Find the accelerations of rod A and wedge B in the arrangement shown in figure if the ratio of the mass of the wedge to that of the rod equals `eta`, and the friction between all contact surfaces is negligible.

Explanation

Let us draw free body diagram of each body, i.e. of rod A and of wedge B and also draw the kinematical diagram for accelerations, after analysing the directions of motion of A and B. Kinematical relationship of acceleration is:
`tan alpha=(w_A)/(w_B)` (1)
Let us write Newton's second law for both bodies in terms of projections having taken positive directions of y and x axes as shown in figure.
`m_Ag-Ncos alpha=m_Aw_A` (2)
and `Nsin alpha=m_Bw_B` (3)
Simultaneously soultions of (1), (2) and (3) yields:
`w_A=(m_Asin alpha)/(m_Asin alpha+m_B cot alpha cos alpha)=(g)/((1+eta cot^2alpha))` and
`w_B=(w_A)/(tan alpha)=(g)/((tan alpha+eta cot alpha))`

Note: We may also solve this problem using conservation of mechanical energy instead of Newton's second law.