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`