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Acceleration with tension

  1. Aug 13, 2013 #1
    I have attached my calculations.
    In the problem do they mean acceleration from K to Y?
    Where did I go wrong in my calculations?
    THANKS IN ADVANCE
     

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  2. jcsd
  3. Aug 13, 2013 #2
    You did not go wrong anywhere, but you did not finish. You have some angle in your formula, while the expected result does not have any.
     
  4. Aug 13, 2013 #3
    I have Tx=(FCOSα)/(2SINα)=(F)/(2TANα)=[F(L2-X2)0.5)]/2X
    I have the wrong answer:(
     
  5. Aug 13, 2013 #4
    You obviously used the wrong sides of the triangle to compute the tangent function. Think about that again. If you come to the same result again, post your thoughts here, and we will discuss.
     
  6. Aug 13, 2013 #5
    Why the wrong sides?
     

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  7. Aug 13, 2013 #6
    Which side is x? How is it related with the angle?
     
  8. Aug 13, 2013 #7
    tanα=x/(L2-X2)0.5
     

    Attached Files:

  9. Aug 13, 2013 #8
    x is the perpendicular distance from the line of F. Is that the case on your diagram?
     
  10. Aug 13, 2013 #9
    Thus it is going to be tanα=(L2-X2)0.5/X? With 2Tsinα=F it seems to give the right answer, however in the "student solution" they refer to that problem as having 2Tcosα=F, Am I right this time?
    Does it matter where I decide to place α ?
     
    Last edited: Aug 13, 2013
  11. Aug 14, 2013 #10
    1. From the start x=L thus is the acceleration infinite?
    2. Is ax changing while F is constant?
     
  12. Aug 14, 2013 #11
    The notation in your working hurts my brain.
     
  13. Aug 14, 2013 #12
    Obviously, there are two angles that you can use, but the end result should be the same.

    Nothing ever gets infinite in the real world. If you get something infinite in a problem, that means that its mathematical model becomes invalid.

    It depends on the angle, so, yes.
     
  14. Aug 14, 2013 #13
    voko, I'm grateful to you for your help!
     
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