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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) For upward motion, the force required is $F_{	ext{up}} = mg(\sin 	heta + \mu \cos 	heta)$.For downward motion, the force required to prevent sliding is $F_{	ext{down}} = mg(\sin 	heta - \mu \cos 	heta)$.According to the problem, $F_{	ext{up}} = 2 F_{	ext{down}}$.Substituting the expressions, we get $mg(\sin 	heta + \mu \cos 	heta) = 2mg(\sin 	heta - \mu \cos 	heta)$.Dividing both sides by $mg$, we have $\sin 	heta + \mu \cos 	heta = 2 \sin 	heta - 2 \mu \cos 	heta$.Rearranging the terms, $3 \mu \cos 	heta = \sin 	heta$, which simplifies to $\mu = \frac{1}{3} 	an 	heta$.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. The coefficient of friction, when the angle of inclination of the plane is $60^{\circ}$ is

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$Assertion$ : Angle of repose is equal to the angle of limiting friction.
$Reason$ : When the body is just at the point of motion, the force of friction in this stage is called limiting friction.

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