Find the contact force between the block and acceleration of the blocks as shown in figure.

Explanation

Considering all the three block as a system to find the common acceleration
`F_("net") = 50-30 = 20N`
`a - ( 20 )/( 10 ) = 2 m //s^(2) `
To find the contact force between B & C we draw F.B.D. of 3 kg block.
`( sum F_("net"))_(x) =ma`
`rArr N_(1) - 30 = 3 ( 2) rArr N_(1) = 36 N `
To find contact force between A & B we draw F.B.D. of 5 kg block

`rArr N_(2) - N_(1) = 5a`
`N_(2) = 5 xx 2 + 36 rarr N_(2) = 46N`