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

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(A) Let $F$ be the maximum value of the force applied such that the block of mass $m = \sqrt{3} \ kg$ does not move on the rough surface.$R$ is the no

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) Let $F$ be the maximum value of the force applied such that the block of mass $m = \sqrt{3} \ kg$ does not move on the rough surface.$R$ is the normal reaction force exerted by the surface on the block.Resolving the forces vertically:$R = F \sin 60^{\circ} + mg$Resolving the forces horizontally,the limiting friction $f = \mu R$ must balance the horizontal component of the applied force $F \cos 60^{\circ}$:$f = F \cos 60^{\circ}$$\mu R = F \cos 60^{\circ}$Substituting the expression for $R$:$\mu (F \sin 60^{\circ} + mg) = F \cos 60^{\circ}$$\mu F \sin 60^{\circ} + \mu mg = F \cos 60^{\circ}$$F (\cos 60^{\circ} - \mu \sin 60^{\circ}) = \mu mg$$F = \frac{\mu mg}{\cos 60^{\circ} - \mu \sin 60^{\circ}}$Given $\mu = \frac{1}{2 \sqrt{3}}$,$m = \sqrt{3} \ kg$,and $g = 10 \ m/s^2$:$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})} = \frac{5}{\frac{1}{2} - \frac{1}{4}} = \frac{5}{\frac{1}{4}} = 20 \ N$Therefore,the maximum value of the force is $20 \ N$.

What is the maximum value of the force $F$ such that the block shown in the arrangement does not move? (in $N$)

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