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

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Question

(C) Let $m$ be the mass of the body,$	heta$ be the angle of inclination,and $\mu$ be the coefficient of friction.The force required to move the body

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $m$ be the mass of the body,$	heta$ be the angle of inclination,and $\mu$ be the coefficient of friction.The force required to move the body up the inclined plane is $F_{up} = mg \sin 	heta + \mu mg \cos 	heta$.The force required to prevent the body from sliding down the inclined plane is $F_{down} = mg \sin 	heta - \mu mg \cos 	heta$.According to the problem,$F_{up} = 2 F_{down}$.Substituting the expressions,we get:$mg \sin 	heta + \mu mg \cos 	heta = 2(mg \sin 	heta - \mu mg \cos 	heta)$Dividing both sides by $mg$:$\sin 	heta + \mu \cos 	heta = 2 \sin 	heta - 2 \mu \cos 	heta$Rearranging the terms to solve for $\mu$:$3 \mu \cos 	heta = \sin 	heta$$\mu = \frac{1}{3} 	an 	heta$Given $	heta = 60^{\circ}$,we have:$\mu = \frac{1}{3} 	an 60^{\circ} = \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. If the angle of inclination of the plane is $60^{\circ}$,then the coefficient of friction is

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