Applied force on the block on a horizontal surface, as shown in the figure, we have


As we know that, force of friction, \(f=\mu R\) From the diagram,
\(\begin{aligned}
& f=\mu\left(W+F \sin 60^{\circ}\right) \\
& F \cos 60^{\circ}=\frac{1}{2 \sqrt{3}}\left(10 \sqrt{3}+F \sin 60^{\circ}\right) \\
& \Rightarrow F \times \frac{1}{2}=\frac{1}{2 \sqrt{3}}\left(10 \sqrt{3}+F \times \frac{\sqrt{3}}{2}\right) \quad\left\{\begin{array}{l}
\because \cos 60^{\circ}=\frac{1}{2} \\
\sin 60^{\circ}=\frac{\sqrt{3}}{2}
\end{array}\right\} \\
& \Rightarrow \frac{F}{2}=\frac{1}{2 \sqrt{3}} \times \sqrt{3}\left(10+\frac{F}{2}\right) \\
& \Rightarrow \frac{F}{2}=\frac{1}{2} \frac{(20+F)}{2} \Rightarrow F=\frac{20+F}{20} \\
& \Rightarrow 2 F=20+F \Rightarrow 2 F-F=20 \\
& \therefore \quad F=20 \mathrm{~N} \\
\end{aligned}\)