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Tension HW problem.

  1. Feb 11, 2006 #1
    alright, i'm having trouble finding a formula to answer this problem.


    a 58N weight is being suspended by 2 cables. the cable on the left side is horizontal, while the other cable is at a 60' angle.

    What is the tension of the right side cable? (T1)

    What is the tension of the left side cable, the horizontal one? (T2)


    I started off by trying to find the sum of all the forces, but I'm already stuck.
    I have;

    {Fx = T2 - T1cos60 = 0
    and
    {Fy = T1sin60 - 58N = 0

    is this right so far, or do i have the cos and the sin switched?
     
  2. jcsd
  3. Feb 11, 2006 #2

    Astronuc

    User Avatar

    Staff: Mentor

    That depends on whether the 60° is with respect to horizontal or vertical.

    If the angle is with respect to the horizontal, then your equations are correct.
     
  4. Feb 12, 2006 #3
    [​IMG]

    that's the problem.

    i've done this;

    {Fx = T2 - T1cos60 = 0
    and
    {Fy = T1sin60 - 58N = 0

    then to find T2 i thought i could do this;

    T1sin60 = 58N
    to
    T1=(sin^-1)60 + 58N

    but the answer is apparently wrong... so i must not have done it right.
     
  5. Feb 12, 2006 #4

    Astronuc

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    Staff: Mentor

    T1sin60 = 58N is correct.

    T1=(sin^-1)60 + 58N is not correct.

    T1 * sin 60° = 58N => T1 = 58N / (sin 60°)

    Try T2 = T1 * cos 60°
     
  6. Feb 13, 2006 #5
    ahhh i totally forgot i could just divide both sides by it

    insted i was doing something else i learned recently.... damnit so many formulas!!! lol

    thank you a lot.
     
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