Calculating Tension (with rope mass)

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SUMMARY

This discussion focuses on calculating tension in a rope with mass, emphasizing that tension varies linearly along the length of the rope. The key equations include F=ma and T=f(y)=mg + (MsubscriptR times g/maximum length) times y. A practical example illustrates that for a uniform rope, tension at the bottom is 20 N, at the middle is 25 N, and at the top is 30 N when a 2 kg mass is hung from a 1 kg rope. The conversation also clarifies that tension in the rope does not depend on individual molecules but rather on the total force distributed across the rope's cross-section.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with basic concepts of tension and force
  • Knowledge of linear equations and their applications in physics
  • Basic grasp of differential equations and their relevance in physics problems
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  • Study the concept of linear tension in ropes and its derivation from Newton's laws
  • Explore the application of differential equations in calculating forces in variable mass systems
  • Learn about Free Body Diagrams and their role in analyzing forces in physics
  • Investigate the molecular structure of materials and how it relates to tension and elasticity
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Students studying physics, particularly those focusing on mechanics, as well as educators looking to explain the principles of tension in ropes with mass.

Rayquesto
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Homework Statement



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?

Homework 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''

The Attempt at a Solution



The relevant 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.
 
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Are you sure it's a differential? I think you can solve that with an integral…maybe not, but that's what it looks like to me.
 
Rayquesto said:
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?


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.
 
ok thank you. I'm spending sometime thinking about what you are saying and by the looks of it, you are referring 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?
 
Rayquesto said:
ok thank you. I'm spending sometime thinking about what you are saying and by the looks of it, you are referring 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?
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.
 
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.
 
Rayquesto said:
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.
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.
 
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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