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I had just started to study Tension deeper when I ran into diagrams like these. I was wondering why the tension force worked on opposite sides rather than at one point? What would happen if they did instead work on only one point? I’ve hardly studied Newton’s third law but I thought the forces came from the same point.

  • $\begingroup$ Can you elaborate on what do you mean when you say that you thought "the forces came from the same point"? $\endgroup$ – Dvij D.C. Sep 5 '19 at 20:51
  • $\begingroup$ I meant that I thought when you pulled on the string the two tensions vectors would begin at the same point, but I saw diagrams that said they begin at x and x+dx $\endgroup$ – Gary Song Sep 6 '19 at 4:54

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In this sort of diagram we are considering a small finite extent of the string at some place which I will call $x_0$ over some fixed distance $\Delta x$: in other words it occupies the interval $(x_0, x_0 + \Delta x).$ Typically such a string is also describing some graph $y(x, t)$ at some time $t$.

This finite extent of the string feels two forces on it, due to the tension in the string. First there is the tension force of the rope to the “left” at $x < x_0$, pulling in some direction given by $-\left .\frac{\partial y}{\partial x}\right|_{x=x_0},$ second there is the tension force of the rope to the “right” at $x > x_0 + \Delta x,$ pulling in some direction given by $+\left .\frac{\partial y}{\partial x}\right|_{x=x_0+\Delta x}$.

The fact that they work from different locations does in theory mean that they will torque, or twist, the chunk of string. In theory they also might move its mass along the $x$ axis. But generally for this sort of problem those effects will be ignored from the beginning, and only if you are very lucky is there some discussion about how they might be neglected in this limit as $\Delta x \to 0.$ Probably the ignore-the-torque argument is justified; probably the ignore-the-longitudinal-motion argument is not. But the basic problem is that if I allow for longitudinal motion of my finite elements in the $x$ axis then I am also modeling longitudinal vibrations along that axis; these then become a strange set of alternate harmonic oscillators which are being coupled to the original string, possibly for the sake of modeling dissipation or so. But if we want a clean model of just the transverse vibrations $y(x)$, then it makes sense to ignore those aspects of reality.

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  • $\begingroup$ Thank you. Are the two tensions pulling left and right equal when delta(x) approaches 0? What do you mean the tensions pulls to the left at x<x_0? $\endgroup$ – Gary Song Sep 6 '19 at 4:56
  • $\begingroup$ I mean that literally this little chunk of rope is like a tiny little human holding one rope in her left hand and one rope in her right hand and getting stretched out by it; she occupies the positions $x$ such that $x_0 < x < x_0 + \Delta x$ and to her left, at positions $x$ such that $x < x_0$ there is some rope, and to her right, at positions $x$ such that $x > x_0 + \Delta y$ there is some other rope. She is pulled to the left by the tension in the left rope, and pulled to the right by the tension in the right rope. $\endgroup$ – CR Drost Sep 6 '19 at 14:58
  • $\begingroup$ I think we assume that the tension magnitudes end up equal but of course the directions end up being unequal, causing a nice term of $\partial^2 y/\partial x^2$ to appear which allows us to have a wave equation $$\ddot y - c^2 y'' = 0.$$ $\endgroup$ – CR Drost Sep 6 '19 at 14:59

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