Q 11-04-023JEE MainAIEEE 2011Medium
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 $\theta$ with the horizontal such that $\tan\theta = 2\mu$, then the ratio $\dfrac{F_1}{F_2}$ is
Answer: (D) $3$
Pushing up, friction acts down the plane: $F_1 = mg(\sin\theta + \mu\cos\theta)$.
Preventing sliding down, friction acts up the plane: $F_2 = mg(\sin\theta - \mu\cos\theta)$.
$$\frac{F_1}{F_2} = \frac{\tan\theta + \mu}{\tan\theta - \mu} = \frac{2\mu + \mu}{2\mu - \mu} = 3$$
Solution by Sreeraj P, M.Sc Physics