A block of mass m is placed on an inclined plane. With what acceleration A towards right should the system move on a horizontal surface so that m does not slide on the surface of inclined plane? Also calculate the force supplied by wedge on the block. Assume all surfaces are smooth.

Explanation

Here also angle of inclination of the plane gt angle of repose
`[ tan^(-1) ((1)/(sqrt(3))) = 30^(@)]`
Hence the block has a tendency to slide the plane free body diagram of the block as seen the frame of refornce of the wedge.
Case(1) for `a_(man)` limiting friction

will act to upward direction
Considering the equilibrium of `M` in the plane of wedge
`mg sin theta = f_(1) + ma_(min) cos theta`
`rArr f_(1) = mg sin theta - ma_(min) cos theta`.....(i)
`f_(1) = mu N`
`rArr mg sin theta - ma_(min) cos theta = mu(ma_(min) sin theta + mg cos theta)`
`a_(min) = g[(sin theta -mu cos theta)/(mu sin theta + cos theta)]`
`= g[( tan theta - mu )/(mu tan theta + 1)] = 10[(sqrt(3)-(1)/(sqrt(3)))/(sqrt(3)(1)/(sqrt(3)+1))]=10/(sqrt(3))ms^(-2)`
Case(2) for `a_(max)` limiting friction will act downward

`ma_(max) cos theta = f_(1) + mg sin theta`
`N = ma_(max) sin theta + mg cos theta`
`f_(1)= mu N`
`rArr a_(max) = (mg sin theta + mu mg cos theta)/(m cos theta - mu m sin theta)`
`= (g(tan theta + mu))/((1 - mu tan theta)) = (10(sqrt(3) + (1)/(sqrt(3))))/(1 - (1)/(sqrt(3)) sqrt(3)) = prop`
So there is no limit on maximum acceleration .