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Tension question

  1. Oct 10, 2007 #1
    since there is a picture for this problem i am providing a link to view the problem

    http://i199.photobucket.com/albums/aa314/anglum/help2.jpg


    i solved for tension of T1 by taking the sin55=90/T1
    solved for T1 and converted to kg and got 11.211 kg which was incorrect

    since the remainder of the problems are dependent on having T1 right i am now stuck

    not sure what to do since the forces on the "knots" is equal to zero i used pythagorean theorem to get my answer since i knew the vertical force on T1 was 90N

    please help

    thank you
     
  2. jcsd
  3. Oct 10, 2007 #2
    i have to start solving all of the problems contained by first gettin T1 value correct?
     
  4. Oct 10, 2007 #3

    PhanthomJay

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    what about the T2 force at this knot? You've got to use Newton 1 in both the x and y directions to solve for the 2 unknown forces with the 2 equations.
    why are you converting a force to a mass unit?

    redo the T1 calc
     
  5. Oct 10, 2007 #4
    i am converting the force to Kg cuz part 1 of the problem asks for tension in T1 in units of kg

    i drew a vertical line down from teh top to the knot ... then knew the force of that had to be 90 N... thus sin55 = 90/T1 correctt???
     
  6. Oct 10, 2007 #5
    if i solve for T1 that way i get 109.8678N and that is incorrect
     
  7. Oct 10, 2007 #6

    PhanthomJay

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    No. You must isolate the knot and note that there is both a T1sin 55 and a T2sin 10 componnent in the vertical direction, one acting up and the other down, the sum total of which algebraically adds to 90 newtons. Look in the x direction as well and apply newton 1 again. You get 2 equations with 2 unknowns, which you can now solve for T1 and T2. I don't know why you would convert the tension to kilos, must be a misprint.
     
  8. Oct 10, 2007 #7
    so my equation to solve looks like this

    T1sin55 + T2sin10 = 90N ?????
     
  9. Oct 10, 2007 #8

    PhanthomJay

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    Watch your signs. T1 component acts up, T2 component acts down.
     
  10. Oct 10, 2007 #9
    ok so itd be T1sin55 - T2sin10 = 90N???? and that is just the vertical tension on that one

    and the horizontal would be T1cos55 - T2cos10 = 0????
     
  11. Oct 10, 2007 #10

    PhanthomJay

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    Yes, good, solve for T1 and T2, then move to the next knot.
     
  12. Oct 10, 2007 #11
    how am i supposed to solve for T1 and T2 ??? combine those equations???
     
  13. Oct 10, 2007 #12

    learningphysics

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    Yes, the equations are right. Solve for T2 in one equation and plug it into the other equation.
     
  14. Oct 10, 2007 #13
    if i solve for T2 in the 2nd equation i get -T2 = -T1cos55/cos10
     
  15. Oct 10, 2007 #14

    learningphysics

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    yup, so T2 = T1cos55/cos10, you can plug in cos55 and cos10...
     
  16. Oct 10, 2007 #15
    ok so then i get where T1 = X

    .81915X - .57357X/.98480 = 90 ?????
     
  17. Oct 10, 2007 #16

    learningphysics

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    you forgot to multiply by sin10
     
  18. Oct 10, 2007 #17
    o sooo

    .8195X - .57357X/.98480 (.173648) = 90??????
     
  19. Oct 10, 2007 #18

    learningphysics

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    looks right.
     
  20. Oct 10, 2007 #19
    so i then get .8159X -.57357x/.98480 = 90/.173648

    then i get

    .8159X - .57357X = (90/.173648) * (.98480)

    .24233X = 510.4118677

    T1 = 2106.267766 that cant be right??!?!
     
  21. Oct 10, 2007 #20
    my math has to be wayyyy off
     
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