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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Correct answer: B. 100 3 ~N

N _2= mg =200 ~N Taking torque about A : aligned & f _ s ~L 60^ + mg ~L 2 60^ = N _2 ~L 60^ \ & f _ s ( 1 2 )+200 ( 1 2 ) 3 2 =200 3 2 \ & f _ s =100 3 ~N aligned

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