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I'm currently going through University Physics by Young & Freedman, and there's one paragraph (13th edition, end of section 4.5, page 123), I just can't wrap my head around - why is the tension in the rope $50\ \mathrm N$ and what is the equation giving us "the magnitude of force acting at that point", when obviously, from what it says in this paragraph, it is not the net force?

the paragraph in question

(Background: I am a mathematics student on a year abroad in the UK who made the questionable decision to take an introductory physics module covering the entire contents of the textbook, assuming that the theory would be covered in the lectures, but apparently the knowledge of A-level physics is assumed and my prior knowledge of physics at the beginning of the term basically boils down to "what is uniform motion?", so at the moment I spend most of the day reading through Y&F and watching Walter Lewin's lectures.)


marked as duplicate by Qmechanic Mar 14 '17 at 17:03

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  • $\begingroup$ The tension everywhere in the rope is 50 N. This means that 50 N is acting on each end of the rope, but the forces on the ends of the rope are in opposite directions, so there is no NET force on the rope. $\endgroup$ – David White Nov 24 '16 at 22:28
  • $\begingroup$ @DavidWhite okay, now what if we have $40\ \mathrm N$ at one end and $60\ \mathrm N$ at the other end on a string of length $\ell$. I know the string is not in equilibrium then, since the net force is non-zero, so it is accelerating, but how would I calculate the tension at any point in the string? $\endgroup$ – Sora. Nov 24 '16 at 23:02
  • 2
    $\begingroup$ More on forces and factors of two: physics.stackexchange.com/q/41291/2451 and links therein. $\endgroup$ – Qmechanic Nov 24 '16 at 23:39
  • $\begingroup$ In your example in the comment above, the tension in the accelerating rope would vary from $60N$ at one end to $40N$ at the other. (The string cannot be massless.) This problem is the same as finding the tension in the coupling chains between carriages in a train. $\endgroup$ – sammy gerbil Nov 27 '16 at 0:25

The problem here is trying to convert a "common sense" idea about what "tension in a rope" means, into mathematics which is simple enough for high-school-level students to understand.

I'm not sure if UK A-level maths includes an introduction to matrices, but if it does the following might help a mathematician understand what's really going on here.

If you interpret "tension" as "some kind of force" then the idea of "some kind of force in the rope" doesn't really make sense. You can only talk about forces acting on the ends of the rope. If you imagine that you cut the rope somewhere along its length, you then have two more "ends" which have equal and opposite forces acting on them.

The "correct" way to deal with all this to replace the idea of "tension in the rope" with the notion of stress. Stress has the dimensions of force/area - the same as pressure, but "stress" and "pressure" are different concepts. In general, stress is a tensor (the next step up in the hierarchy of scalars, vectors, ...) and to find the force on the surface of a body (e.g. a cut through the rope) you take the dot-product of the stress tensor and the unit vector normal to the surface, integrated over the surface area: $$\vec F = \int_S \sigma \cdot \vec n \; dA$$

In general, a tensor in 3D space has $3 \times 3=9$ components (though stress tensors are always symmetric, so there are only 6 independent components) and it can be represented as a $3 \times 3$ matrix like $$\begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{zx} & \sigma_{zy} & \sigma_{zz} \end{bmatrix}.$$ The dot-product operation in the previous equation corresponds to multiplying the matrix by a vector.

If we take the $x$ direction along the length of the rope, only one component of the stress $\sigma_{xx}$ in the rope is non-zero, and the vector corresponding to a cut through the rope is either $\begin{bmatrix} 1 & 0 & 0 \end{bmatrix}^T$ or $\begin{bmatrix} -1 & 0 & 0 \end{bmatrix}^T$ depending on which end of the rope you are considering. So the force vector on the end of the rope is $$\int_S \begin{bmatrix} \sigma_{xx} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} \pm 1 \\ 0 \\ 0 \end{bmatrix} dA = \begin{bmatrix} \pm \sigma_{xx} A \\ 0 \\ 0 \end{bmatrix}$$.

At the level of your textbook, the scalar quantity $\sigma_{xx} A$ is called the "tension $T$ in the rope" (note, this is always positive), but the "force at the end of the rope" is really one component of a force vector, with magnitude $+T$ at one end and $-T$ at the other. That avoids any reference to tensors, but doesn't give a good answer to your question about why the signs of the forces on the end of the rope are the way they are.

  • $\begingroup$ This seems like overkill to me. The problem is 1D - why use matrices? $\endgroup$ – sammy gerbil Nov 26 '16 at 22:53

I am afraid that I do not have access to the textbook.

Imagine a horizontal rope on a frictionless surface under tension.

If the rope was cut at any point it would separate with an end on the left $L$ and an end on the right $R$.
To prevent the separation suppose that external forces were applied to the rope at each of the ends, $\vec F_{\text{External on Rope L}}$ on the left hand end of the rope and $\vec F_{\text{External on Rope R}}$ on the right hand end of the rope.
Newton's third law tells you that the rope would be exerting forces on whatever is producing these external forces
$\vec F_{\text{Rope L on External}} = -\vec F_{\text{External on Rope L}}$
$\vec F_{\text{Rope R on External}} = -\vec F_{\text{External on Rope R}}$.

The magnitudes of all these forces are the same and so when the rope was whole before it was cut one can say the force on the left hand part of the rope due to the right hand part of the rope was equal in magnitude and opposite in direction to force on the right hand part of the rope due to the left hand part of the rope and these are the forces that are called the tension.
Tension only seems to reveal itself when you have an end to the rope because otherwise you have a static equilibrium situation at each position along the rope the net force is zero.

Now suppose that the rope has a mass and it is used to suspend a mass of weight $W$ from a fixed support and the whole system (weight, rope and support) is in static equilibrium.

At the point of contact between the rope and the mass the downward weight of the mass is equal to the upward force on the mass due to the rope which is called the tension in the rope.

A similar analysis will equate force at the top of the rope but this time the downward force on the support due to the rope, the tension in the rope, will be larger than at the bottom of the rope, because the upward force on the rope due to the support has to balance the weight of the mass at the end of the rope and the weight of the rope.

Now choose a segment of the rope.
The upward force on the top of the segment due to the tension in the rope above the segment will be larger than the downward force on the bottom of the segment due to the tension in the rope below the segment.
This is because the difference in these two tensions must equal the weight of the segment.
So in this situation the tension in the rope is not constant.

We now come to an example of simplification in terms of what one has to write which can cause a conceptual problem.

At each point along the rope the upward tension due to the rope above the point has to be equal in magnitude but opposite in direction to the weight of the rope and the mass hanging from it.
The simplification which is often made is to assume that the weight of the mass is much greater than the weight of the rope and this leads to the fact that the tension in the rope is approximately equal to the weight of the mass at the end of the rope and so the tension is approximately constant along the rope.

The shorthand version of this statement is say that the rope is massless.

You can think of a massless rope as transferring a force from one place to another and with the help of a pulley enabling the force to change direction.


The tension in the rope is an intensive property, like density or temperature. It describes the condition of the rope in relation to how strongly it pulls back on things attached to it.

By Newton's 3rd Law, the forces applied to the ends each match the pulling power (tension) of the rope at those points. These forces don't add up to make the tension in the rope, just as when the temperature of the rope is $20^{\circ}C$ at each end, the temperature in the middle is not $40^{\circ}C$.

The answers to Why I think tension should be twice the force in a tug of war look useful.


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