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Tension w/ rope hanging from a ceiling

  1. Feb 25, 2012 #1
    1. The problem statement, all variables and given/known data
    The ends of a massless rope are tied to a ceiling. Two identical 4.21 kg masses are now hung along the length of the rope, dividing the rope into three segments of equal length. Segment 2 (the central segment) is horizontal (parallel to the ceiling). The ends of segments 1 and 3 are attached to the ceiling, making an angle of 50.2° with the ceiling. (The distance between the points at which the rope is hung is greater than the rope segments.) Find the magnitude of the tension in rope 2.

    2. Relevant equations
    Pythagorean and components

    3. The attempt at a solution
    I attached my diagram of how I drew everything. I'm hoping that this is correct. If it is, then I assume rope 2 is the middle section of the rope where it is parallel with the ceiling (my professor doesn't seem to phrase his questions that well, or I'm just not used to it). After solving for the tensions in rope 1 and 3, which are equal to each other, we could move the x-component vector to where I made it in red. These two vectors act in opposite directions, causing the tension of the middle rope to be twice as much. That is my assumption.

    If my assumption is correct, then the tension in rope 2 should be:


    This answer is incorrect, so could someone tell me where I went wrong?

    Attached Files:

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  2. jcsd
  3. Feb 25, 2012 #2
    Why do you multiply by 2?
  4. Feb 25, 2012 #3
    I thought since there is tension from both weights that go in opposite directions (like tug-of-war), then the tension would be twice as great.
  5. Feb 25, 2012 #4
    What about Newton's law?
  6. Feb 25, 2012 #5


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    hi iJamJL! :wink:
    noooo :cry:

    do a free body diagram for a small horizontal bit of string starting at the right-hand end …

    Tcos50.2° to the right, and … to the left? :smile:
  7. Feb 25, 2012 #6
    If you tie a rope to a doorknob and pull on it with a force of 10 N, is the tension in the rope 20 N? I don't think so.
  8. Feb 25, 2012 #7
    Newton's third law says that for every action, there is an equal and opposite reaction. Therefore, if we created those reactions where the masses are, then we would have two equal forces going toward each other (one horizontally to the right of the left mass, and one horizontally to the left of the right mass), resulting in 0. That can't be because tension exists, making the string parallel. I suppose I'm just not thinking in the correct manner. :frown:

    Isn't it the same force, but in the opposite direction?

    What about the left-end rope?
  9. Feb 25, 2012 #8


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    the left-end rope isn't anywhere near the small bit at the right-hand end, is it? :wink:

    what is the force on the left of that small bit?
  10. Feb 25, 2012 #9
    It's equal to the force on the right side, Tcos(50.2°). I saw this before. When we look at the center of the middle section, then the net force would be equal to 0, wouldn't it?
  11. Feb 25, 2012 #10


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    i meant, what's it called?

    it's called the tension :wink:
  12. Feb 25, 2012 #11
    Oh, I knew that, lol. The whole difficulty of the problem for me is the fact that I see two forces of tension that act in opposite directions due to the two weights. I thought we were supposed to find the net force, or net tension, due to those weights. It would either be twice (that's why I multiplied by two: Fnet = F-left + F-right) the force of one, or 0, which didn't make sense at all. The answer seems to be that there is only one force of tension acting on the central section of the rope..but I don't think I fully, 100% understand it.
  13. Feb 25, 2012 #12
    Think about the rope and doorknob example I mentioned. The knob pulls on the rope with 10 N but in the opposite direction from your pull. But the tension in the rope is 10 N not 20 N.
  14. Feb 25, 2012 #13


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    You can't find the tension in the middle section by looking at the whole middle section …

    you can only find it (or any other force) when it's an external force …

    that means you have to "cut" the string, so that you can consider the tension as an external force
  15. Feb 25, 2012 #14


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    In your tug-of-war example.
    Team 1 pull on the rope with a combined force of F Newtons.
    Newtons 3rd law requires the rope to pull back on them with a force of F Newtons.

    Meanwhile Team 2 pulls on their end of the rope with a combined force of F Newtons.
    Newtons 3rd law requires the rope to pull back on them with a force of F Newtons.

    All that is easily achieved by the Tension in the rope being F Newtons - exactly the required force.

    Note: What is impossible is for one team to pull more strongly on the rope that the other team. The team that wins the tug of war is the team that pushes on the ground the hardest!
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