Question

A uniform rod of length $L$ pivoted at the bottom of a pond of depth $L/2$ stays in stable equilibrium as shown in the figure. Find the force acting at the pivoted end of the rod in terms of mass $m$ of the rod.

• A. $(\sqrt{2}-1)mg$
• B. $(\sqrt{2}+1)mg$
• C. $(\sqrt{3}+1)mg$
• D. $(\sqrt{2}+3)mg$

Answer 1

The correct option is A $(\sqrt{2}-1)mg$

Weight of the rod $W=ALdg$, where $d$ is the density of the rod, $A$ is the uniform cross section area of rod. $L$ is the total length of the rod.

Lets say $x$ is the length of submerged part, we have

$x\sin \theta =\frac{L}{2}\Rightarrow x=\left(\frac{L}{2}\right)cosec\theta$

Force of buoyancy $B=A\left(\frac{L}{2}cosec\theta \right)\rho g$

Since the rod is at rest i.e. it is in rotational equilibrium, we have

$W\times \frac{L}{2}\cos \theta =B\times \frac{1}{2}\left(\frac{L}{2}\right)cosec\theta \Rightarrow ALdg\times \frac{L}{2}\cos \theta =A\left(\frac{L}{2}cosec\theta \right)\rho g\times \frac{1}{2}\left(\left(\frac{L}{2}\right)cosec\theta \right)\cos \theta \Rightarrow d=\left(\frac{1}{2}cosec\theta \right)\rho \times \frac{1}{2}cosec\theta \Rightarrow \frac{\rho }{2}=\left(\frac{1}{2}cosec\theta \right)\rho \times \frac{1}{2}cosec\theta \because d=\frac{\rho }{2}\Rightarrow cose{c}^2\theta =2\Rightarrow cosec\theta =\sqrt{2}\Rightarrow \sin \theta =\frac{1}{\sqrt{2}}\Rightarrow \theta ={45}^o$

Pivoted end provides the net downward force $\left(F\right)$ of $B-W$,

$F=B-W\Rightarrow F=B-mg$

Here mass and weight of rod are given by

$m=ALd\text{and}W=mg=ALdg$

$\because B=A\left(\frac{L}{2}\csc \theta \right)\rho g\Rightarrow F=A(\frac{L}{2}\times \sqrt{2})\left(2d\right)g-mg\because \rho =2d\text{and}\theta ={45}^o\Rightarrow F=\sqrt{2}ALdg-mg\Rightarrow F=\sqrt{2}mg-mg\because m=ALd\Rightarrow F=(\sqrt{2}-1)mg$