Question

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

Answer 1

Answer: 20

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

For vertical equilibrium of the block,
$N=mg+F\sin 60^{\circ }=\sqrt{3}g+\sqrt{3}\frac{F}{2}\dots .\left(i\right)$
For no motion, force of friction
$f\ge F\cos 60^{\circ }$
or ${}^{\circ }\mu N\ge F\cos 60^{\circ }$
or $\frac{1}{2\sqrt{3}}\left(\sqrt{3}g+\frac{\sqrt{3}F}{2}\right)\ge \frac{F}{2}$
or $g\ge \frac{F}{2}orF\le 2gor20N$
Therefore, maximum value of F is $20N.$