The maximum value of the force $F$ such that the block shown in the arrangement, does not move is

Let $F$ be the maximum value of force applied when the block of $m=\sqrt{3}kg$ does not move on the rough surface. 

$R=$normal reaction

or $R=F \sin 60°+mg$

$f$= force of friction

$$\begin{gathered} \mu R=F \cos 60° \\ \mu \left(F \sin 60°+mg\right)=F \cos 60° \end{gathered}$$

or $\mu F \sin 60°+\mu mg=F \cos 60°$

or $F=\frac{\mu mg}{\cos 60°-\mu \sin 60°}$

       $=\frac{{\displaystyle \frac{1}{2\sqrt{3}}}\times \sqrt{3}\times 10}{{\displaystyle \frac{1}{2}}-\left({\displaystyle \frac{1}{2\sqrt{3}}}\times {\displaystyle \frac{\sqrt{3}}{2}}\right)}=\frac{5}{{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{4}}}=5\times 4= 20 N$

Maximum value of force = $20 N$