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In the given the wedge is acted upon by ...

In the given the wedge is acted upon by a constant horizontal force 'F'. The wedge is moving on a smooth horizontal surface A ball of mas 'm' is at rest relative to the wedge The ratio of forces exerted on 'm' by the wedge when 'F' is acting and 'F' is withdrawn assuming no friction between the edge and the ball is equal to .

Text Solution

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`(F cos theta = mg sin theta rArr F =mg tan theta)`
If `F =0, N_(2) = mg cos theta, (N_(1))/(N_(2)) = 1 + (Fsin theta)/(mg cos theta)`
`(N_(1))/(N_(2)) =1+ (mg tan theta sin theta)/(mg cos theta) = 1 + tan^(2)● = sec^(2)●`
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