The coefficient of friction between the wedge and the table is `mu`, the friction between the rod and the wedge being negligible.

Answer

`Q=(m_(1)m_(2)g cos(alpha-varphi))/(m_(1) cos^(2) alpha cos varphi +m_(2) sin alpha sin(alpha-varphi)); varphi="arctan "mu`.

Explanation

The equation of motion of the rod will remain unaltered, in the equation of motion of the wedge on should take into account the presence of the force of kinetic friction `T = mu (Q cos alpha +m_(1)g)`,

directed opposite to the acceleration (Fig.). Thus
`-m_(2)g+Q cos alpha =m_(2)a_(y), -Q sin alpha +mu(Q cos alpha+m_(1)g) = m_(1) a_(x)`
After some transformations, noting that `a_(y)= a_(x) tan alpha`, we obtain
`Q=(m_(1)m_(2)g (cos alpha+mu sin alpha))/(m_(1) cos^(2)alpha+m_(2)sin^(2)alpha-mu m_(2)sin alpha cos alpha)`
`a_(x)= (-m_(2)g +mu g cot alpha(m_(2)+m_(1)))/(m_(1)cot alpha+m_(2)tan alpha+mu m_(2)) lt 0`
`a_(y) = (-m_(2)g tan alpha +mu g (m_(2) +m_(1)))/(m_(1)cot alpha +m_(2) tan alpha -mu m_(2)) lt 0`
The solution makes sense when
`m_(1) cot alpha +m_(2) tan alpha gt mu m_(2), m_(2) tan alpha gt mu(m_(2) +m_(1))`
It follows from the second inequality that
`tan alpha gt (mu(m_(2)+m_(1)))/(m_(2))`
If , on the other hand, `tan alpha le (mu(m_(2) +m_(1)))/(m_(2))` then `a_(x)= 0, a_(y)=0` and `Q=(m_(2)g)/(cos alpha)`.