In the figure shown ,find :`(g = 10 m//s^(2))`

(a) acceleration of `1 kg,2 kg` and `3 kg` blocks and
(b) tension `T_(1)` and `T_(2)`.

Explanation

`P_(1) 2T_(1) = 2T_(2)`
` :. T_(1) = T_(2)` …(i)
`1 kg T_(2) = (1) (a)` …(ii)
`2 kg T_(1) - 20 = 2 (a_(r) - (a)/(2))` …(iii)
`3 kg 30 - T_(1) = 3 (a_(r) + (a)/(2))` …(iv)

On solving these equation, we get
`T_(1) = T_(2) = (120)/(11) N`
`a_(1) = a = (120)/(11) m//s^(2)`
`a_(2) = a_(r) - (a)/(2) = - (50)/(11) m//s^(2)`
or `a_(2) = (50)/(11) m//s^(2)` (downwards)
`a_(3) = a_(r) + (a)/(2) = (70)/(11)m//s^(2)` (downwards)