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The nature of wire tension

  1. Oct 3, 2012 #1
    As far as I understand, the wires in classical newtonian mechanics are inextensible, unbreakable and so forth. The part that confuses me is that sometimes there are two tensions in the wire, or rather the tension is in both senses. Is that because there are opposing forces acting on the two ends of the wire, or is that simply because of the electrostatic nature of the force (the particles in the wire all attract each other and resist the extension of the wire).

    And in more general terms, how to figure out the tension/tensions of the wire, and the way they act or balance a physical system?
  2. jcsd
  3. Oct 3, 2012 #2
  4. Oct 3, 2012 #3
    Thank you for the link, it's useful, and helped clarified things a bit.

    What I meant was: consider the following situation: an object of mass m is suspended by an ideal string. So then there's the gravitational force, G = mg pulling the object down, which is transmitted through the string, and it's a force pointing downwards. But the system is in equilibrium, so there's an opposing force equal to the weight created by the support. So that force is equal in modulus but pointing upwards, and is transmitted through the string to the object so that the object is at rest.

    Ok so far? And if so, wouldn't "the tension in the string be oriented in both senses"? (one down for the weight, one up for the balancing). And wouldn't it be equal in magnitude to 2 times the weight?
  5. Oct 3, 2012 #4
    I think I'm beginning to see where you are coming from.

    Think about the mass being supported by a single string, straight up (or down). It is in equilibrium (ie, not moving). At the point where the mass meets the string the forces on that point sum to zero. The mass provides mg Newtons of force stright down, so to hold this in equilibrium the string provides mg Newtons of tension straight up. If instead the mass is supported by two strings, each at some angle to the mass, the total tension (by which I mean the sum of the tensions in each wire) will be greater than mg Newtons, as there is now a lateral component of the forces at the point where the strings meet the mass.

    Does that make it any clearer?
    Last edited: Oct 3, 2012
  6. Oct 3, 2012 #5
    Yes it does. What you're saying is that in an system in equilibrium, every point is at equilibrium, and so for certain points like the one where the mass meets the string, there is a "tension force" to account for the equilibrium. When in fact, the "tension force", is simply the normal, directed through the string.
  7. Oct 3, 2012 #6
    I agree, except the 'normal' is a specific vector at 90 degrees to a plane. These problems are usually solved by using vector diagrams and breaking each force vector (mg and the tension in the wires) into their xy components (although the choice of co-ordinate system is entirely yours) to ensure that the vector forces acting on the mass/string junction sum to zero.
  8. Oct 3, 2012 #7
    I think I understand it now, thanks.
  9. Oct 3, 2012 #8
    It is important to be clear which force acts on which body.

    We mentally separate the system into "free bodies" and consider the force(s) that must be applied to each free body to hold it in equilibrium.

    So in your weight on string case we have two free bodies : the weight and the string.

    The weight has two forces acting on it ;
    The body force of gravity = weight = mass times g, acting downwards
    The pull (from the string) acting upwards on the point of attachment.

    Since the weight is in equilibrium these forces are equal.

    Notice there is no string in this free body.

    The string also has two forces acting on it;

    The upwards reaction at whatever you have hung it from.
    The downwards pull of the weight on the string.

    Again these are in equilibrium and so are equal.

    Notice that I have not mentioned tension.

    This is because all the above forces are external forces.

    That is they are imposed on the free body (string or weight) and can be changed by some external agent, they are not under the control of the free body.

    The tension in the string is an internal force or response to the external loading force to enable it to transfer it from one end of the string to the other.

    It helps greatly to draw these free body force diagrams. Can you do this?
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