question_answer

The minimum force required to start pushing a body up a rough (frictional coefficient \[\mu \]) inclined plane is \[{{F}_{1}}\] while the minimum force needed to prevent it from sliding down is \[{{F}_{2}}.\] If the inclined plane makes an angle \[\theta \]from the horizontal such that \[\tan \theta =2\mu \]then the ratio \[\frac{{{F}_{1}}}{{{F}_{2}}}\] is :                                 [AIEEE 11-05-2011]           

A) 1

B) 2

C) 3

D) 4

Correct Answer: C

Solution :

[c]

\[{{F}_{1}}=mg\sin \theta +\mu mgcos\theta \]

\[{{F}_{2}}=mg\sin \theta -\mu mg\,cos\theta \]

\[\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{\sin \theta +\mu \,cos\theta }{\sin \theta -\mu \,cos\theta }\]

\[\frac{\tan \theta +\mu }{\tan \theta -\mu }=\frac{2\mu +\mu }{2\mu -\mu }=\frac{3\mu }{\mu }=3.\]