91Ó°ÊÓ

The height of the gate is, d = 2 m .

The width of the gate is, w = 4 m .

The torque about the hinge is calculated by finding the force about the hinge and then calculating the torque of this force.

The force on a thin horizontal strip about the hinge is given as:

$dF=\mathrm{ÒÏ}g\left(\frac{d}{2}−y\right)wdy$

Here, y is the distance from the hinge position of the gate, $\mathrm{ÒÏ}$is the density of water which is ${10}^{3}kg/m^3$, g and is the gravitation acceleration due to gravity.

The torque on a thin horizontal strip about the hinge is given as:

$\begin{array}{r}d\mathrm{Ï„}=2dFâ‹\dots \left(\frac{d}{2}−y\right)\\ d\mathrm{Ï„}=2\left(\mathrm{ÒÏ}g\left(\frac{d}{2}−y\right)wdy\right)\left(\frac{d}{2}−y\right)\\ \mathrm{Ï„}=2\mathrm{ÒÏ}gw{∫}_0^{d/2} {\left(\frac{d}{2}−y\right)}^{2}dy\end{array}$

Substitute all the values in the above equation, and we get,

$\begin{array}{r}\mathrm{τ}=2\left({10}^3\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\left(4\mathrm{m}\right){∫}_0^{(2\mathrm{m}/2} {\left(\frac{2}{2}−y\right)}^{2}dy\\ \mathrm{τ}=2\left({10}^3\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\left(4\mathrm{m}\right){∫}_0^{1\mathrm{m}} \left(1+y^2−2y\right)dy\\ \mathrm{τ}=2\left({10}^3\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\left(4\mathrm{m}\right){\left(y+\frac{y^3}{3}−2\left(\frac{y^2}{2}\right)\right)}_0^{1\mathrm{m}}\\ \mathrm{τ}=2\left({10}^3\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\left(4\mathrm{m}\right)\left(1+\frac{(1{)}^3}{3}−(1{)}^2\right)\end{array}$

Solving further as,

$\begin{array}{r}\mathrm{Ï„}=2\left({10}^3\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\left(4\mathrm{m}\right)\left(\frac{1}{3}\mathrm{m}\right)\\ \mathrm{Ï„}=2.61×{10}^4\mathrm{N}â‹\dots \mathrm{m}\end{array}$

Therefore, the torque on a thin horizontal strip about the hinge is $2.61×{10}^{4}N.m$.