Question

A stationary uniform rod of mass $m=1\text{kg}$ and length $l=2\text{m}$ leans against a smooth vertical wall making an angle $\theta ={45}^o$ with rough horizontal floor. Find the frictional force that is exerted by the floor on the rod.

• A. $5\text{N}$
• B. $10\text{N}$
• C. $20\text{N}$
• D. $15\text{N}$

Answer 1

The correct option is A $5\text{N}$

As the rod is stationary so the linear acceleration and angular acceleration of rod is zero.
i.e., $a_{com}=0;\alpha =0$
$\begin{array}{c}{N}_2=f\\ N_1=mg\end{array}\}\because a_{com}=0$


Torque about any point of the rod should also be zero
$\therefore \alpha =0$
${\tau }_A=0\Rightarrow mg\cos \theta \frac{l}{2}+fl\sin \theta =N_1.l\cos \theta$
$\Rightarrow N_1\cos \theta =f\sin \theta +\frac{mg\cos \theta }{2}$
$\Rightarrow f=\frac{mg\cos \theta }{2\sin \theta }=\frac{mg\cot \theta }{2}$
$\Rightarrow f=\frac{1\times 10\times \cot {45}^o}{2}$
$\Rightarrow f=\frac{10}{2}=5\text{N}$
So, option (a) is correct.