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Tension - how is mechanical tension even defined?

  1. Jan 21, 2014 #1
    Tension -- how is mechanical tension even defined?

    Hello, I'm currently a student in AP Physics having a little trouble with the concept of tension, especially that of a pulley system or an Atwood machine.

    First, how is tension even defined? Is it the reaction force to the applied force on the string? I keep getting confused because I have this impression that it is a pair force, but not sure which is defined as tension in the string. For ex, for an Atwood machine, is tension the reaction force of the applied gravitational force due to the weights on each end?

    And also, I'm having some trouble generally understanding the concept that the tensions of the strings on each side of the pulley may be different as well. I have a vague understanding of how the resultant final acceleration of the system may be calculated, but again, am failing to visualize it conceptually.

    If you could take your time to explain or point me towards the source, it would be greatly appreciated. Thanks!
     
  2. jcsd
  3. Jan 21, 2014 #2
    This is a very common problem that people have. That's because tension is really a more complicated quantity than a scalar or a vector, called a second order tensor. Suppose you have a string lying along the x axis, and you apply equal and opposite forces on both ends of magnitude T. One of these forces is pointing in the + x direction, and is given by [itex]T\vec{i}[/itex], and the other is in the - x direction, and is given by [itex]-T\vec{i}[/itex]. So you are right about the duel nature of the tension. Now suppose we focus on a point on the string somewhere between the two ends, we make a cut in the string, and we replace the action of either portion of the string on the other portion of the string by an equivalent force so that we can maintain the equilibrium. If we replace the right portion of the string, we will have to replace it with a force on the left portion of the string equal to [itex]T\vec{i}[/itex]. If we replace the left portion of the string, we will have to replace it with a force on the right portion of the string equal to [itex]-T\vec{i}[/itex]. But, how can be obtain there results automatically, without having to think about it? Suppose we write down the quantity [itex]T\vec{i}\vec{i}[/itex], without temporarily providing explanation of what it means. Now let's focus on the left side of the string where we have made the cut in the string. A unit normal vector to the cross section of the cut from the region to the left where the material is still there to the region on the right where the material is no longer there is given by [itex]+\vec{i}[/itex]. If we take the dot product of [itex]T\vec{i}\vec{i}[/itex] with the normal [itex]+\vec{i}[/itex], we get [tex]T\vec{i}\vec{i}\centerdot\vec{i}=T\vec{i}(\vec{i}\centerdot\vec{i})=
    T\vec{i}[/tex]
    Now let's focus on the right side of the string where we have made the cut in the string. A unit normal vector to the cross section of the cut from the region to the right where the material is still there to the region on the left where the material is no longer there is given by [itex]-\vec{i}[/itex]. If we take the dot product of [itex]T\vec{i}\vec{i}[/itex] with the normal [itex]-\vec{i}[/itex], we get [tex]T\vec{i}\vec{i}\centerdot-\vec{i}=-T\vec{i}(\vec{i}\centerdot\vec{i})=
    -T\vec{i}[/tex]
    So this methodology automatically gives the correct tension force with the correct direction. The quantity [itex]T\vec{i}\vec{i}[/itex] is called the tension tensor.

    I hope this helps.
     
  4. Jan 22, 2014 #3
    Thank you for a detailed answer. I think I am starting to grasp the concept of tensor and its role in equilibrium now.

    However, what would happen if you pushed on the right end of the right portion of the string as -Ti? Then, by necessity, would the tension in the string be +Ti? Another example that strikes my head is a a spinning weight tied to a string; the centripetal acceleration is towards the center of the circular motion but the string still clearly has a tension force on it.
     
    Last edited: Jan 22, 2014
  5. Jan 22, 2014 #4
    A string won't support a compressive load, and will buckle. But, let's suppose it could. Then the tension tensor would be [itex]-T\vec{i}\vec{i}[/itex]. If we dot this with the unit normal in the +x direction, we get [itex]-T\vec{i}\vec{i}\centerdot \vec{i}=-T\vec{i}[/itex]. This says that the force is directed in the negative x direction, as it should be.

    In this case, the tension tensor is given by [itex]T\vec{i_r}\vec{i_r}[/itex], where [itex]\vec{i_r}[/itex] is the unit vector in the radial direction. If we dot this with a unit vector pointing away from the mass (we break the string just inboard of the mass), we get: [itex]T\vec{i_r}\vec{i_r}\centerdot \vec{-i_r}=-T\vec{i_r}[/itex]
     
  6. Jan 23, 2014 #5
    Welcome to the world of internal forces and stresses. The state of internal stress at a location within a body is characterized by the stress tensor [itex]\vec{σ}[/itex], which is a second order tensor. If you focus on a tiny element of flat planar area located somewhere within a body, separating the material on one side of the flat planar element from the material on the other side, you can determine the force exerted by the material on each side on the material on the other side. What you do is erect a normal vector [itex]\vec{n}[/itex] to the flat planar element. The normal can be pointing from the material on side A to the material on side B, or from the material on side B to the material on side A. If it is pointing from the material on side A to the material on side B, and you dot the stress tensor with this normal, you obtain the vector force per unit area [itex]\vec{τ}[/itex] exerted by the material on side B on the material on side A. If it is pointing from the material on side B to the material on side A, and you dot the stress tensor with this normal, you obtain the vector force per unit area [itex]\vec{τ}[/itex] exerted by the material on side A on the material on side B. In either case, you calculate:
    [tex]\vec{τ}=\vec{σ}\centerdot \vec{n}[/tex]
    This is called the Cauchy stress relationship, and is the basis of all continuum engineering mechanics. The vector [itex]\vec{τ}[/itex] is referred to as the traction vector.

    How is this all related to tension in your string? Well, the traction tensor in your string is equal to the stress tensor times the cross sectional area of your string:
    [tex]\vec{T}=\vec{σ}A[/tex]
    In the most general case, the stress tensor, when expressed in component form, has 9 components in a summation involving pairs of unit vectors in the three coordinate directions. However, for the simple case of a thin cylinder comprising your string, only one of these components is non-zero, the component in the x-direction (or, in the case of spinning a weight on a string, the component in the radial direction):
    [tex]\vec{σ}=σ_{xx}\vec{i_x}\vec{i_x}[/tex]
    So,[tex]\vec{T}=σ_{xx}A\vec{i_x}\vec{i_x}[/tex]
    This means that what we call the tension in the string is equal to the product of the stress and the cross sectional area σxxA.

    The reason I'm telling you this is so you know that you don't have to break the string to get the tension force. You can just focus on a specific cross section of the string, and use the Cauchy stress relationship to find the force that the material on one side of the cross section exerts on the material on the other side of the cross section (and vice versa).
     
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