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(D) Let $m$ be the mass of the body and $	heta$ be the angle of inclination.$1$. Force required to just prevent the body from sliding down $(F_1)$:In

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

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(D) Let $m$ be the mass of the body and $	heta$ be the angle of inclination.$1$. Force required to just prevent the body from sliding down $(F_1)$:In this case,the friction force acts upwards along the plane.$F_1 + \mu mg \cos 	heta = mg \sin 	heta$$F_1 = mg \sin 	heta - \mu mg \cos 	heta$$2$. Force required to just move the body up the plane $(F_2)$:In this case,the friction force acts downwards along the plane.$F_2 = mg \sin 	heta + \mu mg \cos 	heta$Given that $F_2 = 2F_1$:$mg \sin 	heta + \mu mg \cos 	heta = 2(mg \sin 	heta - \mu mg \cos 	heta)$$\sin 	heta + \mu \cos 	heta = 2 \sin 	heta - 2 \mu \cos 	heta$$3 \mu \cos 	heta = \sin 	heta$$	an 	heta = 3 \mu$$	heta = 	an^{-1}(3 \mu)$

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

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