of force is used to determine the height$d$above the bottom at which the spring must be mounted in order to produce no torque.

The height above the bottom where the spring must be supplied is expressed as:

$d=\frac{1}{F}\int ydF$

$=\frac{w{\int }_0^h{p}_{0}y+\rho g(h-y)dy}{w\left[p_{0}h+\frac{1}{2}\rho g{h}^2\right]}$

$=\frac{\frac{1}{2}{p}_0{h}^2+\frac{1}{6}\rho g{h}^3}{p_{0}h+\frac{1}{2}\rho g{h}^2}$

$d=\frac{\frac{1}{2}{p}_{0}h+\frac{1}{6}\rho g{h}^2}{p_0+\frac{1}{2}\rho gh}$