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Calculating Tension (with rope mass)

  1. Sep 18, 2011 #1
    1. The problem statement, all variables and given/known data

    I really didn't plan to post this in hopes of finding a particular solution, but rather to learn what to do if a problem contains a pulley, a tension and a rope with mass, or any problem that had mass of rope and a tension. My professor and I tried working out this problem, but we couldn't find a way to do it. My professor told me that if the rope had mass, then the tension would increase as the distance from the object that is creating tension increases. So, the question is How do you calculate tension if there is mass in a rope?

    2. Relevant equations

    F=ma
    T=f(y)= mg + (MsubscriptR times g/maximum length) times y
    *where y=length from mass creating the tension in the first place.

    M d^2y/dt^2 = mg = P(density(rho)) times y times Area times g
    where y=f(t) and m=pyA which means there is a second differential equation involved and my professor couldn't figure it out but guessed that it may have been y=(e^c't) + c''

    3. The attempt at a solution

    The relevent equations are what my professor gave me. I am a little confused as to how the equations connect and where to go next. By the way, I don't really need to know this, but I am highly curious to know how to calculate tension if mass of rope exists. I would also like help in understanding how this relates to masses at the molecular level.
     
  2. jcsd
  3. Sep 19, 2011 #2
    Are you sure it's a differential? I think you can solve that with an integral…maybe not, but thats what it looks like to me.
     
  4. Sep 20, 2011 #3

    PhanthomJay

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    For a uniform rope with mass, the tension variation in the rope is linear, and your professor is correct. No need for any calculus. Draw Free Body diagrams. Suppose you have a rope hanging from a ceiling that has a mass of 1 kg, and you hang a 2 kg mass from the bottom of it. At the bottom end of the rope, using Newton 1, T = mg, where g = 10m/s/s, then T_min = 20 N. At the top of the rope, T_max = 30N. At the middle of the rope, T_mid = 25 N. A linearly varying tension. The same concept can be used with accelerating masses and ropes with mass.
     
  5. Sep 20, 2011 #4
    ok thank you. I'm spending sometime thinking about what you are saying and by the looks of it, you are refering to the tensions not just relative to the rope and the ends, but also the tension due to the molecules in the rope. How does that work? doesn't the entire rope contain mass? Wouldn't that mean that tension would depend upon a constant that contains a differential equation which tells also the weight at a certain point? Well, I see what you are saying when you mean that the tensions vary linearly. So, linearly with a constant that depends upon distance from fixed points?
     
  6. Sep 20, 2011 #5

    PhanthomJay

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    Tension represents a force in the entire rope at a given point along its length. It does not represent the force in individual molecules. If the rope had a tension of 100 N at a certain point, and it was made up of 100 individual smaller strands, each strand of the rope would experience a tension of 1 N. For a uniform rope of mass m hanging vertically, the tension varies linearly from a max at the top end to a min at the bot end.
     
  7. Sep 22, 2011 #6
    I'm going to have to ask my professor about the whole idea of the molecules in the rope and how there are small springs. He sounded very confident that there are tensions in them. I don't know who to believe yet. He seemed very imaginative and correct.
     
  8. Sep 22, 2011 #7

    PhanthomJay

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    Prof is right. The point I was making is that if the tension in a rope is 100 N, and there are a zillion molecules across the ropes cross section, then the tension in each molecule is one-one zillionth of 100 N. The sum total of the tensions in each molecule at the surface area cross section is the total rope tension at that plane.
     
  9. Sep 22, 2011 #8
    I see! Ok OK OKAH! thank you. I get the big picture now. You sound much more clearer now, or rather you make much more sense now.
     
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