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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(4) 100√3 N. Given: M=20 kg, L=5 m, rod at 60° to the wall, so at θ=30° to the floor; g=10 m s⁻². Idea: the wall is smooth, so its reaction N₂ is horizontal; the only other horizontal force is the friction f at the floor. Horizontal equilibrium: f=N₂. Torque about the foot of the rod: Mg·L/2 cos θ=N₂L sin θ N₂=(Mg)/2 cot θ=(20×10)/2 cot 30°=100√3 f=100√3 N≈173 N.

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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