Prism 1 with bar 2 of mass m placed on it gets a horizontal acceleration w directed to the left (figure). At what maximum value of this acceleration will the bar still stationary relative to the prism, if the coefficient of friction between them `k lt cot alpha` ?

Explanation

On the basis of the initial argument of the solution of above, the tendency of bar 2 with respect to 1 will be to move up along the plane.
Let us fix `(x-y)` coordinate system in the frame of ground as shown in figure.
From second law of motion in projection form along y and x-axis:
`mg cos alpha-N=m w sin alpha`
or, `N=m(g cos alpha-w sin alpha)` (1)
`mg sin alpha+f r=mw cos alpha`
or, `f r=m(w cos alpha-g sin alpha)` (2)
but `f r le kN`, so from (1) and (2)
`(w cos alpha-g sin alpha)lek(g cos alpha+w sin alpha)`
or, `w(cos alpha-k sin alpha)le g(k cos alpha+sin alpha)`
or, `w le "g"((cos alpha+sin alpha))/(cos alpha-k sin alpha)`,
So, the sought maximum acceleration of the wedge:
`w_(max)=((k cos alpha+ sin alpha)g)/(cos alpha- k sin alpha)=((k cot alpha+1)g)/(cot alpha-k)` where `cot alpha gt k`