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Adding vectors

  1. Aug 26, 2009 #1
    1. The problem statement, all variables and given/known data
    300 lb force is to be resolved into components along a-a(prime) and b-b(prime).
    a.) Determine the angle by trigonometry knowing that the compnent along line a-a(prime) is to be 240 lb.

    b.) What is the corresponding value of the component b-b(prime)

    Statics1-1.jpg

    in the picture i changed a-a(prime) to x-x(prime) and b-b(prime) to y-y(prime)

    2. Relevant equations
    sin(theata 1)/A=sin(theata 2)


    3. The attempt at a solution

    I dont know how to do this seeing as it is tilted!?!
     
  2. jcsd
  3. Aug 26, 2009 #2

    Doc Al

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    I'm having trouble understanding the problem and the diagram. I see a vector, a horizontal line label x at one end and x' at the other, and a tilted line labeled y' at one end. Can you restate the problem?
     
  4. Aug 26, 2009 #3
    i took a different picture
     
  5. Aug 26, 2009 #4
    This is straight from the book...........

    "The 300-lb force is to be resolved into components along the lines a-a' and b-b'

    a.) Determine the angle alpha by trigonometry knowing that the component along line a-a' is to be 240-lb

    b.0What is the corresponding value of the component along b-b'
     
  6. Aug 26, 2009 #5

    Doc Al

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    Where is it?
     
  7. Aug 26, 2009 #6
  8. Aug 26, 2009 #7
    sorry it took me a while to upload it to photobucket, then onto here sorry
     
  9. Aug 26, 2009 #8
    does this help?
     
  10. Aug 26, 2009 #9

    Doc Al

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    Yes--it's clear now.

    Since you're given the component of the force along a-a', what must alpha be? That's step one. Then you'll have all the angles needed.
     
  11. Aug 26, 2009 #10
    Ok, i thought the angle of a-a' would have been 180 since its a straight line. But then its wrong. Then i thought maybe its 120 but then after working through it again it is wrong.
     
  12. Aug 26, 2009 #11

    Doc Al

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    Answer this: How would you find the x-component of a vector making an angle of alpha with the x-axis? It's the same problem.
     
  13. Aug 26, 2009 #12
    cos(alpha)=a-a'/300

    is that what you are asking?
     
  14. Aug 26, 2009 #13
    which would make alpha=36 deg which is wrong.
     
  15. Aug 26, 2009 #14

    Doc Al

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    I would write it as:
    Fa-a' = F cos(alpha)
    240 = 300 cos(alpha)
     
  16. Aug 26, 2009 #15
    right so alpha=cos^-1(.8)=36 deg, the anwser is 76.1 deg.....which doesn't make sense
     
  17. Aug 26, 2009 #16

    Doc Al

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    cos-1(240/300) = 36.9 deg

    Why do you say that's wrong?
     
  18. Aug 26, 2009 #17
    because in the back of the book it says its 76.1
     
  19. Aug 26, 2009 #18

    Doc Al

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    What's the answer given for b? Is it consistent?
     
  20. Aug 26, 2009 #19
    the answer for b is 336 lb
     
  21. Aug 26, 2009 #20

    Doc Al

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    OK. I understand what they want. The axes are not orthogonal, thus to find the a-a' component you must draw a line parallel to b-b' that intersects the tip of the 300-lb vector. You'll get a triangle, two sides of which are given (240 and 300). You'll be able to use some trig to find alpha. (The answer is correct, now that I understand it. :uhh:)

    (Taking Fcos(alpha) is only good for orthogonal coordinates, not skewed. Sorry about that!)

    And to then find the b-b' component, you'll draw a line parallel to a-a' and get another triangle.
     
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