Question

$F_1$ and $F_2$ are the minimum and maximum forces needed to keep a body on a rough inclined plane stationary. If $\theta$ be the angle of inclination of the surface, so that $\tan \theta =2\mu$. Find the ratio of $F_1$ and $F_2$.

• A. $1:2$
• B. $2:1$
• C. $1:3$
• D. $3:1$

Answer 1

Answer: (C)

Forces acting on the body in the two cases are shown below. The only difference is that in the first case (I) force of friction is acting is upward direction and in the second case (II), it is acting in downward direction.

For equilibrium in case I,
$\mu R+F_1=mg\sin \theta$
$F_1=mg\sin \theta -\mu R$
$=mg\sin \theta -\mu mg\cos \theta$
$[\because R=mg\cos \theta ]$
Similarly, for equilibrium in case II,
$F_2=\mu R+mg\sin \theta$
$=\mu mg\cos \theta +mg\sin \theta$
$[\because R=mg\cos \theta ]$
$\Rightarrow \frac{F_1}{F_2}=\frac{(mg\sin \theta -\mu mg\cos \theta )}{mg\sin \theta +\mu mg\cos \theta }$
$=\frac{\sin \theta -\mu \cos \theta }{\sin \theta +\mu \cos \theta }$
Putting $\mu =\frac{1}{2}\tan \theta$
$\frac{F_1}{F_2}=\frac{\sin \theta -\frac{1}{2}\sin \theta }{\sin \theta +\frac{1}{2}\sin \theta }$
$=\frac{1-\frac{1}{2}}{1+1/2}=\frac{1}{2}\times \frac{2}{3}$
$F_1:F_2=1:3$