Q: 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° 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 h/l and the frictional force f at the bottom of the stick are (g = 10 ms^{-2})

(A) $\frac{h}{l} = \frac{\sqrt{3}}{16} \; , f = \frac{16\sqrt{3}}{3}$

(B) $\frac{h}{l} = \frac{3}{16} \; , f = \frac{16\sqrt{3}}{3}$

(C) $\frac{h}{l} = \frac{3\sqrt{3}}{16} \; , f = \frac{8\sqrt{3}}{3}$

(D) $\frac{h}{l} = \frac{3\sqrt{3}}{16} \; , f = \frac{16\sqrt{3}}{3}$

Ans: D

Solution : $\displaystyle \Sigma F_x = 0 $

$\displaystyle R_1 cos30^o – f = 0 $

$\displaystyle R_1 cos30^o = f $ …(i)

$\displaystyle \Sigma F_y = 0 $

$\displaystyle R_1 sin30^o + R_2 – m g = 0 $

$\displaystyle R_1 sin30^o + R_2 = m g $

$\displaystyle \frac{R_1}{2} + R_1 = m g $ (Since R_{1} = R_{2})

$\displaystyle \frac{3 R_1}{2} = m g $

$\displaystyle R_1 = \frac{2}{3}m g = R_2 $

$\displaystyle R_1 = \frac{2}{3}\times 16 = R_2 $

$\displaystyle R_1 = \frac{32}{3} = R_2 $

Hence , $\displaystyle f = R_1 cos30 = \frac{2}{3}m g \times \frac{\sqrt{3}}{2}$

$\displaystyle f = \frac{2}{3}\times 16 \times \frac{\sqrt{3}}{2}$

$\displaystyle f = \frac{16\sqrt{3}}{3} $

$\displaystyle \Sigma \tau_o = 0 $

$\displaystyle mg \frac{l}{2}cos60^o – R_1(l – x) = 0 $

$\displaystyle mg \frac{l}{2}\times \frac{l}{2} = \frac{2}{3}m g (l – x) $

$\displaystyle \frac{l}{4} = \frac{2}{3} (l – x) $

x = 5l/8

$\displaystyle cos30 = \frac{h}{l-x}$

$\displaystyle \frac{\sqrt{3}}{2} = \frac{h}{l-\frac{5l}{8}}$

$\displaystyle \frac{\sqrt{3}}{2} = \frac{h}{\frac{3l}{8}}$

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