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

Answer 2

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

$F_{1} = mg\sin\theta + \mu mg\cos\theta.......(i)$

(b)

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

$F_{2} = mg\sin\theta - \mu mg\cos\theta.......(i)$

Divide (i) by (ii) we get

$\frac{F_{1}}{F_{2}} = \frac{\sin\theta + \mu\cos\theta}{\sin\theta - \mu\cos\theta} = \frac{\tan\theta + \mu}{\tan\theta - \mu}$

$= \frac{2\mu + \mu}{2\mu - \mu} = 3$ $(\because\tan\theta = 2\mu(Given))$