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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


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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


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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


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


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    That is how I understand the OP:


    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


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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?

    Attached Files:

  14. Jun 19, 2015 #13


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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


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


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    Placing the force arrows is difficult if you don't separate the bodies like shown here in step 3:

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


    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


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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


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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.
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