In the arrangement shown in figure the masses of the wedge `M` and the body `m` are known. The appreciable friction exists only between the wedge and the body `m`, the friction coefficient being equal to `k`. The masses of the pulley and the thread are negligible. Find the acceleration of the body `m` relative to the horizontal surface on which the wedge slides.

Explanation

When a block of mass `m` moves towards right. When `m` moves a distance `x_(0)` downward, the triangular block moves towards right by `x_(0)`. Hence the vertical acceleration of `m` and the horizontal acceleration of `M` are equal (say `a_(0)`), also two blocks move in contact in the horizontal direction.
Block of mass `m`:
`rarr : N = ma` (i)
`f = mu N = mu ma`
`darr : mg - T - f = ma`
`mg - T = f + ma = (mu + 1)ma` (ii)
Triangular block :
`rarr : T - N = Ma`
`T = N + Ma = (M + m)a` (iii)
Adding (ii) and (iii) , we get
`mg = (mu + 1) ma + (M + m)a = (M + mu m + 2m)a`
`a = (mg)/(M + (mu + 2)m)`
Acceleration of `m` :
`a' = sqrt(2) a = (sqrt(2) mg)/(M + (mu + 2) m)`
`= (g sqrt(2))/((M)/(m)) + (mu + 2)`

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