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Problem with Tension

  1. Sep 10, 2010 #1
    1. The problem statement, all variables and given/known data

    There is no one particular problem, but I will give one that I am struggling to understand as an example:

    A block of mass m = 15.0 kg is hanging from three strings. What are the tensions in the strings?

    There is a diagram that goes with this problem. There is one string (C) directly connected to the mass and two strings (A, B) connected to the ceiling that meet at the first string (C). String A meets the ceiling at an angle of 30 degrees, string B at 45 degrees.

    2. Relevant equations

    m = 15.0kg

    3. The attempt at a solution

    This is actually an example in the book, so I have the solution in front of me. I understand how the problem is set up, equations, etc. The sole thing I don't understand is why the strings are pulling the directions they are pulling. For instance, why are strings A and B pulling up on C? Don't all three strings have a pull in both directions? Please make this concept of tension clearer for me. The book does a poor job.
     
  2. jcsd
  3. Sep 10, 2010 #2
    Tension doesn't have direction until you relate it to one end or another. Imagine pulling a length of string between your two hands. You would say that from your point of view the arrows for the external actions are opposite and outwards. At the left end, say, you have the external force going to the left, equilibrated by an internal force going to the right. So when you isolate (in your mind) one end of the string, you have to replace the 'cut' with any force that might be there - in this case, a tension to the right is needed to keep the external action in equilibrium... OK? At the right end, opposite arrows apply. Hence outward facing arrows for applied force are balanced by internal facing arrows for the internal actions. Stand up. You can feel your weight on the ground. Which way is this force? From the point of view of the ground, the force is downwards, but from the point of view of your feet, it is upwards (and if it wasn't there, you would be drawn to the centre of the earth).
     
  4. Sep 11, 2010 #3
    Okay, I had to read this a few times. But I think I get it. You're saying that for every external force on the string, there is an opposite internal force, i. e., tension in the string. But this is only when something is in equilibrium, right?

    For instance, there is another example in the book involving a man pulling a crate with a string. The book says "the thin, massless string simply transmits the applied force from one end to the other with no change in direction or magnitude." In this case, is there no opposite internal force in the string as in the first example?
     
  5. Sep 13, 2010 #4
    Man pulling object with string. Suppose you want to analyse this with a mathematical model. If you want to concentrate on the crate and leave the man out of it and mentally cut the string at (say) its mid-point, you have to replace the cut with the force thus exposed. In other words an arrow representing the force that the man was initiating. This reveals the internal tension in the string. Suppose you want to isolate the man as the free body diagram by cutting the string at its mid-point, then the cut must be replace by a force represented by an arrow going the opposite way to that in the first mathematical model. Bring these two mathematical models into close proximity and you will find the arrows at the cut point are self-equilibrating. If you draw the crate, the string and the man all together, the arrows in the string are by convention drawn pointing towards each other at the ends, representing the string's internal reactions to the applied forces (man pulling, and crate resisting).
     
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