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This question already has an answer here:

inertia ball It is known that when we jerk the bottom string quickly, it will break; and if we pull slowly the top string will break, while the force applied is the same.

My confusion is: according to Newton's first law, an object will remain at rest or at a constant velocity when the net force acting on it is zero. However, in the mass-string system, the mass remained at rest despite the fact that the downward force exerted on the mass, in turn, on the top string exceeds the maximum tension that the string can withstand (as it caused the second string to break). So does this imply that not all the force applied on the bottom string are exerted on the mass (as the tension in the top string is lower than the bottom one)? If so, where did the force go? And why is that when the force is exerted over a longer period of time this disparity between the force applied and the force experienced by the top string seems to vanish?

Essentially, why is the tension in the top string smaller than the bottom one when jerked quickly?

I did some research online and came across answers similar to this one: "By pulling the string slowly, we are putting a strain in the string below and above the weight. Due to the mass of the weight, the strain above the weight is much larger than below. The string snaps wherever the strain is the highest. When a sharp jerk is exerted on the string, the inertia of the weight keeps the strain below the weight. Although there is some strain above the weight, compared to the strain below the weight the strain in the latter is still higher, and the string snaps below the weight."

I understand that it has something to do with inertia, but how exactly does inertia "keeps the strain below the weight"? Please explain with details.

This is my first post, thanks for any help I can get!

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marked as duplicate by sammy gerbil, ZeroTheHero, Kyle Kanos, Jon Custer, Qmechanic May 22 '18 at 15:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The tension in the top string is entirely given by the distance between the ball and the support! Think of that string as a very stiff spring.

To increase that tension, the ball has to move down. Just a bit, not much, but it does have to move to stretch the string.

If you pull slowly, the ball has time to move and does, then the upper string breaks.

If you pull fast, there’s no time for the ball to move and stretch the upper string to its breaking point before the lower string is broken. The extra tension in the bottom string puts a net force on the ball, but that just causes it to accelerate, not move instantly. In the small time before the lower string breaks, it builds up a tiny bit of speed, but not yet any net displacement.

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  • $\begingroup$ Thanks for the reply! Never thought the string as a stiff spring before, really cleared up my confusion a lot! $\endgroup$ – user2249273 May 19 '18 at 2:28
  • $\begingroup$ Can you elaborate on "there is no time for the ball to move..."? and if the upper string does not break, does this mean that the force on the ball is less than the force applied? because according to the spring formula: distance = F/k, so if the force is the same the upper one should be stretched to breaking point faster than the bottom one as there is already gravitational force acting on it. Please explain, thanks. $\endgroup$ – user2249273 May 19 '18 at 2:48
  • $\begingroup$ So does this mean that when a force is applied in a shorter time interval the displacement of the spring will be less? and that we cannot really use the formula F = k d in these cases? $\endgroup$ – user2249273 May 19 '18 at 2:59
  • $\begingroup$ F = k d fine. But what determines d? $\endgroup$ – Bob Jacobsen May 19 '18 at 3:35
  • $\begingroup$ d as the change in the spring's (or the string's) length. So according to the formula, d = F / k, since k is constant for both strings and if the force experienced by either strings is equal, they should have an equal change in length, and since the upper one is already stretched a bit by the weight, wouldn't this mean the upper one will break first as it is closer to the breaking point (maximum length)? $\endgroup$ – user2249273 May 19 '18 at 3:46

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