Where would a string being pulled from both sides snap and why?

Question

Consider a massless string being pulled from both ends with equal forces, increasing gradually and equally on both sides. At the point where the string can no longer withstand the tension, the string snaps (ultimate failure). My question is where (meaning point on the string from an end) will the string snap, and why? Since the string is massless, I think the imperfections in the string is out of the question.

Now considering a real string (having mass), but perfect in the same situation. Where will it snap, and why? I think answer to the first one will apply here as well.

As for the third type (no answer expected, but if you have something to add, you are very welcome), that is string having mass and is imperfect, I know it will snap at the weakest point along its length. I assume that at the weakest point, the tension is somehow greater than at any other point on the string, causing it to give in first.

Answer 1

In the ideal case, the breaking probablitly function $f(x)$ can be defined to have uniform probability distribution in $x$ ($x$ is position). Which basically means that the string can break anywhere in a single experiment with no special $x$ - where is breaks more often.

In a real string, when the bonding electromagnetic force is overcome, the string breaks. Here, $f(x)$ will ofcourse depend inversely on the bonding strength function $B(x)$ (measures the strength of bond at point x). $B(x)$ will depend on the the thickness of the string at point x. This is as far as you can go without the actual string.

Answer 2

I strongly disagree with some points of the first answer (Stephan).

First: the middle of the string is no better than any other point. If the string is to snap at all, any infinite inhomogenity will be relevant for chosing where.

Second: with mass, it is still more wrong - the tension at the ends will be greater than in the middle (the horizontal tension is equal everywhere, but in the ends there is an additional vertical tension of half the mass - the difference in the magnitude of the tension (Pythagoras) will be small of second order for small mass, but if the homogenity is really good, the probability of snapping should rise towards the ends)

Third: here the exact words are important, the OP says, that the tension is greater at the weakest point - well, the force is of course equal (in the massless case :)) but the force per area is greater for a thinner region, so there it will break (if the material homogenous at least).

Answer 3

Consider a massless string

Why massless ? What does this change ?

being pulled from both ends with equal forces, increasing gradually and equally on both sides. At the point where the string can no longer withstand the tension, the string snaps (ultimate failure).

Having mass would not change that.

My question is where (meaning point on the string from an end) will the string snap, and why? Since the string is massless, I think the imperfections in the string is out of the question.

No, having mass does not mean you have no imperfections. Stating the string is homogeneous is all that is required.

The string will snap in the middle, regardless of mass, once the forces are equal and the string has no random imperfections.

Now considering a real string (having mass), but perfect in the same situation.

You are saying it is still a homogeneous string - no random imperfections.

Where will it snap, and why? I think answer to the first one will apply here as well.

Same again. Middle.

As for the third type (no answer expected, but if you have something to add, you are very welcome), that is string having mass and is imperfect

Forget the mass. Now saying it's "imperfect" doesn't mean much. What kind of imperfections ? Are the imperfections localized to one region ? Are they related to tensile strength ? You need to be precise when explaining a problem.

Anyway we'll assume it's some tensile weakness at some point along the string.

I know it will snap at the weakest point along its length.

Basically.

I assume that at the weakest point, the tension is somehow greater than at any other point on the string, causing it to give in first.

The tension is no (initially) greater anywhere.

Typically you'd see the string stretch out, then some fibers (the weakest) break, but the remaining ones would continue to hold until the tension became too great for them. Obviously as the number of fiber's still intact decreases, the tension in the remaining fibers is increased. The string will break when the tension exceeds what the last remaining fibers can hold.

The precise point of final failure won't be the same as the initial point of failure. The first failure will be the weakest fiber, but that obviously won't be the last failure point.

Answer 4

I believe the ends (fixture points) are the more likely failing spots. The first reason is these are the places where it is most hard to avoid microscopic damages to the string texture which will develop into large-scale fractures.

Second, I tried to depict a string under tension in the sketch below approximated by an elastic band/cylinder. It should be clear that near the corners of the fixtures the local tension is in general larger than the one away from there. Details will depend strongly on the inner structure of the string/rubber band, but at the end of the day, only the horizontal component of the net tension must be equal throughout the string/band.