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haridas_mandal

  • 2 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.

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  1. haridas_mandal
    • 2 years ago
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    |dw:1350498047651:dw|

  2. Algebraic!
    • 2 years ago
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    |dw:1350584039429:dw|

  3. haridas_mandal
    • 2 years ago
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    could you explain a bit how the FBD is drawn with Normal Reaction as mg/cos theta

  4. Algebraic!
    • 2 years ago
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    you mean for case 2?

  5. Algebraic!
    • 2 years ago
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    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
    • 2 years ago
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    Oh . yes i have not got the logic clear.

  7. Algebraic!
    • 2 years ago
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    ah whoops, I see why you're asking..

  8. Algebraic!
    • 2 years ago
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    I labelled the theta incorrectly...

  9. Algebraic!
    • 2 years ago
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    |dw:1350585942860:dw|

  10. Algebraic!
    • 2 years ago
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    there now it works...:)

  11. Algebraic!
    • 2 years ago
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    N2 is mg/cos(theta)

  12. Algebraic!
    • 2 years ago
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    |dw:1350586072251:dw|

  13. haridas_mandal
    • 2 years ago
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    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
    • 2 years ago
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    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!
    • 2 years ago
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    m is at rest with respect to the ramp when it's accelerating with the ramp...

  16. Algebraic!
    • 2 years ago
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    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
    • 2 years ago
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    Thanks

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