question_answer

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

A) \[{{\tan }^{-1}}\mu \]

B) \[{{\tan }^{-1}}\left( \mu /2 \right)\]

C) \[{{\tan }^{-1}}2\mu \]

D) \[{{\tan }^{-1}}3\mu \]

Correct Answer: D

Solution :

[d] In case [a]   In case [b] \[mg\sin \theta ={{F}_{1}}-\mu N\] \[\text{N=mg sin}\theta \text{ }\therefore \text{mg sin}\theta \text{+}\mu \text{mg cos}\theta \text{=}{{\text{F}}_{1}}\] \[\text{In second case }\left( b \right)\] \[\mu N+{{F}_{2}}=\text{mg sin}\theta \] \[\Rightarrow \mu \text{mg cos}\theta -{{F}_{2}}=mg\sin \theta \]\[\text{or }{{F}_{2}}=mg\sin \theta -\mu mg\cos \theta \text{ but }{{F}_{1}}=2{{F}_{2}}\] \[\text{therefore mg sin}\theta +\mu mg\cos \theta \]\[=2\left( mg\sin \theta -\mu mg\cos \theta  \right)\]\[mg\sin \theta =3\mu mg\cos \theta \] \[\text{or }\tan \theta =3\mu \text{ or }\theta \text{=ta}{{\text{n}}^{-1}}\left( 3\mu  \right)\]