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Math Help - Cable Tension

  1. #1
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    Cable Tension

    Cable tension question

    Triangle with angle:

    A is 50 degree
    B is 30 degree

    2000lb attached to it.


    My setup:

    T1= cos50
    T2=cos30


    T1sin50 + t2 sin30 = 2000

    solving for T1

    T2= T1 cos50/cos30

    T1 ( sin50 + (cos50/cos30)sin30) = 2000


    Is my set up correct?
    Last edited by NeoSonata; June 5th 2011 at 05:44 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by NeoSonata View Post
    Cable tension question

    Triangle with angle:

    A is 50 degree
    B is 30 degree

    2000lb attached to it.
    What is the orientation of the triangle? Where is the 2000 lb attached to it?

    T1= cos50
    T2=cos30
    These cannot possibly be correct. Did you mean something along the lines of T1 = 2000*cos(50)? etc.

    -Dan
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  3. #3
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    Triangle is facing down w/ 2000lb attached to it.

    A 50 degree __ B 30 degee
    \/
    2000lb


    |t1|cos50 = |t2|cos30



    Equation is
    |t1|sin50 + |t2|sin30 = 2000


    |t2| = |t1| cos50/cos30

    plug this in to the quation above


    |t1| ( sin50 + (cos50/cos30)sin30 ) = 2000

    |t1| = 1758.77

    Is t1 correct?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by NeoSonata View Post
    Triangle is facing down w/ 2000lb attached to it.

    A 50 degree __ B 30 degee
    \/
    2000lb


    |t1|cos50 = |t2|cos30



    Equation is
    |t1|sin50 + |t2|sin30 = 2000


    |t2| = |t1| cos50/cos30

    plug this in to the quation above


    |t1| ( sin50 + (cos50/cos30)sin30 ) = 2000

    |t1| = 1758.77

    Is t1 correct?
    Looks good to me.

    -Dan
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  5. #5
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    e^(i*pi)'s Avatar
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    Are you using lb as a unit of mass or force?

    It will need to be the latter to work properly

    (I come from Europe so I'm used to kg )
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