What is the maximum value of the force F such | vedclass

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(A) For the block to not move,the applied horizontal force must be less than or equal to the limiting friction.$F \cos 60^{\circ} = f_L = \mu N$Vertic

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) For the block to not move,the applied horizontal force must be less than or equal to the limiting friction.$F \cos 60^{\circ} = f_L = \mu N$Vertical forces balance: $N = mg + F \sin 60^{\circ}$Substituting $N$ into the friction equation:$F \cos 60^{\circ} = \mu (mg + F \sin 60^{\circ})$$F \cos 60^{\circ} - \mu F \sin 60^{\circ} = \mu mg$$F (\cos 60^{\circ} - \mu \sin 60^{\circ}) = \mu mg$$F = \frac{\mu mg}{\cos 60^{\circ} - \mu \sin 60^{\circ}}$Given $m = \sqrt{3} \ kg$,$\mu = \frac{1}{2\sqrt{3}}$,$g = 10 \ m/s^2$,$	heta = 60^{\circ}$:$F = \frac{(\frac{1}{2\sqrt{3}}) (\sqrt{3}) (10)}{\cos 60^{\circ} - (\frac{1}{2\sqrt{3}}) \sin 60^{\circ}}$$F = \frac{5}{\frac{1}{2} - (\frac{1}{2\sqrt{3}}) (\frac{\sqrt{3}}{2})} = \frac{5}{\frac{1}{2} - \frac{1}{4}} = \frac{5}{\frac{1}{4}} = 20 \ N$

What is the maximum value of the force $F$ such that the block shown in the arrangement does not move (in $N$)? (Take $g = 10 \ m/s^2$)

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