Question

What is the maximum value of the force $F$ (in newton) such that the block shown in the arrangement, does not move? $\left(g = 10\, m s^{- 2}\right)$

Answer 1

Free body diagram (FBD) of the block (shown by a dot) is shown in figure.

For vertical equilibrium of the block,
$N=mg+F\sin 60 ^{\circ} = \sqrt{3}g+\sqrt{3}\frac{F}{2} \, \ldots .\left(i\right)$
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 $20\, N.$