1. 100 N 2. 100√(3) N 3. 200 N 4. 200√(3) N

Topic: Equilibrium of rigid bodies, rigid body rotation, equations of rotational motion· Rotational Motion

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. 100 N

2. $100\sqrt{3}$ N

3. 200 N

4. $200\sqrt{3}$ N

Doubtly Asked 30 Oct 2026 NEET UG 2025 Practice test neet Save WhatsApp Report

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Answer: 2. $100\sqrt{3}$ N

Key idea: the wall is smooth, so the wall force $N$ is horizontal. For equilibrium, floor friction $f$ equals $N$, and torques about the bottom end must balance.

The rod makes $60^\circ$ with the wall (vertical), length $L=5$ m, $mg=200$ N.

Torque about the foot of the rod: weight acts at $L/2$ with horizontal arm $\tfrac L2\sin60^\circ$; the wall force acts at the top with vertical arm $L\cos60^\circ$.

$$mg\cdot\frac L2\sin60^\circ=N\,L\cos60^\circ\Rightarrow N=\frac{mg}{2}\tan60^\circ=100\sqrt3\ \mathrm N$$

So the friction force is $f=N=100\sqrt3$ N.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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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)

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