Question

A square gate of size 2 m x 2 m is hinged at its midpoint O as shown. Fluid of density $\sigma$ is present to the left of the square. It is held in position by an unknown force F (Given that $\sigma$ is density of fluid)

• A. The torque exerted by the fluid in the upper half on the gate is $\frac{\sigma g}{3}Nm$
• B. The force exerted by the fluid in the lower half on the gate is $\frac{4\sigma g}{3}N$
• C. Total force exerted by the fluid on the gate is $\frac{5\sigma g}{3}N$
• D. The moment of the unknown force is $\frac{4\sigma g}{3}Nm$

Answer 1

Answer: (A), (D)

The torque exerted by the fluid in the upper half on the gate is $\frac{\sigma g}{3}Nm$
The moment of the unknown force is $\frac{4\sigma g}{3}Nm$

The gate will be in equilibrium, if the sum of clockwise moments is equal to the sum of anticlockwise moments taken about hinge O.
(i) The moment of required force F about O is clockwise.
(ii) The moment of force due to fluid in the upper half of the gate about O is clockwise.
(iii) The moment of force due to fluid in the lower half of the gate about O is anticlockwise.
Moment of force F (unknown) about O is F x 1 clockwise.
Moment of the force exerted by fluid above O is given by
${\tau }_1={\int }_o^1\sigma gy\left(2dy\right)(1-y)$
[where $\sigma gy$ is the pressure of the fluid of depth y. Here, 2dy is the area of a layer of thickness dy at y. Also, (1 - y) is the moment - arm about O].
${\tau }_1=2\sigma g{\int }_0^1[ydy-y^{2}dy]$
$=2\sigma g{[\frac{y^2}{2}-\frac{y^3}{3}]}_0^1=2\sigma g(\frac{1}{2}-\frac{1}{3})$
$=\frac{2\sigma g}{6}=\frac{\sigma g}{3}$ clockwise
Similarly, the moment due to the liquid in the lower half (i.e., below O) is
${\tau }_2={\int }_0^1\sigma g(y+1)\left(2dy\right)\left(y\right)$
= $2\sigma g{[\frac{y^3}{3}+\frac{y^2}{2}]}_0^1$
= $2\sigma g[\frac{1}{3}+\frac{1}{2}]$
= $\frac{5\sigma g}{3}$ anticlockwise
$\therefore \tau +{\tau }_1={\tau }_2$
$\Rightarrow \tau +\frac{\sigma g}{3}=\frac{5\sigma g}{3}$
$\Rightarrow \tau =\frac{4\sigma g}{3}Nm$