$\left[g=10 m / s ^{2} ; \sin 60^{\circ}=\frac{\sqrt{3}}{2} ; \cos 60^{\circ}=\frac{1}{2}\right]$
\(\& N=\sin 60^{\circ}+\sqrt{3} g\) \(\ldots(2)\)
From equation
\((1)\;and\;(2)\)
\(\frac{F}{2}=\frac{1}{3 \sqrt{3}}\left(\frac{F \sqrt{3}}{2}+\sqrt{3} g\right)\)
\(\Rightarrow F = g =10\) Newton \(=3 x\)
So \(x=\frac{10}{3}=3.33\)