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  Solved  Paper (Held On 11 May  2011)

A)  1

B)  2

C)  3

D)  4

Correct Answer: C

Solution :

                                \[{{F}_{1}}=mg\sin \theta +\mu mgcos\theta \]                 \[{{F}_{2}}=mg\sin \theta -\mu mg\,cos\theta \] \[\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{\sin \theta +\mu \,cos\theta }{\sin \theta -\mu \,cos\theta }\] \[\frac{\tan \theta +\mu }{\tan \theta -\mu }=\frac{2\mu +\mu }{2\mu -\mu }=\frac{3\mu }{\mu }=3.\]