In the given figure all the surfaces are smooth. The ratio of forces exerted by the wedge on the mass M WHEN F is not applied and when the force F is applied such that M is at rest with respect to the wedge is ---
Answers options-1,1:2,sec^2 theta,cos^2 theta.

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- anonymous

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- anonymous

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- anonymous

could you explain a bit how the FBD is drawn with Normal Reaction as mg/cos theta

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- anonymous

you mean for case 2?

- anonymous

sure... there are only two forces on the block in that case... N and mg...
the horizontal component of N must be F
and the vertical component must be mg

- anonymous

Oh . yes i have not got the logic clear.

- anonymous

ah whoops, I see why you're asking..

- anonymous

I labelled the theta incorrectly...

- anonymous

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- anonymous

there now it works...:)

- anonymous

N2 is mg/cos(theta)

- anonymous

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- anonymous

But what about M at rest ? Is the Normal Reaction force mg/cos theta is sufficient to negate mg sin theta component. I am not so sure whether I have got the concept right. Could you please elaborate. I have got the values right.

- anonymous

After a second thought it struck me that the wedge is pitted against the Mass M and there is as such no component as mg sin theta...am i correct?

- anonymous

m is at rest with respect to the ramp when it's accelerating with the ramp...

- anonymous

yeah I mean, you could decompose mg into vectors along the ramp or into/out of the ramp, but it would be pretty pointless...

- anonymous

Thanks

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