A rod of mass `m_(2)` rests on wedge of mass `m_(1)` (Fig.). Guides allow the rod to move only in the direction of the y-axis and the wedge only in the direction of the x-axis. Find the acceleration of both bodies and the reaction of the wedge. Neglect friction.

Answer

`a_(1)=-"g"=(m_(3))/(m_(2)tan alpha+m_(1)cot alpha), a_(2)=-"g"(m_(2)tan alpha)/(m_(2)tan alpha +m_(1) cot alpha), Q=(m_(1)m_(2) g cos alpha)/(m_(2) sin^(2) alpha +m_(1)cos^(2)alpha)`.

Explanation

Three forces act on the rod: the reaction of the wedge Q, the reaction of the guides F and the gravitational force `m_(2)g` (Fig.). The equation of motion (in the y-direction) is
`-m_(2)g +Q cos alpha = m_(2)a_(2)`
Correspondingly, three forces also act on the wedge: the reaction of the table N, the gravitational force `m_(1)g`, and the reaction of the rod Q

(Fig). The equation of motion (in the x-direction) is
`-Q sin alpha= m_(1)a_(1)`
To obtain the third equation, we compare the displacements of the rod `Deltay=1//2 a_(2)t^(2)` with that of the wedge `Deltax=1//2 a_(1)t^(2)`. Since `Deltay= Deltax tan alpha` it follows that
`a_(2)=a_(1)tan alpha`
Then we solve the system of three equations with three unknowns.