Question

Three blocks A, B,and C of masses $m_1,m_2$, and $m_3$,respectively, are resting one on top of the other as shown in $Fig.6.24$. 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.

Answer 1

This system cannot be in equilibrium because in the horizontal direction, the system has a net external force F. As far as the vertical direction is concerned, all the forces 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)
Body B : There is no external force acting on it vertically; hence, it will not have any acceleration in the vertical direction.
$N_2=N_1+m_{2}g=(m_1+m_2)g$..........(ii)
However, there is on external force F acting 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, $a_B=\frac{F}{m_2}$
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.