Question

A uniform wooden stick of mass $1.6$ kg and length $l$ rests in an inclined manner on a smooth, vertical wall of height $h(<l)$ such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of ${30}^{\circ }$ with the wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio of $h/l$ and the frictional force $f$ at the bottom of the stick are
(g = 10 m$s^{-2}$):

• A. $\frac{h}{l}=\frac{\sqrt{3}}{16},f=\frac{16\sqrt{3}}{3}N$
• B. $\frac{h}{l}=\frac{3}{16},f=\frac{16\sqrt{3}}{3}N$
• C. $\frac{h}{l}=\frac{3\sqrt{3}}{16},f=\frac{8\sqrt{3}}{3}N$
• D. $\frac{h}{l}=\frac{3\sqrt{3}}{16},f=\frac{16\sqrt{3}}{3}N$

Answer 1

The correct option is D $\frac{h}{l}=\frac{3\sqrt{3}}{16},f=\frac{16\sqrt{3}}{3}N$

Balancing torque about the lowest point, we get

$N\frac{h}{sin{60}^0}=mg\frac{l}{2}cos{60}^0$

Also, $N+\frac{N}{2}=mg$

$N=\frac{2mg}{3}$

$\frac{4mg}{3\sqrt{3}}h=mg\frac{l}{4}$

$\frac{h}{l}=\frac{3\sqrt{3}}{16}$

$f=Nsin{60}^0=\frac{mg}{\sqrt{3}}=\frac{16}{\sqrt{3}}$

Answer is option D.