Question

A wooden plank of length $1m$ 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 $0.5m$. The specific gravity of the plank is $\mathrm{0.5.}$ Find the angle $\theta$ that the plank makes with the vertical in the equilibrium position.

• A. $30^{\circ }$
• B. $60^{\circ }$
• C. $45^{\circ }$
• D. $90^{\circ }$

Answer 1

Answer: (C)

The forces acting on the plank are shown in the figure. The height of water level is $1=0.5m.$ The length of the plank is $1.0m=21$. The weight of the plank acts through the centre $B$ of the plank. We have $OB=1$. The buoyant force $F$ acts through the point $A$ which is the middle point of the dipped part $OC$ of the plank.
We have $OA=\frac{OC}{2}=\frac{\ell }{2\cos \theta }$
Let the mass per unit length of the plank be $\rho$. Its weight
$mg=2lpg.$
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, $\theta =45^{\circ }$