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(D) For pushing the body up the inclined plane,the minimum force $F_1$ must overcome both the component of gravity acting down the plane $(mg \sin 	h

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) For pushing the body up the inclined plane,the minimum force $F_1$ must overcome both the component of gravity acting down the plane $(mg \sin 	heta)$ and the maximum static friction force acting down the plane $(\mu mg \cos 	heta)$. Thus,$F_1 = mg \sin 	heta + \mu mg \cos 	heta = mg(\sin 	heta + \mu \cos 	heta)$.For preventing the body from sliding down the inclined plane,the minimum force $F_2$ acts up the plane. The component of gravity acts down the plane $(mg \sin 	heta)$,and the maximum static friction force acts up the plane $(\mu mg \cos 	heta)$. Thus,$F_2 = mg \sin 	heta - \mu mg \cos 	heta = mg(\sin 	heta - \mu \cos 	heta)$.The ratio is given by $\frac{F_1}{F_2} = \frac{mg(\sin 	heta + \mu \cos 	heta)}{mg(\sin 	heta - \mu \cos 	heta)} = \frac{\sin 	heta + \mu \cos 	heta}{\sin 	heta - \mu \cos 	heta}$.Dividing the numerator and denominator by $\cos 	heta$,we get $\frac{F_1}{F_2} = \frac{	an 	heta + \mu}{	an 	heta - \mu}$.Given $	an 	heta = 2 \mu$,substituting this value gives $\frac{F_1}{F_2} = \frac{2 \mu + \mu}{2 \mu - \mu} = \frac{3 \mu}{\mu} = 3$.

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