A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60^\circ 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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Rotational Motion question

2025 · Q151

Rotational Motion question

2025 · Q151

View written solutionFree

Correct answer: D

NEET 2025 Physics - Rotational Motion Question 2 English Explanation

For translational equilibrium

$$\begin{aligned} \& N_1=M g \\ \& N_2=f \end{aligned}$$

For rotational equilibrium

Torque about $A, M g \frac{L}{2} \cos \theta=N_2 L \sin \theta$

$$\begin{aligned} \& \frac{M g}{2} \cot \theta=N_2=f \\ \& \frac{M g}{2} \cot 30^{\circ}=f \\ \& \frac{M g}{2} \sqrt{3}=N_2 \\ \& 100 \sqrt{3}=f \end{aligned}$$

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Rotational Motion question

2025 · Q151

Rotational Motion question

2025 · Q151

View written solutionFree

Correct answer: D

For translational equilibrium

$$\begin{aligned} \& N_1=M g \\ \& N_2=f \end{aligned}$$

For rotational equilibrium

Torque about $A, M g \frac{L}{2} \cos \theta=N_2 L \sin \theta$

$$\begin{aligned} \& \frac{M g}{2} \cot \theta=N_2=f \\ \& \frac{M g}{2} \cot 30^{\circ}=f \\ \& \frac{M g}{2} \sqrt{3}=N_2 \\ \& 100 \sqrt{3}=f \end{aligned}$$