Chapter 14, Problem 14.15P

Review. The lank in Figure P14.15 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 in 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 amount of force water exerts on the hatch.

Answer to Problem 14.15P

The amount of force water exerts on the hatch is $2.94\times {10}^4\text{N}$.

Explanation of Solution

Given info: The depth of water in the tank is $2.00\text{m}$, the height of the rectangular hatch is $1.00\text{m}$ and the width of rectangular hatch is $2.00\text{m}$.

Write the expression for the force exerted on the strip.

$df=P\cdot dA$ (1)

Here,

$dA$ is the area of the small strip on the hatch.

$P$ is the pressure exerted on the strip.

Write the expression for pressure exerted on the strip.

$P=\rho gh$

Here,

$\rho$ is density of the water.

$h$ is the depth of the water in the tank.

$g$ is the acceleration due to gravity.

The area of the cross section of the strip is,

$dA=w\cdot dh$

Here,

$w$ is the width of the rectangular strip.

$dh$ is the strip height.

Substitute $w\cdot dh$ for $dA$ and $\rho gh$ for $P$ in the equation (1).

$dF=\rho gh\left(w\cdot dh\right)$

As, the pressure varies with the depth so the force exerted is also vary.

Write the expression for force exerted on the rectangular height,

${\displaystyle \int dF}={\displaystyle {\int }_{h_1}^{h_1}\rho gh\left(w\cdot dh\right)}$ (2)

Here,

$h_1$ is the height of the rectangular hatch.

$h_2$ is depth of the water in tank.

Substitute $2.00\text{m}$ for $h_2$ and $1.00\text{m}$ for $h_1$ in the equation (2).

$\begin{array}{c}{\displaystyle \int dF}={\displaystyle {\int }_{1.00\text{m}}^{2.00\text{m}}\rho gh\left(w\cdot dh\right)}\\ =\rho gw{\left(\frac{h^2}{2}\right)}_{1.00\text{m}}^{2.00\text{m}}\\ F=\frac{3}{2}\rho gw\end{array}$ (3)

The density of the water is $1000\text{kg}/{\text{m}}^{\text{3}}$ and the acceleration due to gravity is $9.8\text{m}/{\text{s}}^{\text{2}}$.

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $2.00\text{m}$ for $w$ in the equation (3).

$\begin{array}{c}F=\frac{3}{2}\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(2.00\text{m}\right)\\ =2.94\times {10}^4\text{N}\end{array}$

Conclusion:

Therefore, the amount of force water exerts on the hatch is $2.94\times {10}^4\text{N}$.

To determine

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

Answer to Problem 14.15P

The amount of the torque exerted by the water about the hinges is $1.63\times {10}^4\text{N}\cdot \text{m}$.

Explanation of Solution

Given info: The depth of water in the tank is $2.00\text{m}$, the height of the rectangular hatch is $1.00\text{m}$ and the width of rectangular hatch is $2.00\text{m}$.

Write the expression for the torque about the hinge.

$\tau =dF\cdot \left(h-1.00\text{m}\right)$ (4)

Here,

$dF$ is the force exerted on the rectangular hatch.

$\left(h-1.00\text{m}\right)$ is the distance from hinge to strip.

From part (a), the force exerted on the rectangular hatch is,

$dF=\rho gh\left(w\cdot dh\right)$

Substitute $\rho gh\left(w\cdot dh\right)$ for $dF$ in the equation (4).

$\tau =\rho gh\left(w\cdot dh\right)\cdot \left(h-1.00\text{m}\right)$

Write the expression for the torque about the hinge,

$\tau ={\displaystyle \underset{h_1}{\overset{h_2}{\int }}\rho gh\left(w\cdot dh\right)\cdot \left(h-1.00\text{m}\right)}$ (5)

Substitute $1.00\text{m}$ for $h_1$ and $2.00\text{m}$ for $h_2$ in the equation (5).

$\begin{array}{c}\tau =\rho gw{\displaystyle \underset{1.00\text{m}}{\overset{2.00\text{m}}{\int }}\left(h^2-h\right)}dh\\ =\rho gw{\left(\frac{h^3}{3}-\frac{h^2}{2}\right)}_{1.00\text{m}}^{2.00\text{m}}\\ =\frac{5}{6}\rho gw\end{array}$ (6)

The density of the water is $1000\text{kg}/{\text{m}}^{\text{3}}$ and the acceleration due to gravity is $9.8\text{m}/{\text{s}}^{\text{2}}$.

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $2.00\text{m}$ for $w$ in the equation (6).

$\begin{array}{c}\tau =\frac{5}{6}\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(2.00\text{m}\right)\\ =1.63\times {10}^4\text{N}\cdot \text{m}\end{array}$

Conclusion:

Therefore, the amount of the torque exerted by the water about the hinges is $1.63\times {10}^4\text{N}\cdot \text{m}$.