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ByOm Sharma 27 January, 202527 January, 2025

According to the question,

For the equilibrium of the block
$f=m g \sin \theta$
$N=m g \cos \theta$
Required friction $f$ must be less than or equal to limiting friction.
$f \leq \mu N$
$m g \sin \theta \leq \mu m g \cos \theta$
$\tan \theta \leq \mu$
Given $y=\frac{x^{2}}{4}$
Slope $\tan \theta=\frac{d y}{d x}=\frac{1}{4}(2 x)=\frac{x}{2}$
Put in (3)

Just to push the block up friction force will be down the inclined plane
$F_{1}=m g \sin 30+f$
here $f=\mu N$ ,
$N=m g \cos 30^{\circ}$
$F_{1}=\frac{m g}{2}+\mu m g \cos 30$

$F_{1}=\frac{m g}{2}+\frac{\sqrt{3}}{2} \mu m g$
Just to prevent it from sliding down friction force will be up the inclined plane
here $f=\mu N, N=m g \cos 30^{\circ}$
$F_{2}=\frac{m g}{2}-\mu m g \cos 30^{\circ}$
$F_{2}=\frac{m g}{2}-\frac{\sqrt{3}}{2} \mu \mathrm{mg}$
$F_{1}-F_{2}=\sqrt{3} \mu \mathrm{mg}$
$\Rightarrow F_{1}-F_{1}=\sqrt{3}(0.1)(5 \times 10)$
$\left(F_{1}-F_{2}\right)=5 \sqrt{3} \mathrm{~N}$

Om Sharma

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