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3D Physics problem(Forces and Tension)

  1. May 24, 2015 #1
    1. The problem statement, all variables and given/known data

    metal piece = 300N
    As this metal piece is in balance, what is the tension in each cable?
    Picture of the problem in attachment.
    2. Relevant equations


    3. The attempt at a solution


    W = 300*[0 0 1]

    D to A:
    vtda = tda [sin(37), sin(37)*sin(90), -cos(37)]

    D to B:
    vtdb = tdb [sin(37)*sin(90), sin(37), - cos(37)]

    D to C
    vtdc = tdc [cos(30)*sin(37), -sin(30)*sin(37), -cos(37)]

    solve([0 0 0]=300*[0 0 1])+vtda+vtdb+vtdb, {tda,tdb,tdc}

    I have been trying to find the mistake for 3 hours but I really cant find it :/
    When I solve with my TI calculator I get a "false"

    thank you very much
     

    Attached Files:

  2. jcsd
  3. May 24, 2015 #2
    Let the tensions in the 3 cables be ##T_1##, ##T_2## and ##T_3## .
    It is clear that you'll need a system of 3 equations to solve this problem. Luckily, the coordinate system is set-up in such a way that the z axis is the perpendicular from the surface (it is an assumption that I'm making after looking at the diagram, and I'm 99% sure that its the actual case).

    You already know that the metal piece is in equilibrium. This means that the resultant of the x and y components of the tensions equals 0, whereas the resultant of the z components of the tensions equals to the weight of the piece. Can you set up the required 3 equations?

    One tip: always set up the algebraic/vector equation first, and substitute the numbers later. That way, it's always easier to spot any mistakes.
     
    Last edited: May 24, 2015
  4. May 24, 2015 #3

    BvU

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    Hi chief,

    1. What is this funny language you are writing in ?
    2. What are the units of angle in your TI thingy? degrees or radians ?
    3. 37 is approximate, 3/5 for the sine is exact. idem 4/5 for the cos.
    4. z is downwards, your coordinate system is left-handed ? Bad habit!
    5. In the x-direction vtda and vtdc pull in different directions. I don't see that in your expressions
    6. sin(90) = 0. Easier to use the shorter form ? Less insightful, I admit.

    [edit] slow typist
    @PWiz: solve(...) appears to be solving (almost) exactly the three equations you describe. El Jefe does the right thing, he/she doesn't do the thing completely right, though.
     
  5. May 24, 2015 #4
    its in degree
    I am using directional cosine.

    Also is 300*[0 0 1] correct since Z is downward?
     
  6. May 24, 2015 #5
    Yes but ropes are not symmetric around, so z components are not equals.
     
  7. May 24, 2015 #6
    z is downward? how would they be different? only x and y has to be different for each rope.
     
  8. May 24, 2015 #7
    see again angles and write the correct vectors.
     
  9. May 24, 2015 #8
    sorry I really dont understand how Z is different to each of the rope, z is same direction for all of them since its downward. and the angle between each rope and the middle line is equal to every rope.
     
  10. May 24, 2015 #9

    BvU

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    Dear chief,
    You are not correct. Z components follow once you found the components in the xy plane.

    To get going. do pay attention to my remark #5.
     
  11. May 24, 2015 #10
    to make zero x and y components the tension on C must be larger than other two.
     
  12. May 24, 2015 #11
    yes I fixed that on my calculator put still get false as an answer.
     
  13. May 24, 2015 #12

    BvU

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    Theo, chief has allowed the tensions to be different!
     
  14. May 24, 2015 #13
    W = 300*[0 0 1]

    vtda = tda [sin(37), sin(37)*sin(90), cos(37)]

    vtdb = tdb [sin(37)*cos(90), sin(37), cos(37)]

    vtdc = tdc [-cos(30)*sin(37), -cos(60)*sin(37), cos(37)]

    solve([0 0 0]=300*[0 0 1])+vtda+vtdb+vtdb, {tda,tdb,tdc}
     
  15. May 24, 2015 #14
    Your writing notation is rather difficult to follow (it's almost as if you're posting your calculator input). Can you change this?
     
  16. May 24, 2015 #15
    thats how we have learned to write notation on paper, I am not sure how to make this more easy to understand :(
     
  17. May 24, 2015 #16

    BvU

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    Looks good to me. I have no problem solving by hand.
    Don't know the proper TI incantation language.
     
  18. May 24, 2015 #17
    Well you could perhaps call the origin of the coordinate system O and the bottom of the hook D.
    Then you can describe angles much more easily. For example, cos (DAO) = 0.6 , etc.
    I don't know if it's just me, but this is how I learned geometric description in middle school :woot:
     
  19. May 24, 2015 #18

    BvU

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    How about
    solve([0 0 0]=300*[0 0 1]+vtda+vtdb+vtdb), {tda,tdb,tdc} instead of

    solve([0 0 0]=300*[0 0 1])+vtda+vtdb+vtdb, {tda,tdb,tdc}
     
  20. May 24, 2015 #19

    BvU

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    Oh, and if W is pulling down, it's only fair to have vtda, vtdb and vtdc pull upwards. Since 4/5 is positive, I suggest

    solve([0 0 0]=300*[0 0 ##\bf -##1]+vtda+vtdb+vtdb), {tda,tdb,tdc}
     
  21. May 24, 2015 #20

    BvU

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