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 2011, Difficult
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$F _{1}= mg (\sin \theta+\mu \cos \theta)$
$F _{2}= mg (\sin \theta-\mu \cos \theta)$
$\frac{ F _{1}}{ F _{2}}=\frac{\sin \theta+\mu \cos \theta}{\sin \theta-\mu \cos \theta}$
$\frac{ F _{1}}{ F _{2}}=\frac{\tan \theta+\mu}{\tan \theta-\mu}$
$\frac{ F _{1}}{ F _{2}}=\frac{2 \mu+\mu}{2 \mu-\mu}$
$\frac{ F _{1}}{ F _{2}}=3$
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