# 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?

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.

• +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
• If the forces are different (say, in a tug of war where one team is winning), what is the "shared magnitude"? If I'm understanding @MarkEichenlaub 's answer right then there's two forces, not one. – Jonathan Spirit Oct 2 '17 at 22:17
• @JonathanSpirit That is a similar example to the case where you have the Atwood's machine, where you have two different masses hanging on each side. See the figure of my answer here: physics.stackexchange.com/a/387934/46464 Essentially, if you equal the accelerations of the masses and you do the FBD for each body, you will be able to isolate and calculate the tension T. That is the "shared magnitude" in this case. – bruno Feb 22 '18 at 14:27

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.

• 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. – unsym 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. – unsym Nov 24 '10 at 3:02
• @ Mark Eichenlaub ""Tension is not a force, and doesn't have a magnitude and direction"" I do not understand the word "tension" in this context. What does that mean? – Georg Jan 26 '11 at 11:10
• @Georg It is simply a name for when the pressure is negative. – Mark Eichenlaub Jan 26 '11 at 14:20

Free Body Diagram (http://en.wikipedia.org/wiki/Free_body_diagram) is your friend. Cut the rope at any point and replace the missing piece with the internal tension force and see how and it changes along the cable.

If you consider gravity then the rope forms a catenary shape and the tensions is minimum at the sag point (middle) and peak at the ends.

The "tensor" is simply a 3by3 matrix, which when multiplies by the normal vector gives the force. And you can argue that in a box with a uniform (stress) tensor has to have a finite rotational acceleration, so if you take the limit shrinking the box to size zero, you discover the stress tensor must be a symmetric matrix.

• OK but a tensor is not a matrix... – Cedric H. Nov 24 '10 at 12:33

my answer is also zero. because resultant become zero. let a force f1 and f2 pull the rope having equal magnitude. then, magnitude of f1= magnitude of f2 let tension on rope be T1 and T2. then, =(f1+T2)- (T1+f2) (since f1=f2) =f1+T2-T1-f1 =T2-T1 T1=T2 THEN AT THE END OF THE ROPE ARE EQUAL THEN TENSION AT THE MID OF THE ROPE WILL BE ZERO.