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

image of the system

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):


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 $


$\ F- μ mg = Ma_M$

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?

  • $\begingroup$ 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. $\endgroup$ – user42733 Mar 31 '14 at 15:49
  • $\begingroup$ @ParthVader because there's no conceptual question in here. "how do I get to those last two equations?" is not a conceptual question. $\endgroup$ – 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.


Not the answer you're looking for? Browse other questions tagged or ask your own question.