Q 11-04-037JEE MainMedium
A $5$ kg block rests on a horizontal floor ($\mu_s = 0.4$). A force $F$ is applied at $37°$ above the horizontal. The minimum value of $F$ that just starts the block moving is nearly (take $\sin 37° = 0.6$, $\cos 37° = 0.8$, $g = 10\ \text{m/s}^2$)
Answer: (B) $19.2$ N
The upward component reduces the normal force: $N = mg - F\sin 37° = 50 - 0.6F$.
Just moving: $F\cos 37° = \mu_s N$:
$$0.8F = 0.4(50 - 0.6F) \;\Rightarrow\; 1.04F = 20 \;\Rightarrow\; F \approx 19.2\ \text{N}$$
This is less than the $20$ N a horizontal pull would need ($\mu_s mg$).
Solution by Sreeraj P, M.Sc Physics