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Friction problem

  1. Feb 15, 2010 #1
    1. The problem statement, all variables and given/known datamass 1 is 2.0 kg, the static and kinetic coefficients of friction are .30 and .20 respectively.

    a) What is mass 2 if it travels uniformly upward?
    b) Downward?
    c) mass 1 and 2 are at rest?

    2. Relevant equations

    3. The attempt at a solution

    Part c is the easiest. Friction and Tension are going to be the same so [tex]m_{2} = \mu_{s}m_{1}[/tex].

    Part a, I did the same thing about except I used [tex]\mu_{k}m_{1}[/tex].

    Part b, [tex]m_{1}g = \mu_{k}m_{1}g[/tex]
    [tex]m_{2} = m_{1}/\mu_{k}[/tex]

    Is this right?

    Why is it that I can ace Abstract Algebra but this gives me a headache?
  2. jcsd
  3. Feb 16, 2010 #2


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    Hi lockedup! :smile:

    erm :redface: … what is the set-up? :confused:
  4. Feb 16, 2010 #3
    I'm guessing its a triangular slope of some sort with one mass on either side.
  5. Feb 16, 2010 #4


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    Hi inutard! :smile:
    ah, but then we'd have heard about a θ :wink:
  6. Feb 17, 2010 #5
    Good point. Maybe its in the diagram that he is neglecting to show us (As he also mentions tension but he neglected to talk about a rope or string in his question).
  7. Feb 18, 2010 #6
    Damn, I forgot to attach the picture! I'm a she BTW... lol

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  8. Feb 18, 2010 #7


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    (just got up :zzz: …)

    Hi lockedup (f.)! :wink:

    mmm … inutard :smile: guessed right! :biggrin:

    ok then, there's a θ = 37º, which isn't in your equations …

    put it in now …

    what do you get? :smile:
  9. Feb 18, 2010 #8
    Pardon me about the gender mixup! Remember that theres not only frictional force opposing direction of motion but also a force to the bottom-left (depending on the angle) that is caused by the normal force on block m1
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