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Tension in the cables

  1. Jan 24, 2007 #1
    1. The problem statement, all variables and given/known data
    Ok, I've got three cables all pointing in a downward direction towards the orgin. They all connect at the orgin, onto a 750lb weight. Determine the tensions in A,B,C.

    I'm going to put these into (x,y,z) form, so you get the picture.

    A = (4,-2,5)
    B = (-3,-3,6)
    C = (-4,3,4)

    2. Relevant equations

    3. The attempt at a solution

    Ok, I've tried this...not sure if its right, but seems right to me. And...if so, I'm stuck at the point of how to solve the 3 equations and 3 unknowns. I can't remember how to do that. Ok here goes...I hope this isn't complicated.

    R = sum(F) = Ta + Tb + Tc = 0

    Ta = Ta(4i - 2j +5k/sqrt(45))

    Tb = Tb(-3i - 3j +6k/sqrt(54))

    Tc = Tc(-4i + 3j +4k/sqrt(41))

    Skip a step here.....

    R = (0.5963Ta - 0.4082Tb - 0.6247)i + (-0.2981Ta - 0.4082Tb + 0.4685Tc)j + (0.7454Ta + 0.8165Tb +0.6247Tc)k

    Sum(Fx) = 0.5963Ta - 0.4082Tb - 0.6247Tc = 0
    Sum(Fy) = -0.2981Ta - 0.4082Tb + 0.4685Tc = 0
    Sum(Fz) = 0.7454Ta + 0.8165Tb + 0.6247Tc = 0

    I'm stuck here. Also...I'm wondering....in my sum(Fz), I think this is where I'd subtract 750....as it is pointing down in the z direction. Is this correct?

    Thanks in advance
  2. jcsd
  3. Jan 24, 2007 #2
    Is this understandable? And if so, can someone please remind me how to solve those equations
  4. Jan 24, 2007 #3


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    Homework Helper

    What is the direction of the resultant force? Is x, y and z all three components are zero?

    What do you think about 750 lb weight?
  5. Jan 25, 2007 #4
    I'm not positive which way the resultant is pointing. By looking at the diagram, I'd say possibly in the positive z direction. As, the tension forces in the cables are pointing up/out from the orgin, right??

    The tensions are not moving in the x nor y direction. I don't think it would be moving in the z direction either....but is the z component equal to the 750lb weight?
  6. Jan 25, 2007 #5
    So is my form correct up to where I am......and if so, can you please help me remember how to solve this equations for each(Ta,Tb,Tc)? Or, point me in the direction of a website that will help me. Thanks
  7. Jan 26, 2007 #6
    Any help on solving simultaneous equations like this. I can NOT figure it out.....have been working on it for about a day now. It just won't come to me. Thanks....
  8. Jan 26, 2007 #7

    Doc Al

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    Staff: Mentor

    I didn't check your arithmetic, but your approach looks OK except that you left out the 750 lb weight!
    Assuming that gravity acts in the -z direction, then that's where you had better add it your sum of forces calculation.

    One way to solve simultaneous equations is just to start eliminating variables. Pick one equation and isolate one of your variables. For example, find Ta in terms of Tb & Tc from one equation, then substitute that into the other two equations. Now you have two equations and two unknows, Tb and Tc. Repeat the process and you'll have one equation and one unknown. Once you find one unknown, use it to find the others by working backward.

    Tedius? Yes. But it works. (With a bit of practice, you can be even more clever.)

    Here's a page describing several methods: http://www.themathpage.com/alg/simultaneous-equations.htm" [Broken]
    Last edited by a moderator: May 2, 2017
  9. Jan 26, 2007 #8
    Thank You Doc Al.......I've been needing that for some time now. I'll be working hard on this today, to get it done. I'm confident I know how to do it now(or I'm going to give it a good run!!!). Thank you very much for you speedy reply!!!

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