In the above figure, the wedge is acted on by a smooth constant horizontal force F. The wedge is moving on a smooth horizontal surface. A ball of mass m is at rest relative to the wedge. The wedge is not allow to move assuming no friction b/w the wedge and the ball. Show that the ratio of the force exerted on m by the wede when F is acting and F is withdrawn is F / [(M+m) g sin theta cos theta].

  Hi,

when force f is not acting

      normal by wedge on ball , N1 = mg cos Ø .....................(1)

when f is applied,

  total acceleration of wedge & ball,          a   =    F/ (M+m)

 we have to make the fbd of ball with respect to the vedge

 in this there will be three forces mg,  N &  ma (  becoz we are seeing w.r.t. wedge )

   ball is at rest with respect to wedge,

so,   horizontal forces,  N2 sinØ =  ma

                                    N2 =  ma / sinØ

                                    N2 = mF/(M+m) sinØ............(2)

(2)/ (1) .....

 N2 /N1 = F/(M+m)g sin Ø cos Ø