A uniform wooden stick of mass $1.6 \mathrm{~kg}$ and length $l$ rest in an inclined manner on a smooth, vertical wall of height $h(

Question

A uniform wooden stick of mass $1.6 \mathrm{~kg}$ and length $l$ rest 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 $\left(g=10 \mathrm{~ms}^{\circ}\right)$

Aicharya · Accepted Answer

Balancing torque about the lowest point, we get
$\mathrm{N} \frac{\mathrm{h}}{\sin 60^{0}}=\mathrm{mg} \frac{1}{2} \cos 60^{\circ}$
Also, $\mathrm{N}+\frac{\mathrm{N}}{2}=\mathrm{mg}$
$\mathrm{N}=\frac{2 \mathrm{mg}}{3}$
$\frac{4 \mathrm{mg}}{3 \sqrt{3}} \mathrm{~h}=\mathrm{mg} \frac{1}{4}$
$\frac{\mathrm{h}}{1}=\frac{3 \sqrt{3}}{16}$
$\mathrm{f}=\mathrm{N} \sin 60^{0}=\frac{\mathrm{mg}}{\sqrt{3}}=\frac{16}{\sqrt{3}}$
Answer is option D.

A uniform wooden stick of mass \(1.6 \mathrm{~kg}\) and length \(l\) rest in an inclined manner on a smooth,…

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