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.
Best Answer
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.