NEET · Physics · STD 11 - 6. system of particles and rotational motion
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 \(\left.g =10 m / s ^2\right)\)
- A 100 N
- B \(100 \sqrt{3} N\)
- C 200 N
- D \(200 \sqrt{3} N\)
Answer & Solution
Correct Answer
(B) \(100 \sqrt{3} N\)
Step-by-step Solution
Detailed explanation
Taking torque about A :
\(\begin{array}{l} f _{ s } L \cos 60^{\circ}+ mg \frac{ L }{2} \sin 60^{\circ}= N _2 L \sin 60^{\circ} \\ f _{ s }\left(\frac{1}{2}\right)+200\left(\frac{1}{2}\right) \frac{\sqrt{3}}{2}=200 \frac{\sqrt{3}}{2} \\ f _{ s }=100 \sqrt{3} N\end{array}\)
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