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Pulley Question

  1. Jan 25, 2007 #1
    A block of mass M resting on a 19.1° slope is shown. The block has coefficients of friction μs=0.788 and μk=0.482 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 1.86 kg. What is the minimum mass M1 that will stick and not slip?

    [​IMG]

    In terms of mass 1, following eqn can be written:

    m1g sin 19.1 - T - Ff = 0

    which can be rewritten as:

    m1g sin 19.1 - T - m1g cos 19.1 (friction coefficient) = 0
    right? since the object is at rest.
    then i found T by equating it to m2g (T = m2g...see the diagram as to why i have donet this)

    Hence m1g sin 19.1 - m2g - m1g cos 19.1 (0.788) = 0
    and then solved for m1....but im not getting the right answer
    I dont understand why my approach is wrong...where did i go wrong here?
     
  2. jcsd
  3. Jan 25, 2007 #2
    when an object is at rest, the equation representing the coefficients of friction isn't an equality, but an inequality. [tex] f= \mu R [/tex] only when the system is in limiting equilibrium or moving. I appreciate that's what you're trying to find here, but it's an important point to bear in mind.

    And after a quick glance at your approach, it looks fine to me. I think it's probably arithmetic error.... you are using the appropriate coefficient of friction, yes?
     
    Last edited: Jan 25, 2007
  4. Jan 26, 2007 #3
    By "minimum mass M1 that will stick and not slip" isn't it meant "slip" as in move upwards or to the right? Because shouldn't then the sign of the friction force change, and be against the direction of the probable movement, that is opposite the direction of the tension force?
     
  5. Jan 26, 2007 #4
    actually, yes, it should. Sorry I never noticed that, myh bad I was probably off it or something. Thank you for picking that up
     
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