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(A) Let $m$ be the mass of the body,$	heta$ be the inclination,and $f = \mu mg \cos 	heta$ be the maximum static friction.Case $1$: Force $F_1$ requ

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$	heta = 	an^{-1}(3\mu)$

Vedclass · Answer

(A) Let $m$ be the mass of the body,$	heta$ be the inclination,and $f = \mu mg \cos 	heta$ be the maximum static friction.Case $1$: Force $F_1$ required to move the body up the plane.In this case,friction acts downwards along the plane.$F_1 = mg \sin 	heta + f = mg \sin 	heta + \mu mg \cos 	heta$.Case $2$: Force $F_2$ required to prevent the body from sliding down.In this case,friction acts upwards along the plane.$F_2 = mg \sin 	heta - f = mg \sin 	heta - \mu mg \cos 	heta$.Given that $F_1 = 2F_2$,we have:$mg \sin 	heta + \mu mg \cos 	heta = 2(mg \sin 	heta - \mu mg \cos 	heta)$.Dividing by $mg$:$\sin 	heta + \mu \cos 	heta = 2 \sin 	heta - 2 \mu \cos 	heta$.Rearranging the terms:$3 \mu \cos 	heta = \sin 	heta$.$	an 	heta = 3 \mu$.Therefore,$	heta = 	an^{-1}(3\mu)$.

The force required just to move a body up an inclined plane is double the force required just to prevent the body from sliding down. If $\mu$ is the coefficient of friction,the inclination of the plane to the horizontal is:

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