anonymous
  • anonymous
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.
Physics
schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1350498047651:dw|
anonymous
  • anonymous
|dw:1350584039429:dw|
anonymous
  • anonymous
could you explain a bit how the FBD is drawn with Normal Reaction as mg/cos theta

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anonymous
  • anonymous
you mean for case 2?
anonymous
  • 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
  • anonymous
Oh . yes i have not got the logic clear.
anonymous
  • anonymous
ah whoops, I see why you're asking..
anonymous
  • anonymous
I labelled the theta incorrectly...
anonymous
  • anonymous
|dw:1350585942860:dw|
anonymous
  • anonymous
there now it works...:)
anonymous
  • anonymous
N2 is mg/cos(theta)
anonymous
  • anonymous
|dw:1350586072251:dw|
anonymous
  • 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
  • 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
  • anonymous
m is at rest with respect to the ramp when it's accelerating with the ramp...
anonymous
  • 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
  • anonymous
Thanks

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