# 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?

## closed as off-topic by Brandon Enright, David Z♦Mar 31 '14 at 15:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

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• 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.