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

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

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$20$

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(A) For the block not to move,the horizontal component of the applied force $F$ must be less than or equal to the limiting friction force $f_L$.The horizontal component of the force is $F \cos 	heta$.The normal reaction $N$ on the block is $N = mg + F \sin 	heta$.The limiting friction force is $f_L = \mu N = \mu(mg + F \sin 	heta)$.Equating the horizontal force to the limiting friction for the maximum value of $F$:$F \cos 	heta = \mu(mg + F \sin 	heta)$$F \cos 	heta = \mu mg + \mu F \sin 	heta$$F(\cos 	heta - \mu \sin 	heta) = \mu mg$$F = \frac{\mu mg}{\cos 	heta - \mu \sin 	heta}$Given values: $m = \sqrt{3} \ kg$,$\mu = \frac{1}{2\sqrt{3}}$,$	heta = 60^\circ$,$g = 10 \ m/s^2$.$F = \frac{(\frac{1}{2\sqrt{3}}) 	imes \sqrt{3} 	imes 10}{\cos 60^\circ - (\frac{1}{2\sqrt{3}}) \sin 60^\circ}$$F = \frac{5}{\frac{1}{2} - (\frac{1}{2\sqrt{3}}) 	imes \frac{\sqrt{3}}{2}}$$F = \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 $N$)? (Given: $m = \sqrt{3} \ kg$,$\mu = \frac{1}{2\sqrt{3}}$,$\theta = 60^\circ$)

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