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How to find the angle between the resultant and the X axis?

  1. Oct 25, 2010 #1
    1. The problem statement, all variables and given/known data
    How do you find the angle between the resultant and the x-axis? Please refer to the link in the picture. I just don't know how to get theta which is 10.26 degrees.


    2. Relevant equations
    https://www.physicsforums.com/attachment.php?attachmentid=29314&d=1287763774


    3. The attempt at a solution
     
  2. jcsd
  3. Oct 25, 2010 #2

    cepheid

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    I don't understand. Doesn't the diagram show you exactly how to do it? :confused:
     
  4. Oct 25, 2010 #3

    Integral

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    Use algebra to isolate sin( [itex] \theta [/itex]) then apply arcsin.
     
  5. Oct 25, 2010 #4
    The answer is wrong.
     
  6. Oct 25, 2010 #5
    Can you please elaborate?
     
  7. Oct 25, 2010 #6

    cepheid

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    No, 10.26 degrees is correct.

    Solve for sin(theta) from the equation that you get from the sine law. Once you know sin(theta), then you know theta, because you can apply the inverse sine (a.k.a. arcsine) on your calculator just as Integral mentioned.
     
  8. Oct 25, 2010 #7
    Yeah and i keep getting 10.16.
     
  9. Oct 25, 2010 #8

    cepheid

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    Can you post your calculation steps? Maybe we can see where you went wrong.
     
  10. Oct 25, 2010 #9
    15.02/sin(154)
    = 34.2632

    6.1/34.2632
    = .1780

    Sin^- .1780 = 10.26

    10.26 + 25 = 35.26 as my final answer
     
  11. Oct 25, 2010 #10
    deaninator is correct; the answer is wrong.

    Break the vectors down to their x- and y-components and solve it that way. You'll see that the resultant vector makes an angle of 40.75[itex]^\circ[/tex] with the x axis.

    The problem stems from the fact that the wrong angle is being found in the attachment. The angle you need to find is the one below the resultant vector, not the one above.

    Using the sine law, you would find this angle by:

    [tex]\frac{9.3}{sin \theta} = \frac{15.02}{sin 154^\circ}[/tex]
     
  12. Oct 25, 2010 #11
    then you add 25 degrees right?
     
  13. Oct 25, 2010 #12
    nvm sorry
     
  14. Oct 25, 2010 #13
    Hey do you know a site where I can find where I learn how to break down vectors to their x- and y- components?
     
  15. Oct 25, 2010 #14

    cepheid

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  16. Oct 25, 2010 #15
    cool website... i should find more of such kinds..
     
  17. Oct 26, 2010 #16
    Yes.

    25 degrees is the angle between the positive x-axis and the vector on the bottom of the parallelogram (the 6.1 magnitude vector).

    [itex]\theta[/tex] is the angle between that vector and the resultant vector.

    The sum of the 2 angles is the angle between the resultant and the x-axis.
     
  18. Oct 26, 2010 #17
  19. Oct 26, 2010 #18

    D H

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    Alternatively, they found the right angle but added it to the wrong angle. They should have calculated 51 degrees less 10.26 degrees = 40.74 degrees. (The reason they didn't get 40.75 degrees is because that 10.26 degrees should be 10.25 degrees.)
     
  20. Oct 31, 2010 #19
    Can you tell me how you got to to 40.74 please?
     
  21. Nov 1, 2010 #20
    Referring back to the diagrams in the attachment, you found the angle [itex]\theta[/tex] to be 10.26[itex]^\circ[/tex]

    That is the angle between the 9.3 magnitude vector (from the first diagram) and the resultant vector R. (Note that the angle between the 6.1 magnitude vector and the resultant vector is not the same.)

    Since the 9.3 magnitude vector makes an angle of 51[itex]^\circ[/tex] with the x-axis and the resultant vector is between that vector and the x-axis, then the resultant vector is 10.26[itex]^\circ[/tex] closer to the x-axis. Therefore the resultant vector makes an angle of 51[itex]^\circ[/tex] - 10.26[itex]^\circ[/tex] = 40.74[itex]^\circ[/tex] with the x-axis.
     
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