Sum of forces divided by total mass I was watching a video in YouTube and the guy was using a very strange way to calculate the acceleration of a system 
Is this always valid? If so, can you please provide the proof?
Note that he multiplies the mass of the disk by the constant in its inertia.
 A: $\let\a=\alpha \def\Wp{W_{\mathrm{1p}}} \def\half{{\textstyle{1 \over 2}}}$
Unfortunately you don't inform us as to your degree of confidence with
Newtonian mechanics. So it's difficult to calibrate an answer.
Surely the proposed solution is right but far from a simple application
of 2nd law. You have to decompose the problem into three sub-parts:


*

*mass on the incline with attached rope

*the pulley with the ropes

*mass vertically falling with vertical rope.
1) Mass $m_1$ is subjected to two active forces (parallel to plane):
component $\Wp$ of its weight and tension $T_1$ of the rope. Its
acceleration upwards is given by
$$m_1\,a = T_1 - \Wp.\tag1$$
2) Mass $m_2$ has the same acceleration, but downwards. If $T_2$ is
rope's tension there
$$m_2\,a = m_2\,g - T_2.\tag2$$
3) As to disk we must apply equation for angular momentum (I don't
know its usual name in english): 
$$I\,\a = r\,T_2 - r\,T_1 \tag3$$
where $\a$ is angular acceleration = $a/r$, ($r$ = disk radius),
$I=c\,M\,r^2$, $c=1/2$ for a homogeneous disk.
Adding eqs. (1) and (2), and substituting for $T_1-T_2$ its value
taken from (3)
$$(m_1 + m_2)\,a = m_2\,g - \Wp - {I \a \over r} = 
  m_2\,g - \half\,m_1\,g - \half\,M\a\,r = 
  m_2\,g - \half\,m_1\,g - \half\,M\,a$$
$$a = {m_2\,g - \half\,m_1\,g \over m_1 + m_2 + \half\,M}.$$
This example should help you to solve similar problems.
