Question

In Figure, water stands at depth $\mathit{D}\mathbf{=}\mathbf{35}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}$behind the vertical upstream face of a dam of width $\mathit{W}\mathbf{=}\mathbf{314}\mathbf{}\mathit{m}$ .(a)Find the net horizontal force on the dam from the gauge pressure of the water.(b)Find the net torque due to that force about a horizontal line through parallel to the (long) width of the dam. This torque tends to rotate the dam around that line, which would cause the dam to fail. (c)Find the moment arm of the torque.

Short answer

1. The net horizontal force on the dam from the gauge pressure of the water is$188x{10}^{9}N$

1. The net torque due to the net horizontal force about a horizontal line through O parallel to the long width of the dam is $2.20x{10}^{10}N.m$
2. The moment arm of the torque is$11.7m$

Step-by-step solution

The given data

1. Depth of water,$D=35.0m$
2. Width of the dam,$W=314m$

Understanding the concept of pressure, force and torque

Using the formula of pressure, we can find the net horizontal force on the dam from the gauge pressure of the water. Using this force into the formula of torque, we can find the net torque due to the net horizontal force about a horizontal line through O parallel to the long width of the dam.

Formulae:

Force applied on a body in terms of pressure, $F=PA$ (i)

Pressure applied on a fluid surface, $P=\mathrm{\rho }gh$ (ii)

Torque applied on a body, $\tau =Fr$ (iii)

a) Calculation of net horizontal force on dam

Using equation (i), the force on a body can be written as:

$$\begin{gathered} F=\int dF \\ =\int PA \end{gathered}$$

As,$P=\mathrm{\rho }gy\mathit{}\mathit{\&}\mathit{}dA\mathit{=}Wdy$,

role="math" localid="1657196238814" $$\begin{gathered} F={\int }_0^D\mathrm{\rho }gyWdy \\ =\frac{1}{2}\mathrm{\rho }gW{D}^2\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(a\right) \\ =\frac{1}{2}\left(1.0\times {10}^3\frac{\mathit{k}\mathit{g}}{{\mathit{m}}^{\mathit{3}}}\right)\left(9.8\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\left(314m\right){(35.0m)}^2=1.88\times {10}^{9}N \end{gathered}$$

Therefore, the net horizontal force on the dam from the gauge pressure of the water is$1.88\times {10}^{9}N$

b) Calculation of net torque

As,$r=(D-y)\&dF=\mathrm{\rho }gyWdy$using equation (iii), we get

role="math" localid="1657196905901" $$\begin{gathered} \tau =d\tau ={\int }_0^D\mathrm{\rho }gyW(D-y)dy \\ =\mathrm{\rho }gW\left(\frac{1}{2}{D}^3-\frac{1}{3}{D}^3\right) \\ =\frac{1}{6}\mathrm{\rho }gW{D}^3...................................................................\left(b\right) \\ =\frac{1}{6}\left(1.0\times {10}^3\frac{kg}{m^3}\right)\left(9.8\frac{m}{s^2}\right)(314.0m){(35.0)}^3 \\ =2.20\times {10}^{10}N.m \end{gathered}$$

Therefore, the net torque due to the net horizontal force about a horizontal line through

O parallel to the long width of the dam is $2.20\times {10}^{10}N.m$

c) Calculation of moment arm of the torque

$11.7m$From equation (iii), we get$r=\frac{\tau }{F}$

Using equations (a) and (b), we get

$$\begin{gathered} r=\frac{{\displaystyle \frac{1}{6}}\mathrm{\rho }gW{D}^3}{{\displaystyle \frac{1}{2}}\mathrm{\rho }gW{D}^2} \\ =\frac{D}{3} \\ =\frac{(35.0m)}{3} \\ =11.667m \\ =11.7m \end{gathered}$$

Therefore, the moment arm of the torque is