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Pulley free body question - what am I doing wrong?

  1. Mar 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Note that the FAQ is not clear if I am required to use LaTeX for formatting formulas. I have not used LaTeX, but I will retype my question if I am violating the rules. Please let me know :-)

    Non-extensible cable attached to ceiling. Cable goes down, around a pulley and then back up, then over another pulley and down. At the end of the cable is mass 1. Hanging from the first pulley is mass 2. I am trying to find a formula for the acceleration of M1 as a function of the mass of 1 and 2.

    2. Relevant equations

    I have set g = -9.8m/s^2 (positive acceleration is up, negative is down)

    I have a free-body diagram for mass 1:

    Total force on mass 1: f1 = f1(up) + f1(down)
    f1(up) is the force imparted by the cable pulling up
    f1(down) = m1 * g

    And for mass 2:

    Total force on mass 2: f2 = 2 * f2(up) + f2(down)
    f2(up) is the force imparted by each of the two cables lifting on the pulley above mass 2.
    f2(down) = m2 * g

    I set also f2(up) = f1(up) since the same cable that holds mass 1 also goes to the pulley for mass 2. This is my first question - is this assumption of the forces upward correct?

    I also say that because of the pulley arrangement the acceleration of mass 1 (a1) = 2 * the acceleration of mass 2 (a2). I say this because the velocity of mass 1 is double the velocity of mass 2 (due to the pulley arrangement).

    So we have:

    a1 = 2 * a2
    f2(up) = f1(up)
    a1 = (f1(up) + f1(down)) / m1
    a2 = (2 * f2(up) + f2(down)) / m2

    Is this correct so far?

    3. The attempt at a solution

    Then I try to solve for a1 in terms of m1 and m2.

    First, solve for f1(up) in terms of a1:

    a2 = (2 * f2(up) + f2(down)) / m2
    a2 * m2 = 2 * f2(up) + f2(down)
    a2 * m2 = 2 * f1(up) + m2 * g
    (a2 * m2) - (m2 * g) = 2 * f1(up)
    ((a2 * m2) - (m2 * g)) / 2 = f1(up)
    (((a1 / 2) * m2) - (m2 * g)) / 2 = f1(up)

    Now plug this in to the original a1 equation:

    a1 = (f1(up) + f1(down)) / m1
    a1 = ((((a1 / 2) * m2) - (m2 * g)) / 2 + f1(down)) / m1
    a1 = ((((a1 / 2) * m2) - (m2 * g)) / 2 + m1 * g) / m1
    a1 * m1 = (((a1 / 2) * m2) - (m2 * g)) / 2 + m1 * g
    2 * a1 * m1 = (a1 / 2) * m2 - (m2 * g) + 2 * m1 * g
    2 * a1 * m1 - (a1 / 2) * m2 = 2 * m1 * g - (m2 * g)
    a1 * (2 * m1 - m2 / 2) = g * (2 * m1 - m2)
    a1 = (g * (2 * m1 - m2)) / (2 * m1 - m2 / 2)

    Now, as a sanity check, I try some examples:

    Example 1:

    m1 = 1 kg
    m2 = 2 kg
    (I would expect this to show no acceleration)

    a1 = (-9.8 (2 * 1 - 2)) / (2 * 1 - 2 / 2) = 0
    good!

    Example 2:

    m1 = 1.1 kg
    m2 = 2 kg
    (I would expect this to show a slow negative acceleration as m1 falls to the floor)

    (-9.8 (2 * 1.1 - 2)) / (2 * 1.1 - 2 / 2) = -1.96 / 1.2 = -1.63 m/s^2
    good!

    Example 3:

    m1 = 0.9 kg
    m2 = 2 kg
    (I would expect this to show a slow positive acceleration as m1 rises to the ceiling)

    (-9.8 (2 * 0.9 - 2)) / (2 * 0.9 - 2 / 2) = 1.96 / 0.8 = 2.45 m/s^2
    good!

    Example 4:

    m1 = 0.1 kg
    m2 = 2 kg
    (I would expect this to show a fast positive acceleration as m1 rises quickly to the ceiling)

    (-9.8 (2 * 0.1 - 2)) / (2 * 0.1 - 2 / 2) = 17.64 / -0.8 = -22.05

    Huh?

    I have been through this several (many!) times and cannot figure out what I'm doing wrong. Can anyone offer some insight?

    Additionally, with m1 = 0.5 kg the equation produces zero in the denominator, which is probably not right :-/

    Thanks very much in advance!
     
    Last edited: Mar 29, 2013
  2. jcsd
  3. Mar 29, 2013 #2
    I set up the problem a little differently . I said..

    m1g - f(up) = m1a
    m2g-2F(up) = m2(a/2)

    I solved for a and got

    a = 2g(m2-2m1)/(m2-4m1)

    for m1 = 1 and m2 = 2 a = 0
    for m1 = 1.1 and m2 = 2 a = -1.635
    for m1=.9 and m2 = 2 a = 2.45
     
  4. Mar 30, 2013 #3

    ehild

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    Homework Helper
    Gold Member

    You mixed up the signs. The block 1 and 2 move in opposite directions so a1=-2a2.

    It is a bit difficult to follow your notations. The upward forces come from the tension in the string, and it is the same everywhere if the pulleys and the string itself are massless. We usually denote the tension by T. Taking upward the positive direction, you have the equations

    m1a1=T-m1g
    m2a2=2T-m2g
    and a1=-2a2

    ehild
     

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  5. Mar 30, 2013 #4
    I tried to use your notation. F(up) is the same as T.
    I am using F = ma for both masses to get the two equations.
    For m2, F = m2g - 2T
    for m1 F = m1g - T

    m1g - T = m1a
    m2g-2T = m2(a/2)

    solve for T in both equations, T = T can solve for a in terms of m1 and m2.


    I solved for a and got

    a = 2g(m2-2m1)/(m2-4m1)

    for m1 = 1 and m2 = 2 a = 0
    for m1 = 1.1 and m2 = 2 a = -1.635
    for m1=.9 and m2 = 2 a = 2.45
    Report
     
  6. Mar 30, 2013 #5

    ehild

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    Homework Helper
    Gold Member

    If a is the acceleration of m1, the acceleration of m2 can not be a/2, as they move in opposite directions (one upward, the other downward).


    ehild
     
  7. Mar 30, 2013 #6
    Hi barryj,

    You have the same problem I do where for certain values of m1 and m2 you get a zero in the denominator.
     
    Last edited: Mar 30, 2013
  8. Mar 30, 2013 #7
    Yes, you're right! The wrong sign on a was the key.

    Now I get:

    a1 = 2g(2m1 - m2) / (m1*p^2 + m2)

    Plugging in -9.8m/s^2 for g gives:

    m1 = 2 kg
    m2 = 2 kg
    a1 = -3.92 m/s^2

    m1 = 1 kg
    m2 = 2 kg
    a1 = 0 m/s^2

    m1 = 0.9 kg
    m2 = 2 kg
    a1 = 0.7 m/s^2

    m1 = 0.5 kg
    m2 = 2 kg
    a1 = 4.9 m/s^2

    m1 = 0.1 kg
    m2 = 2 kg
    a1 = 14.72 m/s^2

    And no more possibility of zero in the denominator (unless both m1 and m2 are zero).

    Thanks!!!
     
  9. Mar 30, 2013 #8
    Hmmmmm... Good points from both of you.

    I'll have to think about this. I assumed I was correct since the sanity checks were sane,
     
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