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Very confused about concept of tension

  1. Mar 23, 2013 #1
    I'm having some very basic conceptual problems about tension. If you have a pulley where two masses are hanging on opposite ends of an ideal string, why is the upwards force that the one mass experiences the same as the upwards force that the other experiences? People have tried to explain this to me by saying that it's logical because it's the same string and it's massless, but I just don't get it. What I DO get is: (a) that the acceleration of the masses will be the same and (b) that if the masses are equal, they will be in equilibrium, and the tension will be equal to the weight. Is there perhaps a more mathematical answer to my original question (e.g. reciprocity or something?) Also, exactly how is tension defined? Is it by definition a scalar quantity, or is it a vector? If it's a vector, how is the direction defined?
     
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  3. Mar 23, 2013 #2

    SteamKing

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    Tension is a type of force. As a force, it has the properties of a vector, i.e., magnitude and direction. Strings by nature can carry only tension (you can't push a string). The direction of the tension will therefore act along the length of the string.

    Tension is what makes tug of war. In your example, the string is draped over a pulley, and both masses are tugging at each other. If you want to know why the tension is the same in both parts of the string, draw a free body diagram of each mass. What forces act on each mass?
     
  4. Mar 23, 2013 #3

    CWatters

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    Think about two little bits of the string in the middle somewhere. Each bit is subject to forces on it's ends.

    The total force on the red bit must be the same as the green bit or one would accelerate toward or away from the other either stretching or compressing the rope. If the rope were stretched or compressed the forces would change restoring the balance.

    Consider what happens if the two bits of rope are at the ends.
     

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  5. Mar 23, 2013 #4
    Tension is a force that arises due to shift in position of particles form their equilibrium positions in the the thread.

    Consider the situation when you pull the massless thread between your hands horizontally. Now consider a constituent molecule. Because the thread has been stretched, the particles are all displaced from their original relative positions. Thus the molecule will experience attractive forces from either side (which are equal in magnitude as the molecule is at rest relative to other molecules). Thus every particle experiences force in such a manner by symmetry. This is what I believe happens in case of it being used in a pulley too.

    At the ends of the thread, the tension is balanced by the pull from your hand. Tension is a form of force and hence is a vector. It's direction is such that it tries to bring back the displaced particles to equilibrium position. Please go to the following link if I'm not clear.
    http://en.wikipedia.org/wiki/Tension_(physics [Broken])
     
    Last edited by a moderator: May 6, 2017
  6. Mar 23, 2013 #5
    Another (and more fundamental) way of looking at this is treat tension as a second order tensor. This sounds like it would make things more difficult, but it actually makes things simpler. Suppose that the rope is laid out along the x axis, and let [itex]\vec i_x[/itex] be a unit vector in the positive x direction. The Tension Tensor [itex]\vec \tau[/itex] can be represented mathematically by
    [tex]\vec \tau=T\vec{ i_x}\vec{ i_x}[/tex] where T is the scalar magnitude of the tension. Note that there are a pair of unit vectors placed in juxtapostion with eachother, with no apparent operation implied between the two of them. We'll see how this plays out shortly, so please bear with me.

    Consider some arbitrary point x along the rope. The tensile force (vector) exerted by the portion of the rope located at x+ on the portion of the rope located at x- is equal to the Tension tensor dotted with a unit vector pointing in the direction from x- to x+, namely [itex]+\vec i_x[/itex]. So,

    Tension force exerted by portion of rope located at x+ on portion of rope located at x- is equal to [itex]\vec \tau\centerdot \vec{ i_x}=T\vec{ i_x}(\vec{ i_x}\centerdot\vec{ i_x})=T\vec{ i_x}[/itex]

    That is, it is equal to the scalar magnitude of the tension times a unit vector in the + x direction.

    Next lets consider the tensile force (vector) exerted by the portion of the rope located at x- on the portion of the rope located at x+. This is just the Tension tensor dotted with a unit vector pointing from x+ to x-, namely [itex]-\vec i_x[/itex]. So,

    Tension force exerted by portion of rope located at x- on portion of rope located at x+ is equal to [itex]\vec \tau\centerdot(- \vec{ i_x})=T\vec{ i_x}(\vec{ i_x}\centerdot(\vec{- i_x}))=-T\vec{ i_x}[/itex]

    That is, it is equal to the scalar magnitude of the tension times a unit vector in the - x direction.
     
  7. Mar 23, 2013 #6
    I'm going to latch onto the picture of the string that CWatters attached, because it looks as if it could be useful, and I like visual representations. (We haven't done tensors yet, but it looks like a good explanation and I'll definitely come back to it.) I understand that the net force on each bit of string must be the same, otherwise the string as an entirety would compress or stretch. That makes sense. So, say each of the coloured bits were at an end of a piece of string with masses attached (say with weights 5N and 3N respectively) hanging over a pulley wheel as I described. The downwards force the first bit of string experiences would be 5N, and 3N downwards for the second bit. The first bit would then have to experience a 3N upwards force, and 5N upwards for the second in order for the net force on each bit to be the same (looking at the direction of the forces from left to right along the string, rather than up and down). Is this correct? What in this situation would represent tension?
     
  8. Mar 23, 2013 #7
    The tension T throughout the string would be uniform, including at the ends. Call it T. Now do a force balance on each of the two masses as a free body. This will result in two equations in two unknowns, the string tension T, and the acceleration of both masses a. Solve for T and a.
     
  9. Mar 24, 2013 #8
    I know how to draw free-body diagrams and solve for T and a, and I know that T is uniform because I've been told so on numerous occasions. My question is conceptual. I want to know WHY T is uniform. I am looking for a proof, or at least a better way of understanding it.
     
  10. Mar 24, 2013 #9

    ehild

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    You can imagine the string as a stretched spring. If you select a piece of it as shown in the picture, the other parts act at the ends with forces T1 and T2: These are the tensions in the string at the positions of the ends of the selected piece. If the string has mass, and the mass of the piece is m, the piece will accelerate with a=(T1-T2)/m. If the string is massless, the smallest difference of tension at both ends would cause infinite acceleration. So the tension must be the same everywhere in a massless spring.

    ehild
     

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  11. Mar 24, 2013 #10

    SteamKing

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    Imagine a string which is massless and tied at one end to a hook in the ceiling. The other end of the string is tied to a single weight, and the weight is then suspended from the ceiling by the string. The tension in the string is the force which is counteracting the force of gravity which tends to pull the weight toward the center of the earth. Since the string is massless, it is not affected by gravity. The tension in the line is equal to and opposite of the weight tied to the bottom end of the string. If you were to make a free body diagram of this arrangement, it would not matter where you cut the string: the tension would be the same regardless of how much string were attached to the weight.

    If, however, the string did have some finite mass per length, then it would be affected by gravity. In analyzing a free body diagram of the weight and a certain amount of attached string, the position of the cut would produce a different set of forces, which would include the weight by itself plus the weight of whatever length of string is attached to the weight.

    A massless string is an idealization. If the weight were suspended by a chain instead, then the tension in the chain varies with position and depends on how many links of chain are attached to the weight below the point at which the chain is cut for free body analysis.
     
  12. Mar 24, 2013 #11
    Here is another way of saying what Echild said. Imagine a straight string, with a tension T1 on one end and a tension T2 on the other end. If the mass of the string is m, a force balance on the string gives:

    T1 - T2 = ma

    As the mass of the string approaches zero, the difference between T1 and T2 approaches zero, irrespective of the acceleration.
     
  13. Mar 24, 2013 #12
    Thanks -- that makes perfect sense. :smile:

    But now here's a problem I have with applying this. Imagine an ideal string hanging over a pulley with two different masses attached to either end of the string. Now consider the forces acting on the bit of string that is directly in contact with the first mass: weight and an opposing tension force (not, in this case, two opposing tension forces). It seems to me that according to Chestermiller's explanation, the tension on that bit should be equal in magnitude to the weight, because since the string is massless, the net force on the string should be zero.

    Fnet = (0)a = tension - weight

    But that's not the answer, because we can make the same argument for the other side of the pulley and get a different value for the tension. What's going wrong here?
     
  14. Mar 24, 2013 #13

    ehild

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    If there are different masses attached at the ends of the string, the masses accelerate. The heavier mass moves downward and the lighter one moves upward. If the acceleration of the heavier mass is a, Newton second equation says that ma=mg-T(tension).

    ehild
     
  15. Mar 24, 2013 #14
    The tension is not equal to the weight of either mass. It is less than the weight of the heavier mass and more than the weight of the lighter mass. That way the heavier mass can accelerate downward, and the lighter mass can accelerate upward (both masses moving with the same acceleration magnitude).
     
  16. Mar 25, 2013 #15
    I get that. I understand that tension is not just equal to weight, otherwise it wouldn't be uniform. I'm just confused about the fact that I know the string is accelerating with an acceleration that's equal to the acceleration of the masses, but the string itself is massless. Fnet = (0)a = T1 - T2 as Chestermiller and ehild explained, therefore T1 = T2 = T -- I understand that. Now I'm looking at the bit of string that's directly in contact with the mass. I am correct when I say that the bit of string doesn't feel a downwards tension force, but instead a force equal to the weight of the mass? Then Fnet = (0)a = mg - T. The problem is that mg clearly doesn't equal T.
     
  17. Mar 25, 2013 #16

    ehild

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    Imagine the mass is connected to the string and you hold the other end. When in rest, the string feels mg force, the tension is mg, and you also exert mg force. What force would you feel in a lift in free fall?
    You release the string so string and mass fall free. What force does the string exert on the mass now? Is the string tight or slack?

    The string feels the force the mass pulls it with. As the whole thing is accelerating, it feels mg-ma force at that end where the mass is attached. Then Fnet=mg-ma-T =0 ---> T=mg-ma.

    ehild
     
  18. Mar 25, 2013 #17
    Only your weight.

    The tension should be zero and the string should slacken.

    Wait, why mg-ma? Why not just mg?
     
  19. Mar 25, 2013 #18
    Here's the problem: You wrote down the equation for Fnet without precisely identifying what free body you are looking at. Please precisely identify the free body you are looking at. For example, is it the segment of string adjacent to the mass plus the mass? or, is it the segment of string adjacent to the mass sans the mass, in which case the very end of the string at the junction with the mass needs to be considered.
     
  20. Mar 25, 2013 #19

    ehild

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    In a free-falling lift, you fell weightless.You certainly saw astronauts in the Space Station floating in the air ...

    In free fall the mass does not pull the string, the string does not pull the mass: if the acceleration is a=g (downward) then the tension is T=0: ma=mg-T, T=0.

    If the mass is in rest, a=0, mg-T=0--> T=mg.

    The mass is attached to the string and feels an upward force T from the string. According to Newton's Third Law, the mass exerts the same force T with opposite sign to the string. The mass accelerates, and the acceleration is determined by the net force: ma=mg-T. So T= mg-ma.

    ehild
     
  21. Mar 31, 2013 #20
    Sorry for the late reply, I had some login problems.

    Okay, I thought you were still referring to the free-falling string. I understand that bit now.

    I know these questions must be really annoying, but I've got one more. Imagine a pulley with a mass exerting a force of 5N on the left end and a mass exerting a force of 3N on the right end. Okay, so I know the tension is NOT simply 5N-3N. But here is what I'm thinking: The 3N mass is exerting a force on the string as it accelerates. And the bit of string that experiences that force is tugging at the bit of string above it with the same force, according to NIII. (And that bit tugs at the bit above it, continuing along the whole string.) The same argument can be made for the mass at the other end of the string. So shouldn't any bit of string feel a force of 3N in one direction and 5N in the other?

    Thanks for all the help so far!
     
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