Question

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 $\theta$ (in degrees) that the plank makes with the vertical in the equilibrium position (exclude the case $\theta =0$).

Answer 1

weight of the plank $W=Aldg$, where $d$ is the density of the plank, $A$ is the uniform cross section area of plank. $l$ is the total length of the plank.
Force of buoyancy $B=\left(A0.5\sec \theta \right)\rho g$
where $0.5\sec \theta$ is the submerged length.
Since the plank is at rest i.e. it is in rotational equilibrium, we have
$W\times \frac{l}{2}\sin \theta =B\times \frac{0.5\sec \theta \sin \theta }{2}\Rightarrow Aldg\times \frac{l}{2}\sin \theta =\left(A0.5\sec \theta \right)\rho g\times \frac{0.5\sec \theta \sin \theta }{2}$
$\Rightarrow Al\left(0.5\rho \right)g\times \frac{l}{2}\sin \theta =\left(A0.5\sec \theta \right)\rho g\times \frac{0.5\sec \theta \sin \theta }{2}\because d=0.5\rho$
$\Rightarrow {\sec }^2\theta =2\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}\Rightarrow \theta ={45}^{\circ }$