A uniform rod of mass M and length L leans against a frictionless wall, with quarter of its length hanging over a corner as shown. Friction at corner is sufficient to keep the rod at rest. Then the ratio of magnitude of normal reaction on rod by wall and the magnitude of normal reaction on rod by corner is


(A) 1 2   sin   θ              (B) 2 sin   θ                (C) 1 2   cos   θ                (D) 2 cos   θ

Dear Student,
forces are normal by corner N normal by wall N' weight of the rod and frictional force along the rod in upward direction .Mg=Ncosθ+fsinθN'+Nsinθ=fcosθMg-Ncosθ=fsinθtanθ=Mg-NcosθN'+Nsinθtanθ=Mg-NcosθNN'+NsinθNbalance the torque about the corner.Mg4*L8cosθ+N'3L4sinθ=3Mg4*3L8cosθN'3L4sinθ=MgL4cosθ3N'tanθ=Mgtanθ=3N'tanθ-NcosθNN'+NsinθNratio N'N=αtanθ=3αtanθ-cosθα+sinθsinθ.tanθ=2αtanθ-cosθα=sinθ.tanθ+cosθ2tanθ=12tanθcosθ=12sinθoption ARegards
 

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