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

Question

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\ \mathrm{m/s^2}):

Vishal Gupta · Accepted Answer

B. Take torques about the foot. The wall pushes horizontally with N; the rod's weight acts at its midpoint. With the rod at 60^\circ to the wall (i.e. 30^\circ to the floor): N \cdot L\sin30^\circ = mg\cdot\frac{L}{2}\cos30^\circ \Rightarrow N = mg\frac{\cos30^\circ}{2\sin30^\circ} = 200	imes\frac{\sqrt3/2}{1} = 100\sqrt3 N. Friction at the floor balances N: 100\sqrt3 N.

System of Particles and Rotational Motion

More from this chapter in past papers