Q. 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 :
AIEEEAIEEE 2011Laws of Motion
Solution:
$F_1$ = mg sin$\theta$ + $\mu$mg cos$\theta$
$F_2$ = mg sin$\theta$ - $\mu$mg cos$\theta$
$\frac{F_{1}}{F_{2}} = \frac{sin\, \theta + \mu\, cos \,\theta}{sin\, \theta - \mu \, cos \,\theta }$
$\frac{tan\theta+\mu}{tan\theta-\mu} = \frac{2\mu+\mu}{2\mu-\mu} = \frac{3\mu}{\mu} = 3.$
