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Finding the tensionin the cords

  1. Nov 21, 2009 #1
    Capture.JPG


    what i have done is used my equations for equilibrium on the structure, the problem being that i only have 2 that i can effectively use here and i have 4 unknowns

    [tex]\Sigma[/tex]Fy= Ay + By + Cy + Dy - P = 0
    [tex]\Sigma[/tex]Mc= D*L - B*L - 2*A*L = 0


    what else can i do here, i tried "cutting" the beam at different places but this exposes another moment and another force so that doesnt help either.
     
  2. jcsd
  3. Nov 21, 2009 #2

    Q_Goest

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    Hi Dell.
    2 hints in the wording of the question: The bar is "rigid" and there are "four identical wires".

    Consider the bar to be infinately rigid and that the wires, being identical, also stretch in proportion. In other words, the wires produce a force as if they were springs. If these were springs, would that help?

    Also, consider the way these wires are in tension. Let's look at A, B, C and D in series and rewrite the tension in each of these in tension instead of in the way they provided.

    Ta = P/10
    Tb = P/5 = 2P/10
    Tc = 3P/10
    Td = 2P/5 = 4P/10

    Notice anything?
     
  4. Nov 22, 2009 #3
    1, 2, 3, 4, that i saw before, but thats only because they gave me the answers

    what does rigid and identical wires help me? since the bar is rigid it has no deformation and if i take the wires as springs F=kx, and i can say that Ka=Kb=Kc=Kd correct??

    but from here what do i do?
     
  5. Nov 22, 2009 #4

    Redbelly98

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    To rewrite what Q_Goest said slightly differently,

    Ta = 0.1
    Tb = 0.2
    Tc = 0.3
    Td = 0.4​

    Now do you see a pattern?

    Since the bar is straight and rigid, the amount by which the wires stretch must be of the form

    Δy = mx + b​

    Correct.
     
  6. Nov 22, 2009 #5
    okay that i didnt think of
    y=mx+b

    very logical, like you said the bar is rigid,

    [tex]\Delta[/tex]A = m*0 + b
    [tex]\Delta[/tex]B = m*L + b
    [tex]\Delta[/tex]C = m*2L + b
    [tex]\Delta[/tex]D = m*3L + b

    now what do i so with this, how do i connect this to the tension,
    can i say F=k*y using the y's above

    then i have 6 equations and 7 unknowns A,B,C,D, K,m,b
     
  7. Nov 22, 2009 #6

    D H

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    You don't need to know the specific values of the spring constant k, the displacement b, or that scale factor m.

    Suppose you know the tension in strings A and B. Given just the facts that the bar is rigid and the strings are identical, you should be able to determine the tensions in strings C and D. Can you do that?

    Addendum
    Do this, you will have only two unknowns (e.g., TA and TB). Use these in the force and torque balance equations and voila! you will get the answer.
     
  8. Nov 22, 2009 #7
    A + B + C + D - P = 0
    D*L - B*L - 2*A*L = 0

    if A and B are given i can solve it easily, but they arent,
    what am i not seeing here?
     
  9. Nov 22, 2009 #8

    D H

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    You did not write C and D in terms of A and B. That the rod is rigid and the strings are identical means you can do just that.
     
  10. Nov 22, 2009 #9
    A + B + C + D - P = 0
    D*L - B*L - 2*A*L = 0

    D = B+2A
    C = P-3A-2B

    now what
     
  11. Nov 22, 2009 #10
    F=k[tex]\Delta[/tex]Y

    and since the function is linear

    Ta=Ta
    Tb=2Ta
    Tc=3Tc
    Td=4Ta

    Ta(1+2+3+4)=p

    Ta=P/10

    and so on,

    is this a way to solve it,

    do i not need to define [tex]\Delta[/tex]Y like i did before ??
     
  12. Nov 22, 2009 #11
    obviously i do, i need that to prove that the force is linear
     
  13. Nov 22, 2009 #12

    D H

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    This happens to give the correct answer here. Hang the weight anywhere other than below points C and this will be wrong. There is however, a generic way to express the tensions in strings C and D in terms of the tensions in strings A and B.
     
  14. Nov 22, 2009 #13
    i got it thanks

    since [tex]\Delta[/tex]Y is linear, F=KY is also linear, therefore since the forces are equally placed,

    Ta=Ta
    Tb=2Ta
    Tc=3Ta
    Td=4Ta
    [tex]\Sigma[/tex]=10Ta=P

    Ta=P/10
    Tb=2P/10
    Tc=3P/10
    Td=4P/10
     
  15. Nov 22, 2009 #14

    D H

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    No, you don't have it! You got lucky here. You do not know ahead of time that Tb=2Ta. For example, if the weight was suspended from the middle of the rod, the tensions in all four strings would be equal.
     
  16. Nov 22, 2009 #15
    i know, i solved the next question where the weight was at 2/3L but i just recalculated the equations and managed to get it right,

    i found 2 equations with P in them, then found the ratio between mL and b, from there i can find the relations of Ta->Tb->Tc... then Ta+Tb+...=P

    from there its easy
     
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