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 normal reaction of the floor on the stick. The ratio $\frac{h}{l}$ and the frictional force f at the bottom of the stick are $(g = 10 m s^{- 2})$

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$\frac{h}{l} = \frac{3}{16}, f = \frac{16 \sqrt{3}}{3} N$

$\frac{h}{l} = \frac{3 \sqrt{3}}{16}, f = \frac{16 \sqrt{3}}{3} N$

$\frac{h}{l} = \frac{\sqrt{3}}{16}, f = \frac{16 \sqrt{3}}{3} N$

$\frac{h}{l} = \frac{3 \sqrt{3}}{16}, f = \frac{8 \sqrt{3}}{3} N$

answer is D.

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Detailed Solution

$\begin{array}{l} V e r t i c a l d i r e c t i o n, N + \frac{R}{2} = m g = 1.6 \times 10 \\ \Rightarrow N = 16 - \frac{R}{2} \\ H o r i z o n t a l D i r e c t i o n, f = \frac{\sqrt{3} R}{2} \\ \Rightarrow R = \frac{2 f}{\sqrt{3}} \\ About A, T_{C W} = T_{A C W} \\ \Rightarrow (R) \frac{h}{\cos 30} = (m g) \frac{l}{2} \cos 60 \\ \Rightarrow R [\frac{2 h}{\sqrt{3}}] = (16) \frac{l}{2} \times \frac{1}{2} \\ \Rightarrow \frac{h}{l} = \frac{2 \sqrt{3}}{R} \end{array}$

If the reaction of the floor on stick is taken as normal reaction, then N=R

$\Rightarrow 16 - \frac{R}{2} = R \Rightarrow R = \frac{32}{3} N$
f = $\frac{\sqrt{3}}{2} R = \frac{\sqrt{3}}{2} [\frac{32}{3}] = \frac{16 \sqrt{3}}{3} N$
$\frac{h}{l} = \frac{2 \sqrt{3}}{R} = \frac{2 \sqrt{3}}{32 / 3} = \frac{3 \sqrt{3}}{16}$