In the arrangement shown in show that tension in the string between masses `m_(2)` and `m_(3)` is `T = (2 m_(1) m_(3) g)/(m_(1) + m_(2) + m_(3))`

Explanation

Let `T_(1)` be the tension in the string between `m_(1)` and
`m_(2)` and `T` be the tension in the string between `m_(2)`
and `m_(3)` If a is acceleration of the system in the
direction shown then
`m_(1) a =T_(1) - m_(1) g` ….(i)
`(m_(2) + m_(3)) a = (m_(2) + m_(3)) g - T_(1)` ....(ii)
and `m_(3) a = m_(3) g - T` ...(iii)
Adding (i) and(ii) `(m_(1)+ m_(2) +m_(3) a = (m_(2) + m_(3) - m_(1)) g`
`a = (m_(2) + m_(3) - m_(1)) g// (m_(1) + m_(2) + m_(3))`
put in (iii) `(m_(3) (m_(2) + m_(3) -m_(1)) g)/(m_(1) + m_(2) + m_(3)) =m_(3) g - T`
`T = m_(3) g-(m_(3) (m_(2) + m_(3) -m_(1)) g)/(m_(1) + m_(2) + m_(3))`
`T = 2 m_(1) m_(3) g//(m_(1) + m_(2) + m_(3))` .