Tension in a string in equilibrium (Y&F University Physics section 4.5)

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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.

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Sora.

Undergraduate, currently studying mathematics and Japanese at the Ruhr-University of Bochum. Fluent in German and English, advanced knowledge of Japanese, basic knowledge of Russian and Spanish. Interested in programming (C, C++ and Java). Using LaTeX for some years now, having used it to share my lecture notes of single- and multivariable analysis and calculus with my fellow students, part of the Introduction to LaTeX for mathematicians staff at university. Amateur musician, mainly playing the drums and piano.

Updated on June 14, 2022

Comments

  • Sora.
    Sora. over 1 year

    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.)

    • David White
      David White almost 7 years
      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.
    • Sora.
      Sora. almost 7 years
      @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?
    • Qmechanic
      Qmechanic almost 7 years
      More on forces and factors of two: physics.stackexchange.com/q/41291/2451 and links therein.
    • sammy gerbil
      sammy gerbil almost 7 years
      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.
  • sammy gerbil
    sammy gerbil almost 7 years
    This seems like overkill to me. The problem is 1D - why use matrices?