# My understanding of tension is a little loose....

• etotheipi

#### etotheipi

A video on the MIT open courseware site, https://ocw.mit.edu/courses/physics/8-01sc-classical-mechanics-fall-2016/week-2-Newtons-laws/7.1-pushing-pulling-and-tension/, defines tension as the magnitude of the force between any two adjacent segments (A and B) of string - that is, ##T = |\vec{F_{AB}}| = |\vec{F_{BA}}|##. For simplicity, I'll just try to understand a massless string/spring first where tension is uniform! This seems consistent with ##T = k|\Delta x|##. My previous understanding is that the tension was just the magnitude of the force exerted on any object in contact with any segment of the string, most commonly (but not necessarily) the ends, the direction of which can be worked out by a little common sense and a FBD (i.e. extension or compression...).

However I've also seen it referred to as a rank-1 tensor - I haven't really covered tensors so this might be a little out of my depth - though I understand this to be sort of like a vector. But if my string is curved around a pulley somewhere, I could separate it into lots of little slices and the tension forces would all point in slightly different directions, so one vector for "tension" doesn't seem good enough!

I wonder whether someone could shed some light on this? Thanks!

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Technically speaking, a scalar is a tensor of order 0, and a vector is a tensor of order 1. Tension force is a vector. When a rope is wrapped around an ideal pulley, the magnitude of the tension force is the same, but the tension force is a vector with changing directions at separate points. So for the rope around the pulley, you could say that the tension in the rope is say 100 Newtons , and you are talking about the magnitude of the tension. Or you could say the tension in the rope is 100 Newtons acting down on each side of the pulley, so now it’s a vector for this statement. I would forget about calling it a tensor.

• etotheipi
Technically speaking, a scalar is a tensor of order 0, and a vector is a tensor of order 1. Tension force is a vector. When a rope is wrapped around an ideal pulley, the magnitude of the tension force is the same, but the tension force is a vector with changing directions at separate points. So for the rope around the pulley, you could say that the tension in the rope is say 100 Newtons , and you are talking about the magnitude of the tension. Or you could say the tension in the rope is 100 Newtons acting down on each side of the pulley, so now it’s a vector for this statement. I would forget about calling it a tensor.

Thank you, it appears the same name applies to both concepts and that is fine.

In that case, ##\vec{T} = T(x) \hat{n}## where ##\hat{n}## is a unit vector in the direction of the tension and ##x## is the length along a massive spring. And if the spring happens to be light, then ##T(x) = k|\Delta x|##.

• etotheipi
• vanhees71 and etotheipi
I thought it might be helpful to look over my post #12 of the following thread: https://www.physicsforums.com/threads/so-tension-is-not-a-force.960204/#post-6089855

Thank you for the link, I've read the first parts a few times over and they're starting to make a little more sense. Of course I'll need to do a lot more reading to get anywhere near comfortable with simple tensors.

For a rope or a rod under uniaxial tension load, the stress tensor reduces simply to: $$\boldsymbol{\sigma}=\frac{T}{A}\mathbf{i_x}\mathbf{i_x}$$where T is the tension and A is the cross sectional area. If we dot this with a unit vector perpendicular to the rope cross section, in the positive x direction ##\mathbf{n}=\mathbf{i_x}##, we have $$\boldsymbol{\tau}=\boldsymbol{\sigma}\centerdot \mathbf{n}=\frac{T}{A}\mathbf{i_x}$$

In a construction like this, can ##T## then be negative? Because I see no reason why not here, given ##\frac{T}{A}## is just the component of the stress vector in the positive ##x## direction! However, I've learned so far that ##T = k|x|## for a light spring which is necessarily positive.

Thank you for the link, I've read the first parts a few times over and they're starting to make a little more sense. Of course I'll need to do a lot more reading to get anywhere near comfortable with simple tensors.

In a construction like this, can ##T## then be negative? Because I see no reason why not here, given ##\frac{T}{A}## is just the component of the stress vector in the positive ##x## direction! However, I've learned so far that ##T = k|x|## for a light spring which is necessarily positive.
T can be negative if the body is under compression.

• • vanhees71 and etotheipi
T can be negative if the body is under compression.

Right, so in that case is ##T=kx## a more apt expression?

Right, so in that case is ##T=kx## a more apt expression?
The analysis I presented has nothing to do with the amount of deformation or displacement. That comes into play in the relationship between the stress tensor and the strain tensor.

• etotheipi
The analysis I presented has nothing to do with the amount of deformation or displacement. That comes into play in the relationship between the stress tensor and the strain tensor.

I think for the second time this week I'm having a little trouble with notation. My interpretation of that part of your post is ##\vec{\sigma} = \frac{\vec{T}}{A}##, with ##\vec{T} = T_x \hat{i}##, whilst ##T_{x} = \pm T = \pm k|x|## depending on whether the spring is under extension or compression, assuming all the necessary assumptions so that the spring is such as it would be during an introductory mechanics course. Is this what you meant?

I think the full stress tensor/strain tensor relationship might be a bit too much for me to understand at the moment, however I'll have a look around and see if I can make some sense of all of this!

I think for the second time this week I'm having a little trouble with notation. My interpretation of that part of your post is ##\vec{\sigma} = \frac{\vec{T}}{A}##, with ##\vec{T} = T_x \hat{i}##
No. The stress tensor is exactly as I wrote it. It is a second-order tensor, not a vector. Please re-read what I wrote in the other thread again (and more carefully). This approach using dyadic notation is like nothing you've seen before.

• etotheipi
No. The stress tensor is exactly as I wrote it. It is a second-order tensor, not a vector. Please re-read what I wrote in the other thread again (and more carefully). This approach using dyadic notation is like nothing you've seen before.

Right, I apologise, I meant ##\boldsymbol{\tau}## instead of sigma.

etotheipi said:
My understanding of tension is a little loose...

Good one. . . . . .

• etotheipi