Repeat both the parts of the above question, If the train moves with an acceleration `a = g//2` up the plane.

Explanation

`T sin theta - mg sin 30^(@) = ma = (mg)/(2)`

`:. T sin theta = mg` …(i)
`T cos theta = mg cos 30^(@)`
`T cos theta = (sqrt3 mg)/(2)` …(ii)
Solving Eqs, (i) and (ii), we get
`theta - tan^(-1) ((2)/(sqrt3)`
and `T = (sqrt7)/(2) mg`
`= 5sqrt7 N`