A square gate of size 1 m × 1 m is hinged at its mid-point. A fluid...

A square gate of size

1 \mathrm{ m} \times 1 \mathrm{ m}

is hinged at its mid-point. A fluid of density

\rho

fills the space to the left of the gate. The force F required to hold the gate stationary is A square gate of size 1 m × 1 m is hinged at its mid-point. A fluid of density fills the space to the left of the gate.

A square gate of size 1 m × 1 m is hinged at its mid-point. A fluid of density fills the space to the left of the gate.

Options

A $\frac{\rho g}{3}$
B $\frac{\rho g}{2}$
C $\frac{\rho g}{6}$
D $\frac{\rho g}{8}$

Correct answer

C. $\frac{\rho g}{6}$

Step-by-step solution

The net force acting on the gate element of width dy at a depth y from the surface of the fluid, is

\begin{aligned}\mathrm{dy}&=\left(\mathrm{p}_{0}+\rho_{\mathrm{g}} \mathrm{y}-\mathrm{p}_{0}\right) \times1 \mathrm{dy}\\ &=\rho \mathrm{gydy} \end{aligned}

Torque about the hinge is

\mathrm{d} \tau=\operatorname{pgydy} \times \left(\frac{1}{2}-\mathrm{y}\right)Net torque experience. by the gate is

\begin{aligned} \tau_{\text{net }} &=\int \mathrm{d} \tau+\mathrm{F} \times \frac{1}{2}\\ &=\int_{0}^{1} \rho g \mathrm{ydy}\left(\frac{1}{2}-\mathrm{y}\right)+\mathrm{F} \times \frac{1}{2}=0 \\ \Rightarrow & \mathrm{F}=\frac{\rho \mathrm{g}}{6} \end{aligned}

i.e., The force

Frequired to

hold the gate stationary is \frac{p g}{a}

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