# Difficult optimal cable cost analysis

1. Feb 1, 2013

### algar32

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

http://snag.gy/z4LzZ.jpg

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.

Thanks

2. Feb 1, 2013

### tiny-tim

hi algar32!
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

3. Feb 1, 2013

### algar32

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.

4. Feb 1, 2013

### SteamKing

Staff Emeritus
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.

5. Feb 1, 2013

### algar32

Last edited: Feb 1, 2013
6. Feb 2, 2013

### tiny-tim

??

can you type it out, please?​

7. Feb 2, 2013

### voko

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.

8. Feb 2, 2013

### algar32

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)?

Thanks.

9. Feb 2, 2013

### voko

You don't solve for x and z. You solve just for b as a function of x and z.

10. Feb 2, 2013

### algar32

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.

11. Feb 2, 2013

### algar32

I am unsure how to do this. Thanks.

12. Feb 2, 2013

### voko

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.

13. Feb 2, 2013

### voko

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.

14. Feb 2, 2013

### algar32

Thanks. Here is my work:

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

(3c/2)-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?

15. Feb 3, 2013

### voko

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.

16. Feb 3, 2013

### rollingstein

One interesting outcome is x and z are always related by

x=-z

Unless having a tensionless chain is ok.

17. Feb 3, 2013

### voko

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).

18. Feb 3, 2013

### rollingstein

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)

(x,z)=(2.243,0.449)

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
19. Feb 3, 2013

### rollingstein

Not even sure how one could do those minimizations by hand. It'd require solving simultanously two quadratics in x and z.

20. Feb 3, 2013

### voko

That result is correct. Of course, the report requested must prove that this really is the point resulting in a minimal cost.