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Stick being pulled by a string from the top

  1. Feb 6, 2014 #1
    1. The problem statement, all variables and given/known data
    We have a stick which is vertical, its length is 2m and its weight 45.9kg. A string is pulling the top of the stick from the floor at a distance of 2m from the left of the bottom of the bar. A force of 100N is pulling from the right at the middle of the stick. Which has to be the friction coefficient for the friction force acting on the bottom of the stick? The stick has to be in rest.

    https://imagizer.imageshack.us/v2/800x600q90/607/kb5l.jpg [Broken]

    2. Relevant equations

    F=ma
    p=mg
    R=u*N

    u=(friction coefficient)
    N=(normal reaction)
    p=weight
    T=tension

    3. The attempt at a solution

    y axis: p+Ty=N -> N=p+Tcos(Θ)

    x axis: F=R+Tx ->F=uN+Tsin(Θ) -> u=([itex]\frac{F-Tsin(Θ)}{p+Tcos(Θ)}[/itex])
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 6, 2014 #2

    BvU

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    So ?
    Make a drawing.
    [itex]\sum \vec F = 0[/itex] is one condition for non-movement. Any others ?
     
    Last edited: Feb 6, 2014
  4. Feb 6, 2014 #3
    I can't solve the equation, I don't know the value of T
     
  5. Feb 6, 2014 #4

    BvU

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    I'm even worse off: I don't know what T is!
     
  6. Feb 6, 2014 #5
    T is the tension of the string
     
  7. Feb 6, 2014 #6

    BvU

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    I kind of guessed that. If you don't know it, keep it and you'll find an expression for the friction coeffient in terms of T. More can't be asked from you. In the mean time I can ask you: what is theta ?
    Helper's lives are lengthened by completeness in 1. and 2. Also in 3.

    Drawing ?
     
  8. Feb 6, 2014 #7
    Oh the drawing I forgot I'm sorry :) I'll upload it in a minute
     
  9. Feb 6, 2014 #8
  10. Feb 6, 2014 #9

    BvU

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    Sure helps. You even added it as a picture in post #1. Makes the thread unintelligible for later readers, but never mind. So ##\cos \theta = \sin \theta = \frac {1}{2}\sqrt 2##.
    I don't think there's anything I can improve on what you've come up with so far.

    However, what about T ? You know the thing isn't moving. Back to the "Any others" in post #2.
     
  11. Feb 6, 2014 #10

    collinsmark

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    You've formed one equation (still two unknowns though).

    Perhaps you should take a moment*.

    *(By that I don't mean "take a moment to relax and contemplate the situation," but rather I mean take a moment. Given that nothing is rotating either, it means that sum of all torques [moments] around any given point must also be zero. You can find a new equation by taking a moment around any point, however, some points may be better than others. :wink:).

    [Edit: I see now that BvU is back. :smile:]
     
    Last edited: Feb 6, 2014
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