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Sliding Block and Pulley

  1. Apr 13, 2008 #1
    1. The problem statement, all variables and given/known data
    A 5.50 kg block (m1) is connected by means of a massless rope to a 3.15 kg block (m2). The pulley is frictionless. Calculate the maximum value for the coefficient of static friction, if the 5.50 kg block is to begin sliding.

    The diagram is attatched to the post.

    2. Relevant equations
    Friction coefficient = (Frictional Force)/(Normal Force)
    Acceleration = (m2*9.8)/(m1 + m2)
    Of course, Force = mass*acceleration

    3. The attempt at a solution
    Acceleration = (m2*9.8)/(m1 + m2)
    Acceleration = 3.5688 m/s^2
    3.5688*5.5 = 19.6284 N = frictional force
    Normal force = 5.5*9.8 = 53.9 N

    19.6284/53.9 = .36416

    I know that the answer isn't right, what am I doing wrong?
     

    Attached Files:

    Last edited: Apr 13, 2008
  2. jcsd
  3. Apr 13, 2008 #2

    Doc Al

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    Attach the diagram as an image. (That site requires authorization to view.)
     
  4. Apr 13, 2008 #3
    Ok, I did.
     
  5. Apr 13, 2008 #4

    Doc Al

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    Note that the block is just about to barely begin sliding. So what can you take as the acceleration? (Hint: Note that you are finding the coefficient of static friction.)
     
  6. Apr 13, 2008 #5
    I already gave the acceleration, 3.5688 m/s^2, and doesn't the force of static friction balance the force of tension, which is 3.5688*5.5?
     
  7. Apr 13, 2008 #6

    Doc Al

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    I know. My point was that this acceleration is wrong. :wink: How did you arrive at this value?
    Exactly! So what's the acceleration if the forces are balanced?
     
  8. Apr 13, 2008 #7
    The weight of the hanging mass exerts a net force on both masses. (Assuming that the surface is frictionless)

    (3.15*9.8)/(3.15+5.5) = 3.5688 m/s^2

    Also, the acceleration needs to be zero, so then i did 5.5*3.5688 to find the frictional force.
     
    Last edited: Apr 13, 2008
  9. Apr 13, 2008 #8

    Doc Al

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    You found the acceleration of the masses assuming no friction. But you know there's friction, so this is not relevant.

    Doesn't this contradict what you just did above?

    The acceleration is zero. So what's the net force on m1?
     
  10. Apr 13, 2008 #9
    there is none because force is equal to mass * acceleration. so then what am i doing wrong?
     
  11. Apr 13, 2008 #10

    Doc Al

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    The net force is zero. So add up the forces on m1 and set equal to zero. What horizontal forces act on m1?

    (This is a static equalibrium problem.)
     
  12. Apr 13, 2008 #11
    Ok, I got it right...

    I figured that the tension in the rope would be equal to the weight of m2 in equilibrium.
    Then, I set it to equal the force of friction.
    I divided the force of friction by the mass of m1 to get .573.

    Thanks for the help!
     
  13. Apr 13, 2008 #12

    Doc Al

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    Good! You're welcome.
     
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