The minimum force required to move a body up | vedclass

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(C) The minimum force required to move a body up a rough inclined plane is given by $F_1 = mg(\sin 	heta + \mu \cos 	heta)$.The minimum force requir

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$30$

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(C) The minimum force required to move a body up a rough inclined plane is given by $F_1 = mg(\sin 	heta + \mu \cos 	heta)$.The minimum force required to prevent the body from sliding down the rough inclined plane is given by $F_2 = mg(\sin 	heta - \mu \cos 	heta)$ if the force is applied parallel to the plane. However,if we consider the force required to just prevent sliding (static equilibrium),it is $F_2 = mg(\sin 	heta - \mu \cos 	heta)$. Given the standard interpretation for this problem type where $F_2$ is the force to prevent sliding down,we use $F_2 = mg(\sin 	heta - \mu \cos 	heta)$.Given $F_1 = 3F_2$,we have $mg(\sin 	heta + \mu \cos 	heta) = 3mg(\sin 	heta - \mu \cos 	heta)$.Dividing by $mg$,we get $\sin 	heta + \mu \cos 	heta = 3\sin 	heta - 3\mu \cos 	heta$.Rearranging terms,$4\mu \cos 	heta = 2\sin 	heta$.Thus,$	an 	heta = 2\mu$.Substituting $\mu = \frac{1}{2\sqrt{3}}$,we get $	an 	heta = 2 	imes \frac{1}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$.Therefore,$	heta = 30^{\circ}$.

The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and the inclined plane is $\frac{1}{2 \sqrt{3}}$,then the angle of the inclined plane is (in $^{\circ}$)

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