Chapter 14, Problem 15P

(a)

To determine

The force that exerted by water on the hatch.

Answer to Problem 15P

The force that exerted by water on the hatch is $\underset{\_}{29.4\text{kN}}$ to the right.

Explanation of Solution

Observe Figure P14.15 given in the question. Figure 1 shown below gives the value of height and width of the hatch and also the depth of water stored in the tank.

The system experiences downward pressure from the atmosphere and an outward pressure from the water. So the net pressure will be the difference between these two pressures. The excess pressure is equal to $\rho gh$.

Consider a small portion of hatch between the heights $h$ and $h+dh$.

Write the expression for the force experienced at the hatch.

    $dF=PdA$                                                                                                                   (I)

Here, $dF$ is the force experienced at the hatch, $P$ is the pressure, and $dA$ is the area of the elemental portion considered between $h$ and $h+dh$ of the hatch.

Write the expression for the pressure.

    $P=\rho gh$                                                                                                                    (II)

Here, $g$ is the acceleration due to gravity, and $h$ is the height of hatch.

Write the expression for area of elemental portion considered.

    $dA=wdh$                                                                                                                 (III)

Here, $w$ is the width of the hatch.

Use expression (II), and (III) in (I) to find $dF$.

    $dF=\left(\rho gh\right)wdh$                                                                                                     (IV)

Integrate expression (III) to find the total force.

    $\begin{array}{c}F={\displaystyle \underset{h_1}{\overset{h_2}{\int }}dF}\\ ={\displaystyle \underset{h_1}{\overset{h_2}{\int }}\rho gwhdh}\\ =\rho gw{\left(\frac{h^2}{2}\right)}_{h_1}^{h_2}\end{array}$                                                                                                      (V)

Here $F$ is the total force, $h_1$ and $h_2$ are the limits of integration.

Conclusion:

Substitute $1.00\text{m}$ for $h_1$, $2.00\text{m}$ for $h_2$, $2.00\text{m}$ for $w$, $1000{\text{kg/m}}^3$ for $\rho$, and $9.80{\text{m/s}}^2$ for $g$ in equation (IV) to find $F$.

    $\begin{array}{c}F=\left(1000{\text{kg/m}}^3\right)\left(9.80{\text{m/s}}^2\right)\left(2.00\text{m}\right)\left(\frac{{\left(2.00\text{m}\right)}^2}{2}-\frac{{\left(1.00\text{m}\right)}^2}{2}\right)\\ =29.4\text{kN}\end{array}$

Therefore, the force that is exerted by water on the hatch is $\underset{\_}{29.4\text{kN}}$ to the right.

(b)

To determine

The torque exerted by the water about the hinges.

Answer to Problem 15P

Magnitude of the torque exerted by the water about the hinges is $\underset{\_}{16.3\text{kN}\cdot \text{m}}$.

Explanation of Solution

Write the expression for the torque exerted by water about hinges between $h$ and $h+dh$ of the hatch.

    $d\tau =dFr$                                                                                                                 (VI)

Here, $d\tau$ is the torque exerted by water about hinges between $h$ and $h+dh$ of the hatch, $r$ is the lever arm of force $dF$.

Use expression (IV) in expression (VI).

    $\begin{array}{c}d\tau =dFr\\ =\left(\rho gh\right)rwdh\end{array}$                                                                                                   (VII)

From Figure 1 it is clear that the lever arm of force $dF$ is the distance $\left(h-1.00\text{m}\right)$ from hinge to strip.

Therefore rewrite expression (VII).

    $d\tau =\left(\rho gh\right)\left(h-1.00\text{m}\right)wdh$                                                                              (VIII)

Integrate expression (VIII) from height $h_1$ to $h_2$ to find the torque.

    $\begin{array}{c}\tau ={\displaystyle \underset{h_1}{\overset{h_2}{\int }}\left(\rho gh\right)\left(h-1.00\text{m}\right)wdh}\\ ={\displaystyle \underset{h_1}{\overset{h_2}{\int }}\rho g{h}^2-\left(1.00\text{m}\right)}\rho ghwdh\\ ={\left[\rho g\frac{h^3}{2}-\left(1.00\text{m}\right)\rho g\left(\frac{h^2}{2}\right)w\right]}_{h_1}^{h_2}\\ =\rho g\left(\frac{h_2^3}{3}-\frac{h_1^3}{3}\right)-\left(1.00\text{m}\right)\rho gw\left(\left(\frac{h_2^2}{2}-\frac{h_1^2}{2}\right)\right)\end{array}$                                                      (IX)

Conclusion:

Substitute $1000{\text{kg/m}}^3$ for $\rho$, $9.80{\text{m/s}}^2$ for $g$, $2.00\text{m}$ for $w$, $2.00\text{m}$ for $h_2$, and $1.00\text{m}$ for $h_1$ in equation (IX) to find $\tau$.

  $\begin{array}{c}\tau =\left(1000{\text{kg/m}}^2\right)\left(9.80{\text{m/s}}^2\right)\left(\frac{{\left(2.00\text{m}\right)}^3}{3}-\frac{{\left(1.00\text{m}\right)}^2}{3}\right)-\left(1.00\text{m}\right)\left(1000{\text{kg/m}}^3\right)\left(9.80{\text{m/s}}^2\right)\left(1.00\text{m}\right)\left(\left(\frac{{\left(2.00\text{m}\right)}^2}{2}-\frac{{\left(1.00\text{m}\right)}^2}{2}\right)\right)\\ =16.3\text{kN}\cdot \text{m}\end{array}$

The positive torque implies that it is directed counterclockwise.

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