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[Friction] Angled bar pressing down against a plate

  1. Nov 1, 2015 #1
    1. The problem statement, all variables and given/known data
    VIfUiu2.png
    This is my problem in all its glory.

    2. Relevant equations
    f = uN

    3. The attempt at a solution
    For part a I simply did that tan(alpha) = f/N and therefore alpha = arctan(f/N).
    This simplifies to alpha = arctan(u). I believe this part is correct.

    The second part of this question is what gets me. I set up the sum of the moments for the bar.
    My equation is: W*sin(alpha)*L/2 - N*sin(alpha)*L + f*cos(alpha)*L = 0

    Algebra gave me that N = - W*sin(alpha) / [2*(u*cos(alpha) - sin(alpha)]

    I made a FBD for the plate and the equations I got were that F - f = 0 and Ngroundonplate - N = 0. (The plain N being the normal force of the bar on the plate)

    Since f = uN, substitution of N gave me f = - u*W*sin(alpha) / [2*(u*cos(alpha) - sin(alpha)] and then dividing top and bottom by cos(alpha) I got that f = INFINITY.

    If this makes sense, can anyone explain how? If it's wrong can anyone point me in the right direction?
     
  2. jcsd
  3. Nov 1, 2015 #2

    haruspex

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    Which way is f acting on the bar?
     
  4. Nov 1, 2015 #3
    If the workpiece is moving to the left, the friction force should be towards the right, shouldn't it?
     
  5. Nov 1, 2015 #4

    haruspex

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    Is that the friction force on the workpiece or the friction force on the bar?
     
  6. Nov 1, 2015 #5
    The friction on the workpiece.
     
  7. Nov 1, 2015 #6

    haruspex

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    Ok, but the equation I quoted is in regard to forces on the bar, no?
     
  8. Nov 1, 2015 #7
    Ahh, I see now! I had set up that sum of moments for part a and I neglected to alter it for part b. Thanks a bunch!
    After altering it I get F = uW/4. This makes much more sense!
     
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