A uniform wooden stick of mass $1.6\,\text{kg}$ 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 $h/l$ and the frictional force $f$ at the bottom of the stick are ($g = 10\,\text{m}\,\text{s}^{-2}$)

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

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Answer: D. $\frac{h}{l} = \frac{3\sqrt{3}}{16}, f = \frac{16\sqrt{3}}{3}\,\text{N}$

System of Particles and Rotational Motion A uniform wooden stick of mass $1.6\,\text{kg}$ 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 $h/l$ and the frictional force $f$ at the bottom of the stick are ($g = 10\,\text{m}\,\text{s}^{-2}$) Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

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