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

• etotheipi
In summary, tension is defined as the magnitude of the force between any two adjacent segments of string or rope, and can be represented as a vector or scalar depending on the context. In the case of a rope or rod under uniaxial tension load, the stress tensor simplifies to a scalar value of T/A, where T is the tension and A is the cross-sectional area. T can be negative if the body is under compression, and the relationship between the stress tensor and the strain tensor takes into account the amount of deformation or displacement.
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
PhanthomJay said:
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|##.

Chestermiller said:
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.

Chestermiller said:
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.

etotheipi said:
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
Chestermiller said:
T can be negative if the body is under compression.

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

etotheipi said:
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
Chestermiller said:
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!

etotheipi said:
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
Chestermiller said:
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

## 1. What is tension and how is it related to science?

Tension is a force that occurs when an object is pulled or stretched. In science, tension is an important concept in understanding the behavior of materials and the forces acting on them.

## 2. How does tension affect the behavior of materials?

Tension can cause materials to stretch or deform, depending on their properties and the amount of force applied. It can also cause materials to break or fail if the tension exceeds their strength.

## 3. Can tension be measured and quantified?

Yes, tension can be measured using instruments such as force gauges or strain gauges. It is typically measured in units of force, such as newtons or pounds.

## 4. What are some real-life examples of tension?

Tension can be observed in everyday situations, such as when you stretch a rubber band or pull on a rope. It is also present in structures like bridges and buildings, where tension forces help distribute weight and maintain stability.

## 5. How does tension play a role in scientific experiments?

Tension is often a key factor in scientific experiments, particularly in the study of materials and their properties. Scientists may use tension to test the strength or elasticity of a material, or to observe how it responds to different forces.

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