The masses `m_(1) m_(2)` and `m_(3)` of the three bodies shown in fig . Are 5 , 2 and 3 kg respectively Calculate the valuse of tension `T_(1) T_(2)` and `T_(3)` when (i) the whole system is going upward with an acceleration of `2 m//s^(2)` (ii) the whole system is stationary ` (g=9.8 m//s^(2))` .
.

Explanation

All the three bodies are moving together with an an upward acc . Of `2 m//s^(2)` The force pulling the system upwards is `T_(1)` and downward pull of gravity is
` (m_(1) + m_(3)) g`
According to Newton s 2 nd law of motion
`T_(1) - (m_(1) + m_(2) + m_(3) ) g = (m_(1) + m_(3) ) a`
or `T_(1) = (m_(1) + m _(2) + m_(3) ) (a + g )`
` = (5+2+3) (2+9.8) = 118 N `
Similarly , for motion of `m_(2)` and `m_(3)` ., we write
`T_(2) = (m_(2) + m_(3) ( a+g ) = (2+3) (2 + 9.8)`
`= 590 N`
and for motion of `m_(3)`
`T_(3) = m_(3) (a+g ) = 3 (2+ 9.8 ) = 35 .4 N`
(ii) When the whole system is stationary
`a= 0 `
Using the same equations as above with a = 0 ,
` T_(1) = (m_(1) + m_(2) + m_(3)) g = 10 xx 9.8 = 98 N`
`T_(2) = (m_(2) + m_(3)) g = 5 xx 9.8 = 49 N`
`T_(3) = m_(3) xx g = 3 xx 9.8 = 29 . 4 N `