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(A) For the block not to move,the applied horizontal force must be less than or equal to the limiting friction force.The horizontal component of the a

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$20$

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(A) For the block not to move,the applied horizontal force must be less than or equal to the limiting friction force.The horizontal component of the applied force is $F_x = F \cos 60^{\circ} = F/2$.The vertical forces acting on the block are the weight $W = mg = \sqrt{3}g$ (taking $g = 10 \ m/s^2$,$W = 10\sqrt{3} \ N$) and the vertical component of the applied force $F_y = F \sin 60^{\circ} = F\sqrt{3}/2$.The normal reaction $R$ is $R = W + F \sin 60^{\circ} = 10\sqrt{3} + F\sqrt{3}/2$.The limiting friction force is $f_L = \mu R = \frac{1}{2\sqrt{3}} (10\sqrt{3} + \frac{F\sqrt{3}}{2})$.For the block to remain stationary,$F \cos 60^{\circ} \le f_L$.Setting $F \cos 60^{\circ} = f_L$ to find the maximum force:$\frac{F}{2} = \frac{1}{2\sqrt{3}} (10\sqrt{3} + \frac{F\sqrt{3}}{2})$$\frac{F}{2} = \frac{10\sqrt{3}}{2\sqrt{3}} + \frac{F\sqrt{3}}{4\sqrt{3}}$$\frac{F}{2} = 5 + \frac{F}{4}$$\frac{F}{2} - \frac{F}{4} = 5$$\frac{F}{4} = 5 \implies F = 20 \ N$.

What is the maximum force $F$ (in $N$) that can be applied such that the block does not move? (Given $\mu = 1/(2\sqrt{3})$ and $m = \sqrt{3} \ kg$)

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