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Two blocks and pulley systemhelp on small part

  1. Dec 3, 2003 #1
    You got to see the picture 1st.

    A block of mass m1 = 2 kg rests on a table with which it has a coefficient of friction µ = 0.66. A string attached to the block passes over a pulley to a block of mass m3 = 4 kg. The pulley is a uniform disk of mass m2 = 0.5 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley.

    --------
    a) With what acceleration does the mass m3 fall?
    b) What is the tension in the horizontal string, T1?
    c) What is the tension in the vertical string, T3?
    =======================

    Okay, shouldn't be that HARD of a problem. But this is what I've done so far.

    mass 1 (right is positive)
    -------
    Summation of forces in x direction
    T1-u*m1*g=m1a

    Summation of forces in y direction
    N-m1g=0
    N=m1g

    mass 3 (up is positive)
    ------
    Summation of forces in y direction
    T3-m3*g=m3*a

    My biggest question is the following equation correct:

    mass 2
    ------
    (T1-T3)R=I*alpha
    simplifies to
    (T1-T3)R=.5*m2*R^2*(a/R)
    or
    (T1-T3)=.5*m2*a => is this right for mass 2 for net torque

    From here I manipulated the equations until I got a=...
    This is what I have so far, but I'm not getting the acceleration value right, unless I made an algebra error.
    But the equations look right so far, right?
     

    Attached Files:

  2. jcsd
  3. Dec 3, 2003 #2

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    It looks like you've got your torque acting in the wrong direction. Try
    [tex]\tau_{net}=T_3r-T_1r[/tex]
    as your torque formula instead.
     
  4. Dec 3, 2003 #3
    Ok, I did what NateTG said, and it gave me a different accerlation of -15.008 m/sec^2 for m3. But its not the right answer. Was there something else I'm supposed to do?
     
  5. Dec 3, 2003 #4

    Doc Al

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    Staff: Mentor

    To avoid premature insanity, pick a consistent set of sign conventions. You know that m1 moves to the right, the pulley moves clockwise, and m3 goes down. So... use those directions as the positive direction in all your equations.

    The way you currently wrote your equations, what you call "a" in one equation equals "-a" in another. Yikes!
     
  6. Dec 3, 2003 #5
    Thanks for the tip DocAl! I figured out the answer. Yes!!!!!

    I learn something new everytime I post here. :smile:
     
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