As in the diagram three blocks connected together lie on a horizontal frictionless table and pulled to the right with a force F=50 N if `m_(1)=5kg m_(2)=10` kg and `m_(3)=15 kg` find the tensions `T_(1)` and `T_(2)`

Explanation

Given `f= 50 N`
`m_(1) = 5kg `
All the blocks move with common acceleration a under the force F=50 N
`:. F =-(m_(1) + m_(2) + m_(3)) a`
or a `(F )/(m_(1)+m_(2)+m_(3)) = (50)/(5+10+15)`
To dtermine `T_(1)` Refer to the free - body diagram for `m_(1)` shown in the diagram
Clearly the tension `T_(1) ` produces acceleration
`T_(1) = m_(1) a = 5xx (5)/(3) = (25)/(3) = 8.33 n`

to determine `T_(2)` Refer to the free-body diagram for `m_(3)` shown in the diagram (b) force F acts towards right and tension `T_(2)` acts towards left
`:. F- T_(2) = m_(3)a`
`50 = T_(2) = 15 xx (5)/(3)`
`T_(2) =25 N`