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Math Help - Tension on rope

  1. #1
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    Tension on rope

    Ropes 3m and 5m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 5kg. The ropes, fastened at different heights make angles of 52 degrees and 40 degrees with the horizontal. Find the tension in each wire and the magnitude of each tension.

    I started with an example in the book and then realized that the lengths of the rope where the same. This is what I did:

    T1 = -T1cos(52) + T1sin(52)
    T2 = T2cos(40) + T2sin(40)

    T1 + T2 = 5

    (-T1cos(52) + T2cos(40))i +( T1sin(52)+ T2sin(40))j = 5

    -T1cos(52) + T2cos(40) = 0
    T1sin(52)+ T2sin(40) = 5

    T2 = (T1cos52)/cos40

    T1sin(52)+ ((T1cos52)/cos40)sin(40) = 5

    T1 = 3.83kg

    T2 = ((3.83)cos52)/cos40
    T2 = 3.1kg

    and this is where I realized I forgot about the different lengths of the ropes. Can anyone help? I don't have any idea how to go about this problem.



    Thanks
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  2. #2
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    Quote Originally Posted by kenshinofkin View Post
    Ropes 3m and 5m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 5kg. The ropes, fastened at different heights make angles of 52 degrees and 40 degrees with the horizontal. Find the tension in each wire and the magnitude of each tension.

    I started with an example in the book and then realized that the lengths of the rope where the same. This is what I did:

    T1 = -T1cos(52) + T1sin(52)
    T2 = T2cos(40) + T2sin(40)

    T1 + T2 = 5

    (-T1cos(52) + T2cos(40))i +( T1sin(52)+ T2sin(40))j = 5

    -T1cos(52) + T2cos(40) = 0
    T1sin(52)+ T2sin(40) = 5

    T2 = (T1cos52)/cos40

    T1sin(52)+ ((T1cos52)/cos40)sin(40) = 5

    T1 = 3.83kg

    T2 = ((3.83)cos52)/cos40
    T2 = 3.1kg

    and this is where I realized I forgot about the different lengths of the ropes. Can anyone help? I don't have any idea how to go about this problem.



    Thanks
    The length of the rope is irrelevant. Lami's theorem gives the answer quickly:

    Lami's theorem - Wikipedia, the free encyclopedia.

    By the way, if you take the weight force as 5, the unit of force is kg wt, not kg.
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  3. #3
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    Quote Originally Posted by kenshinofkin View Post
    Ropes 3m and 5m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 5kg. The ropes, fastened at different heights make angles of 52 degrees and 40 degrees with the horizontal. Find the tension in each wire and the magnitude of each tension.

    I started with an example in the book and then realized that the lengths of the rope where the same. This is what I did:

    T1 = -T1cos(52) + T1sin(52)
    T2 = T2cos(40) + T2sin(40)

    T1 + T2 = 5

    (-T1cos(52) + T2cos(40))i +( T1sin(52)+ T2sin(40))j = 5

    -T1cos(52) + T2cos(40) = 0
    T1sin(52)+ T2sin(40) = 5

    T2 = (T1cos52)/cos40

    T1sin(52)+ ((T1cos52)/cos40)sin(40) = 5

    T1 = 3.83kg

    T2 = ((3.83)cos52)/cos40
    T2 = 3.1kg

    and this is where I realized I forgot about the different lengths of the ropes. Can anyone help? I don't have any idea how to go about this problem.



    Thanks
    I didn't check the numbers (or the algebra) but the setup looks good. By the way, tension is a force. What's the unit for that?

    -Dan
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  4. #4
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    Quote Originally Posted by kenshinofkin View Post
    Ropes 3m and 5m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 5kg. The ropes, fastened at different heights make angles of 52 degrees and 40 degrees with the horizontal. Find the tension in each wire and the magnitude of each tension.

    I started with an example in the book and then realized that the lengths of the rope where the same. This is what I did:

    T1 = -T1cos(52) + T1sin(52)
    T2 = T2cos(40) + T2sin(40)

    T1 + T2 = 5

    (-T1cos(52) + T2cos(40))i +( T1sin(52)+ T2sin(40))j = 5

    -T1cos(52) + T2cos(40) = 0
    T1sin(52)+ T2sin(40) = 5

    T2 = (T1cos52)/cos40

    T1sin(52)+ ((T1cos52)/cos40)sin(40) = 5

    T1 = 3.83kg

    T2 = ((3.83)cos52)/cos40
    T2 = 3.1kg

    and this is where I realized I forgot about the different lengths of the ropes. Can anyone help? I don't have any idea how to go about this problem.



    Thanks
    Assuming the system is in equilibrium :-

    F_y = T_1 sin52 + T_2 sin40 = 5g

    F_x = T_1 cos52 = T_2 cos40

    Solve these simultaneously and you you should get the tension. The length of the rope is not relevant to the calculation of tension.
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  5. #5
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    The length of the ropes doesn't matter for the tension but what about the magnitude of each tension? Would I multiply the vectors for each of them by their length?

    So

    T1 = -2.36i + 3.02j
    T2 = 2.37i + 2j

    would be

    T1 = 3(-2.36i + 3.02j)
    T2 = 5(2.37i + 2j)
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by kenshinofkin View Post
    The length of the ropes doesn't matter for the tension but what about the magnitude of each tension? Would I multiply the vectors for each of them by their length?

    So

    T1 = -2.36i + 3.02j
    T2 = 2.37i + 2j

    would be

    T1 = 3(-2.36i + 3.02j)
    T2 = 5(2.37i + 2j)
    The tensions do not depend on the length of the ropes. Period. End. Never. Finito. Nunca, nada, jamas. Nein. Nyet.

    Think about it. You have a tension in Newtons. That means a tension is a force, because it has the units of force. You want to multiply it by a length, which will give it a unit of Nm, a unit of energy.

    The only thing that the tensions have to do with the ropes is that they are directed along them. We occasionally will use the length of the rope to find the angle that the tension is at, but never for the magnitude.

    -Dan
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  7. #7
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    Quote Originally Posted by topsquark View Post
    The tensions do not depend on the length of the ropes. Period. End. Never. Finito. Nunca, nada, jamas. Nein. Nyet.

    [snip]
    -Dan
    You forgot bupkis
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