Question

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

Answer 1

$F_1=mg(sin\theta +\mu cos\theta )$ [as body is just in position to move up, friction force is downwards]
$F_2=mg(sin\theta -\mu cos\theta )$ [as body is just in position to slide down, friction is upwards]

$\therefore \frac{F_1}{F_2}=\frac{sin\theta +\mu cos\theta }{sin\theta -\mu cos\theta }=\frac{tan\theta +\mu }{tan\theta -\mu }=\frac{2\mu +\mu }{2\mu -\mu }=3$