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.

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

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)

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$1.88\times {10}^{9}N$

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 $2.20\times {10}^{10}N.m$

$11.7m$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