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Coefficient of Friction at constant speed

  1. Jul 24, 2006 #1
    Hello, please help me out on this:
    Two equall masses lie on each side of a rough wedge, connected by a string, the wedge makes anlges 53, and 47, to the horizontal, find the coefficient of friction for the masses to move at a constant speed.

    MY THINKING:
    since masses are equal:
    mgsin(53) = mgcos(53)(u) + mgcos(47)(U) + mgsin(47) ---> these are all the opposing forces, and force of fricton of each mass, since the slop is different for them.
    the mg(s) cancels out. Leaving:
    sin(53)-sin(47) = (cos53+cos47)(u)
    then i got u = 0.052
    which is wrong...
     
  2. jcsd
  3. Jul 24, 2006 #2
    I do not know what a wedge* is, but I imagine the situation which was more probable with the given data.

    Is [tex]\mu_k = 0.098[/tex]? :wink:
    If it is, then I will explain my thoughts. :smile:

    *My english needs improvements. :approve:
     
  4. Jul 25, 2006 #3

    Doc Al

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    Staff: Mentor

    Your thinking looks OK to me. Why do you think it's wrong?
     
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