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Static cylinder confusion

  1. Dec 4, 2013 #1
    Hi everyone,

    If I have a static cylinder on an inclined rough plane at an angle alpha supported by a chord under tension, which leaves the surface of the cylinder at a tangent upwards and parallel to the plane. Does the weight component trying to pull it downwards equal the sum of the friction + tension? And does the friction oppose the translational motion, rotational, or both?
    :confused:

    Any insights are appreciated, thx.
     
    Last edited: Dec 4, 2013
  2. jcsd
  3. Dec 4, 2013 #2

    CWatters

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    Draw a force diagram. If the cylinder is not rotating what can you say about the forces/torques?
     
  4. Dec 4, 2013 #3

    tiny-tim

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    hi devious b! :smile:

    there's only three forces: the weight, the reaction force, and the tension

    so they must all go through the same point :wink:

    (because … ?)
     
  5. Dec 4, 2013 #4
    My problem is the force diagram. Does the friction F act upward opposing the motion down and the torque caused by the tension?
     
  6. Dec 4, 2013 #5

    tiny-tim

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    if you follow my hint, it'll automatically give you the answer to that :smile:
     
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