A block of mass m is placed on wedge of mass M (Fig.). Find the accelerations of the block and the wedge in the reference system fixed to the table, and the reaction. Friction is to be neglected.
Analyse the limiting case when the wedge remains stationary.

Answer

`a_(x)="g"(M sin alpha cos alpha)/( M + m sin^(2)alpha) , a_(y) = ((M+m) sin^(2)alpha)/(M+m sin^(2)alpha)`
`b_(x)= -"g"(m sin alpha cos alpha)/(M+m sin^(2)alpha); Q=(mMg)/(M+m sin^(2)alpha)`.

Explanation

The problem reduces to the solution of a system of four equations:
`-Q sin alpha = Mb_(x) , " " -mg + Q cos alpha = m a_(y)`
`Q sin alpha = ma_(x), " " a_(y) = (-a_(x) +b_(x)) tan alpha`
The case of a fixed wedge may be considered separately, but it may also be obtained from the general case, if we put `m lt lt M`.We obtain
`a_(x)=g sin alpha cos alpha, a_(y^(-)) - g sin^(2)alpha`
`a = sqrt(a_(x)^(2) +a_(y)^(2)) = g sin alpha, Q= mg cos alpha`