Two masses of 5 kg and 3 kg are suspended with help of massless inextensible strings as shown in figure. Calculate `T_(1)` and `T_(2)` when whole system is going upwards with acceleration = 2 `m/s^(2)` 2 (use g = 9.8 `ms^(-2)`).

Explanation

Given `m_(1) = 5 kg m_(2) = 3kg`
`g= 9.8m//s^(2) ` and `a=m//s^(2)`

FBD force body diagram for block of 5 kg and 3 kg are shown in figure
For block of mass 5 kg
`T_(1) -T_(2) - 5g = 5a`
`T_(1) - T_(2) =5g + 5a`
`T_(1) -T_(2) = 5 (g + a)`
For block of mass 3 kg
`T_(2) -3g = 3a`
`T_(2) =3g + 3a`
From equation (i )
`T_(1) -T_(2) =5 (g + a)`
`T_(2) =T_(2) + 5(g + a)`
`=8(g + a)`
`= 8(9.8 + 2)`
`=94.4 N`