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Tension of two ropes given mass and angle

  1. Mar 18, 2009 #1
    1. The problem statement, all variables and given/known data

    A 1400 kg steel beam is supported by two ropes.

    2. Relevant equations

    Tx = mgsin(ANGLE) ??????????????????
    Ty = mgcos(ANGLE) ????????????????

    g = 9.8?

    Tension of one wire = [((Tx)^2) + ((Ty)^2)]^(1/2) ????????

    3. The attempt at a solution


    I wasn't too sure whether or not i should use the angle: 20 or 70.
    So i did it both ways they were both wrong, but i did it both ways...

    [((1400(9.8)sin(20))^2) + ((1400(9.8)cos(20))^2)]&(1/2)

    And then i did:

    [((1400(9.8)sin(70))^2) + ((1400(9.8)cos(70))^2)]&(1/2)

    Am i approaching this problem correctly? I have a test Thursday and i feel pretty F***** right now. Any help would be appreciated.
     

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  3. Mar 18, 2009 #2

    LowlyPion

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    I can't see your picture as yet.

    But resolving angles is no sweat. You always want the component in the direction of the axes, and however your angle is defined - relative to the horizontal or the vertical - you want the angle function determined by your drawing. Side opposite is sine, and side adjacent is cosine.

    As to your method I think you need to think about what you are doing.

    Most importantly remember for statics that the ∑ F in the x are 0 and the ∑ F in the y are 0.

    Merely take the components of your tensions in the x and they add to 0. Now the components in the Y are supporting the beam. so those components of tension must add to the weight.

    ∑ Ty = T1y + T2y = m*g

    For the x ∑ Tx = 0 then T1x = - T2x
     
  4. Mar 18, 2009 #3
    So the the total Ty should equal 13720N correct?

    m*g
    1400*9.8 = 13720N ??

    How would i go about finding the tensions of the individual ropes?

    I was thinking that:

    13720(sin20)) = gives you the tension of rope 1 with an angle of 20 degrees, and 13720(sin30)) = gives you the tension of rope 2 with an angle of 30 degrees.

    Am i approaching this correctly?
     
    Last edited: Mar 18, 2009
  5. Mar 18, 2009 #4

    LowlyPion

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    If the angles are with the horizontal then yes, the sum of T1*Sin(20) and T2*Sin(30) will yield the m*g.

    Now write an equation for the x components.

    Then all you need to do is solve for T1 and T2 directly.

    2 equations and 2 unknowns means the end of the voyage is in sight.
     
  6. Mar 18, 2009 #5
    T1sin(20) + T2cos(30) = 0 for the x-components?

    I'm a little confused on how to go further from here...

    The angles i believe are with the vertical axis, they are in the picture as rope 1 being 20 degrees to the left of the vertical y axis and rope 2, 30 degrees to the right of the vertical y axis.
     
  7. Mar 18, 2009 #6

    LowlyPion

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    Sorry I can't see your picture. I previously said if the angles were with the horizontal. So if they are with the vertical then simply swap sin and cos. Sin is the x components then and cos is vertical.
     
  8. Mar 18, 2009 #7
  9. Mar 18, 2009 #8

    LowlyPion

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    Got it. So ...

    T1*Cos(20) and T2*Cos(30) = m*g.

    And

    T1*Sin(20) = -T2*Sin(30)
     
  10. Mar 18, 2009 #9
    Shouldn't it be:

    T1*Sin(20) - T2*Sin(30) = 0

    then...
    T1*Sin(20) = T2*Sin(30)

    then i did....

    T1 = [(T2*Sin(30))/(Sin(20))]

    using...
    T1*Cos(20) + T2*Cos(30) = m*g.

    Replace T1, with what i found earlier and...
    [(T2*Sin(30))/(Sin(20))]*Cos(20) + T2*Cos(30) = m*g.

    Pull out T2.....

    T2([(Sin(30))/(Sin(20))]*[Cos(20)] + Cos(30)) = 13720 N

    T2 = 13720 N / {([(Sin(30))/(Sin(20))]*[Cos(20)] + Cos(30))}

    T2 = 6125.64 N ?????????


    I know its wrong, i am just not sure where i am messing up.
     
    Last edited: Mar 18, 2009
  11. Mar 18, 2009 #10

    LowlyPion

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    I think you are already anticipating the direction of the Tension.

    It's certainly true that |T1*Sin(20)| - |T2*Sin(30)| = 0

    But the x components are opposed and hence not interchangeable if you take their directionality into account as

    T1*Sin(20) = T2*Sin(30) might imply.

    Ultimately you are really only interested here in |T1| = |T2|*Sin30/Sin20 for solving for T1 and T2, so I suppose it doesn't matter here.
     
  12. Mar 18, 2009 #11
    I'm a little confused. Because i think what your saying is that i am doing it correctly? But when i solved out my answer for T2 the online system says I am wrong.
     
  13. Mar 18, 2009 #12

    LowlyPion

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    As for your calculation

    I'd put it in numbers. It's easier for me to follow.

    T1 = 1.462 T2

    Then

    1.462 T2 * Cos20 + T2 * Cos 30 = 1.374T2 + .866 T2 = 1400*9.8 = 13720

    T2 = 6125 (Which is what you got.)
    T1 = 1.462*6125 = 8955
     
  14. Mar 18, 2009 #13

    LowlyPion

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    Yes it looks like you are ok. Maybe you need to reverse the values for T1 and T2 in the answering system?
     
  15. Mar 18, 2009 #14
    I'm an idiot. I was putting the T2 answer in the T1 box. LOLLL
     
  16. Mar 18, 2009 #15
    I really appreciate your help. Thank You.
     
  17. Mar 18, 2009 #16

    LowlyPion

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    Sure.

    Cheers.
     
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