Where would a string break if an instantaneous UNEQUAL tensile force is applied on the ends?

Question

Similar to a question asked previously except this time with unequal tensile forces at the ends: Where would a string being pulled from both sides snap and why? Assume the string is homogenous.

Would it be possible to predict where the string will fail? i.e Can I say with certainty whether the string will fail at a point closer to the end with greater force applied or vice versa.

Not sure if the way I've posed this problem is accurate, but for reference it stems from a game where players pull on both sides of a string and whoever has the longer piece wins. I am neglecting the "pinching force" by the players hands and assuming a perfect tensile force is applied.

Answer 1

It is not possible for a massless string to have unequal forces applied, so this question requires that we consider only the case of a massive string.

Because the string is massive it will accelerate, so in the steady state we can consider it to be a string held vertically in a pseudo-gravitational field of magnitude $g=\Delta F/m$. In such a configuration the maximum stress is at the top of the string and the stress decreases linearly down to the bottom. So the string would break at the top.

However, that assumes a static loading, not a dynamic loading. If the loading is increased from 0 to the maximum in a time which is much shorter than the length of the string divided by the speed of sound in the string, then for a brief time there will be a very high stress right at the points of application with the stress going to zero in the middle. In this configuration it is possible for the string to break on both ends while the middle remains intact.

Answer 2

Typically a string is assumed to be massless in beginning physics class problems. This is an approximation of course. Its purpose is to make the problem simpler so you can focus on the forces on the objects the string is attached to.

The approximation breaks if you apply different different forces on the ends of a massless string. There would be a non-$0$ total force on the string. $a = F/m = $ doesn't work.

So you have a string with a mass. A uniform string has a uniform mass per unit length. A new approximation for dealing with this might be a chain, where each link has the same mass.

Let us make the problem easy. Suppose one of the forces is $0$. Then you are pulling one link $1$, link $1$ pulls on link $2$, etc. If you draw a free body diagram, there are two forces on each link.

All of the links have the same acceleration. So all have the same total force, call it $F$. That means the sum of the two forces is the same. They are opposite, but not equal.

Starting at the far end. The force on the far end of the last link is $0$. The other force must be $F$. The force on the far end of the next link is $F$. The other force must be $2F$. In this way the forces get bigger and bigger. The forces on the first link are the biggest, and this is the link that will break.

You can see an interesting problem something like this one here: I Rented A Helicopter To Settle A Physics Debate

Answer 3

In your realistic game, the answer is: No, you cannot predict where the string will break. It solely depends on where the manufacturing weaknesses are located. This assumes that "instantaneous" is slow compared to the speed of sound in the string material, and that the string is light compared to the applied forces. In this case, there is no such thing as "unequal tensile forces", the actual force with which the string tugs on each players hand are equal. The lucky player wins.

Now, let's relax the "light compared to the applied forces", i.e. assume that the mass of the string is relevant. Something like heavy chain links bound together with pieces of flimsy string. In this case, the unequal forces mean that you have a net force that acts to accelerate the chain. Now, each link is accelerated by the force difference between the two strings that link to it. So, as you go from the end with the lower force towards the high force end, tension of the chain rises. As such, the chain is likely to break at the side with the stronger player. The weaker player wins.

Not satisfied yet? Let's also relax the "slow compared to the speed of sound" condition: Assume that the "instantaneous" is so fast that each player starts a tension wave at their end of the string. This tension wave travels towards the other end of the string until it hit meets the other tension wave. Now, each player has put only the amount of tension on their end of the string that matches their own force. However, when the two tension waves meet, the string is suddenly tensioned with the sum of these two forces. If the string is going to break at all, this is where. The location of this point only depends on the timing of the two players, as the speed of the tension waves depend only on the speed of sound in the string material. So, if one player is slightly late with applying their force, their wave does not have the same time to travel down the string, and thus the string will break closer to them. In this case, the quicker player wins.