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50N from left, 50N from right. Tension in string still 50N?

  1. Jun 19, 2015 #1
    Let's say I go to the store and purchase a rope. On the package of string, it says that the string can only withstand 100 Newtons before the rope snaps. I go home and test the tension in the rope. I tell a friend to pull on the rope at 99 Newtons while I pull on the rope with 99 Newtons. Did I exceed the recommended tension? If the labeling on the package is correct, will the rope snap because there is some force within the string that causing the rope to experience more than 99 Newtons in the horizontal directions (excluding gravity)? Not sure what stress and strain are yet, but could I be confusing tension with some other concept that I have yet to learn about. I feel that tension in the string should somehow be as strong as the sum of the forces acting on it.
     
  2. jcsd
  3. Jun 19, 2015 #2

    A.T.

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    The sum of forces acting on the string is zero, if you both pull with equal but opposite forces.
     
  4. Jun 19, 2015 #3
    EDIT: Never mind.
     
    Last edited: Jun 19, 2015
  5. Jun 19, 2015 #4

    A.T.

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    The tension is 99N if both pull with 99N. Not sure what is "double" here?
     
  6. Jun 19, 2015 #5
    Here's what I think.

    If you are holding one end of the string and standing still, your friend pulling at 99N will exert 99N of force on the string. If you then start pulling at 99N in the opposite direction, the total force on the string is 198N. The forces are equal and opposite so the string won't move anywhere but it may well snap as you are over the 100N limit.
     
  7. Jun 19, 2015 #6

    A.T.

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    Nope.
     
  8. Jun 19, 2015 #7
    EDIT: Hmm, I'm confused myself now. After all, the scenario is static.
     
  9. Jun 19, 2015 #8

    A.T.

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    That is how I understand the OP:

    balanced%20forces.gif

    From:
    http://physics101samimuller.blogspot.de/2014/12/tension-force.html [Broken]

    No. The sum of forces is zero, but the tension is 99N
     
    Last edited by a moderator: May 7, 2017
  10. Jun 19, 2015 #9
    Yeah, I edited my post. Given how the scenario is static, whether a wall is holding it in place, or something else pulling with the same force, comes down to the same thing.
     
  11. Jun 19, 2015 #10
    Thank you. If I may, suppose my friend, standing on the left side of the string, continues to pull with a force of 99 N while I, on the right side of the string increase my pull to 100 N. Does the string have 99 Newtons on Tension, 100 Newtons, or a combination? Does my friend on the left feel 99 N of force pulling or 100? Do I feel 100 or 99 N?
     
  12. Jun 19, 2015 #11

    A.T.

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    It will have a tension gradient, going from 100N on your side, to 99N on his side. An the whole thing isn't static anymore, because the forces don't balance.

    The string pulls on you with same but opposite force as you pull on the string. Same for your friend.
     
  13. Jun 19, 2015 #12
    Would the attached image provide an accurate representation of the description you provided?

    If so, then what, may I ask, does the rope feel in the middle? Suppose you choose a point exactly in the middle of the rope and measure its tension. Would it be 100 N, 99 N, or 100 N in one direction and 99 in the other direction?
     

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  14. Jun 19, 2015 #13

    tony873004

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    If you pull with 99 N, your friend has no choice but to pull with 99 N. That's Newton's 3rd Law. Your friend could be a doorknob. Tie the other end of the rope to a doorknob. If you pull with 99 N, the doorknob must pull with 99 N.
    This reminds me of the tug-of-war question. Which team wins? Most would say the team that pulls hardest. No! According to Newton's 3rd Law, they must both pull with the same force. The team that wins is the team that pushes the ground hardest.
     
  15. Jun 19, 2015 #14
    Of course Tony. If the doorknob can only pull with 99 N and I pull with 100 N, the doorknob will have to accelerate at f=ma. That would be 1m/s^2 if the doorknob is 1kg.

    And to expand on this using two people to vary the pulling forces,

    If we suppose friction on the ground is included in the force of the pulls and the puller on the left pulls with 99 N of force and the puller on the right pulls with 100 N, there will be a net force on the rope of 1 N, right? So the rope would accelerate at F/m = a. Suppose the rope is 1 kg. The rope will accelerate at 1N/1kg = 1 m/s^2 relative to the ground. If the puller on the left, supposing he continues to grasp the rope, will have no choice but to "slide" or accelerate in some way at 1m/s^2 along with the left end of the rope, yes?
     
    Last edited: Jun 19, 2015
  16. Jun 19, 2015 #15
    These tension forces can be a bit confusing.
     
  17. Jun 20, 2015 #16

    A.T.

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    Only for a static situation or if we consider the rope to be mass-less.
     
  18. Jun 20, 2015 #17

    A.T.

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    Placing the force arrows is difficult if you don't separate the bodies like shown here in step 3:
    http://adaptivemap.ma.psu.edu/websi...and_machines/analysisofframesandmachines.html

    Assuming a uniform mass distribution in the rope it would be 99.5N.

    Yes.

    Initially he could just extend his arm, but beyond that he will fall over or slide if he fails to provide 100N.
     
  19. Jun 20, 2015 #18

    NascentOxygen

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    To determine the acceleration, you need to use the combined masses of the rope and the person/object fastened to its other end since these move as one.
     
  20. Jun 20, 2015 #19
    A.T, thanks, I didn't see post #11 last night. It's very informative.
     
  21. Jun 20, 2015 #20

    Merlin3189

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    I think this has been dismissed? Though I find these discussions a bit hard to follow sometimes.

    But what I wanted to say was, before the second sentence, before you start pulling on the rope, what happens to it? I guess it will start to uncoil as he moves away from you, as he will have to if he is to apply 99N, but if it's a heavy rough rope perhaps eventually the friction of the ground on the rope will exceed 99N and stop him. (If it's not heavy enough, he pulls out all the rope and continues accelerating into the distance, so that you can never exert your pull because the rope has gone!)

    Now when you start pulling on the other end in the opposite direction, it depends how long the rope is. If he had pulled out less than half before the friction stopped him, then the same will happen for you and you will be stopped before you have pulled out half the rope. Each of you will be pulling with 99N against the static friction of your portion of rope and there will be a slack pile of rope remaining in the middle.
    If he had pulled out exactly half, then the situation would be the same, but without any slack in the middle: you would still each be pulling against the static friction of half the rope and there would be no tension exactly at the centre. The tension in the rope will be zero at the centre and rise gradually away from the centre to reach a maximum of 99N at each end. ( If the ends of the rope are lifted from the ground, there may be a variation in the rate of change of tension, but the tension at centre and end are unaffected .)

    If he had pulled out more than half the rope, then you will start pulling on "his" part of the rope before you stop. The zero tension point will disappear and there will be tension in all parts of the rope, with the minimum position moving towards the centre. You will stop as this minimum reaches the centre and the situation is then symmetric with each of you pulling with 99N, the maximum tension at the ends, reducing gradually thanks to friction where the rope is still in contact with the ground and to gravity where you have lifted the rope from the ground. The tension in the centre will always be less than 99N in this way of looking at it.

    If you start taking account of the elasticity of the rope and the mass of you and your friend, then you may get some oscillations which could vary the tension above and below the static values. But then you would have to have some spare rope, so that you can let the rope slip to limit your tug to 99N when it tries to accelerate your mass, and it gets a bit more complicated once you start looking dynamically.

    That is rather more than I intended to say, when I wondered what would happen if we paused after the first sentence!

    The "total force on the string" seems to be lacking the weight, any forces from the ground and any force due to air pressure. If you add up the sum of the magnitudes of all the forces on the string, I guess it could add up to more than 198N. Maybe "total force" is not a very useful concept?


    Well I hope we are sticking with the static situation. But I don't see why "mass-less"? If the rope is heavy, but suspended and static, then the situation is symmetric and the force equal (though not opposite) at each end. If the rope rests at least partially on the floor, then your friend has the option of leaving some of the pulling to friction! But gravity will not preferentially aid him, unless you are higher than he.

    As far as "no choice" goes, I think this may be a barber not shaving himself. Unless you (& your friend) are allowed to release the rope, or at any rate let it slide through your hands, I can't see how you can maintain exactly 99N at all times. Is this the a pachyderm in the forum.
     
  22. Jun 20, 2015 #21

    A.T.

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    No we didn't.
     
  23. Jun 20, 2015 #22
    Hi Merlin,

    Perhaps you went over my head a little bit, lol. Not entirely sure what you mean by "pulling out half of the rope."

    Thanks
     
  24. Jun 20, 2015 #23

    Merlin3189

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    Ok. Well in that case I'd have to say that whether the rope is massless or not, you can't keep to 99N unless you and your friend are also massless.

    If we can have a light inextensible string attached to two opposing 99N forces with no mass and no other forces, then the original question gets even more trivial.
    If so, I'd take a different tack in explaining tension - what is it, where does it come from, why is total tension not meaningful, etc.
     
  25. Jun 20, 2015 #24
    Suppose I tied the rope to a box of 1,000,000 kg (M) and, while I experience enough friction to remain static, the box is on a frictionless surface.

    If I pull the rope with 1N, and keep pulling on the rope so that it remains 100% taut, the box will accelerate toward me at F/M = a = 1/1,000,000 = .000001m/s^2. It will accelerate very slowly, but it will still accelerate. I find it amazing that no matter how much mass an object has, it will still accelerate when acted upon by another force.

    However, when a box of just 100kg (m) is sitting on a table with some friction, it won't budge when applying 100N (or 100x the force that I applied on the gigantic mass in the previous scenario).

    Does this difference in the way the two boxes interact with my pulling force mean that the 100kg box (m) is pulling against my 100N pull with more opposing force than the 1,000,000 kg box (M)?

    Does the M kg box even have a resisting/opposing force? If I continue applying the small force of 1N to the M kg box, it will continue to accelerate, so there must not be any force to match my force and balance out the net force to 0, right?

    So does mass itself does not provide a pulling force at all? Box M may feel to me like it's pulling against my pull, but it really isn't since it's actually accelerating very slowly - even though I can't detect it?

    Scenario 1: If I try to evaluate all the opposing forces to find the net force, suppose I'm pulling 100N on the rope, the rope is pulling 100N on me, the other end of the rope attached to Box M is pulling on it with 100N and the box M is pulling on the rope with 100N, is this correct? If so, then why does the box M in the frictionless state not oppose my pull with 100N? It seems all the forces should cancel each other out for a net force of 0. The rope and I should be a net force of 0 and the rope and the box M should be another net force. Two net forces of 0 should mean no acceleration, I thought.

    Scenario 2: Now, if I pull the box m (with enough friction to remain static), I still count a net force of 0. My 100 N applied to the rope, the rope applying 100N on me, the rope applying 100 N on the box, the box applying 100 N on the rope, the surface applying 100 N on the block, the block applying 100 N on the surface. 3 net forces of zero.

    In both scenarios, all of the force pairs seem to balance each other out to a net force of 0, unless I'm over counting or undercounting a force somewhere..
     
  26. Jun 20, 2015 #25

    Merlin3189

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    Ocata: OP has the friend start pulling on the string (sorry I picked up rope from another post without noticing) and I wondered what would happen. If the string were laid out in a straight line, then (assuming it to be inextensible, but not light) it would immediately start to accelerate like a solid body. Assuming a small delay before you try to pull on your end, then it might be out of your reach before you could catch hold.
    So I thought, just have it be long and loosely piled in a heap so that even while your friend was racing away, your end would stay still for a second or so.

    If perchance the string is light (or lightish) then applying 99N to it is going to accelerate it very quickly - a 2m string of 100N working load strength would be less than 0.1kg and so would accelerate away at about 1000m/sec/sec or more. Pausing for even 0.1 sec before grabbing the string to exert your pull, the string is at least 5m away from you and travelling at 100m/sec! Even if you were running before your friend started pulling, you'd not catch it.
    Perhaps your friend starts by pulling the string towards you and you catch the end as it flies past. If you are massless, then you're simply going to reduce the rope's acceleration to zero and move with your friend and the rope at whatever speed they have reached. If you have mass, and as AT wants we look at dynamics, then it all gets more complex. If the string were inextensible then the problem becomes impossible: one of you has to have infinite acceleration. So now we have to consider elasticity and length of the string and although the problem may be tractable, we are almost certainly going to break the constant 99N rule of the original problem.


    The problem with the OP question is the difficulty of applying a single force of 99N to a string, unless that string is a long, heavy rope. I thought that using a long, heavy rope might make the problem more tractable.
    Additionally allowing it to slide along a rough floor provides damping to the spring mass system and maybe allows us to gloss over the dynamic problems and look at the steady state results..
     
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