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Q. A wooden plank of length $1 m$ and uniform cross-section is hinged at one end to the bottom of a tank as shown in figure. The tank is filled with water up to a height of $0.5 m$. The specific gravity of the plank is $0.5 .$ Find the angle $\theta$ that the plank makes with the vertical in the equilibrium position.Physics Question Image

Mechanical Properties of Fluids

Solution:

The forces acting on the plank are shown in the figure. The height of water level is $1=0.5 m .$ The length of the plank is $1.0 m =21$. The weight of the plank acts through the centre $B$ of the plank. We have $O B=1$. The buoyant force $F$ acts through the point $A$ which is the middle point of the dipped part $OC$ of the plank.
We have $O A=\frac{O C}{2}=\frac{\ell}{2 \cos \theta}$
Let the mass per unit length of the plank be $\rho$. Its weight
$mg =2 lpg .$
The mass of the part $OC$ of the plank $=\left(\frac{\ell}{\cos \theta}\right) \rho .$
The mass of water displaced
$=\frac{1}{0.5} \frac{\ell}{\cos \theta} \rho=\frac{2 \ell \rho}{\cos \theta}$
The buoyant force $F$ is, therefore, $F=\frac{2 \ell \rho g}{\cos \theta}$
Now, for equilibrium, the torque of $mg$ about $O$ should balance the torque of $F$ about $O$
So , $mg ( OB ) \sin \theta= F ( OA ) \sin \theta$
or, $(2 \ell \rho) \ell=\left(\frac{2 \ell \rho}{\cos \theta}\right)\left(\frac{\ell}{2 \cos \theta}\right)$
or, $\cos ^{2} \theta=\frac{1}{2}$ or, $\cos \theta=\frac{1}{\sqrt{2}},$ or, $\theta=45^{\circ}$