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In a purely theoretical scenario (no gravity, etc.), consider a string with n evenly spaced points of equal weakness along the length of it. If equal tension is applied from both ends, where will it break? Will it break at each of the points of weakness thus dissecting it into n+1 pieces, or will it only snap at the outermost points of weakness, or will it snap at the centre point given n is odd? I feel like it will divide evenly at the instant the threshold force is attained.

However, if the force is first propagated through the outermost points, then those will break first immediately ending the scenario. However, I am not sure if that is how force propagation works.

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  • $\begingroup$ Have you attempted a force analysis on the problem? I don't know that it's answerable in this state. $\endgroup$
    – Triatticus
    Commented Nov 7, 2023 at 17:36

2 Answers 2

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As you pull from both ends, the string begins to accelerate and stretch slightly. The stretching force propagates through the string at the speed of sound - there is a period of time where the middle of the string doesn't even know you're pulling yet. It's not until the waves meet in the middle do we actual get proper tension where the string is fully resisting your pulling force rather than stretching. When this happens, the tension force propagates from the center, not the ends, so the string snaps in the center.

Think of the scenario where you pull on a string attached to the wall. As you start to pull, the string does not instantaneously snap in your hands, as the wall hasn't even felt a pulling force yet - the rope in your hands has no idea if the other end is anchored or free. Only when the wave reaches the wall does it fully pull back, applying enough tension force to break the string. This force propagates from the wall side, so the string snaps near the wall.

This answer assumes that the pulling force applied is just barely enough to break the string. If you yank on a long string very hard, it's possible that the string fails before the wave propagates all the way to the other end.

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Force propagates through the string at the speed of sound (in the string). So your final assumption would be correct. The outermost weak points will feel the pull before the more distant weak points feel the pull. Imagine picking up a very long string and pulling on the ends. You will lift the ends up before the middle even leaves the ground.

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  • $\begingroup$ But a one-sided pull will merely accelerate the string, not break it. You need a pull back from the other direction to have tension. Before the force propagates, each end of the string has no idea whether the other end is free to move or will induce tension because it's anchored. $\endgroup$ Commented Nov 7, 2023 at 18:08
  • $\begingroup$ @NuclearHoagie the original question is that equal tension is applied to both ends of the string. In the case where it is only pulled by one end, then it won't break unless the friction of the string on the ground is greater than the weak point. $\endgroup$ Commented Nov 7, 2023 at 18:53
  • $\begingroup$ @NuclearHoagie I just read your answer below. So I guess it comes down to just how weak are the "weak points" in the string. If they are very weak, so that the string breaks before it stretches, then my answer would hold. If they are stronger, so that the string stretches before it breaks, then your answer would be correct. $\endgroup$ Commented Nov 7, 2023 at 18:57
  • $\begingroup$ I agree, my answer supposes that the final pulling force your arm applies is just barely enough to break the string, in which case fully 100% of the force must go into breaking the string, and cannot go anywhere else like stretching it. But if you yank hard enough on a long string, the string could break very quickly close to you. $\endgroup$ Commented Nov 7, 2023 at 19:09

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