The minimum force required to start pushing a body up a rough (having co-efficient of friction $\mu$ ) inclined plane is $\vec{F}_1$ while the minimum force needed to prevent it from sliding is $\vec{F}_2$. If the inclined plane makes an angle $\theta$ with the horizontal such that $\tan \theta=2 \mu$, then the ratio $F_1 / F_2$ is

• A

  4
• B

  1
• C

  2
• Correct answer
  D

  3

Solution

When body is being prevented from sliding down $\begin{aligned} & F_{\min }=F_2=m g \sin \theta-\mu m g \cos \theta \\ & F_2=m g \sin \theta-\mu m g \cos \theta\end{aligned}$ When body is pushed up friction acts downward $\mathrm{F}_1=\mathrm{mg} \sin \theta+\mu \mathrm{mg} \cos \theta$ $\left.\begin{array}{l}F_1-F_2=2 \mu \mathrm{mg} \cos \theta \\ F_1+F_2=2 m g \sin \theta\end{array}\right\} \quad \frac{F_1+F_2}{F_1-F_2}=\frac{\tan \theta}{\mu}=2: 1 \Rightarrow \frac{2 F_1}{2 F_2}=\frac{2+1}{2-1}$ or, $\frac{F_1}{F_2}=3: 1$

Hence, the correct answer is option: 'D'.