The force required to move a body up a rough | vedclass

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(C) For upward motion,the force required is $F_{up} = mg(\sin 	heta + \mu \cos 	heta)$.For downward motion,the force required to prevent sliding is

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}}$

Vedclass · Answer

(C) For upward motion,the force required is $F_{up} = mg(\sin 	heta + \mu \cos 	heta)$.For downward motion,the force required to prevent sliding is $F_{down} = mg(\sin 	heta - \mu \cos 	heta)$.According to the problem,$F_{up} = 2 F_{down}$.Substituting the expressions,we get:$mg(\sin 	heta + \mu \cos 	heta) = 2 mg(\sin 	heta - \mu \cos 	heta)$$\sin 	heta + \mu \cos 	heta = 2 \sin 	heta - 2 \mu \cos 	heta$$3 \mu \cos 	heta = \sin 	heta$$\mu = \frac{1}{3} 	an 	heta$Given $	heta = 60^{\circ}$,we have $	an 60^{\circ} = \sqrt{3}$.Therefore,$\mu = \frac{1}{3} 	imes \sqrt{3} = \frac{1}{\sqrt{3}}$.

The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. The coefficient of friction,when the angle of inclination of the plane is $60^{\circ}$,is

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