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A very difficult question about a pulley system with friction

  1. Sep 27, 2011 #1
    1. The problem statement, all variables and given/known data
    See the picture


    2. Relevant equations



    3. The attempt at a solution
    This is a very difficult, can anyone guide me step by step to finish this question.

    First, let start with part(a)

    In order to find the ratio, it seems we need to have the motion equations first.
    I get 2 motion equations.
    But I have 2 question now.
    (1) Are the equation I obtained correct and useful?
    (2)If it is what can I do next?(In my opinion, it would be integration, but I wonder what are the intervals.)
     

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  2. jcsd
  3. Sep 28, 2011 #2
    I'm a little confused, is the second jpg. the problem? If so why do you have theta in your try at the solution?
     
  4. Sep 28, 2011 #3
    It is because I don't know how to do:tongue:
    I thought that theta will be eliminated finally.

    So do you have any ideas?
     
  5. Sep 28, 2011 #4
    Also, the answer is shocking.
    It is [itex]\frac{T_{1}}{T_{2}}[/itex]=[itex]e^{\mu\pi}[/itex]

    That why I have no idea
     
  6. Sep 28, 2011 #5
    Maybe the original poster knows?
     
  7. Sep 28, 2011 #6

    PAllen

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    I assume the second jpg is the problem and the first is the attempt at solution. The angles were introduced by the OP and are irrelevant.

    One key point: there is nothing in the problem jpg about the disk rotating - only sliding is going on here.

    Think about friction and gravity. Maybe start by assuming the disk is frictionless.

    Show us where this takes you.
     
  8. Sep 28, 2011 #7
    The second picture is the question.
    The first picture is my attempt to solve it.


    Now if we assume the pulley is frictionless, it means we can omit it and focus on the two masses.

    I obatin the following equations (see attachment).
    Do these equation useful?
     

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  9. Sep 28, 2011 #8
    He gave me another tips.
    This question required integration.
     
  10. Sep 28, 2011 #9

    PAllen

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    Close so far, but you have: force - force = acceleration
    in your equations. Fix this, and you'll have the frictionless case down.

    Then, consider that you just 'know' the acceleration resulting from mass difference and friction. It should then be simple to extend your frictionless case to be expressed using this 'given' acceleration.

    Even if integration is desired for the ultimate answer in terms of friction coefficient based on the surface properties of the string and disk, you can first consider a coefficient for the string suspended over the disk as a complete system. This has some total coefficient of friction. If you double the masses, what do you expect will happen to the total friction? So then you can easily use the prior to answer all questions in terms of a 'total system friction coefficient'.

    Finally, the only step needing integration is expressing this total system coefficient in terms of standard coefficient that simply relates normal force to friction per unit length of string. On this, you need to figure the normal force between the string and disk at each point. Once you get here, I cannot help you. It has been too many years for me - I would have to review to figure out that force distribution, and I'm not inclined to do that.
     
  11. Sep 28, 2011 #10
    I have corrected the F=ma error and I have reached this step.

    It will increase.

    Do you mean in the integral, we have mass element [itex]dm[/itex] and friction element [itex]df[/itex]?
     

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  12. Sep 28, 2011 #11

    PAllen

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    Well, you can say more than 'it will increase'. If at every point, friction is proportional to normal force, and all forces in the system are doubled, then, without further ado, you can say what about the total friction?

    Now, I would normally stop here, and assume you measure the system friction coefficient. But if you are required to calcluate it, you need to figure out how the normal force is distributed along the disk. The integral should be just d theta, over pi radians. But what you integrate is normal force as f(theta), and friction coefficient, which is constant. It is f(theta) that I can't see offhand how to derive.
     
  13. Sep 29, 2011 #12
    Total friction also double?

    That means you also have no idea about how to get the answer?
    My professor said he will give us the answer one month later because this is a very difficult and interesting question.
     
  14. Sep 29, 2011 #13

    PAllen

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    yes
    Correct, I can't help further without spending more time than I can afford on this. Hopefully someone else can.

    NOTE: You have a few mistakes in your last set of equations.

    hint 1: Your equations for tension should not include friction. That's built into m1*a or m2*a. The result will be much simpler.

    hint 2: Acceleration (there is only one acceleration, as you've noted) can be obtained by noting that the sum of the net forces on each weight (involving acceleration) must equal the balance of gravitation forces and friction.

    The overall results should be quite simple (at least in terms of total system coefficient of friction).
     
  15. Sep 29, 2011 #14

    NascentOxygen

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    True. But it seems to be drawn to resemble a pulley, and we are told to consider the disc to be massless....

    Maybe.

    OP: Did you originate the word "pulley" in the subject line, or was it in the heading of the questions where you sourced this?
     
    Last edited: Sep 29, 2011
  16. Sep 29, 2011 #15

    Sorry, I don't understand what you mean. Are you asking where did I get this question?
     
  17. Sep 29, 2011 #16

    NascentOxygen

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    Was this in a chapter titled Motion and Pulleys? Or did you think it looked like a pulley so put that in your subject header, though the chapter didn't mention pulleys?
     
  18. Sep 29, 2011 #17
    My professor just gave this question to us during the tutorial class. All the question are related to writing equations.

    I don't know if he extract it from the chapter of pulley. (We haven't leart pulley yet)
     
  19. Sep 29, 2011 #18

    PAllen

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    I did notice that wording. But then, either:

    1) masslessness is irrelevant, but friction is important and the problem is fully specified (and lives up to being hard if you are to integrate from a simple coefficient of friction).
    or
    2) if pulley is assumed (and assumed to have no rolling friction), then masslessness is relevant but friction is irrelevant and problem is completely trivial.
    or
    3) if both rolling and sliding are going on, the problem is underspecified and unsolvable.

    It seems to me (1) is the only assumption that makes the problem both interesting and solvable.
     
  20. Sep 29, 2011 #19

    NascentOxygen

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    There's a red herring somewhere. But the fact that athrun200 says the supplied answer involves e(coeff frict) points to this not being a pulley. A fixed shaft then.

    So, athrun200, can you do integral calculus?
     
  21. Sep 30, 2011 #20
    Yes, I am taking a course about multivariable calculus and it covers integral calculus.
    Does this question requires it?
     
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