The minimum force required to start pushing a body up a rough (frictional coefficient $\mu$) inclined plane is F₁ while the minimum force needed to prevent it from sliding down is F₂. If the inclined plane makes an angle $\theta$ with the horizontal such that $\tan \theta =2\mu$, then the ratio $\frac{F_1}{F_2}$ is

 

The minimum force required to start pushing a body up a rough inclined plane is 

$F_1=mg\sin \theta +\mu mg\cos \theta \dots \left(1\right)$

Minimum force needed to prevent the body from sliding down the inclined plane is

$$\begin{gathered} F_2=mg\sin \theta -\mu mg\cos \theta \dots \left(2\right) \\ Divide \left(1\right) by \left(2\right), we get \\ \frac{F_1}{F_2}=\frac{\sin \theta +\mu \cos \theta }{\sin \theta -\mu \cos \theta } \\ =\frac{\tan \theta +\mu }{\tan \theta -\mu } \\ =\frac{2\mu +\mu }{2\mu -\mu } [\because \tan \theta =2\mu (given\left)\right] \\ =3 \end{gathered}$$