Question

Review: The tank in Figure P14.15 is filled with water of depth $\mathit{d}$. At the bottom of one sidewall is a rectangular hatch of height $\mathit{h}$and width $\mathit{w}$that is hinged at the top of the hatch. (a) Determine the magnitude of the force the water exerts on the hatch. (b) Find the magnitude of the torque exerted by the water about the hinges.

Short answer

(a) The magnitude of the force the water exerts on the hatch is $F=\frac{1}{2}\rho gwh[2d-h]$.

(b) The magnitude of the torque exerted by the water about the hinges is $\tau =\frac{\rho gw}{2}\left[d{h}^2-\frac{1}{3}{h}^3\right]$.

Step-by-step solution

Pressure:

The pressure in a fluid is the force per unit area exerted by the fluid on a surface:

$\mathit{P}\mathbf{=}\frac{\mathbf{F}}{\mathbf{A}}$

Here,$\mathit{P}$ can also be given as,

$\mathit{P}\mathbf{=}\mathit{\rho }\mathit{g}\mathit{h}$

Where,$\mathit{P}$is the pressure, $\mathit{\rho }$is the density,$\mathit{g}$ is the acceleration due to gravity, $\mathit{h}$ is the height,$\mathit{A}$ is the area, and, $\mathit{F}$ is the force exerted on the fluid.

(a) Determine The magnitude of the force the water exerts on the hatch:

The air outside and water inside both exert atmospheric pressure, so only the excess water $\rho gh$pressure counts for the net force.

At a distance $y$from the top of the water, take a strip of hatch between depth $y$and $y+dy$. It feels force as given below.

$\begin{array}{rcl}dF& =& PdA\\ & =& PWdy\\ & =& \left(\rho gyw\right)dy\end{array}$

The total force is define as below.

$\begin{array}{rcl}F& =& \int dF\\ & =& \underset{d-h}{\overset{d}{\int }}\rho gwydy\\ & =& \frac{1}{2}{\overline{)\rho gw{d}^2}}_{d-h}^d\\ & =& \frac{1}{2}\rho gw\left[d^2-{(d-h)}^2\right]\end{array}$

$\begin{array}{rcl}F& =& \frac{1}{2}\rho gw\left[d^2-d^2+2dh-h^2\right]\\ & =& \frac{1}{2}\rho gwh\left[2d-h\right]\end{array}$

Hence, the magnitude of the force the water exerts on the hatch is $F=\frac{1}{2}\rho gwh\left[2d-h\right]$.

(b) Determine The magnitude of the torque exerted by the water about the hinges:

The lever arm of $dF$is the distance $y-(d-h)$from hinge to strip:

$\begin{array}{rcl}\tau & =& \int d\tau \\ & =& \underset{d-h}{\overset{d}{\int }}\rho gwy(y-(d-h))dy\end{array}$

$\begin{array}{rcl}\tau & =& \rho gw\underset{d-h}{\overset{d}{\int }}[y^2-y(d-h)]dy\\ & =& \rho gw{\left[\frac{y^3}{3}-\frac{(d-h)y^2}{2}\right]}_{d-h}^d\\ & =& \frac{\rho gw}{6}\left[2d^3-3{(d-h)}^3-3(d-h)d^2+3{(d-h)}^3\right]\\ & =& \frac{\rho gw}{6}\left[2d^3-3(d-h)d^2+{(d-h)}^3\right]\end{array}$

$\begin{array}{rcl}\tau & =& \rho gw{\left[\frac{y^3}{3}-\frac{(d-h)y^2}{2}\right]}_{d-h}^d\\ & =& \frac{\rho gw}{6}\left[2d^3-3{(d-h)}^3-3(d-h)d^2+3{(d-h)}^3\right]\\ & =& \frac{\rho gw}{6}\left[2d^3-3(d-h)d^2+{(d-h)}^3\right]\end{array}$

$\begin{array}{rcl}\tau & =& \frac{\rho gw}{6}\left[2d^3-3d^3+3d^{2}h+d^3-3d^{2}h+3d{h}^2-h^3\right]\\ & =& \frac{\rho gw}{6}\left[3d{h}^2-h^3\right]\\ & =& \frac{\rho gw}{2}\left[d{h}^2-\frac{1}{3}{h}^3\right]\end{array}$

Hence, the magnitude of the torque exerted by the water about the hinges is $\tau =\frac{\rho gw}{2}\left[d{h}^2-\frac{1}{3}{h}^3\right]$.