# Rope tension question

If two ends of a rope are pulled with forces of equal magnitude and opposite direction, the tension at the center of the rope must be zero. True or false?

The answer is false. I chose true though and I'm not understanding why. Forces act at the center of mass of the object, so if there are two forces of equal and opposite magnitude, then they should cancel out resulting in zero tension, no?

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The tension of the rope is the shared magnitude of the two forces. Imagine cutting the rope at a point and inserting a spring scale in its place. The reading will show the tension. A rope with zero tension would be hanging loosely or laying on the ground, neglecting the rope's mass.

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+1 for the physical interpretation with a spring scale, but shouldn't the tension be equal to the shared magnitude of the forces, not equal to the sum of the magnitudes? – Mark Eichenlaub Nov 23 '10 at 5:39
Yeah, I think Mark is right. That, or I'm always explaining it incorrectly to students. Please, don't say I screwed up so bad. :p – Raskolnikov Nov 23 '10 at 9:35
Whoops, corrected! :) – Vortico Nov 23 '10 at 22:44

Here's a slightly more mathematical answer to go with Vortico's excellent physical answer.

Tension is not a force, and doesn't have a magnitude and direction. This is confusing, because we frequently talk about the "force of tension", which is a different thing from the tension itself.

The tension is actually an example of a mathematical object called a "rank-two tensor". A rank two tensor is a function whose input and output are both vectors. The input is a direction, and the output is the force exerted on the piece of rope lying in that direction. So for a vertical rope, you can take a unit vector pointing down as the input to the tension tensor, and it gives back a force of, say, 20N up. That would mean that at any point in the rope, the rope is pulling up on the parts beneath it with 20N of force. Then if you input a unit vector point up, you'll get back a force of 20N down. Any part of the rope pulls down on whatever is above it (more rope, or maybe a ceiling) with 20N of force. We then say that "the tension in the rope is 20N", but what we mean is actually the entire story above about inputting directions and getting out forces. Every little piece of rope has two 20N forces on it from the surrounding rope, and also exerts two 20N forces on the surrounding rope.

This is a little pedantic, since the tension is just a number and the whole bit about the tensor does not seem very important. It becomes more important when we talk about surface tension, perhaps in a metal sheet. In that case we could imagine that the tension really is different in different directions. Physically, if you cut a slit in the sheet, and then measured how much force it takes to hold the slit closed, you could find that the force required (per unit length of the slit) depends on the slit's direction. You would need the full power of the rank-two tensor to describe that tension.

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Treating tension as a tensor is interesting. However, is it possible to define such tensor? A tensor takes a input vector and transform it into a output force vector. When you apply a force to the rope, its direction change, and in your example, the tensor should also change. Also when you push the rope, the tension should be zero. – hwlau Nov 23 '10 at 11:09
I'm not quite sure what you're driving at. Are you thinking that the tensor should be some intrinsic property of the rope? It's just a description of the forces one piece of the rope exerts on whatever is at its endpoints. If the direction of the rope changes, the tensor changes. If the rope is bent, the tension tensor is different at different points in the rope. If the tension is zero, the tension is the zero tensor. – Mark Eichenlaub Nov 23 '10 at 15:56
@Mark: You are correct. I am just wondering whether there is a need for the tensor idea for this simple problem, if the tensor will always change when the input and output force change. – hwlau Nov 24 '10 at 3:02
I see. I guess no, there's no need. But I thought it might be helpful. – Mark Eichenlaub Nov 24 '10 at 6:09
@Georg It is simply a name for when the pressure is negative. – Mark Eichenlaub Jan 26 '11 at 14:20