**Motion of two bodies
one resting on the other**

When a body A of mass m is resting on a body B of mass M then two conditions are possible:

**1) A force F is applied to the upper
body, and then following four situations are possible:**

**i) When there is no friction:**

(a) The body A will move on body B with acceleration (F/m): \({{a}_{A}}=\frac{F}{m}\).

(b)
The body B will remain at rest: a_{B} = 0

(c) If L is the length of B as shown in figure, A will fall from B after time t:

\(t=\sqrt{\frac{2L}{a}}=\sqrt{\frac{2mL}{F}}\,\,\,\left( as\,s=\frac{1}{2}a{{t}^{2}}\,\,and\,\,a=\frac{F}{m} \right)\).

**ii) If friction is present between A
and B only and applied force is less than limiting friction (F < F _{l}):**

F
= Applied force on the upper body; F_{l} = Limiting friction between A
and B; F_{k} = Kinetic friction between A and B.

a)
The body A will not slide on body B till F < F_{l} i.e.., F < μ_{s}mg

b) Combined system (m + M) will move together with common acceleration: \({{a}_{A}}={{a}_{B}}=\frac{F}{m+M}\).

**iii) If friction is present between A
and B only and applied force is greater than limiting friction (F > F _{l}):**

In this condition the two bodies will move in the same direction but with different acceleration. Here force of kinetic friction μ_{k}mg will oppose the motion of A while cause the motion of B. F – F_{k} = ma_{A} ⇒ \({{a}_{A}}=\frac{\left( F-{{F}_{k}} \right)}{m}=\frac{\left( F-{{\mu }_{k}}mg \right)}{m}\).

**iv) If there is friction between B and floor:**

Where,
F’_{l} = μ’ (M + m)g = limiting friction between B and floor, F_{k}
= Kinetic friction between A and B and B will move only if F_{k} > F’_{l} and then F_{k} – F’_{l} = Ma_{s}.

However, if b doesn’t move then static friction
will work between body B and the floor. I.e.., friction force is equal to
applied force not F’_{l}.