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In the given figure, the wedge is acted upon by a 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 The ratio of forces exerted on 'm' by the wedge when 'F' is acting and 'F' is withdrawn assuming no friction between the edge and the ball, is equal to : |
Answer» Solution : `N_(1)=mg COS THETA+F SIN theta` `N_(2)=mg cos theta (N_(1))/(N_(2))=1+(F sin theta)/(mg cos theta)` `(N_(1))/(N_(2))=1+(cancel(m)cancel(G) tan theta sin theta)/(cancel(m)cancel(g)cos theta)` `(N_(1))/(N_(2))=1+Tan^(2)theta=sec^(2)theta` |
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