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Homework Help: Wire tensions in the diagram?

  1. Sep 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi, id like to know how to find the tensions in the diagram below.


    I keep getting the wrong answer, the right answers are 181 and (i dont know the other)

    How do I get the answers?

    2. Relevant equations


    3. The attempt at a solution

    I keep getting 686 for one of the answers
  2. jcsd
  3. Sep 14, 2010 #2
    Draw a force triangle, Ft3 must be equal to the sum Ft1+Ft2. You know the directions of Ft1 and Ft2 and you know the length and direction of Ft3. So you get a triangle of which you know the angles and the length of one side, so you can find the length of the other sides.
  4. Sep 14, 2010 #3


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    Gold Member

    Hello ConstableZiM,

    Yes, one of the answers is ~181 N. I don't know where you got the 686 thought. Show us your work and maybe we can point you in the right direction. :smile:

    [Edit: gerben beat me to the post.]
  5. Sep 14, 2010 #4
    Ok, so this is what happened, I got the answer from the board, but the steps were not given... So I know one of the answers.

    Ive been trying to figure it out for a long time now... heres what I did.

    T2Sin30+T1Sin40 = 196
    T1 = T2 + 196
    So, T2Sin30 + T2 Sin40 - 196 Sin40 = 196
    T2 = 281.8 <<< My latest attempt... Which is closer, but still wrong...

    I dont really know what the length of T3 is gerben, I know the force is -196 N... This is the first time I've ever taken physics... So I'm still not used to the concepts and stuff...
  6. Sep 14, 2010 #5
    The length of the force vector is just the amount of force (so the length of Ft3 is mg).
    The direction of the force vector is the direction in which the force pulls.
  7. Sep 14, 2010 #6
    THANKS! I get it now... I now know where the problem is, I substituted T1 into the equation from t1+t2+t3=0... It should have worked though, if that equation is true... I used the T1 value from the T2cos40=T1cos30 to substitute into T1 now... Why didn't it work with the other equation? Does the t1+t2+t3=0 equation only work with the vertical forces?

    Again, thanks for the help.
  8. Sep 14, 2010 #7
    The T1, T2 and T3 forces are not in the same direction, so you should not just add their lengths. You should add the 'components' of the forces in the same direction:

    Notice that T3 pulls only vertically down, while T1 and T2 both pull a bit upwards but also a bit sidewards.

    The amount that T1 pulls upwards (T1sin30) plus the amount that T2 pulls upwards (T2sin40) should be equal to the amount that T3 pulls downwards (mg). So, this gave you equation: T1Sin30+T2Sin40 = 196.

    The amount that T1 pulls to the left (T1cos30) should be equal to the amount that T2 pulls to the right (T2cos40). So, this gave you equation: T1cos30 = T2cos40.
  9. Sep 14, 2010 #8
    Thanks, I literally almost lost my head over this... Reeeaally helped me...
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