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Problem in mechanics involving a pulley

  1. Feb 24, 2006 #1
    Okay
    I chanced upon this problem and have lost some sleep since then.
    There is a pulley system with a single stationary pulley with 2 masses m1 and m2 suspended using a string which is inelastic and massles.The coefficient of static FRICTION between the PULLEY and the STRING is mu.What are the accelerations of the masses?

    [One way that occurs to me is to find the normal reaction between the pulley and the thread at a point which makes theta with the centre of the pulley and then integrate between limits 0 and pi.Though i am not sure how to go about calculating the normal force at a point(guess using tension in the string and resolving it at that point).]
     
  2. jcsd
  3. Feb 24, 2006 #2
    I used the method i specified and got the following answer
    |a|=g(x2-x1)/(x2+x1)
    where x2=m2(1-mu)
    x1=m1(1+mu)
    assuming m2 sufficiently greater than m1 to set the system in motion.
     
  4. Feb 24, 2006 #3

    Hootenanny

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    If the string is massless then there can be no friction. Friction is given by [itex]F_{max} = \mu R[/itex]; if there is no mass, then there is no normal reaction force ([itex]R[/itex]) and hence no friction.
     
  5. Feb 24, 2006 #4

    BobG

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    So, the only way you can ever get a pulley to turn is to use very heavy rope or cable? A pulley with a light weight string and heavy masses will cause the string to slide over the pulley instead of turn the pulley?

    In any event, it's unclear what the OP is looking for. Are you looking for the maximum acceleration before the string slides over the pulley instead of turns it? Or are you looking for the friction due to the bearings within the pulley? (sounds more like the first, since for the second, you'd be more concerned about the coefficient of kinetic friction).

    The force of gravity provides a downward force on the two masses regardless of where the string is on the pulley, so the normal force, using horizontal to the ground as the reference point, would be the sine of the angle (ranging from 0 to 90 - it's easier to just add your two masses together, since the sine from 180 to 90 is the same as the sine from 0 to 90).
     
  6. Feb 25, 2006 #5

    Hootenanny

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    Ahh yes I see now.
     
  7. Feb 25, 2006 #6
    I am looking for the acceleration produced in each block..which happens to be constant..
    My physics coach hinted at using torque but i am not very sure how..and yes ..desite the the string being massless..the gravitational force causes the masses attached to the string to produce a reaction between the string and pulley.
    And BOb if you could elaborate..[I am not sure you could just say the normal at a point would be sum of masses *sine(angle)].
     
  8. Feb 25, 2006 #7

    BobG

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    Conceptually, the force from each mass will be distributed over the length of the string. Looking at two points, one at 120 deg and one at 60 degrees (with mass 1 being on the 120 deg side and mass 2 being on the 60 degree side), and looking just at the forces on those two individual points (however the forces are distributed), you'd have:

    [tex]f1 sin (120) + f2 sin (60)[/tex]

    The sine of 120 is equal to the sine of 60, so you can use sine 60 for both. The distributive property gives you the sum of the forces times the angle.

    The forces have to be distributed along the length of the contact area between the string and the pulley. In other words, the sum of the forces will wind up equalling the sum of the masses times the acceleration of gravity (it had better or else you'll have to find a way to explain what happened to the difference).
     
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