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
Down the plane: $mg\sin\theta = 30$ N. Maximum friction up the plane: $\mu_s mg\cos\theta = 0.5 \times 40 = 20$ N.
Friction alone cannot hold the block (tan 37° = 0.75 > 0.5), so the extra force needed is
$$F_{min} = mg(\sin\theta - \mu_s\cos\theta) = 30 - 20 = 10\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics