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Physics book defines tension

  1. Aug 6, 2008 #1
    This really isn't a specific problem in my studies but a general one:

    My physics book defines tension as applying equal magnitude forces, but opposite in direction to the ends of a body. It is clear that this would cause the object to be in equilibrium. It also goes on to say tension is a scalar quantity equal to the magnitude of the forces applied. If this is the case, why does the book ask for the tension in objects that have forces applied to the ends of an object that are not equal in magnitude, and thus cause the object to accelerate? They also show tension as a vector quantity...a lot.

    Is it just bad use of words on the book's part?
     
  2. jcsd
  3. Aug 6, 2008 #2

    Kurdt

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    Re: Tension

    If you use a rope to pull an object, the rope will exert a force on that object and the magnitude of that force is the tension in the rope. I think the confusion arises since the force exerted by ropes is always called (or implied to be) tension as well.
     
  4. Aug 6, 2008 #3
    Re: Tension

    If I had a heavy rope in a verticle position and applied a force on each end such that the top force was greater than the bottom force and the rope accelerates upward, how can you find the tension in the rope? Wouldn't it be impossible to find the tension since the forces acting on one point are not equal in magnitude but opposite in direction?
     
  5. Aug 6, 2008 #4

    PhanthomJay

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    Re: Tension

    For simplicity, let's take gravity out of the picture and imagine a heavy horizontal rope (that is, a horizontal rope with mass) resting on a frictionless horizontal service with an applied rightward horizontal force on the right end, and a lesser applied leftward horizontal force on the left end, such that the rope accelerates to the right under the net force. The tension force in the rope will them be variable along its length, varying from a minimum value at the left end (equal to the applied leftward force at the left end), to a maximum value at the right end (equal to the applied rightward force at that end). At any one point in the rope, the internal tensions acting at that point would be equal on either side of the point, within the confines of the calculus, that is to say, the internal tension to the right of the point would be just infinitesimally greater than the internal tension to the left of that point.
     
  6. Aug 6, 2008 #5
    Re: Tension

    Thank you PhanthomJay! =)
     
  7. Aug 7, 2008 #6
    Re: Tension

    Here is another question: Can an idealized point particle with a mass not equal to zero be in tension if it has equal magnitude forces that are opposite in direction acting on it? If that is the case, how can that particle be in tension if it is accelerating?
     
  8. Aug 8, 2008 #7

    PhanthomJay

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    Re: Tension

    I'm not totally sure of your question, but when you start talking about tension in 'idealized point particles with mass', you're getting into the calculus which I'd rather not tackle (I'm just an engineer, after all; I leave calculus to the experts or trust my computer). So let's talk instead about rigid bodies, then we can look internally at a small section of that body. I'll use some numbers because letters confuse me.
    Imagine, if you will , a 2 meter long uniform rigid horizontal rod of mass 10 kg, subject to equal and opposite forces at either end, of 100N apiece. The tension in the rod is 100N throughout, and the tension at any section or point in the rod is also 100N, and the rod does not move, or it moves with constant velocity, but in any case, it does not accelerate.
    Now take that same rod and apply a 100N force pointing to the right, at the right end, and a 25N force pointing to the left, at the left end, and now since the unbalanced force is 75N to the right, the rod will accelerate per newton 2 with an acceleration of 7.5m/s^2. Now let's examine the tension in the rod. As I noted earlier, it will be linearly variable, 25N at the left end, 100N at the right end, 62.5N in the middle, etc. You should prove this to yourself with a free body diagram cut through some section. The tension on any point within the rod will be variable, depending on which point you are looking at, and there will a differential tension at that point causing the acceleration (oops, we're into the calculus). The 'point' here is that the rod is in a variable state of tension while it is accelerating. I'm not sure if this is a clear explanation.
     
  9. Aug 8, 2008 #8
    Re: Tension

    I understand what you are saying. I am just thinking too hard about the problem.

    It makes sense that tension varies along a continuous body that is accelerating. When you do the calculus the mass of the point in question is considered to be zero so the tension on the left end of that point is equal to the tension on the right. Even though we would think the point is in equilibrium it can accelerate according to Newton's second law.

    Where I get confused is that I know a rigid body in reality is not a continuum but discrete. In that case, the above approach seems to fail because the discrete points of the material do indeed have mass.
     
  10. Aug 8, 2008 #9

    PhanthomJay

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    Re: Tension

    You're digging real deep now. First, I would like to make a correction in regard to your comment on the continuous body. The particle does not have zero mass, it has infinitesimal mass, dm. The particle is not in equilibrium.or else it couldn't accelerate. The tension on the right side of the particle is infinitessimally greater than the tension on the left side of the particle. Let me try to explain using a discrete particle at the center of the rod, with a length of 0.0001 meter, quite small ,indeed. The mass of that particle using my number of 10 kg for the 2 meter rod (5kg/m), is 0.0005 kg. The tension on the right of the particle is 62.5N as I have noted. The tension on the left of the particle is unknown, call it X. Now apply Newton 2 to a FBD of that particle, to whit:
    F_net =ma
    62.5 -X = (0.0005)(7.5)
    X = 62.49625
    So the tension is slightly less on the left of the particle, than it is on the right. This applies to any particle on that rod. Clear as mud??
     
  11. Aug 8, 2008 #10
    Re: Tension

    You are showing that smaller and smaller segments have net forces that approach zero but have constant acceleration.

    If you take the limit of the length of the segment to approach zero the value of m approaches zero. Isn't it possible to make the conclusion that at a length of zero, the segment is now a point which has a mass of zero a constant acceleration of 7.5 and a net force of zero? I say you could since Newton's second law is not violated with these numbers as inputs.
     
  12. Aug 8, 2008 #11

    PhanthomJay

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    Re: Tension

    That's the reasoning behind the calculus...the net force approaches zero, and the mass of the particle approaches zero, but neither is zero. At a particle at the midpoint on the rod, the internal forces acting on that particle are 62.5N to the right, and 62.4999999999999999....N to the left. And so forth.
     
  13. Aug 8, 2008 #12
    Re: Tension

    I need to brush up on my understanding of some calculus concepts then. Thanks again for bearing with me PhathomJay!

    =)
     
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