Question

Consider the arrangement as shown in the figure. The lift is moving with acceleration `(g)//(4)` in vertically downward direction. Find the contact force (a) between the blocks (b) between the lower block and the floor of lift.

Answer 1

`mg-N_(1)=ma_(0)=(mg)/(4)rArrN_(1)=(3mg)/(4)`

`2mg+N_(1)-N_(2)=2ma_(0)=2mxx(g)/(4)=(mg)/(2)`
`2mg+(3mg)/(4)-N_(2)=(mg)/(2)`
`N_(2)=(9mg)/(4)`
OR
(m + 2m)g - `N_(2) = ( m + 2m) a_(0) = 3m xx (g)/(4)`
`N_(2) = (9mg)/(4)`