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Q.
What is the maximum value of the force $F$ such that the block shown in the arrangement, does not move?
IIT JEEIIT JEE 2003Laws of Motion
Solution:
Free body diagram (FBD) of the block (shown by a dot) is shown in figure
For vertical equilibrium of the block
$N = mg + F \sin 60^{\circ}$
$=\sqrt{3}g+\sqrt{3}\frac{F}{2} ...(i)$
For no motion, force of friction
$f \ge \cos 60^{\circ}$ or $\mu N\ge F \cos 60^{\circ}$
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} $
or $F \le 2g$ or $20\,N$
Therefore, maximum value of $F$ is $20 \,N$.
