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Block on a wedge

  1. Mar 11, 2015 #1
    Suppose we want to find the constraint equation relating the acceleration of a block on a wedge to the acceleration of the wedge (see attached images). We can write:
    $$h - y = (x - X) \tan{\theta}$$
    What if ##X## was measured to the centre of mass of the wedge for instance? Wouldn't the correct equation be:
    $$h - y = (X - x) \tan{\theta}$$
     

    Attached Files:

  2. jcsd
  3. Mar 11, 2015 #2

    Doc Al

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    Staff: Mentor

    I see how you got this and it makes sense.

    I do not see how you got that. You still need the sides of the same triangle, you've just redefined what X is. So you need to rewrite your first equation using your new X. For example, if ## X_{cm} = X + a##, then your formula would become ##h - y = (x - X_{cm} + a) \tan{\theta}##.
     
  4. Mar 11, 2015 #3
    I just realized the error in my reasoning. This makes sense now. Since ##X## and ##X_{cm}## differ by a constant, their second derivatives should be equal. We chose ##X## to be the distance to the edge of the wedge to simplify the trigonometry required, right?
     
    Last edited: Mar 11, 2015
  5. Mar 11, 2015 #4

    Doc Al

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    Staff: Mentor

    Right.
     
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