The minimum force required along the plane to start pushing a body up a rough (frictional coefficient $μ$ ) inclined plane is $F_{1}$ while the maximum force along the plane needed to prevent it from sliding down is $F_{2}$ . If the inclined plane makes an angle $θ$ with the horizontal such that $\tan θ = 2 μ$ , then the ratio $\frac{F_{1}}{F_{2}}$ is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 3.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$F_{1} = m g \sin θ + μ m g c o s θ$

$F_{2} = m g \sin θ - μ m g c o s θ$

$∴ \frac{F_{1}}{F_{2}} = \frac{\tan θ + μ}{\tan θ - μ} = 3$

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test