Find the force of interaction between the bodies as shown in fig. Blocks are in contact.

Explanation

All the bodies move together along horizontal direction with an acceleration,
`a=([F_("horizontal")]_("net"))/(m_("total")) = (F cos theta)/(m_(1)+m_(2)+m_(3))`

By Newton's second law for block `m_(1)` ,
`N_(1)=m_(1)a = m_(1)((F cos theta)/(m_(1)+m_(2)+m_(3)))`
and for block `m_(2)`, we have `N_(2)-N_(1) = m_(2)a`
or `N_(2) = N_(1) + m_(2)a = N_(1)+m_(2)((F cos theta)/(m_(1)+m_(2)+m_(3)))`
and for block `m_(2)` we have `N-(2)-N_(1)=m_(2)a`
or `N_(2)=N_(1)+m_(2)a = N_(1)+m_(2)((F cos theta)/(m_(1)+m_(2)+m_(3)))`
Here the force of interaction is of compressive nature.
Hence, `N_(2)=((m_(1)+m_(2))f cos theta)/((m_(1)+m_(2)+m_(3)))`
We can caluculate the value of `N_(2)` by taking `m_(1) and m_(2)` together as system. From FBD of `'m_(1)+m_(2)'` as fig.
or `N_(2)=(m_(1)+m_(2))a=((m_(1)+m_(2))f cos theta)/((m_(1)+m_(2)+m_(3)))`.