Hi, I undersand that tension is the force exerted on an object by a string or rope when pulled. Where I get confused is when I'm asked to find the tension at a point along a rope or tension in a section of rope...would the tension in/of a section of rope be the net force acting on it, or the force it exerts on either end (if this is true, I assume the force acting at each end is necessarily always the same?)? Sorry if I'm not making myself clear...here's an example which might show what I mean... A verticle rope is attatched at both ends (to walls). Find the tension T at distance y above the lower end of the rope. The mass of the section below y is M and the mass above is m. Hope I make sense! Thanks, Matt
The tension in a rope is the gross force that one small section of rope exerts on its neighboring section in one direction or the other. It is not the net force that the neighbors from both directions exert upon that small section. If you have a tug-of-war, a small section of rope will be under a large tension. But it will be undergoing negligible net force.
I think you may be being asked to prepare to draw a normal force diagram. For a uniform vertically hanging rope, that would vary triangularly from 0 at the bottom to the weight of the rope at the top.
Many people have suffered the same difficulty that you are presently experiencing. This is because new initiates to physics are not taught that tension in a rope is really a tensorial quantity. It should really be expressed as: T = T i i where T is the scalar magnitude of the tension, T is the tension tensor, and i is a unit vector in the "positive direction" along the rope. Note that the two unit vectors in the above equation are placed in juxtaposition with one another, with no operation implied between them. Two vectors in juxtaposition like this are called a dyad. Please bear with me, and I will show you how this representation can be used. Let s be a specific location along the rope. The way the traction tensor works is that, if you want to know the force exerted by the portion of the rope at >s on the portion of the rope at <s, you dot the tension tensor with a unit vector directed from <s to >s (i.e., in the positive s direction i): F = T [itex]\cdot[/itex]i = T i i [itex]\cdot[/itex]i = T i (i [itex]\cdot[/itex]i) = T i Note that you have dotted the right-hand member of the dyad with the unit vector in the plus s direction i. This has had the effect of mapping the tension tensor T into the tension vector F. Suppose you want to know the force exerted by the portion of the rope at <s on the portion of the rope at >s. In this case, you dot the tension tensor with a unit vector drawn from >s to <s (i.e., in the negative s direction -i). In this case, you find that: F = T [itex]\cdot[/itex](-i) = -T i In summary, what this notation does is provide you with an automatic foolproof method of determining the force exerted at a given location by the portion of the rope on one side of the given location on the portion of the rope on the other side of the given location. Once you get used to this approach, you should have very little trouble later on learning about the stress tensor and how to apply it.
Natv1: The mass is uniformly distributed. The algebraic sum of the forces on one side of a section (one definition of 'normal force') is a linear distribution in the shape of a triangle when plotted as a graph. Thanks to Chestermiller for the rigorous explanation.
Here is another way to understand tension, as a negative pressure. In physics we can generalize the notion of a force to any type of system configuration change. A torque is a force that does work when an object rotates, a stress is a force that does work when an object undergoes a strain deformation, and a pressure/tension (scalar stress) does work when a system changes volume (3d) area (2d e.g. surface tension) or length (1d). In the example you give of the hanging massive rope, ask what is the (negative) work done if you allow one point on the rope to expand to a segment of length dL. The answer is that the rope below that point will drop that distance in the constant g potential. [tex]T = -dE/dL = gM[/tex] If for example you are considering a vibrating string, ether longitudinal or transversal vibrations , the string must stretch to displace beyond the equilibrium state. Figure out the potential energy of the elastic string dU = TdL and kinetic energy and you can work out its dynamics.
You need to be careful about your units here. It's Stress that has the same dimensions as Pressure. Tension has the same dimensions as Force.
In each case we're talking force per boundary measure. Pressure = (negative) volumetric tension has units force per area. Surface tension has units force per length. Linear tension, force per point unit or just force. The important point is that a tension is not (translational) force but rather a generalized (scaling) force.
http://content.answcdn.com/main/content/img/oxford/Oxford_Food_Fitness/0198631472.isometrics.1.jpg Try these isometric exercises. Try the "tension" one on the right of the picture. You can feel tension on your fingers.