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 $F_1$/$F_2$.

Answer 1

frictional force is $f$

$f=\mu mg\cos \theta$

$F_1=\mu mg\sin \theta +f=mg(\sin \theta +\mu \cos \theta )$

$F_2+f=mg\sin \theta$

$F_2=mg\sin \theta -\mu mg\cos \theta =mg(\sin \theta -\mu \cos \theta )$

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