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Tushar Watts Grade: 12
        


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 Show that the ratio of the force exerted on m by the wede when F is acting and F is withdrawn is F / [(M+m) g sin theta cos theta].

8 years ago

Answers : (1)

Pratham Ashish
17 Points
										

  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


 


 


                                            

8 years ago
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