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  4. 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 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)

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Q. A uniform wooden stick of mass $1.6\, kg$ and length $\ell$ rests in an inclined manner on a smooth, vertical wall of height $h (<\ell)$ 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 $30^{\circ}$ 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 $\frac{ h }{\ell}$ and the frictional force $f$ at the bottom of the stick are $(g = 10\,ms^{-2})$

JEE AdvancedJEE Advanced 2016

Solution:

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$N \sin 30^{\circ}+ N =1.6\, g$
$\frac{3 N }{2}=1.6 \,g$
$ \Rightarrow N =\frac{3.2}{3} g \ldots$(i)
$f = N \cos 30^{\circ}=\frac{3.2}{3} g \times \frac{\sqrt{3}}{2}=\frac{16 \sqrt{3}}{2} N\ldots$(ii)
$\tau_{ A }=1.6 \times \frac{\ell}{2} \sin (90+60)- N ( x ) \sin 90^{\circ}=0$
$\Rightarrow 1.6 g \times \frac{\ell}{2} \times \frac{1}{2}= Nx =\frac{32}{3} x$
$4 \ell=\frac{32}{3} x $
$4 \ell=\frac{32}{3} \times \frac{ h }{\sqrt{3}} \times 2$
$\frac{ h }{\ell}=\frac{3 \sqrt{3}}{16}$