Explanation

(a) Free body diagram (FBD) of the block (shown by a dot) is shown in figure.
[image:5_image252.emf]
For vertical equilibrium of the block,
$N = mg + F\sin{60{^\circ} = \sqrt{3}}g + \sqrt{3}\frac{F}{2}\ \ldots.(i)$
For no motion, force of friction
$f \geq F\cos{60{^\circ}}$
$or\ \ {^\circ}\mu\ N \geq F\cos{60{^\circ}}$
$or\frac{1}{2\sqrt{3}}\left( \sqrt{3}\ g + \frac{\sqrt{3}\ F}{2} \right) \geq \frac{F}{2}$
$or\ \ g \geq \frac{F}{2}\ \ or\ \ F \leq 2\ g\ \ or\ 20\ N$
Therefore, maximum value of F is 20 N.