Three block A,B,and C of mass `m_(1),m_(2)` and `m_(3)`, respectively are resting one on top of the other as shown in fig. A horizontal force F is applied on block B. Assuming all the surfaces are frictionless, calculate (1) acceleration of block A, block B, and block C, (2) normal reactions between A and B, B and C, and between C and ground .

Explanation

This system cannot be in equilibrium beacause in the horizontal direction, the system has a net external force F. As far as the vertical direction is concerned, all the force are internal action and reaction forces. These forces are equal and opposite, hence, they will cancel each other out

Body A : No external force is acting on it, hence, it will remain stationary in equilibrium. So acceleration at block A, `a_(A)=0` and
`N_(1)=m_(1)g`...(i) ltbr Body B: there is not external force acting on it vertically, hence, it will not have any acceletation in the vertical direction.
`N_(2)=N_(1)+m_(2)g=(m_(1)+m_(2))g` ...(ii)
However, there is one external force F action on it. so it will have some acceleration a in the horizontal direction such that
`F+m_(2)a_(B)` ...(iii)
which gives acceleration of block B , `a_(B) = (F)/(m_(2))`
Body C and earth: Same comments as in the case of body A above.
`N_(3)=N_(2)+m_(3)g` ...(iv)
and `N_(3)=(m_(1)+m_(2)+m_(3))g` ...(v)
No external force acts on block C in the horizontal direction, Hence, `a_( C)` is equal to 0.