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 res…

Question

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 m/s^2):

1. A. 100 N
2. B. 100√3 N (Correct answer)
3. C. 200 N
4. D. 200√3 N

Correct Answer

Option B — 100√3 N

Detailed Solution & Explanation

The correct answer is $100\sqrt{3}$ N.

Key Points

• The rod makes $60^\circ$ with the wall, i.e. $30^\circ$ with the floor. Let $N_1$ be the normal from the floor, $N_2$ the normal from the smooth wall, and $f$ the friction.
• Translational equilibrium: $N_1=Mg=200$ N and $N_2=f$ (the wall is smooth, so only $N_2$ is horizontal at the top).
• Rotational equilibrium about the floor contact: $Mg\cdot\dfrac{L}{2}\cos\theta=N_2\,L\sin\theta\Rightarrow N_2=\dfrac{Mg}{2}\cot\theta$ with $\theta=30^\circ$.
• $f=N_2=\dfrac{200}{2}\cot30^\circ=100\sqrt{3}$ N.