Two blocks are placed at rest on a smooth fixed inclined place. A force F acts on block of mass `m_1` and is parallel to the inclined plane as shown in figure. Both blocks move up the incline. Then
(i) Draw free body diagram blocks of mass `m_1` and blocks mass `m_2`
(ii) Find acceleration of blocks of mass `m_1` and blocks mass `m_2`
(iii) Find normal reaction between the blocks of mass `m_1` and `m_2`

Explanation

(i) Let f be the normal reaction between the blocks.
FBD of mass `m_1`
FBD of mass `m_2`
(ii) From FBD of mass `m_1`:
`F-(m_1gsintheta+f)=m_1a`
From FBD of mass `m_2`
`f-m_2gsintheta=m_2a`
Adding the equations (i) and (ii):
`(m_1+m_2)a=F-(m_1+m_2)gsintheta`
`a=(F-(m_2+m_2)gsintheta)/((m_1+m_2))`
(iii) From equation (2)
`f=m_2a+m_2gsintheta`
`f=m_2[(F-(m_1+m_2)gsintheta)/((m_1+m_2))]+m_2gsintheta`
`f=m_2[(F-(m_1+m_2)gsintheta+(m_1+m_2)gsintheta)/((m_1+m_2))]`
`f=(m_2F)/(m_1+m_2)`