A uniform wooden stick of mass $1.6 \mathrm{~kg}$ 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 focr. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the st $ck$. The ratio $h / l$ and the frictional force $f$ at the bottom of the stick are $\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)$
IIT 2016, Advanced
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$(d)$ Considering the normal reaction of the floor and wall to be $N$ and with reference to the figure.
$Image$
By vertical equilibrium.
$N+N \sin 30^{\circ}=1.6 g \Rightarrow N=\frac{3.2 g}{3} \ldots$
By horizontal equilibrium
$f=N \cos 30^{\circ}=\frac{\sqrt{3}}{2} N=\frac{16 \sqrt{3}}{3} \operatorname{From}(i)$
Taking torque about A we get
$1.6 g \times A B=N \times x$
$1.6 g \times \frac{l}{2} \cos 60^{\circ}=\frac{3.2 g}{3} \times x \therefore \frac{3 l}{8}=x \ldots .$
But $\cos 30^{\circ}=\frac{h}{x} \therefore x=\frac{h}{\cos 30^{\circ}} \ldots (iii)$
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