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What can you tell me about the tension in a string?

  1. Nov 25, 2006 #1
    What can you tell me about the tension in a string? Why is it equal over the string? Why is the direction of the tension over the ends of the string opposite?
     
  2. jcsd
  3. Nov 25, 2006 #2

    arildno

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    1. Usually, we make the approximation considering the string to be MASSLESS.
    But, since the string can't have infinite acceleration of any kind, it follows from Newton's second law that the sum of forces acting upon the string is zero, F=0.
    2. Furthermore, any particular piece of a massless string must also be massless, since mass is always non-negative. Hence, if the sum of forces acting upon the PIECE of the string is called f, then we also have f=0

    Do you agree so far?
     
  4. Nov 25, 2006 #3
    Yes, I am seeing that the key to all this the the string to be massless.
     
  5. Nov 25, 2006 #4

    arildno

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    Okay!

    Now, suppose you've got a massless string fixed in the ceiling, an object of mass M hanging in the other end.
    Now, imagine an arbitrary partition of the rope into two pieces, the lower end L (attached to the object), the higher end H (attached to the ceiling).

    Now, ask yourself:
    What forces act upon the piece L?
    Clearly, the rope supports the weight by a force Mg, (working upwards), but therefore, the object, by Newton's 3.law exerts a force -Mg on L.

    However, L is in equilibrium (apart from being massless), hence the piece H must provide a force Mg on L in order for L to remain at rest. (How would this argument be changed if the rope had mass of its own?)
    Similarly, by Newton's 3. law, L exerts a force -Mg on H, which the ceiling needs to counteract in order for H to remain at rest.

    But, therefore:
    The tension force in the rope is of equal magnitude throughout the rope, and the attached object experiences an UPWARDS tension force equal to its weight in magnitude, whereas the ceiling experiences a DOWNWARDS tension force equal to the object's weight?

    Are we still in agreement?
     
  6. Nov 25, 2006 #5

    Danger

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    Arildno, that's just about the best explanation of the situation that I've ever seen. :cool:
     
  7. Nov 25, 2006 #6

    arildno

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    I hope you are not being ironic..:blushing:

    I'm not finished, though..we still need to consider the cases where we can have variable tension in a massless string.
     
  8. Nov 25, 2006 #7
    Yes, we are. You may continue...
     
  9. Nov 26, 2006 #8

    arildno

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    Well, Daniel Craig was a worthy choice. :smile:

    Now, we'll look at a case where we may have varying tension in a massless string:
    Let a string go through a narrow tube of length L, so that if you try to pull the string through, there is a resistive force from the tube acting upon the string which you need to overcome.
    For simplicity, we'll let the resistive force acting upon a SEGMENT of the string be strictly proportional to the length of the said segment.

    Let x be a variable denoting a position along the string, so that the region x<0 represents the free end of string, 0<=x<=L the segment of the string contained within the tube, and x>L the freed piece of the string, ending in your hand, which supplies a pulling force P on the string.

    Now, consider how the tension must be in each of these regions of the string:
    At the very tip of the string in the free end x<0, there is no external force acting upon the string. Hence, any string segment in this region starting at the tip cannot be subject to any tension force from the rest of the string, since that would yield an unbalanced Newton's 2.law for the given segment.
    Therefore, the tension T(x)=0 for x<0.

    Now, let us look at a segment of length X beginning at x=0, i.e, enclosed in the tube. This segment experiences a resistive force -kX from the tube wall, so clearly, the portion of the string x>X must provide a tension force kX on our segment 0<=x<=X.
    Thus, the tension VARIES as we proceed through the tube, and we have:
    T(x)=kx, 0<=x<=L

    Finally, in the freed region x>L ending with your hand, the tension must be constant, so that we get here T(x)=kL, x>L.
    In particular, we see that our pulling force P=kL, which simply express that P must balance the ENTIRE resistive force from the tube.

    I'll end this discussion later on, by looking at a typical case encountered, namely when the string itself is not just straight, but might be bent around a pulley.
     
  10. Nov 28, 2006 #9

    arildno

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    Okay, even if Pepon hasn't returned yet, I'll end this discussion by looking at the typical pulley situation, using as little fancy maths as possible.

    Consider a massless string slung around a circular pulley, and assume that there is tension in the string.

    Consider a circular arc-segment of that string, for simplicity with its midpoint at the "top" of the pulley, so that the tangent at the midpoint is horizontal, and the normal vertical.

    Now, tension forces acts upon this segment from the rest of the string, and at each end of the string, that force acts outwards from the string segment (by Newton's 3.law, the segment itself provides forces on the rest of the string towards the segment).

    First of all, irrespective of the magnitudes of these tensile forces, BOTH of them provides a downwards force component onto our segment.

    Therefore, unless the pulley provided a NORMAL force acting vertically upon the segment, the segment would gain an infinite downwards acceleration, since it is massless. That this normal force ought to be proportional to the magnitude of the tension forces, i.e, the tension, should be evident.
    Furthermore, it should be evident that it is the actual presence of the pulley, and its normal force that allows/necessitates the segment to bend into shape in the first place.


    Now, let us look at the required balance of forces in the horizontal direction:
    If the tension is constant, i.e, the magnitude of the two tension forces being equal, it is evident by symmetry that the horizontal force components cancel each other. Thus, there CANNOT exist any external non-zero, horizontal force (like friction) acting upon the segment (since that would unbalance Newton's 2.law).

    Conversely, assume that there isn't any non-zero, ezternal force acting upon the segment. Then, again by symmetry, the tension across the segment must be constant; otherwise, the horizontal compononents of the tension forces would not cancel, in the case of the smooth, circular pulley.


    But this means that we have derived the precise condition for when the tension is constant in a massless string:

    The tension in a massless string is constant if and only if there is no external, tangential force acting upon the string or any of its segments.

    If this condition is fulfilled, it follows that a taut, massless string merely transmits, and possibly, redirects any force applied at one end point to the other end point.
    Which was to be shown..:smile:
     
  11. Nov 28, 2006 #10

    vanesch

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    Just to add a small point after arildno's excellent contribution, I would like to point out why "the tension at the ends of the string is opposite".
    The reason is that a tension is, strictly speaking, not a force, but a tensor (a second-order tensor). A second-order tensor is a mathematical beast to which you give a vector, and get another vector in return.
    In all generality, the tensor describing the tension in a body takes in the normal vector to a small surface (with the length of the vector the surface of the small surface), and returns the force acting on that small surface (in other words, the momentum change that this small surface would undergo if suddenly the rest of the matter were cut away on the negative side of it).

    If we represent the normal vector by n and the tensor by sigma, we have:

    F = sigma.n

    Now, in a string, the only cut that makes sense is a perpendicular cut to the string itself as we don't really consider its width, and that leaves us with only two possible "normal vectors" to feed the tensor: the normal vector pointing in one direction of the string, and the normal vector opposite to it, pointing in the other direction.
    So there will be just a "normal vector" S, and -S which make sense.

    ---------------------------------------------------
    -S <===x===> S

    Now, for the normal vector S, we have F1 = sigma.S and F1 is the force that would be exerted on the piece of string if we cut away the left part of the string. If the string is under "pull", it would then clearly move to the right. This is the force that will be excerted by the string on whatever it is attached to on the left side (because then, indeed, the "left part of the string" at that point is missing).

    For the normal vector (-S) we have F2 = sigma.(-S) = - sigma.S = -F1. It is the force that would be exerted on the piece of string if we cut away the right part of the string this time. And that's the force that will be exerted by the string on whatever will be attached to it on the right (because, this time, the "right part of the string" is missing).

    This is the origin of the minus sign between the two forces: it is because they are forces obtained from an identical stress situation (sigma) but with opposite normal vectors S and -S.
     
  12. Nov 28, 2006 #11

    arildno

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    A rather subtle point deserves a separate mention:

    The condition of masslessness decouples the kinematics and the dynamics of the string:
    Since in general, the kinematic quantity "acceleration a" is related to the dynamic quantity "force" F through the relation a=F/m.
    However, in the case of masslessness, we get the INDEFINITE relation a=0/0! (I.e, no relation whatsoever..)

    This means that we must PRESCRIBE the motion of the string to a sufficient extent, for example require that the string does not oscillate.

    We DO have one constraint of motion for the string, namely that its length is constant.
    This shows that usually, it makes most sense to regard the string as PART of a physical system that also include some objects with mass. Their force laws, along with the length constraint on the string along with a few provisions on how the string is allowed to move is often sufficient to find the motion of the entire system, including that of the string. The typical example of that is the "trivial" Atwood machine problem.
     
  13. Nov 28, 2006 #12
    Strings/ropes, cannot support any axial compressive loads; therefore, the only force on a string must be tensile.

    Typically, the load on the string is much larger than the weight of the string. As a consequence, the effect of the strings mass is negligible in increasing the tension as you go down the string.

    Notice, I never said the string is massless, I just said the string's mass is much smaller than the applied load and is negligible. Why the distinction? Because for all I care, the string could be 1,000lbs in weight. But if its supporting a 500,000lb load, it appears to have small mass. Try and pick up that "massless" string. :wink:

    Because Newton's laws come in equal and opposite pairs, the tension at the top and bottom must be the same.

    I don't understand why were talking about ropes inside poles with resistive forces (I am touching my left ear with my right hand).
     
    Last edited: Nov 28, 2006
  14. Nov 28, 2006 #13
    I don't understand. Stress is a tensor, but force is a vector, by definition.

    However, you are always right, so I will sit back and wait for your explanation as to what Im doing wrong. :smile:
     
    Last edited: Nov 28, 2006
  15. Nov 29, 2006 #14

    Gokul43201

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    How does this contradict the line you just quoted?

    But to differ slightly from vanesch, I'd add that tension is not a tensor, but one of the positive components of the stress tensor that lie along its principal diagonal.
     
    Last edited: Nov 29, 2006
  16. Nov 29, 2006 #15
    Because he said force (tension) is is not a vector, but a tensor.
     
  17. Nov 29, 2006 #16

    Gokul43201

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    He's not refering to the tensile force when he speaks of tension, but say the [itex]\sigma_{zz} [/itex] component of the stress tensor (if z points along the length of the rope).
     
  18. Nov 29, 2006 #17
    Oh, ok. Typically, then he should say "Normal Stress", not "Tension."

    Or:
    Tensile stress
    Tension force.
     
  19. Nov 29, 2006 #18

    vanesch

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    The point I wanted simply to make, is the origin of the "flip of sign" of the "force of tension" on both sides of the string, which was one of the questions of the OP. I was pointing out that what was constant along the string was the "situation of stress" which is described by the stress tensor (forgive me any erroneous nomenclature), and that this is a beast that gives you a force when you give it a direction in which you want to know the force.
    Now, as on both sides of the string, one wants to know the force in OPPOSITE DIRECTIONS, you have to feed two opposite normal vectors to the stress tensor, and hence it will spit out two opposite forces.

    So it is not (as I understood the problem of the OP, but I might be wrong) that "somewhere along the string, the "force of tension flips sign". But maybe I was addressing a non-existent problem...
     
  20. Nov 29, 2006 #19
    I see what you're saying now. Thanks. :wink:
     
  21. Nov 29, 2006 #20

    arildno

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    Hmm..when it comes to determine "when is the approximation of masslessness a good one?", not only the weight of the string must be negligible to some other force present, but also, the string's mass-acceleration.
     
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