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Forces acting on a pulley question

  1. Aug 26, 2015 #1
    1. The problem statement, all variables and given/known data

    Student 1 (50-kg), sits on a chair with metal runners, at rest. Student 2 (85-kg), sits on an identical chair. Both students keep their feet off the floor. A rope runs from student 1's hands around a light pulley and then over her shoulder to the hands of a teacher standing on the floor behind her. The low friction axle of the pulley is attached to a second rope held by student 2. All ropes run parallel to the chair runners.

    a) If student 1 pulls on the end of the rope, will her chair or 2’s chair slide on the floor?

    b) If instead the teacher pulls on his rope end, which chair slides? Why this one?

    c) If student 2 pulls on his rope which chair slides? Why?

    d) Now the teacher ties his end of the rope to student 1’s chair. Student 1 pulls on the end of the rope in her hands. Which chair slides and why.

    2. Relevant equations



    3. The attempt at a solution
    a) Student 1's chair will slide forward then student 2's chair will slide forward as well.
    b) Both student 1 and 2's chairs move because there is tension in the rope that pulls on student 1 and that pulley is connected to student 2 as well.
    c) Both student 1 and 2 slide forward because when student 2 pulls on his rope he pulls the pulley which increases the tension in the rope of student 1
    d) Student A will move first followed by 2

    Im not quite sure how to approach this problem and was just trying to reason through it :P Could someone tell me if I'm on the right track? Thank you!
     

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  3. Aug 26, 2015 #2

    haruspex

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    It is not made clear, but I think you are supposed to assume (1) that there is friction between runners and floor, and (2) that the rope is only pulled hard enough to achieve some motion.
    What do you know about frictional forces?
     
  4. Aug 26, 2015 #3
    Assuming that the mass of the chair does not count, the force due to friction would be equal to Ff = μN = ( μ)(g)( mass of the person) correct?
     
  5. Aug 26, 2015 #4

    haruspex

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    What is mu here, the static or kinetic coefficient?
     
  6. Aug 26, 2015 #5
    I believe that it is the static
     
  7. Aug 26, 2015 #6

    haruspex

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    Yes. A common misunderstanding about static friction is saying that the frictional force is equal to the coefficient multiplied by the normal force. That's true for kinetic friction but not for static. Can you post a correct formulation of the relationship?
     
  8. Aug 26, 2015 #7
    μs = fs max/N
     
  9. Aug 26, 2015 #8

    haruspex

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    Right, it only gives you the maximum value of the frictional force.
    So if the rope is pulled only just hard enough to start something moving, will both chairs move?
     
  10. Aug 26, 2015 #9
    So will only the lighter of the two chairs move in that case?
     
  11. Aug 26, 2015 #10

    haruspex

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    If it were not for the pulley, yes. But you do need to take the pulley into account.
    Assuming the same coefficient of friction for each, consider what tension in the rope will move the first chair, and what tension will move the second chair.
     
  12. Aug 26, 2015 #11
    ok i think I am starting to get it now, trying question a) again my attempt at the answer is:
    Assuming that the tension (T1) in the string held by student 1 and the teacher is the same at all points. Also the tension (T2) in the second rope is twice the tension of T1. When student 1 pulls on the rope, they only have to exert a tension of slightly more than 85/2 to move student 2’s chair. Student 1 will not move until T1 is greater than 50, therefore student 2’s chair moves first.
     
  13. Aug 26, 2015 #12

    haruspex

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    Right answer, but you have not expressed the logic correctly.
    As you wrote, the tension is the same throughout the rope, but then you wrote "the tension (T2) in the second rope is twice the tension of T1". What should you have said instead?
     
  14. Aug 26, 2015 #13
    hmmm I'm not quite sure what I did wrong there, is it because we do not know how much greater the tension in the shorter string actually is? So I cannot definitely say that it is twice the tension of T1?
     
  15. Aug 26, 2015 #14

    haruspex

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    I'm very sorry - doing too many things at once. What you wrote is fine.
     
  16. Aug 26, 2015 #15
    ok thank you for your help :)
     
  17. Aug 26, 2015 #16

    haruspex

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    One more thought - you also have to assume the chairs have mass under 15kg.
     
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