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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
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Verified Answer
The correct answer is:
$\frac{\rho g}{6}$
The net force acting on the gate element of width dy at a depth y from the surface of the fluid, is
$$
\begin{aligned}
\mathrm{dy} &=\left(\mathrm{p}_{0}+\rho_{\mathrm{g}} \mathrm{y}-\mathrm{p}_{0}\right) \times 1 \mathrm{dy} \\
&=\rho \mathrm{gydy}
\end{aligned}
$$
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 $F$ required to hold the gate
stationary is $\frac{p g}{a}$
$$
\begin{aligned}
\mathrm{dy} &=\left(\mathrm{p}_{0}+\rho_{\mathrm{g}} \mathrm{y}-\mathrm{p}_{0}\right) \times 1 \mathrm{dy} \\
&=\rho \mathrm{gydy}
\end{aligned}
$$
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 $F$ required to hold the gate
stationary is $\frac{p g}{a}$
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