Question

A wooden plank of length 2$l$ 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 $l$ m. The specific gravity of the plank is 0.5. Find the angle $\theta$ that the plank makes with the vertlcal in the equilibrium position. (Exclude the case $\theta$ = 0)

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

Answer 1

The correct option is A ${45}^{\circ }$

Answer is A.
The forces acting on the plank are shown in the figure. The height of water level is l. The length of the plank is 2l. The weight of the plank acts through the center B of the plank. We have OB = l. The buoyant force F acts through the point A which is the middle point of the dipped part OC of the plank.
We have $OA=\frac{OC}{2}=\frac{l}{2cos\theta }$.
Let the mass per unit length of the plank be p.
Its weight mg = 2 lpg.
The mass of the part OC of the plank = $\left(\frac{l}{cos\theta }\right)p$.
The mass of water displaced = $\frac{1}{0.5}\frac{l}{cos\theta }p=\frac{2lp}{cos\theta }$
The buoyant force F is, therefore, $\frac{2lpg}{cos\theta }$.
Now, for equilibrium, the torque of mg about O should balance the torque of F about O.
So, mg(OB) sin = F(OA) sin
or, $\left(2lp\right)l=\left(\frac{2lp}{cos\theta }\right)\left(\frac{l}{2cos\theta }\right)$
or, $co{s}^2\theta =\frac{1}{2}$
or, $cos\theta =\frac{1}{\sqrt{2}}$
or, $\theta ={45}^{\circ }$.
Hence, the angle that the plank makes with the vertical in the equilibrium position is 45 degrees.