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Newton`s 3rd Law Tension Solved Question

  1. Sep 15, 2007 #1
    1. The problem statement, all variables and given/known data

    [​IMG]



    2. Relevant equations
    They are listed in the image above, but it`s all rooted in the basic Tension equation:
    F-Ft=ma



    3. The attempt at a solution
    This is a different question I am asking. The problem IS solved as you can see. But I am having trouble understanding the logic behind it.

    The writers say that mass 1 has twice the acceleration as mass 2. They totally lost me there. Why is that the case? And how do I interpret it from the image? And then they say that when the pulley moves a distance d, m1 moves 2d. Once again, I`m clueless as to why that is. And when they say the pulley moves, they mean the pulley is moving rather than rotating, right? But either way, I lost the point. Why will m1 move distance of 2d? Wouldn`t that entail it going ahead of the pulley, or will the rope pulling it make it come in contact with the pulley?

    Then they say, the tension in Ft1 is half that of Ft2. Why?? How did they arrive at that observation? I cannot interpret the reason from regarding the image alone. Why would the total force on the pulley being zero make Ft1 have less tension??

    Somebody please help me.
     
    Last edited: Sep 15, 2007
  2. jcsd
  3. Sep 15, 2007 #2

    Doc Al

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    Yes, when they talk about the pulley moving, they mean its center moving sideways. Note that the pulley and mass 2 are connected, so they move together.

    The best way to understand how the rope/pulley arrangement constrains the motion of the masses is to take a piece of string and play around with it. Since the string is doubled over the pulley, mass 1 must move twice as fast as mass 2.

    Another way to look at it is to examine the rope wrapped around the pulley. Think of it as having two parts: the top side and the bottom side. If the pulley moves 1 foot to the right, the top side lengthens by 1 foot. Since the rope's length is fixed, that means the bottom section must shorten by 1 foot, making m1 closer to the pulley by 1 foot. So... the pulley has moved to the right by 1 foot and m1 is closer to the pulley by 1 foot... thus m1 must have moved a net distance of 2 feet to the right. Make sense?

    Sooner or later m1 will bang into the pulley--and then the party's over for this little set up! :smile: (Pretend that it never gets that far.)

    Each rope segment attached to the pulley exerts a force on the pulley. The left rope attaches twice, so it exerts its tension twice. That force is balanced by the single right rope segment, so it must exert twice the tension.

    Realize that these effects are connected. Since the rope/pulley constrains the system so that m2 only moves half as far as m1, it must exert twice the force to conserve energy.
     
  4. Sep 15, 2007 #3
    Doc Al, I can`t thank you enough for the attention you gave to this. I`m still milling your explanation around in my head, but you addressed everything very well for me.

    So, basically m1 moves twice as much because it:
    1) Moves the same distance as the pulley.
    2) The tension from Ft1 pulls on the rope attached to m1 and the length of the bottom part of the rope diminishes.

    Correct?
     
  5. Sep 15, 2007 #4

    Doc Al

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    I think you're getting it, but allow me to restate it a bit. M1 moves twice as fast as M2 (which moves at the same speed as the pulley) since:
    (1) M1 gets closer to the pulley at the same rate as the pulley moves
    (2) the pulley itself moves

    When you add these two speeds you'll see that M1 must move twice as fast as M2.
     
  6. Sep 15, 2007 #5

    arildno

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    Dearly Missed

    1) is NOT correct.

    Look at the top segment of the rope, ending right in front of the pulley (a straight line segment plus a quarter circle:
    Suppose the pulley moves some distance D forward.
    Then, the top segment of the rope still consists of a quarter circle and a straight line sgment.
    Clearly, since the arc length of the quarter circle does not change, and the top segment is fixed to the wall, its end unmoving, it means that the straight line segment of the top rope piece must have increased with a length D. Otherwise, the pulley wouldn't be able to move a distance D forward, contradicting the hypothesis.

    Now, the lengthening of the top segment cannot be in the form of a stringlike stretching; a rope is considered to be of FIXED length.

    Hence, that increase in length of the top piece of the rope can only come about if the lower rope straight line segment has SHORTENED by a length D.

    Now, if m1 had only moved the same distance as the pulley, then the relative distince between them, would remain constant.
    But that relative distance is, precisely, determined by the length of the lower rope's straight line segment!

    Hence, m1 must move twice the distance D as the pulley, in order to decrease the lower rope's straight line part with length D.

    Thus, the acceleration of m1 must be twice that of the pulley, and therefore twice that of m2, since the pulley and m2 has the same acceleration.
     
  7. Sep 15, 2007 #6

    Doc Al

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    Say what????

    This is exactly what I stated. Reread what I wrote:
    (1) M1 gets closer to the pulley at the same rate as the pulley moves​
     
  8. Sep 15, 2007 #7

    arildno

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    1) was a reference to Sabelle's misstatement "1) Moves the same distance as the pulley."

    I need time to write my reply, yours wasn't up when I began it..:smile:
     
  9. Sep 15, 2007 #8

    Doc Al

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    D'oh! :uhh: (Next time please use quotes!)

    That's happened to me a few too many times. :redface:
     
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