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Weight tension problem

  1. Aug 6, 2016 #1
    1. The problem statement, all variables and given/known data
    See attached picture.

    2. Relevant equations
    Change in work = force x distance moved

    3. The attempt at a solution

    So I took the ratio AC/AB = 1/ $\sqrt{2} $ which I set equal to T/50 where T is the tension we want to find. Solving this gave me T=50/ $\sqrt{2} $.
    I just wanted to know whether this reasoning is valid and I would also like alternative solutions.
     

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  2. jcsd
  3. Aug 6, 2016 #2

    David Lewis

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    The formula for work is irrelevant. We know sum of forces acting on the point where the wires are joined = zero.
     
  4. Aug 6, 2016 #3

    haruspex

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    The instructions say to use Virtual Work.
    You do not state what your reasoning was, so it is impossible to answer that. Yes, it gives the right answer, but does not seem to comply with the instructions.
     
  5. Aug 7, 2016 #4
    Do you have any hints on how to solve this using virtual work?
     
  6. Aug 7, 2016 #5

    haruspex

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    Do you understand the principle and method of virtual work? Se e.g. https://en.m.wikipedia.org/wiki/Virtual_work.
     
  7. Aug 7, 2016 #6
    Well I think it's when you move a body by a virtual displacement, which will enable you to compute all the possible sums of the work done by each component in the mechanical system. I tried to do this in the problem above:

    We can take the weight W and imagine it moving from it's current position to the top of the line (i.e. until it lies on the same plane as the line). By moving it we do work against gravity, which is equal to the Gp = vertical distance moved x 50 x 9.81 = 5√2/2 x 50 x 9.81.

    However I don't see how I can relate the tension of the wires to this (e.g. what is the 'distance' they move?).
     
  8. Aug 7, 2016 #7

    haruspex

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    You should use arbitrarily small virtual displacements, calculus-style.
    The work done by a tension is the component of the displacement in the direction of the tension. This will equal the shortening in the string.
     
  9. Aug 7, 2016 #8
    Does the direction of the virtual displacement matter? Or can I just say that the weight would move an arbitrary small value downwards? This would imply a lengthening in the string right?
     
  10. Aug 7, 2016 #9

    haruspex

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    That's right, the direction does not matter. But in general it is taken as a vector, and the virtual work done should tend to zero faster than the displacement no matter what the direction of the vector. So typically consider two directions of movement.

    Edit: if there is some symmetry you may be able to get away with only considering one direction. Not able to view the image right now.
     
    Last edited: Aug 7, 2016
  11. Aug 7, 2016 #10
    So here is my attempt:

    If we lift the weight W by a small vertical distance Δh, W will gain P.E. of Δh x mg in the process. Since energy is conserved, the work done by the wire (i.e. the shortening of the wire by a small length, which we call Δl must be equal to the gain in P.E. by the weight.)

    From this we obtain the equation:

    2Δl = 2Δh x mg

    I need help on how to continue...( I apologise for the continued questions, but I am new to this and I just started reading the F. lectures, so you must understand me struggling with these concepts...)

    Thank you.
     
  12. Aug 7, 2016 #11

    haruspex

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    Not sure why you have those factors of 2 in there. If it's because there are two wires, it should not be on the right hand side.
    Next you have to figure out what the change in length of the wires is as the weight moves up Δh. It helps to think about components of movement. What is the component of that movement in the direction of the wire?
     
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