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Finding the coordinate of a point by Law of Cosines/Sines

  1. Sep 21, 2015 #1
    1. The problem statement, all variables and given/known data
    A model for the suspension of a vehicle is shown where the spring has stiffness k = 178 N.mm and an unstretched length of 347 mm.

    Here is the picture: http://i.imgur.com/1dTVs12.jpg

    Part a asked to determine the value of P and the force supported by member AB so that the suspension was in equilibrium, and I got that right.

    However part b asks "If P = 0, determine the x and y coordinates of point A, namely xA and yA."

    2. Relevant equations
    Law of sines
    Law of cosines
    Basic trig

    3. The attempt at a solution
    I used the law of cosines to find all the angles
    A = 15.737 degrees
    B = 138.128 degrees
    C = 26.135 degrees

    I don't know where to go from here to find the coordinates, I feel like this is probably really easy but I'm over thinking it.

    Thank you so much for your help!
     
  2. jcsd
  3. Sep 21, 2015 #2

    SteamKing

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    You've calculated three angles, the distances AB and BC are shown in the diagram, and points B and C are fixed to the vehicle.

    If nothing else, you could make a sketch and find the third distance.

    In case you've forgotten how to apply the Law of Sines, here is a picture:

    law_of_sines.png
     
  4. Sep 21, 2015 #3

    RUber

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    Are you considering point B to be your origin?
    In that case, you can draw a right triangle with hypotenuse of length 229mm, and determine the angles (B-90), 90, (180-B). From there, the calculations are 229mm(sine ) and 229mm(cosine). Then be sure to orient your vectors in the proper direction.
     
  5. Sep 21, 2015 #4
    I did that, and I got the same lengths, I don't need to find the lengths I need to find where the coordinates are : /
     
  6. Sep 21, 2015 #5
    B is my origin because it is fixed in place and when the forces aren't exerted on AB or the spring the hinge swings open with B fixed in place. I dont get what you mean when you say
     
  7. Sep 21, 2015 #6

    SteamKing

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    You use the lengths of the sides and a little geometry to determine the coordinates. The (x,y) coordinate of point A is not just going to appear. You definitely should make a sketch of your triangle before working out the location of point A.
     
  8. Sep 21, 2015 #7

    RUber

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    You have the angles:
    B = 138.128 degrees.
    If you draw the x-axis, you will see that the right triangle formed with B, A, and the axis will have one angle measuring (138.128-90)degrees.
    The hypotenuse is equal to 229mm according to your drawing.
    Then use sine = opposite / hypotenuse to solve for the y component (negative) and cosine = adjacent/hypotenuse to solve for the x component.
     
  9. Sep 21, 2015 #8
    Sin(138.128)/347=Sin(26.135)/c
    c = -2869.62

    Sin(138.128)/347=Sin(15.737)/a
    a = 98.8653
     
  10. Sep 21, 2015 #9
    Sin(138.128)/347=Sin(26.135)/c
    c = -2869.62

    Sin(138.128)/347=Sin(15.737)/a
    a = 98.8653
     
  11. Sep 21, 2015 #10

    RUber

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    That number for c doesn't seem reasonable, and doesn't follow from the math. You are still using the law of sines to tell you what you already know...only now you are doing the calculations incorrectly. (Check degrees and not radians).
     
  12. Sep 21, 2015 #11
    Whoops it was radians, my bad.

    a = 141.001
    b = 228.998

    Which is exactly the information the problem already provides.
     
  13. Sep 21, 2015 #12

    RUber

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    Draw a right triangle.
    upload_2015-9-21_19-57-43.png
     
  14. Sep 21, 2015 #13

    RUber

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    What should the measure of the angle at B be?
    Call that angle m.
    sin(m) = opp/hyp
    cos(m) = adj/hyp
    Does that make more sense?
     
  15. Sep 21, 2015 #14
  16. Sep 21, 2015 #15

    RUber

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    Why would 48.128 be down there? Wasn't that part of angle B? 90+48.128 = 138.128.
     
  17. Sep 21, 2015 #16
    I don't know, I was trying to do what you mentioned before about (B-90), 90, (180-B)
     
  18. Sep 21, 2015 #17

    RUber

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    You shouldn't just blindly put the angles in. Use what you know. If the angle at B was 138.128, and you just keep the part that falls below the x-axis, how much is left?
    That part is still at point B in the diagram.
     
  19. Sep 21, 2015 #18
    I didn't think I put them in randomly... Did we cut the triangle in half somewhere? I'm still not sure how we got this right triangle.

    EDIT: So the part that is still at point B would be 138.128-90 right? Do I just have the numbers switched?
     
  20. Sep 21, 2015 #19
    This homework is due in an hour and is 10 out of the 30 points :frown: please help
     
  21. Sep 21, 2015 #20

    RUber

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    Your edit is right. We didn't move anything, just ignored the first 90 degrees of the angle to make a right triangle below.
     
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