A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of $

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 $g = 10	ext{ }m/s^{2}$ )

Accepted Answer

For translational equilibrium$$\begin{matrix} & N_{1} = Mg \ & N_{2} = f \end{matrix}$$For rotational equilibriumTorque about $A,Mg\frac{L}{2}cos	heta = N_{2}Lsin	heta$$$\begin{matrix} & \frac{Mg}{2}cot	heta = N_{2} = f \ & \frac{Mg}{2}cot30^{\circ} = f \ & \frac{Mg}{2}\sqrt{3} = N_{2} \ & 100\sqrt{3} = f \end{matrix}$$

Question

Options

Solution

More Rotational Motion Questions