Mass `m_(1)" & "m_(2)` are placed on a rough inclined plane as shown in figure. Find out the acceleration of the blocks and contact force in between these surface.

Explanation

As we know if `mu_(1) gt mu_(2)` both will travel together so `a_(1)=a_(2)=a`
F.B.D

Which is equivalent to
`a=(3gsin37^(@)-(f_(1)+f_(2)))/3`
`a=(18-7.2)/2=3.6m//sec^(2)`

Now F.B.D of 1 kg block is

`gsin37^(@)+N-4=(1)a`
`N=3.6+4-6=1.6` Newton