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 $\theta$ with the horizontal such that $tan\theta =2\mu$, then the ratio $\frac{F_1}{F_2}$ is

• A. 4
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (D)

3

The minimum force required to start pushing a body up a rough inclined plane is

$F_1=mgsin\theta +\mu gcos\theta$ ......(i)

Minimum force needed to prevent the body from sliding down the inclined plane is

$F_2=mgsin\theta -\mu gcos\theta$ ...........(ii)

Divide (i) by (ii), we get

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