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Double Pulley Problem

  1. Sep 22, 2008 #1
    1. The problem statement, all variables and given/known data

    An object of mass m1 on a frictionless horizontal table is connected to an object of mass m2 through a very light pulley P1 and a light fixed pulley P2.

    [​IMG]

    (a) If a1 and a2 are the accelerations of m1 and m2, respectively, what is the relation between these accelerations?

    (b) Express the tensions in the strings in terms of g and the masses m1 and m2.

    (c) Express the accelerations a1 and a2 in terms of g and the masses m1 and m2.

    2. Relevant equations

    [tex]\Sigma[/tex]F = ma

    3. The attempt at a solution

    The way that I did it was I made two force diagrams. The first force diagram of m1 I made just has force going to the right, force T1 (tension of rope). Then I have a second force diagram of m2 with a force T2 going up and a force of mg (or weight of m2) going down. but the thing is that I keep on getting one a (acceleration), but the answer has a1 and a2. Can someone please tell me what I am doing wrong with my force diagrams. I think if i can have the correct force diagram, then I will be able to do the problem (hopefully!).
     
  2. jcsd
  3. Sep 22, 2008 #2

    LowlyPion

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    The movable pulley gives you a mechanical advantage of 2, but it means the tension translated to m1 is only 1/2 from T2, and hence so too will the acceleration of m1 be different.

    Stated as Tensions that's T2 = 2T1
     
    Last edited: Sep 23, 2008
  4. Sep 23, 2008 #3
    Okay, so if T2= 2T1, then the force diagram looks like this (this is what i got):
    The force diagram of m1 has T1 going to the right, m1g going down, and the normal vector n going up.
    The force diagram of m2 has T2 going up and m2g going down.

    So heres the way I did it, but its wrong for some reason, and i dont no why:
    [tex]\Sigma[/tex]Fm1= T1 = m1a1
    [tex]\Sigma[/tex]Fm2= T2 - m2g=m2a2.

    So what i did to find the relationship between the accelerations was substitute T2 = 2T1.
    2m1a1 - m2g= m2a2
    a1= m2(a2 + g)/2m1
    But it says that i am wrong, and there is not suppose to be the gravity constant in the relationship between the accelerations. Can someone please tell me what am i doing wrong?
     
  5. Sep 23, 2008 #4

    LowlyPion

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    You can't avoid g.

    Edit: But in part (a) the relationship of the accelerations falls out of the pulley doesn't it?
     
    Last edited: Sep 23, 2008
  6. Sep 23, 2008 #5
    I mean for part (a)
     
  7. Sep 23, 2008 #6
    Should i determine part (b) or (c) before I do (a)
     
  8. Sep 23, 2008 #7

    LowlyPion

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    Sorry, I edited my previous post when I realized you were talking about a).

    Here it is again. But in part (a) the relationship of the accelerations falls out of the pulley doesn't it?
     
  9. Sep 23, 2008 #8
    What do you mean? That a1=a2? Im really confused...
     
  10. Sep 23, 2008 #9

    LowlyPion

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    Visualize how far you have to pull the rope at the m1 side of the pulley to move the weight on the m2 side. Don't you have to move a distance of 2 for every 1 you move on the m2 side? Isn't that where the mechanical advantage comes from? You pull only 1/2 as much, but over twice the distance to move things on the m2 side? For any velocity at m2 then there is 2 times the velocity at m1.
     
  11. Sep 23, 2008 #10
    Okay i get that for any velocity at m2 there is 2 times the velocity at m1, so a1= 2a2. But Im confused on how to set of the force equations.Did i write the force equations right? Can you give me a hint on how to set it up?

    Force eqns:
    [tex]\Sigma[/tex] Fm1= T1 = m1a1
    [tex]\Sigma[/tex] Fm2= T2 - m2g=m2a2.
     
    Last edited: Sep 23, 2008
  12. Sep 23, 2008 #11

    LowlyPion

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    Ok, one equation is wrong. T2 is minus and M2*g is plus as you need to preserve the same sign convention between the two equations if you want to solve them.
     
  13. Sep 23, 2008 #12

    LowlyPion

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    If 2x1 = x2
    and 2v1 = v2
    What is a1 again?
     
  14. Sep 23, 2008 #13
    a1=2a2 ...lol
     
    Last edited: Sep 23, 2008
  15. Sep 23, 2008 #14
    how is that eqn wrong. m2g is the force going down (hence, negative) and T2 is going up, isnt the force eqn i wrote right? Also can you give me any more hints on writing each tension T in terms of m1, m2 and gravity?
     
  16. Sep 23, 2008 #15

    LowlyPion

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    You need to make Tension the same sign in whatever direction you choose.

    You chose to the right in equation 1 and it translates to the left in equation 2. As the problem stands it is net down. So T must be < m2g
     
  17. Sep 23, 2008 #16
    Okay i am still drawing a blank on this problem... So lets go back from the beginning. My text book does not explain mechanical advantage, so could you please explain a little more in depth on mechanical advantage on having two pulleys in relation to the two masses?
     
  18. Sep 23, 2008 #17

    LowlyPion

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    Maybe this will help?
    http://en.wikipedia.org/wiki/Pulley#Types_of_systems
     
  19. Sep 23, 2008 #18
    It helped a little, but still kind of foggy. Is there any more hints you could give me to help me solve the problem? I am completely stuck.
     
  20. Sep 24, 2008 #19
    Can anyone give me more hints to answer this problem?
     
  21. Oct 11, 2008 #20
    I think, we set the T's equal , solve for a1 which will give us a2. I don't think the a's fall out when solving for the tension, but I think we can find the a's.... I hope.
     
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