Find the acceleration of rod A and wedge B in the arrangement shown in fig if the ratio of the mass of wedge to that of the rod equals `eta` and the friction between the contact surfaces are negligible.

Explanation

Let in any time the rod displaces down by y, the corresponding horizontal displacement of wedge will be x from the figure we have
`y=x tan theta` …(i)
Differentiating both sides of Eq. (i) we get
`(dy)/(dt)=(dx)/(dt) tan theta`
or `v_("rod")=v_("wedge") tan theta` ...(ii)
and `a_("rod") = a_("wedge") tan theta` ...(iii)
Let mass of the rod is m, then mass of wedge `M=etam`
By newton second law
For vertical motion of rod
`mg-N cos theta = ma_("rod")` ....(iv)
For horizontal motion of wedge,
`N sin theta = Ma_("wedge")`
`=M((a_("rod"))/tantheta)`....(v)
Solving Eqns (iv) and (v) we get
`a_("rod")=g/([1+M/mcot^(2)theta])`
`=g/([1+etacot^(2) theta])`
and `a_("wedge")=a_("rod")/tan theta=g/([tan theta+eta cot theta])`