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 :

• A. 3
• B. 4
• C. 1
• D. 2

Answer 1

Answer: (A)

3

From the Free Body Diagrams, we have

$F_1=mgSin\theta +\mu mgCos\theta$ ...$\left[1\right]$

$F_2=mgSin\theta -\mu mgCos\theta$ ...$\left[2\right]$

On adding equations 1 and 2, we have

$Sin\theta =\frac{F_1+F_2}{2mg}$ ...$\left[3\right]$

On subtracting equations 1 and 2, we have

$\mu Cos\theta =\frac{F_1-F_2}{2mg}$ ...$\left[4\right]$

Dividing equations 3 and 4, we have

$\frac{Sin\theta }{\mu Cos\theta }=\frac{F_1+F_2}{F_1-F_2}$

$\frac{Tan\theta }{\mu }=\frac{F_1+F_2}{F_1-F_2}$

Given, $Tan\theta =2\mu$

$\frac{2\mu }{\mu }=\frac{F_1+F_2}{F_1-F_2}$

$2\times \left(F_1-F_2\right)=F_1+F_2$

$3F_2=F_1$

$\frac{F_1}{F_2}=3$

Hence, the answer is OPTION A.