JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Laws of Motion

The applied force $F$ pushes the block downward and to the left at $60^{\circ }$ to the horizontal, so its vertical component presses the block into the surface.

Vertical equilibrium:
$N=mg+F\sin 60^{\circ }=\sqrt{3}(10)+\frac{\sqrt{3}F}{2}=10\sqrt{3}+\frac{\sqrt{3}F}{2}$

Horizontal equilibrium at the critical (about-to-slide) condition, friction is limiting:
$F\cos 60^{\circ }=\mu N$

$\frac{F}{2}=\frac{1}{3\sqrt{3}}\left(10\sqrt{3}+\frac{\sqrt{3}F}{2}\right)$

$\frac{F}{2}=\frac{10\sqrt{3}}{3\sqrt{3}}+\frac{\sqrt{3}F}{2\cdot 3\sqrt{3}}=\frac{10}{3}+\frac{F}{6}$

$\frac{F}{2}-\frac{F}{6}=\frac{10}{3}\Rightarrow \frac{3F-F}{6}=\frac{10}{3}\Rightarrow \frac{F}{3}=\frac{10}{3}$

$F=10$ N

Given $F=3x$: $x=\frac{10}{3}=3.33\approx 3$ (rounded to the nearest integer)

Note on the physics: increasing $F$ raises the driving component $F\cos 60^{\circ }=F/2$ but also raises the friction limit, through $\mu F\sin 60^{\circ }=F/6$. Sliding is possible only because $\cos 60^{\circ }>\mu \sin 60^{\circ }$, i.e. $\mu <\cot 60^{\circ }=0.577$. Here $\mu =\frac{1}{3\sqrt{3}}\approx 0.192$, so a finite critical force exists. For $\mu \ge \cot 60^{\circ }$ the block could never be pushed, no matter how large $F$ is.