A light semi-cylindrical gate of radius R and | vedclass

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(D) The gate is a semi-cylinder of radius $R$ and length $l$. The pressure at a depth $h$ from the free surface is $P = \rho g h$.Consider an elementa

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(D) The gate is a semi-cylinder of radius $R$ and length $l$. The pressure at a depth $h$ from the free surface is $P = ho g h$.Consider an elemental strip of the gate at an angle $	heta$ with the vertical,having width $R d	heta$ and length $l$. The depth of this strip from the free surface is $h = R(1 - \cos	heta)$.The pressure on this strip is $P = ho g R(1 - \cos	heta)$.The force on this elemental strip is $dF_p = P \cdot dA = ho g R(1 - \cos	heta) \cdot (l R d	heta) = ho g R^2 l (1 - \cos	heta) d	heta$.The torque $d	au$ due to this force about the pivot $O$ is $d	au = dF_p \cdot R \sin	heta = ho g R^3 l (1 - \cos	heta) \sin	heta d	heta$.To find the total torque $	au$ about $O$,we integrate from $	heta = 0$ to $\pi$:$	au = \int_{0}^{\pi} ho g R^3 l (\sin	heta - \sin	heta \cos	heta) d	heta = ho g R^3 l [-\cos	heta - \frac{\sin^2	heta}{2}]_{0}^{\pi} = ho g R^3 l [(-(-1) - 0) - (-1 - 0)] = 2 ho g R^3 l$.The force $F$ is applied at the bottom edge,which is at a distance $R$ from the pivot $O$. Thus,the torque due to $F$ is $	au_F = F \cdot R$.Equating the torques: $F \cdot R = 2 ho g R^3 l$,which gives $F = 2 ho g R^2 l$.

$A$ light semi-cylindrical gate of radius $R$ and length $l$ is pivoted at its midpoint $O$ of the diameter as shown in the figure,holding liquid of density $\rho$. The force $F$ required to prevent the rotation of the gate is equal to:

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