The minimum value of F so that the block is i | vedclass

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(C) For the block to be in equilibrium,the applied force $F$ must balance the component of gravity along the incline and the frictional force. The ang

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$mg/2$

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(C) For the block to be in equilibrium,the applied force $F$ must balance the component of gravity along the incline and the frictional force. The angle of the incline is $\alpha = 60^{\circ}$ and the coefficient of friction is $\mu = 1/\sqrt{3} = 	an 30^{\circ}$.Resolving forces parallel and perpendicular to the incline:Parallel to the incline: $F \cos 	heta = mg \sin 60^{\circ} - f$Perpendicular to the incline: $N + F \sin 	heta = mg \cos 60^{\circ}$For minimum $F$,the friction $f$ must be at its maximum value,$f_{max} = \mu N = \frac{1}{\sqrt{3}} N$.Substituting $N = mg \cos 60^{\circ} - F \sin 	heta$ into the parallel force equation:$F \cos 	heta = mg \sin 60^{\circ} - \frac{1}{\sqrt{3}} (mg \cos 60^{\circ} - F \sin 	heta)$$F \cos 	heta = mg \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{3}} (mg \frac{1}{2} - F \sin 	heta)$$F \cos 	heta = \frac{\sqrt{3}}{2} mg - \frac{1}{2\sqrt{3}} mg + \frac{1}{\sqrt{3}} F \sin 	heta$$F (\cos 	heta - \frac{1}{\sqrt{3}} \sin 	heta) = mg (\frac{3-1}{2\sqrt{3}}) = mg \frac{2}{2\sqrt{3}} = \frac{mg}{\sqrt{3}}$$F (\frac{\sqrt{3} \cos 	heta - \sin 	heta}{\sqrt{3}}) = \frac{mg}{\sqrt{3}}$$F (2 (\frac{\sqrt{3}}{2} \cos 	heta - \frac{1}{2} \sin 	heta)) = mg$$F (2 \sin(60^{\circ} - 	heta)) = mg$$F = \frac{mg}{2 \sin(60^{\circ} - 	heta)}$To minimize $F$,$\sin(60^{\circ} - 	heta)$ must be maximum,i.e.,$1$. Thus,$F_{min} = mg/2$.

The minimum value of $F$ so that the block is in equilibrium is

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