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/s² )

We are given a uniform rod of mass \(20\;\text{kg}\) and length���\(5\;\text{m}\) that leans against a smooth vertical wall. The rod makes an angle of \(60^\circ\) with the wall, so the angle with the horizontal floor is \(30^\circ\).

First, find the weight of the rod:

$\mathrm{W}=\mathrm{m}\times \mathrm{g}=20\times 10=200N$

The weight acts at the centre of the rod, i.e., at a distance \(\dfrac{L}{2}=2.5\;\text{m}\) from either end.

Let \(f\) be the friction force exerted by the rough floor, \(N_f\) the normal reaction of the floor, and \(N_w\) the normal reaction of the smooth wall (horizontal).

Equilibrium of forces gives

$\mathrm{N\_f}=200N$ (vertical equilibrium)

$\mathrm{N\_w}=f$ (horizontal equilibrium, because the wall is smooth).

Take moments about the point where the rod touches the floor (the pivot). The friction force and the floor normal pass through this point, so they produce no torque.

The torque due to the weight is

The clockwise torque from the wall’s normal force \(N_w\) is

Setting the sum of torques to zero (counter‑clockwise positive):

$\mathrm{N\_w}\times 2.5=250\sqrt{3}$

Hence

$\mathrm{N\_w}=250\sqrt{3}/2.5=100\sqrt{3}N$

Because \(N_w = f\), the friction force exerted by the floor is

$f=100\sqrt{3}N$

Thus the required friction force is \(100\sqrt{3}\;\text{N}\).