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$^2$)

Question

Dr. Abubacker · Accepted Answer

[Solution by Claude - not from source] Friction $f=N_{wall}$. Torque about the foot: $N_{wall}L\cos60^\circ=mg\frac L2\sin60^\circ\Rightarrow N_{wall}=\frac{mg	an60^\circ}{2}=\frac{200\sqrt3}{2}=100\sqrt3$ N. Answer (2).

Q. 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$^2$)

Explanation

Reviewed by Dr. Abubacker

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