Simple Atwood Machine as System of Particles
The system shown in the figure is known as simple Atwood machine. Initially the masses are held at rest and then let free. Assuming mass `m_(2)` more than the mass `m_(1)`, find acceleration of mass center and tension in the string supporting the pulley.

Explanation

We know that accelarations `a_(1)` and `a_(2)` are given by the following equations.
`a_(2) = (m_(2) - m_(1))/(m_(2) + m_(1))g downarrow` and `a_(1)=(m_(2) - m_(1))/(m_(2) + m_(1))g uparrow`
Making use of eq. , we can find acceleration `a_(c )` of the mass center. We denote upward direction positive and downward direction negative signs respectively.
`Mvec(a)_(c ) = Sigma(m_(i)bar(a)_(i))rarr" "(m_(1) + m_(2))a_(c) = m_(1)a_(1) - m_(2)a_(2)`
Substituting values of accelerations `a_(1)` and `a_(2)`, we obtain `a_(C ) = 2m_(1)m_ (2) - (m_(1)^(2) + m_(2)^(2))/(m_(1)+m_(2)^(2))`

To find tension T in the string supporting the pulley, we again use eq.(9)
`Sigma vec(F)_(i) = Mvec(a)_(c )rarr " " T - m_(1) g - m_(2) g = (m_(1) + m_(2))a_(c)`
Substituting expression obtained for `a_(c )`, we have
`T = (4m_(1) m_(2))/(m_(1) + m_(2)) g`