In the system shown in figure pully is smooth. Strings is massless and inextensible.Find acceleration of the system a, tension `T_(1)` and`T_(2)` `(g=10m//s^(2))`

Explanation

Here net pulling force will be
Weight of `4kg` and `6kg` blocks on one side -weight of `2kg` blocks on one the other side - therefore.
`a=("Net pulling force")/("Total mass")`
`=((6xx10)+(4xx10)-(2)(10))/(6+4+2)`
`=(20)/(3)m//s^(2)`
for `T_(1)`, Let us consider `FBD` of 2kg block. Writing equation of motion, we get.
`T_(1)-20=2_(a)` or `T_(1)+20xx2(20)/(3)=(100)/(3)N`
For `T_(2)`,We may consider `FBD` of` 6kg` block. Writing equation of motion, we get.
`60 - T_(2) = 6_(a)`
`T_(2) =60 - 6_(a) = 60 -6 [(20)/(3)]`
`= (60)/(3)N`