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

• A

  $100 N$
• Correct answer
  B

  $100 \sqrt{3} \mathrm{~N}$
• C

  $200 N$
• D

  $200 \sqrt{3} \mathrm{~N}$

Solution

$\mathrm{N}_2=\mathrm{mg}=200 \mathrm{~N}$
Taking torque about A : $\begin{aligned} & \mathrm{f}_{\mathrm{s}} \mathrm{~L} \cos 60^{\circ}+\mathrm{mg} \frac{\mathrm{~L}}{2} \sin 60^{\circ}=\mathrm{N}_2 \mathrm{~L} \sin 60^{\circ} \\ & \mathrm{f}_{\mathrm{s}}\left(\frac{1}{2}\right)+200\left(\frac{1}{2}\right) \frac{\sqrt{3}}{2}=200 \frac{\sqrt{3}}{2} \\ & \mathrm{f}_{\mathrm{s}}=100 \sqrt{3} \mathrm{~N} \end{aligned}$

Hence, the correct answer is option: 'B'.