Friction influencing the motion of a mass and non-inertial frame of reference [closed]

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. 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?

• I don't understand why this was put on hold. The user is not asking us to solve the question, he is merely confused about a part in the question. It's obvious that he/she has put in effort to solve it. – user42733 Mar 31 '14 at 15:49
• @ParthVader because there's no conceptual question in here. "how do I get to those last two equations?" is not a conceptual question. – David Z Apr 1 '14 at 23:49

There is a pseudo force acting on $m$ because of the accelerating $M$ which will be equal to $$F'=ma$$ where $a$ is the acceleration of $M$. To find $a$, $$F=(M+m)a$$ Thus,$$F'=\frac{mF}{M+m}$$
But, there is a frictional force of $\mu mg$ on the opposite side and the net force is $0.25g$ on $m$ (which you already calculated). So,$$0.25g=F'-\mu mg$$ I think you can solve the rest.