Q 11-04-050JEE MainMedium
A $5$ kg block rests on a rough plane inclined at $37°$ to the horizontal ($\mu_s = 0.5$). Find the minimum force, in newtons, that must be applied along the plane (upwards) to keep the block from sliding down. (Take $\sin 37° = 0.6$, $\cos 37° = 0.8$, $g = 10\ \text{m/s}^2$.)
Numerical value type. Enter your answer.
Answer: 10
Concept. Friction (up the plane) cannot exceed $\mu_smg\cos\theta$; the applied force supplies the rest.
Step 1. Forces. $mg\sin\theta = 30$ N, $\mu_smg\cos\theta = 20$ N.
Step 2. Minimum force. $30 - 20 = 10$ N.
Answer. $10$.
Solution by Sreeraj P, M.Sc Physics