## haridas_mandal 3 years ago 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.

1. haridas_mandal

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2. Algebraic!

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3. haridas_mandal

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

4. Algebraic!

you mean for case 2?

5. Algebraic!

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

6. haridas_mandal

Oh . yes i have not got the logic clear.

7. Algebraic!

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

8. Algebraic!

I labelled the theta incorrectly...

9. Algebraic!

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10. Algebraic!

there now it works...:)

11. Algebraic!

N2 is mg/cos(theta)

12. Algebraic!

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13. haridas_mandal

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.

14. haridas_mandal

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?

15. Algebraic!

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

16. Algebraic!

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

17. haridas_mandal

Thanks