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3 Pulley Acceleration Question

  1. Dec 5, 2008 #1
    1. The problem statement, all variables and given/known data

    Three objects m1. m2 and m3 are suspended
    from three massless and frictionless pulleys as
    shown in the diagram. ml is 6 kg, m2 is 2 kg
    and m3 is 3 kg. Find the tension of the string
    when the system is set in motion (i.e. all
    masses are moving). Answer in N.

    2. Relevant equations

    Based on the free body diagrams.

    3. The attempt at a solution

    We had covered this earlier in the course, and I believed that I understood the material. When I tried to do the problem though, I realized that I have no idea where to start since I can't get the righter answer. I set up the basic equations like for m1: ma = mg - T, but that can't be right since you would only need two equations then.

    I believe where I am going wrong is the acceleration, can anyone shed some light on this for me? Thanks very much :)
     

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    Last edited: Dec 5, 2008
  2. jcsd
  3. Dec 5, 2008 #2

    Doc Al

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    There are three masses, thus three force equations. Hint: Figure out the constraint equation that relates the three accelerations.
     
  4. Dec 5, 2008 #3
    I don't know what you mean by constraint equations, but the three force equations I came up with were:

    m3a= T-mg

    m2a= 2T-mg

    ma= mg-T
     
  5. Dec 5, 2008 #4

    Doc Al

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    They can't all have the same acceleration.

    Try this. Use a1, a2, and a3 as the acceleration of each block.

    Since the blocks are connected by a string, their accelerations are not independent. Hint: Assume that m1 & m3 both accelerate downward, thus m2 must accelerate upward. Now see if you can relate the three accelerations. (That relationship is what I call the constraint equation. All pulley problems have one. This one's a bit tricky.)
     
  6. Dec 5, 2008 #5
    Before I try that, I just want to clear one small thing up.
    Say, I have only one pulley (I'm not sure if it matters if it's a heavy one or not), and I put two masses on each string. Now, if one falls down, would the two have the same velocity but in opposite directions? If so, can this be applied to acceleration as well?

    And if all of the above is true (?), the reason why I can't do that for this question is because there are multiple pullies?

    And yeah, I think this was what my professor was getting at in class, that each acceleration is independent (we did an example where a1=2a2). I feel like an idiot for still not grasping that concept.
     
  7. Dec 5, 2008 #6

    Doc Al

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    Yes, exactly. The "constraint equation" for a single pulley is just: a1 = -a2.

    Right again.

    Well, they are not independent, but I know what you mean. They are certainly not all the same.
    Don't. Getting the constraint right is the trickiest part of dealing with pulleys. (Related to that is getting the signs right.) Play around with it a bit. Here's a hint: What if m3 = m1? How would the accelerations relate in that simpler (but similar) case?
     
  8. Dec 5, 2008 #7
    To tell you the truth, I am still not sure what you mean.
    If m1=m3, then there should only be 2 equations=
    m1a1 = m3g-T
    m3a1 = m3g-T ,
    wouldn't these two cancel out, leaving:

    m2a2 = 2T-mg?

    I don't know, but I'm thinking that if m1=/m3, then m2 is:

    m2(a1+a3)=2T-mg?
    But this can't be right either, since a1 and a3 are not in the same direction. I'm still confused. :uhh:
     
  9. Dec 6, 2008 #8

    Doc Al

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    What I meant with my example of m1 = m3 is this. Since the masses on each side are equal, they will have the same acceleration. If they both move down by 1 m (say), how much does the middle mass move up? (a1 & a3 are in the same direction.)
     
  10. Dec 6, 2008 #9
    Since you said earlier about the velocity, distance, and acceleration ratio, is this correct?

    m1a1 = mg-T1
    m3a3 = mg-T1
    m2a2 = 2T1 - mg

    For the second mass in the middle, since it has two tensions, it has a slack value of 2, so if I set the displacement as y for each one while assuming that 1 goes down while 3 goes down as well.

    2y2 = y1 + y3
    2v2 = v1 + v3
    2a2 = a1 + a3 ? <---- If this is my 4th equation, can I solve using the equations I stated earlier for each component?

    I also have one more question about a different topic, should I post it in a different thread?
     
  11. Dec 6, 2008 #10

    rl.bhat

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    I think it is wrong.
    It should be
    m1a1 = m1g-T1
    m3a3 = m3g-T1
    m2a2 = 2T1 - m2g
    and finally, m1a1 +m3a3 = m2a2.
     
  12. Dec 6, 2008 #11
    Yes, I understand your corrections to the first 3 equations since I forgot to put the mass numbers in, but I don't understand the final one?
     
  13. Dec 6, 2008 #12

    rl.bhat

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    Net downward force = net upward force.
     
  14. Dec 6, 2008 #13
    So that is basically my 4th equation, but you include the masses to make the forces instead of the accelerations?

    Could you also use just my 4th equation that deals with only acceleration to solve?
     
  15. Dec 6, 2008 #14

    rl.bhat

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    In the problem you are asked to find the tension in the string, which you can get directly from the above equation.
     
  16. Dec 6, 2008 #15
    Yes ok, I will work it out now.
    If I have further questions, I will post here again tomorrow.
     
  17. Dec 7, 2008 #16

    Doc Al

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    As rl.bhat already pointed out, these are incorrect because you just use "m" on the right hand side. Also, since there's only one tension, just call it T instead of T1 (less symbol clutter). I'd write them like so:
    m1a1 = m1g-T
    m3a3 = m3g-T
    m2a2 = 2T - m2g

    Your 4th equation is exactly correct (considering magnitudes only). Good!
     
  18. Dec 7, 2008 #17

    Doc Al

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    :confused: Why would you think that true?
     
  19. Dec 7, 2008 #18
    So setting my 4th equation using the masses for force would be incorrect?

    Also, I did the question using only the accelerations as the 4th equation, and I think I got the right answer :)
     
  20. Dec 7, 2008 #19

    Doc Al

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    Yes, that would be incorrect. (It makes no sense.)

    Let's hope so! :wink:
     
  21. Dec 7, 2008 #20

    rl.bhat

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    Sorry, it is not true.
     
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