Answer
The horizontal force that should be applied has a magnitude of $141N$.
Work Step by Step
1) The upper block is under the influence of 2 horizontal forces:
- The horizontal force applied to the block $F$ as mentioned in the exercise
- Static frictional force $f_s$, which opposes $F$.
The upper block starts to slide when $F=47N$, which means $f_s^{max}=47N$.
2) The below block is under the influence of 3 horizontal forces:
- The horizontal force applied to the block $F$ as mentioned in the exercise
- Static frictional force $f_{s1}$ from the surface and static frictional force $f_{s2}$ from the upper block.
$f_{s2}$ is the same as $f_s$ from 1). Therefore, $f_{s2}^{max}=\mu_smg=47N$
$f_{s1}$ is a bit different: on the vertical side, the below block is pulled down by both its weight and the weight of the upper block (both have similar values since they have the same mass). Therefore, $F_N=2mg$
$$f_{s1}^{max}=2\mu_smg=2f_{s2}^{max}=94N$$
$F$ will only be able to cause the block to slide when it surpasses all the frictional forces, which equals $$F=f_{s1}^{max}+f_{s2}^{max}=141N$$
