The concept of tension

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Hello, I have read over many textbooks and websites (including posts on this forum) relating to the concept of tension but I am still a bit confused. Many sources seems to only define tension rather than explain it. They seem to assume that you know it the way you know 1+1 =2.

Here is the part where I am confused about. A rope can be seen as a coupling between railroad cars. Suppose I have a locomotive dragging a cart along with a coupling. The locomotive exerts a force of 100 N through the coupling to the cart. Now, by Newton's 3rd Law of motion, there must be a reaction force. The reaction force is where the cart is pulling back on the locomotive with the same force of 100 N but in the opposite direction. Now, the coupling have two forces pulling on it.

Force 1: locomotive pulling on cart (100 N)
Force 2: cart pulling on locomotive (100 N)
Now, shouldn't the coupling be able to withstand 200 N because there are two forces at work here?

All the reading I have done seems to suggest that the coupling only need to withstand 100 N but how could that be right? Doing so seems to ignore Newton's 3rd law. Thanks.
 

Answers and Replies

  • #2
russ_watters
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When you stand on a scale, what are the forces involved? What does the scale read?
 
  • #3
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It's easier to think in terms of energy. If you stretch a spring, say, with a force of 2N for 2m you have done 4 Nm of work. When the tension is released you get back the work. The fact that the spring pulls on both ends with a force of 2N doesn't mean you can get he work back twice. I'm sure you can think of an experiment to test this.
 
  • #4
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A standard bathroom scale will measure your weight by telling you the normal force that the scale is pushing upon you as a reaction force to your force gravity. I don't see the connection though. The scale only measures how compressed the spring is.
 
  • #5
vanesch
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Here is the part where I am confused about. A rope can be seen as a coupling between railroad cars. Suppose I have a locomotive dragging a cart along with a coupling. The locomotive exerts a force of 100 N through the coupling to the cart. Now, by Newton's 3rd Law of motion, there must be a reaction force. The reaction force is where the cart is pulling back on the locomotive with the same force of 100 N but in the opposite direction. Now, the coupling have two forces pulling on it.
I understand your confusion, having had similar conceptual problems once. Tension is indeed a rather subtle concept, and often confused with force, because both are related.

Tension is a physical quantity which describes a mechanical state of matter in a certain sense. It is useful because it results in "forces at its boundaries", and because there is a definite relationship between this mechanical state of matter and its "response" (which is deformation).

Mathematically, tension is not described by a vector (as force is), but rather by a 2-tensor. A 2-tensor is a machine that EATS a vector, and SPITS OUT another vector. When we work in 3 dimensions, a 2-tensor can be represented by a 3x3 matrix, the eating of a vector can be represented by the matrix multiplication of the tension matrix with the 3 coordinates of the vector, and the result is again a column of 3 numbers, which will be the coordinates of the vector that is spit out.

The operational definition of the tension-tensor (hehe) is as follows:
it is the tensor which spits out a FORCE VECTOR when given to eat a UNIT VECTOR times a surface.
To understand this, imagine the following situation: in the middle of a piece of matter with a certain "stress", we decide to cut away a bit of matter, along a small surface of area S, and normal (unit) vector n. That is, where there was matter in front of S (in the positive direction of n), we have removed it.
As such, the matter at this surface S will now start to accelerate (because not withheld anymore by the matter we cut away), and we have to APPLY A FORCE to that naked surface S to restore the original situation of equilibrium (that was previously taken care of by the matter we just cut away).
Well, the unit vector times S, S.n, is the vector we have to give to the tensor, and the force that restores tranquility is what is spit out.

An example: consider water under pressure P. Then the tensor looks like:

-P 0 0
0 -P 0
0 0 -P

Imagine that we cut away some water and leave a small surface S naked in the positive x-direction at some point. The water would squirt out of the surface in the positive x-direction if we didn't do anything! So in order to restore the equilibrium, we would have to apply (with a magical finger or something) a force on that surface which "pushes against the desire of the water to come out", and which would have to be S.P. (surface times pressure), but we would have to push in the MINUS x-axis on the water (as the water tends to go in the +x direction where there is now "open room" after our cutting away of the matter).
Well, that's exactly what the tension (or stress) tensor does:
The surface S has a unit vector (1,0,0) (a unit vector in the +x direction, the direction in which we removed the matter from the surface).

If we multiply the tensor with S.(1,0,0), we get out the vector (-P.S,0,0), which is exactly the force we have to apply to the surface S to have equilibrium.

So a material under uniform pressure (such as water) has an above stress tensor.

Your binding piece has the following stress tensor (assuming the axis between the locomotive and the cart is the x-axis):

s 0 0
0 0 0
0 0 0

Where s is the tension, that is, your 100N divided by the cross section S of the binding piece (so, s = 100N/S).

Let us imagine this:


(locomotive) ====piece====== (cart)

--------------------> +x axis


If we cut away the locomotive, then at the surface, on the left, the unit vector is (-1, 0,0). The force that has to act upon the piece to "replace the locomotive" is then given by the tensor times S.(-1,0,0), which gives us:

(-s.S,0,0) which is nothing else but (-100N,0,0). Indeed the locomotive can be replaced by a force of 100N acting on the left.

If we cut away the cart, then the unit vector is (1,0,0). The tensor acting on this gives us a force, which replaces the action of the cart, equal to (+s.S,0,0), or (+100N,0,0), which is a force of 100N acting on the right.

So the stress situation of the piece is well-described by the above stress tensor.
 
  • #6
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Awesome answer.
 
  • #7
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My math skills are not quite advanced so it will take me some times to digest that heh. Thanks though.

I was just working on an elevator problem and it got me thinking. The elevator exerts a force downward due to the force of gravity. The rope only have to hold the elevator in place so it doesn't fall. Therefore, the rope only have to withstand -Fg. The action force is the elevator pulling down on the rope and the reaction force is the rope pulling up on the elevator. Is the reaction force is what defined as "tension"?
 
  • #8
vanesch
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I was just working on an elevator problem and it got me thinking. The elevator exerts a force downward due to the force of gravity. The rope only have to hold the elevator in place so it doesn't fall. Therefore, the rope only have to withstand -Fg. The action force is the elevator pulling down on the rope and the reaction force is the rope pulling up on the elevator. Is the reaction force is what defined as "tension"?
Well, tension is a "microscopic" quantity, within the material, so normally it is expressed in "force per area" (just like pressure, which is nothing else but a specific kind of tension). What makes a material react, is the tension and the kind of material. Some piece of metal will break when, at some point, the tension goes beyond the limit allowed for that kind of material (there's a lot to say about this! It's what you learn in material science classes).
If you have a thick cable, then, for a given stress situation, you will have a large force on the end surface. If you have a thin cable, then for the same stress situation, you will have a much smaller force on the end surface. But the breaking will occur at about similar stress situations in the material (force divided by cross section).
It is THIS fact, namely, that materials react according to the local stress situation (independent of how exactly the overall global situation is), that makes stress such a useful concept.

For reasonably small stresses for instance, a given material obeys a generalization of Hooke's law: the "relative deformation" tensor, or strain tensor, is linearly related to the stress tensor, and this totally independent on what's the form or size of the overall mechanical piece. In isotropic materials, the linear relationship becomes simply a proportionality, and the constant is called Young's modulus. Macroscopically, it gives rise to Hooke's law.

So, again, tension is NOT a force. It is caused by forces at the boundaries (or internally) of a piece of matter, but it is a different animal. A force is a vector, and stress is a 2-tensor. However, in simple, one-dimensional problems, one has tendency to replace the stress tensor by a kind of "stress force" (which is nothing else but the value of the relevant element of the stress tensor, times the cross section), but when taking this too literally, one runs into all kinds of problems (as you do, and as I did when I was a student). It is because fundamentally, tension is a 2-tensor, and NOT a vector. Even in 1 dimension (where there is this problem with the sign).
 
  • #9
russ_watters
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A standard bathroom scale will measure your weight by telling you the normal force that the scale is pushing upon you as a reaction force to your force gravity. I don't see the connection though. The scale only measures how compressed the spring is.
Whether it is tension or compression, the issue is the same: you have a force due to your weight (say, 100lb) and the reaction of the ground pushing up on the scale (also 100lb), yet the scale shows 100lb, not 200lb. If the sum of the two forces were not zero (they are in opposite directions), that would imply acceleration would exist.

A force needs to be opposed by another force, whether it is from inertia and acceleration or just balanced static forces.
 
  • #10
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Maybe everything actually weighs half the amount we think it does and the scales are doubling the value !
 
  • #11
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Hey, thanks for all the explainations. That clear it up for me.
 

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