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 $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Professor PC Thomas Classes · Answer

(d) : Given that : Mass of rod, $m=20 \mathrm{~kg}$, length of rod, $l=5 \mathrm{~m}$ Weight of the uniform $\operatorname{rod}=m g$ $ =20 	imes 10=200 \mathrm{~N} $ \ For vertical direction, $W=N_2$ For horizontal direction, $f=N_1$ (where $f$ is the frictional force) Since rod is stationary, so net torque is zero (about $B$ ) ⇒ Distance from wall to the centre of mass of $\operatorname{rod}(L / 2)$ is  \ $ D=\frac{L}{2} \cos 30^{\circ}=\frac{1}{2}\left(\frac{\sqrt{3}}{2}ight)=\frac{\sqrt{3}}{4} $ \ ⇒ Now, distance $O A=L \sin 30^{\circ}=\frac{L}{2}$.\ Taking torque about point $B$ to be zero. \ $ \begin{aligned} & \Rightarrow \quad m g 	imes \frac{L}{2} \cos 30^{\circ}=N_1 	imes L \sin 30^{\circ} \ & \Rightarrow \quad m g 	imes \frac{1}{2} 	imes \frac{\sqrt{3}}{2}=f 	imes \frac{1}{2} \quad\left[	ext { As, } N_1=fight] \ & \Rightarrow \quad 200 	imes \frac{\sqrt{3}}{4}=\frac{f}{2} 	ext { or } f=100 \sqrt{3} \mathrm{~N} \end{aligned} $