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Difficult optimal cable cost analysis

  1. Feb 1, 2013 #1
    1. The problem statement, all variables and given/known data


    I was going through some old course work from when I was an undergrad and was unable to complete this one. If you could just provide me with the proper approach/steps it would be much appreciated. I know how to solve all of the tensions, but the cost analysis approach is confusing me.

  2. jcsd
  3. Feb 1, 2013 #2


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    hi algar32! :wink:
    you should get an expression for the length, and an expression for the tension

    multiply them together, that gives you the cost (as a function of x y and z, the coordinates of B)

    then minimise it, probably using calculus …

    show us what you get :smile:
  4. Feb 1, 2013 #3
    How do I get an expression for length and an expression for tension? That is the part I am struggling with. Also, if I make the tension expression won't it have 2 variables in it? How can I handle that if I am trying to rref it? Any further steps/work would be appreciated.Thanks.
  5. Feb 1, 2013 #4


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    Well, the expression for the length is going to be a function of the unknown x and z position of the point B. You may not know what else if you don't write some equations.
  6. Feb 1, 2013 #5
    Last edited: Feb 1, 2013
  7. Feb 2, 2013 #6


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    ?? :confused:

    can you type it out, please?​
  8. Feb 2, 2013 #7
    You are making it more difficult than it has to be. You know that the forces of tension in the cables act along the cables, so ## \vec{T}_{AB} = b \vec {AB} = b(x, 4, z) ##, ## \vec{T}_{AC} = c \vec {AC} = c(-2, 4, -2) ##, ## \vec{T}_{AD} = d \vec {AD} = d(-3, 4, 3) ##, where ##b##, ##c## and ##d## are some positive constants. Letting ## \vec{T} = (0, -mg, 0) ##, we must have ## \vec{T}_{AD} + \vec{T}_{AC} + \vec{T}_{AB} + \vec{T} = 0 ##. Write this equation component-wise, and use Gauss's elimination to determine b in terms of x and z.
  9. Feb 2, 2013 #8
    Okay thanks... Is this what you mean?

    b(x,4,z) + c(-2,4,2)+d(-3,4,3)+(0,-981,0)=0

    I don't see how I can solve that for x and z? Could you show this process? Was I supposed to put in any number (let's say 1) for b, c and d?

    By doing this I will have the most cost effective rope based on the equation givin in the problem (length*tension=cost)?

  10. Feb 2, 2013 #9
    You don't solve for x and z. You solve just for b as a function of x and z.
  11. Feb 2, 2013 #10
    You say to use gaussian elimination, so I have just been setting up a few matrices and plugging them into the calculator and using rref (reduced row ef). I am still not getting a value for x and z. Would it be possible to show me what matrix I should be using? thanks.
  12. Feb 2, 2013 #11
    I am unsure how to do this. Thanks.
  13. Feb 2, 2013 #12
    b(x,4,z) + c(-2,4,2)+d(-3,4,3)+(0,-981,0)=0 is a system of three linear equations, where the unknowns are b, c and d (assume x and z are known). Use Gauss's elimination to find b, it will be a function of x and z.
  14. Feb 2, 2013 #13
    Just to be clear. This has to be done manually, calculators won't do it. I recommend rewriting the equation as c(-2,4,2) + d(-3,4,3) + b(x,4,z) = (0,981,0), so that the b-terms are in the right-most column of the matrix, that will simplify finding b, requiring just two Gaussian steps.
  15. Feb 2, 2013 #14
    Thanks. Here is my work:

    x 4 z | 0 b
    -2 4 2 | -981 c
    -3 4 3 | 0 d


    x 4 z | 0 b
    0 2 0 | -1471.5 c
    -3 4 3 | 0 d

    b - 2c

    x 0 z | 2943 b
    0 2 0 | -1471.5 c
    -3 4 3 | 0 d

    (x + z) b = 2943

    What would my final answer be to optimize the costs?
  16. Feb 3, 2013 #15
    Even assuming that the system is written down properly, I do not see how your steps would result in anything. I think you should review systems of linear equations.

    Re writing down the system, observe that the equation c(-2,4,2) + d(-3,4,3) + b(x,4,z) = (0,981,0) is a vector equation; the entities in brackets are vectors, while the coefficients in front of them are numbers. A vector equation means that the same equation holds for each of its components. For example, taking the first component (X-direction), we obtain -2c - 3d + xb = 0. This is a linear equation for c, d, and b (x is assumed known). Writing the equations for the two other components, we obtain a system of three linear equations.
  17. Feb 3, 2013 #16


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    One interesting outcome is x and z are always related by


    Unless having a tensionless chain is ok.
  18. Feb 3, 2013 #17
    I just noticed that in #8 and later the equation is not correct. It should be c(-2,4,-2) + d(-3,4,3) + b(x,4,z) = (0,981,0).
  19. Feb 3, 2013 #18


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    Yup. That was key. Now the weird x=-z constraint disappears.

    We get:


    Length^2 = x^2 + z^2 +16

    Therefore we minimise the product:

    (x^2 + z^2 +16)/(5*x+z+12)


    Not sure if this is correct.
    Of course, I cheated using Wolfram. Analytically is another challenge. I'd love to see that minimization done by hand.
    Last edited: Feb 3, 2013
  20. Feb 3, 2013 #19


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    Not even sure how one could do those minimizations by hand. It'd require solving simultanously two quadratics in x and z.
  21. Feb 3, 2013 #20
    That result is correct. Of course, the report requested must prove that this really is the point resulting in a minimal cost.
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