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Why is the net force on block A twice that on block B?

  1. Jul 9, 2015 #1
    www.webassign.net/serpop/p7-47.gif

    In the figure that I've posted in the link above, it is assumed that both blocks are initially held at rest and then released. Why does block A move with twice the speed of block B?
     
  2. jcsd
  3. Jul 9, 2015 #2

    jtbell

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    It's because of the Law of Conservation of String. (a deep and subtle result of String Theory :cool:)

    Suppose block B moves upwards a distance d. What happens to the lengths of the middle and right-hand sections of the string? What has to happen to the left-hand section of the string in order to keep the total length constant?
     
    Last edited: Jul 9, 2015
  4. Jul 9, 2015 #3
    Forget about speed for a minute. If you pull block A down one centimeter, what happens to block B? It goes up, right? How far does it go up?

    edit - jtbell beat me to the punch
     
  5. Jul 9, 2015 #4
    I'm still not quite sure.. to me, it seems that if I were to pull block A down by 1 cm, then the middle string will want to pull up by 1 cm.
     
  6. Jul 9, 2015 #5

    jtbell

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    If the middle string shortens by 1 cm, doesn't the right-hand string also have to shorten by 1 cm? What would happen to the total length of the string if that were true?
     
  7. Jul 9, 2015 #6
    Hmm. So if I were to imagine that the middle and right string were instead one string with a block attached, then this string would move up by 1 cm upon pulling block A down by 1 cm.

    If this new string were bent around a pulley and attached to a platform as in the figure, such that its length is the same, then upon pulling block A down by 1 cm, the middle string must move up 1/2 cm and also for the right string?
     
  8. Jul 9, 2015 #7
    Asumming the pulleys have no mass (so the tension is the same across all the string), then there is only one tension pulling block A upwards while there are two pulling block B.
     
  9. Jul 9, 2015 #8
    I understand that, and I actually tried using Newton's second law to show what the acceleration of block A would be. I modeled each block separately as a particle under a net force in the vertical direction. The horizontal components of force are all zero so,

    For block A,

    T = m(g-aA)

    where aA is the acceleration of block A

    For block B,

    T = ½m(aB+g)

    Where aB is the acceleration of block B

    This yields,

    aB = g-2aA

    and I'm not sure whether that is correct. I certainly know that I haven't been able to take any information from it...
     
  10. Jul 10, 2015 #9

    jtbell

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    Assuming that the two masses are equal, this is correct. This gives you one equation for the two unknowns aA and aB. You can get a second equation from the geometrical condition that the total length of string is constant. You're close to it in your second statement in post #6. This gives you two equations in two unknowns.
     
  11. Jul 10, 2015 #10
    I got to the same result as you! . And it is strange, if the speed A at any given moment is twice as the speed at b. Then aA=2aB. I also don't get it.... It seemed such an easy problem. We are probably wrong on something. From the calculations we also get that (g-aA)/(g+aB)= 1/2. I don't know if that is relevant, but it is interesting
     
  12. Jul 10, 2015 #11

    jtbell

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    Yep.
    This is also valid. It's actually the same as Kyle's last equation from post #8:
    which you can verify by multiplying out your equation to get rid of the fraction on the left side. So you have two equations in the two unknowns aA and aB. What's the first thing that you should think when you see two equations in two unknowns? :biggrin:

    (or three equations in three unknowns, or four equations in four unknowns, etc.)
     
    Last edited: Jul 10, 2015
  13. Jul 10, 2015 #12
    So we now have a system of equations and can solve for both accelerations and compare them, right?..

    I'm still wrapping my head around the string conservation idea too.. The only way it makes sense to me is by imagining that the middle and right string instead is just one string that has been "divided" into two strings. If the left string is pulled down by 1 cm, then the combination of the movement of the middle and right string must be 1 cm also. Is this where I'm wrong?

    Edit:

    Since Thom's equation is just a rearrangement of my equation, wouldn't we just end up with 1 = 1?

    Edit #2:

    I realize that you may be referring to the equation

    aA = 2aB

    In the context of solving the system

    And I was reffering to

    (g-aA) / (g+aB) = ½

    My apologies for the confusion. I'm trying to show that

    aA = 2aB

    Which is why I'm trying to understand the string conservation idea.
     
    Last edited: Jul 10, 2015
  14. Jul 10, 2015 #13
    Two equations two unknowns should be easy to solve right ? But in this case i couldn't. Can you show us the result for aA and aB? The system of equations ends up undetermined in my case. Help !
     
  15. Jul 10, 2015 #14
    Very nice :biggrin:
     
  16. Jul 10, 2015 #15
    You have said this a few times. It confuses me, because the middle and right strings ARE just one string. There is only one string, it is attached to A on one end and the ceiling on the other end.
     
  17. Jul 10, 2015 #16

    jtbell

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    The second and third equations in my post #11 are indeed rearrangements of each other, so solving them together simply gives 1 = 1 as Kyle noted. Oops. :blushing:

    Try solving the first and second equations together, or the first and third. If you get stuck, show what you did.
     
  18. Jul 10, 2015 #17
    We need one more equation to solve this....
     
  19. Jul 10, 2015 #18

    Doc Al

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    Here's a simple way to derive the acceleration constraint from the "conservation of string". (Although just playing with a piece of string should be enough for you to figure it out.)

    Let ##x_a## be the distance below the ceiling of mass a; let ##x_b## be the distance below the ceiling of mass b. Using those distances, express the length of the string:
    ##L = x_a + 2x_b##.

    Now take the time derivative of both sides. (Since the length doesn't change, its derivative is zero.) Do it once to get the velocities: ##0 = v_a + 2v_b##; Do it again to get the acceleration: ##0 = a_a + 2a_b##.

    That tells you that ##a_a = -2a_b##. (If all you care about is the magnitude, then ##a_a = 2a_b##.)

    Make sense?
     
  20. Jul 10, 2015 #19
    Good explanation Doc, But can that be true as well as aB = g-2aA ?
     
  21. Jul 10, 2015 #20

    jtbell

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    Yes. This allows you to find the values of aB and aA, by solving the two equations together.
     
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