Prism 1 of mass `m_1` and width angle `alpha` (see figure) rests on a horizontal surface. Bar 2 of mass `m_2` is placed on the prism. Assuming the friction to be negligible, find the acceleration of the prism.

Explanation

Let us draw the force diagram of each body, and on this basic we observe that the prism moves towards right say with an acceleration `vecw_1` and the bar 2 of mass `m_2` moves down the plane with respect to 1, say with acceleration `vecw_(21)`, then, `vecw_2=vecw_(21)+vecw_1` (figure).
Let us write Newton's second law for both bodies in projection form along positive `y_2` and `x_1` axes as shown in the figure.
`m_2g cos alpha-N=m_2w_(2(y_2))=m_2[w_(21(y_2))+w_(1(y_2))]=m_2[0+w_1 sin alpha]`
or, `m_2g cos alpha-N=m_2w_1sin alpha` (1)
and `Nsin alpha=m_1w_1` (2)
Solving (1) and (2), we get
`w_1=(m_2g alphacosalpha)/(m_1+m_2sin^2alpha)=(gsin alphacosalpha)/((m_1//m_2)+sin^2alpha)`