Chapter 15, Problem 13P

Review. The tank in Figure P15.13 is filled with water of depth d = 2.00 m. At the bottom of one sidewall is a rectangular hatch of height h = 1.00 m and width w = 2.00 m 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.

To determine

The magnitude of the force the water exerts on the hatch.

Answer to Problem 13P

The magnitude of the force the water exerts on the hatch is $29.4\text{ kN}$.

Explanation of Solution

The air outside and the water inside the tank exert atmospheric pressure so that only excess water pressure counts for the total force.

The diagram is shown below.

Consider a strip of hatch between depth $h$ and $h+dh$ .

Write the equation for the pressure due to the water at depth $h$ .

  $P=\rho gh$        (I)

Here, $P$ is the pressure due to the water at depth $h$, $\rho$ is the density of the water and $g$ is the acceleration due to gravity.

Write the equation for the pressure in terms of force.

  $P=\frac{F}{A}$

Here, $F$ is the force and $A$ is the area.

Rewrite the above equation for $F$.

  $F=PA$

Use the above equation to write the expression for the force exerted on the considered strip.

  $dF=PdA$        (II)

Here, $dF$ is the force exerted on the strip and $dA$ is the area of the strip.

Write the equation for $dA$.

  $dA=\left(2.00\text{ m}\right)dh$        (III)

Put equations (I) and (III) in equation (II).

  $dF=\rho gh\left(2.00\text{ m}\right)dh$        (IV)

Write the equation for the total force.

  $F={\displaystyle \underset{h=1.00\text{ m}}{\overset{2.00\text{ m}}{\int }}dF}$

Put equation (IV) in the above equation and rearrange it.

  $\begin{array}{c}F={\displaystyle \underset{h=1.00\text{ m}}{\overset{2.00\text{ m}}{\int }}\rho gh\left(2.00\text{ m}\right)dh}\\ =\rho g\left(2.00\text{ m}\right){\displaystyle \underset{h=1.00\text{ m}}{\overset{2.00\text{ m}}{\int }}hdh}\\ =\rho g\left(2.00\text{ m}\right){\left[\frac{h^2}{2}\right]}_{1.00\text{ m}}^{2.00\text{ m}}\\ =\rho g\left(2.00\text{ m}\right)\frac{3}{2}\end{array}$        (V)

Conclusion:

The density of water is $1000{\text{ kg/m}}^3$ and the value of $g$ is $9.80{\text{ m/s}}^2$.

Substitute $1000{\text{ kg/m}}^3$ for $\rho$ and $9.80{\text{ m/s}}^2$ for $g$ in equation (V) to find $F$.

  $\begin{array}{c}F=\left(1000{\text{ kg/m}}^3\right)\left(9.80{\text{ m/s}}^2\right)\left(2.00\text{ m}\right)\frac{3}{2}\\ =29.4\times {10}^3\text{ N}\frac{1\text{ kN}}{1000\text{ N}}\\ =29.4\text{ kN}\end{array}$

Therefore, the magnitude of the force the water exerts on the hatch is $29.4\text{ kN}$.

To determine

The magnitude of the torque exerted by the water about the hinges.

Answer to Problem 13P

The magnitude of the torque exerted by the water about the hinges is $16.3\text{ kN}\cdot \text{m}$.

Explanation of Solution

Write the equation for the total torque.

  $\tau ={\displaystyle \underset{h=1.00\text{ m}}{\overset{2.00\text{ m}}{\int }}d\tau }$        (VI)

Here, $\tau$ is the total torque exerted by the water about the hinges and $d\tau$ is the torque exerted on the considered strip.

The lever arm of the force $dF$ is the distance $\left(h-1.00\text{ m}\right)$ from the hinge to the strip.

Refer to the diagram and write the equation for $d\tau$.

  $d\tau =\left(h-1.00\text{ m}\right)dF$

Put the above equation in equation (VI).

  $\tau ={\displaystyle \underset{h=1.00\text{ m}}{\overset{2.00\text{ m}}{\int }}\left(h-1.00\text{ m}\right)dF}$

Put equation (IV) in the above equation and rearrange it.

  $\begin{array}{c}\tau ={\displaystyle \underset{h=1.00\text{ m}}{\overset{2.00\text{ m}}{\int }}\left(h-1.00\text{ m}\right)\rho gh\left(2.00\text{ m}\right)dh}\\ =\rho g\left(2.00\text{ m}\right){\displaystyle \underset{h=1.00\text{ m}}{\overset{2.00\text{ m}}{\int }}{h}^{2}dh}-\rho g\left(2.00\text{ m}\right)\left(1.00\text{ m}\right){\displaystyle \underset{h=1.00\text{ m}}{\overset{2.00\text{ m}}{\int }}hdh}\end{array}$

Find the value of the above integral.

  $\begin{array}{c}\tau =\rho g\left(2.00\text{ m}\right){\left[\frac{h^3}{3}\right]}_{1.00\text{ m}}^{2.00\text{ m}}-\rho g\left(2.00\text{ m}\right)\left(1.00\text{ m}\right){\left[\frac{h^2}{2}\right]}_{1.00\text{ m}}^{2.00\text{ m}}\\ =\rho g\left(2.00\text{ m}\right)\left(\frac{7}{3}{\text{ m}}^3\right)-\rho g\left(2.00\text{ m}\right)\left(1.00\text{ m}\right)\left(\frac{3}{2}{\text{ m}}^2\right)\\ =\rho g\left(2.00\text{ m}\right)\left(\frac{5}{6}{\text{ m}}^3\right)\\ =5\rho g\left(\frac{1}{3}{\text{ m}}^4\right)\end{array}$        (VII)

Conclusion:

Substitute $1000{\text{ kg/m}}^3$ for $\rho$ and $9.80{\text{ m/s}}^2$ for $g$ in equation (VII) to find $\tau$.

  $\begin{array}{c}F=5\left(1000{\text{ kg/m}}^3\right)\left(9.80{\text{ m/s}}^2\right)\left(\frac{1}{3}{\text{ m}}^3\right)\\ =16.3\times {10}^3\text{ N}\frac{1\text{ kN}}{1000\text{ N}}\\ =16.3\text{ kN}\end{array}$

Therefore, the magnitude of the torque exerted by the water about the hinges is $16.3\text{ kN}\cdot \text{m}$.