Friction influencing the motion of a mass and non-inertial frame of reference
So... recently a Newtonian-mechanics related exercise has raised in me many questions basically ragarding which forces influence a given system and how, I know they might sound bit too "noobish" but I just had to get them out of my chest (yes it really gave me something to think about in the last day).
perhaps you should take a look at the above mentioned exercise to better understand what I am trying to figure out
So, I'm trying to figure out how this system works... the exercise says (please forgive me for all the grammatical errors you will eventually find):
Problem
a cube with a mass $\ M=50Kg$ is standing on a plain surface and can move on it without friction. 766
On that cube is standing another cube with a mass $\ m=10Kg$, at a given distance $\ d=0.5 m$ from the upper-left corner of the cube. Initially, when everything is stationary, a force $\ F=100N$ is applied to the bigger cube horizontally (I assume that the force $\ F$ is constant); at the moment $\ t=2s$ the smaller cube falls. Calculate the friction coefficient between the two cubes.
Variables used:
$\ μ$= friction coefficient
$\ g$= gravitational acceleration on earth =$\ 9.81 m/s^2$
$\ a_M$= acceleration of the bigger cube ($\ M$)
$\ a_m$= acceleration of the smaller cube ($\ m$)
$\ F_f$= friction force
So, I assume that the bigger cube carries the smaller one which moves with a constant, negative acceleration until it falls (please correct me if I'm wrong), with equation of motion
$\ x(t)= x_0 + v_0t + 1/2 at^2$
so $\ 0.5m = - 1/2 at^2$ ------> $\ a= -0.25 m/s^2 $
now...
$\ F- μ mg = Ma_M$
in other terms (and again please correct me if I'm wrong): $\ F$ is partially dissipated by the friction between the cubes, and the bigger cube moves under the influence of a force which equals to the applied force $\ F$ minus the friction force $\ F_f= μ mg$
now the universe collapses: the solution says:
$\ μmg = ma_m$
$\ a_r = a_M - a_m = [ μg (m + M) -F] / M$ which leads to $\ μ=0.15 $
But... what leads to those last two equations? 512