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A General Question about Tension

  1. Feb 25, 2013 #1
    I understand that a weight held up by a weightless string causes tension (T) in the string. And it makes perfect sense that T = mg when the string is perfectly vertical. It is well known that once the string is put at an angle, T = mg/cosθ, but this makes no sense to me at all. How can we get more force out of mg than mg itself? It's almost like force is not conserved here, which I don't even know if has to be.

    I think that a possible solution is that there is a horizontal normal force holding the string to keep the string at an angle. But, this force seems to only exist because of the mg, so how can mg ≤ mg/cosθ?
     
  2. jcsd
  3. Feb 25, 2013 #2
    Force being conserved is not a rule xD
    Energy is conserved.
    If the string is placed at an angle, the weight will no longer be stationary and neither will it have a constant speed, so it is accelerating. you can no longer equate forces in a plane to 0.
     
  4. Feb 25, 2013 #3

    SteamKing

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    Forces are not conserved. However, for equilibrium to be obtained, the sum of the forces and the sum of the moments about a particular point must both equal zero.
     
  5. Feb 25, 2013 #4
    Are you talking about a static situation, or is the weight moving? If static, then something else has to be holding the weight up!
     
  6. Feb 25, 2013 #5
    Damn, sorry, I should have been more clear. The string is being held up by something such as another frictionless string.
     
  7. Feb 25, 2013 #6
    In what case exactly is T =mg/cosθ ? Shouldn't it be mg x cosθ ?
     
  8. Feb 26, 2013 #7
    We are not getting more Force out of mg than mg itself. We are getting less force out of the original tension T than the original tension T itself. It makes sense since part of the weight of the mass (mg) is being balanced by tension while the other part is being balanced by another string which you just mentioned in post no. 5.
     

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  9. Feb 26, 2013 #8
    The hyponuse being the string, and mg is the Fy force.
     
  10. Feb 26, 2013 #9
    Your diagram is not exactly what I had in mind. Instead, a frictionless rope attached horizontally to another rope supporting a weight. The top half of the diagram looks like this __/ The bottom half has the slanted string extending vertically downward from the joint of the two ropes holding a weight. It seems like there isn't any extra force causing T to be greater than mg.
     
  11. Feb 26, 2013 #10

    sophiecentaur

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    That diagram is not strictly accurate because it implies the 'vertical' string is, in fact vertical. If there is any tension in the other string (T1>0) then the other string will no longer be vertical and you then have (as always) vertical and horizontal components in both tensions, cancelling each other out. So T2 will be greater than mg.
    Whatever setup you invent, to be 'real' there must be equilibrium along both axes.
     
  12. Feb 26, 2013 #11
    Ah ok. In this care, the tension in the string can be greater than the mass because it is not only supporting the mass, but also pulling on the other string as well. The tension depends on the angles of both of the strings.

    If string 1 is at some angle theta (from horizontal) and string 2 is totally horizontal, that means that string 1 is fully supporting the weight and also fully pulling against string 2, so the vector sum of the forces is
    T1 = sqrt(mg^2 + T2^2)
    also,
    tan(theta) = mg/T2
    If we know theta, we can solve for T1 and T2
    T1 = sqrt(mg^2 + mg^2*cot^2(theta)) = mg*csc(theta)
     
  13. Feb 26, 2013 #12
    But it still seems that string 2 is pulling on string 1 only because of mg. In other words, if we detach the mass, then there wouldn't be a force from string 1 pulling on string 2. This extra force beyond mg seems to come from nowhere.
     
  14. Feb 26, 2013 #13
    That's not the system I mean. String 1 is the string with the angle. String 2 is the horizontal string in this diagram __/ The corner where the strings both meet, string 1 keeps going exactly vertically downwards; I don't show the vertical continuation of string 1.
     
  15. Feb 26, 2013 #14

    sophiecentaur

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    We are talking in a vacuum about these set-ups. Either draw a diagram and post it or draw it on paper for yourself and include all horizontal and vertical forces. The solution for each arrangement will be different in detail but, if you are doing things right, you will get forces balanced in all directions, if the mass is not accelerating.
    There are no 'loopholes' in this. Follow the rules and you get no anomalies.
    Your problem may be that you are making assumptions about strings being vertical when they can't be.
     
  16. Feb 26, 2013 #15
    Do you still need help with this?
    If i understand correctly ... ur calculations would be the same if the mass was at the intersect of the two strings. and there would be two separate tensions in the angled string.
    Take [itex]T_{1}[/itex] as the tension in the string that is at an angle [itex]θ[/itex] to the horizontal and [itex]T_{2}[/itex] as the tension in the horizontal string leaving [itex]mg[/itex] as the tension in the second half of the first string.
    There you have your two unknowns [itex]T_{1}[/itex] and [itex]T_{2}[/itex] to be solved resolving vertically and horizontaly. There is no dividing by cosθ :)

    Edit: this ur system? (attached)
     

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    Last edited: Feb 26, 2013
  17. Feb 26, 2013 #16
    Your diagram is the right system.

    I really do understand how to get the unknowns of this system; this is all clear to me. However, at a different level of my understanding, it seems as though there is a horizontal force coming from nowhere even though I completely understand what the right answer is.

    I am trying to understand how T2 can be greater than mg, when it seems as though mg is the only reason for there being a T2 force and a horizontal force. My reason for thinking this is that if we cut the weight off of the bottom string holding it, the forces on the strings disappear. So it seems the reason that the forces are there in the first place is only because of mg. I always get these questions right because I just follow the proper proceedure in doing them, but the logic of it does not make sense to me.
     
  18. Feb 26, 2013 #17
    Chimpz has the right diagram in post #15. Please read my reply to it.
     
  19. Feb 27, 2013 #18

    sophiecentaur

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    So your only question is about where 'extra' forces seem to come from. Asking that question, as already mentioned, implies that there must be some conservation law for forces. But there isn't, so there is no point in asking where the forces 'come from'.
    If you were actually constructing that setup, you would need to do some work in pulling the suspension string sideways from vertical. That work would be raising the level of the suspended mass from where it was originally hanging (that's where energy conservation would come in - does that help you in any way?). But problems like this, involving forces in equilibrium do not consider how the situation was arrived at.
    I remember, at A level, a very long time ago, in 'Applied Maths', we solved some such problems using the principle of 'virtual work', in which you distort a framework, infinitessimally, and calculate the work done. You turn this into an equation, which solves the network for you. Don't ask me to repeat it - the last time I did it would have been 50 years ago.
     
  20. Feb 28, 2013 #19
    You know, a lever can be used to lift a heavy weight with a light force. Where does the extra force come from?
    See this: http://en.wikipedia.org/wiki/Mechanical_advantage

    It is not force that is conserved, but force over distance. If you remove the string 1, the weight will drop. Due to string 2, it must move a larger distance to drop the same amount of height. If you think about it, this effect exactly accounts for the mechanical advantage.
     
  21. Feb 28, 2013 #20
    Wow, that's interesting. I need to think about this for a while, and hopefully it will become clear.
     
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