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Homework Help: Tension in a rope on frictionless pulley

  1. Dec 18, 2007 #1
    1. The problem statement, all variables and given/known data

    A 4kg object is palced on a table with a coefficient of friction of 0.2, and attached to a rope. This rope is run through a frictionless pulley, and attached to an 8kg weight. What is the tension in the rope?

    2. Relevant equations

    F=ma
    F of friction = (coefficient of friction)(Force normal)

    3. The attempt at a solution

    Force normal of weight = 8(9.8) = 78.4
    Force normal of object = 4(9.8) = 39.2
    Force of friction on object = 0.2(39.2)

    I don't know where to go from here, and I have an exam tomorrow. Help would be greatly appreciated, obviously :P

    Thanks, I hear you guys are the best for quick physics help ;)
     
  2. jcsd
  3. Dec 18, 2007 #2

    Doc Al

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    Apply Newton's 2nd law to each mass, then combine the two equations.
     
  4. Dec 18, 2007 #3
    So add both forces from F=ma together?
     
  5. Dec 18, 2007 #4

    Doc Al

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    Not sure what you mean. What I mean is: Apply F=ma to each mass separately. You'll end up with two equations that you will solve together.

    What forces act on each mass?
     
  6. Dec 18, 2007 #5
    Well, the weight is pulling the object because it is hanging off the table, through the pulley...
     
  7. Dec 18, 2007 #6

    Doc Al

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    What pulls on the object is the tension in the string. (Of course that tension also pulls up on the weight.)
     
  8. Dec 18, 2007 #7
    Ahh, I think I get it (but physics has confused me from the start, so bear with me). The tension is essentially a net force?
     
  9. Dec 18, 2007 #8

    Doc Al

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    No. "Net force" just means the total force on an object. The tension is just one of several forces.
     
  10. Dec 18, 2007 #9
    Alright, I'll be honest, I still don't get it :(

    This was a quick question I came up with to demonstrate what I need to know. What would be the tension? Then I could figure out how to actually get it.
     
  11. Dec 18, 2007 #10

    Doc Al

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    Start by applying Newton's 2nd law to the hanging mass. What forces act on it?
     
  12. Dec 18, 2007 #11
    Gravity times its own weight, and the object on the table (which I presume can act as friction). Anything I'm missing?

    The object on the table is also prone to friction, and the weight that is hanging pulls back on it.
     
  13. Dec 18, 2007 #12

    Doc Al

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    Stick to the forces acting on the hanging mass. Yes, its weight is one force. What's the other? (Forget about the object on the table for the moment. We'll analyze that one next.)
     
  14. Dec 18, 2007 #13
    I can't think of any, unless you mean the rope/tension itself.
     
  15. Dec 18, 2007 #14

    Doc Al

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    Of course I mean the rope tension!

    There are two forces acting on the hanging mass: The rope tension, which acts up; the weight, which acts down.

    Now apply Newton's 2nd law.
     
  16. Dec 18, 2007 #15
    Haha, thought that would be too easy :P

    The force down = 8(9.8) = 78.4 N

    Not sure how to deal with the rope tension, tbh...
     
  17. Dec 18, 2007 #16

    Doc Al

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    The rope tension is unknown, so just call it T. (You'll end up solving for it.)
     
  18. Dec 18, 2007 #17
    Oh, ok.

    So the net force is F down - T
     
  19. Dec 18, 2007 #18

    Doc Al

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    The net force on the hanging mass is [itex]T - m_2 g[/itex], where [itex]m_2[/itex] equals 8 kg.
     
  20. Dec 18, 2007 #19
    Alright, so essentially just the reverse: T - 78.4.
     
  21. Dec 18, 2007 #20

    Doc Al

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    OK, now apply Newton's 2nd law.
     
  22. Dec 18, 2007 #21
    I thought I already did...

    Would I be right to say that it's 7.84 = T - 78.4?

    Just because the coefficient of friction is 0.2, force normal on the object is 39.2, and thus friction is 7.84.
     
    Last edited: Dec 18, 2007
  23. Dec 18, 2007 #22

    Doc Al

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    Apply Newton's 2nd law to the hanging mass only. Friction has nothing to do with it. Call the acceleration "a".
     
  24. Dec 18, 2007 #23
    Well, mass * a = 8 * 9.8 = 78.4. I really don't understand what you want me to do after that...
     
  25. Dec 18, 2007 #24

    Doc Al

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    The acceleration is not 9.8 m/s^2. That's the acceleration of a freely falling object, not something attached to a string. Call the acceleration "a".
     
  26. Dec 18, 2007 #25
    I truly and honestly don't know what to do.
     
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