# Why can we assume, when dealing with tension, that the mass of a rope is $0$ but still assume that there are forces on it?

I started studying tension and I can't understand the following concept:

If I have two objects $A,B$ collegated by a rope, where $A$ applies a certain force $\vec{F}$ on it, I would have that the tension in the medium is $$T=\sum \vec{F}=\vec{F}_{AR}+\vec{F}_{BR}.$$

So far ,so good.

Now, If we assume the rope to be massless I would have that $$T=\sum \vec{F}=\vec{F}_{AR}+\vec{F}_{BR}=0 ,$$ but now what doesn't click in my mind is that:if we assume the rope to be massless how can we have in the first place forces such as $\vec{F}_{AR}$and $\vec{F}_{BR}$?

Is this something that we just take the assumption of and then see mathematically where it leads us?

I know of course that this assumption is made to simplify reality but I just find this as extremely abstract because even though we assume the mass of the rope to be massless we still assume that there are forces acting on it.

Isn't this a little bit of a paradox?

• When we say "massless", what we really mean is that the mass is finite but so small that it doesn't matter for the problem and so we simply neglect it. – CuriousOne Jan 8 '16 at 20:33
• @CuriousOne that should probably be an answer – David Z Jan 8 '16 at 21:09
• If you want to account for the mass (really weight) of the rope, then you would just integrate those equations over the length of the rope. – Sponge Bob Jan 8 '16 at 23:02

## 2 Answers

By your opening statement, it sounds like you're confusing the concepts of tension and net force. If the rope can be taken to be massless, then that means that its mass is small enough that it can be taken to be practically zero. For this to be the case, the net force on the rope also has to be zero, otherwise the acceleration would go to infinity when you made the mass go to zero. Another way of saying this is that you can assume the mass is an extremely small value and calculate the result, and the result doesn't significantly change when you vary that value. This is what we really mean when we say it's massless. It's massless for all practical purposes. The mass is positive but is small enough to ignore.

When you learn about relativity, by the way, you'll find out that a massless object cannot in fact stand still! So this is like a first-year-physics meaning for the word "massless." It'll change as you take more classes (or read more books; it's unclear whether you're asking from the context of a course or from self-study, but it sounds like, either way, you're working with the subject matter of a typical freshman physics course, so I'll write this answer in that context. I apologize if I've misapprehended where you're at.).

Net force is not the same thing as tension! For the case of a rope, which in the freshman-physics approximation we can take to be one-dimensional, tension is a scalar and not a vector. (More generally, for 3-d objects, "tension" generalizes to "stress," which is a more complicated thing called a second-rank tensor. Don't worry about that for now. I say this just to make you aware that, when I'm calling tension a scalar, I'm cheating a little bit in a way that's hard to explain without going into a lot of detail!).

Suppose the rope is lying from left to right, and object A is at its left end and is pulling on it leftward with 10 N of force. Suppose object B is at its right end and is pulling on it rightward with 10 N of force. Then we have $F_{AR}=-10\hat{x}$ Newtons and $F_{BR}=+10\hat{x}$ Newtons, where $\hat{x}$ is a unit vector pointing to the right. Now you can see that your initial equation is actually correct (if you erase the "T = " part): The sum of those two forces, the net force on the rope, is indeed 0. So it's not a problem that we're ignoring the mass. The net force is zero, so the acceleration is zero, regardless of the fact that the mass is really, really small.

Tension is a separate concept. Tension is not a vector. The action of tension on a specific object is a force, and therefore a vector, but the tension itself can act in different directions especially if there are pulleys involved. Note the tension in the rope is pulling rightward on A (which is to the left of the rope) but it's pulling leftward on B (which is to the right of the rope). So which direction, left or right, would you assign to the tension vector inside the rope, irrespective of whether we're looking at A or B? The answer is neither; it's not a vector at all. In freshman physics, you can take tension to be a scalar. In this case that scalar is 10 Newtons.

In fact you can, if you wish, take this case to be the definition of what you mean by the word tension. The rope is pulling with 10 N of force on both of the things that it's pulling on, in equal and opposite amounts.

Note also: This is not the same "equal and opposite" you get in Newton's third law (another thing which is often confused with these concepts). The notions of tension, net force, and equal and opposite reaction forces are completely different things. They can be easily confused until you're used to them.

One thing I can tell you is this: Once you fully get these concepts, you're basically 90% of the way through freshman physics. It's tricky and confusing until you get it. Then it becomes obvious.

• One last clarification:what do we mean then when we say that the tension at any point in the rope is the magnitude of the force acting at a point?If there are two forces involved,not necessarily equal and in opposite directions,then I wouldn't have same tension at every point.Is tension now the magnitude of the net force of the forces involved ? – Mr. Y Jan 9 '16 at 8:00
• To find the tension in the rope at one point, imagine splitting it into a left half and a right half at that point. The left half pulls leftward on the right half, and the right half pulls rightward on the left half. The magnitude of both those forces is the same (Newton 3) and is equal to the tension. The tension can vary from point to point (for example if it's hanging vertically and it's not massless). But Newton's third law causes the tension at a single point to be well defined. – elifino Jan 10 '16 at 17:52
• The concept of stress takes this one level further: It recognizes that, in an extended 3D object, you could have drawn that imaginary cutting surface at any orientation, and the force isn't necessarily going to be normal to that surface (it can have what's called a shear component). You end up with a 3x3 array of numbers called a tensor. Don't worry if this doesn't yet 100% make sense; I'm just giving you a preview of where this ends up if you keep studying it at higher levels. – elifino Jan 10 '16 at 17:55

It is just a limiting situation in which the tensile strength and stiffness of the material are very high while the density of the material is very low. There are many materials in the real world that actually approach this kind of behavior. Ordinary ropes, strings, and wires are often good examples to this. Of course, a high performance example is Kevlar. Actually, all that is really required is that the mass of the rope be small compared to the masses it is attached to and moving around.