A uniform rod of mass $20$ kg and length $5$ m leans against a smooth vertical wall making an angle of $60°$ with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is
(take $g = 10\ \text{m/s}^2$)
Answer: (B) $100\sqrt{3}$ N
The smooth wall gives only a horizontal normal force $N_w$. Horizontal equilibrium: friction $f = N_w$.
The rod makes $60°$ with the wall. Its top is at height $L\cos 60°$, and its centre is a horizontal distance $\dfrac{L}{2}\sin 60°$ from the foot.
Torques about the foot of the rod:
$$N_w \cdot L\cos 60° = mg\cdot\frac{L}{2}\sin 60°$$
$$N_w = \frac{mg}{2}\tan 60° = \frac{200}{2}\sqrt{3} = 100\sqrt{3}\ \text{N}$$
So the friction force is $f = 100\sqrt{3}$ N.
Solution by Sreeraj P, M.Sc Physics