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 Show that the ratio of the force exerted on m by the wede when F is acting and F is withdrawn is F / [(M+m) g sin theta cos theta].```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

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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