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Dynamics question

  1. Mar 16, 2009 #1
    1. The problem statement, all variables and given/known data
    th_untitled1.jpg

    Find out the acceleration of the rod A and the wedge B in the arrangement if the ratio of mass of the wedge to that of the rod equals [tex]\eta[/tex], and the friction between all contact surfaces is negligible


    2. Relevant equations
    Mass of rod = m, Mass of wedge = M

    mg - [tex]N cos \alpha[/tex] = ma, downwards

    3. The attempt at a solution
    Normal reaction is perpendicular the wedge surface and I don't know if it has a component that can accelerate the wedge to the right. If the force is moving to the left, surely the wedge can't be accelerated to the left.

    I need to know which force accelerates the wedge to the right. I'm thinking that the normal reaction has the component [tex]N sin \alpha[/tex] directed to the left.
    If I knew which component of force was directed to the right, I could solve the remaining part of it.
     
    Last edited: Mar 16, 2009
  2. jcsd
  3. Mar 16, 2009 #2
    Someone, please?
     
  4. Mar 16, 2009 #3

    tiny-tim

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    Welcome to PF!

    Hi sArGe99! Welcome to PF! :smile:

    conservation of energy? :wink:
     
  5. Mar 16, 2009 #4
    Yes. I did try that out. Don't think I have got enough information to use that.
     
  6. Mar 16, 2009 #5
    Normal reaction acts perpendicular to the wedge surface. Is that right, in this case?
    I think that is where I might have gone wrong.
     
  7. Mar 16, 2009 #6

    tiny-tim

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    Yes you have …

    try again (use x and x', where x is horizontal displacement), and show us what you get :smile:
     
  8. Mar 16, 2009 #7
    I think the kinematic relationship is Distance moved by Rod / Distance moved by wedge = tan (alpha)?
     
  9. Mar 17, 2009 #8

    tiny-tim

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    just woke up :zzz: …
    You know it is …

    get on with it!
     
  10. Mar 17, 2009 #9
    Oh.. I could do with this one, I guess.
    It can be solved using force equations, right? I actually wanted to try that method out first.
     
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