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Force along u and v axes

  1. Dec 23, 2015 #1
    1. The problem statement, all variables and given/known data
    in this question , why cant i find Fu = 600cos 30= 520N ?
    why cant I find Fv = 600cos 120 = -300N ?

    2. Relevant equations


    3. The attempt at a solution
    Fu = 600cos 30= 520N ?


    Fv = 600cos 120 = -300N ?
     

    Attached Files:

  2. jcsd
  3. Dec 23, 2015 #2

    SteamKing

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    Well, for one thing, it appears that Fu and Fv are not the sides of a right triangle. It's hard to tell from the attached image.

    For solving oblique triangles, the law of sines or the law of cosines must be used. For more information, see the article below:

    https://en.wikipedia.org/wiki/Solution_of_triangles
     
  4. Dec 24, 2015 #3
     
  5. Dec 24, 2015 #4
    Why can't I do in this way??
     
  6. Dec 24, 2015 #5

    SteamKing

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    As has already been explained, the forces Fu and Fv are not the sides of a right triangle.

    The triangle which contains Fu and Fv has angles of 120°, 30°, and 30°. Do you see any right angles there?

    Haven't you learned the difference yet between an oblique triangle and a right triangle?

    The solution in your text clearly shows the use of the Law of Sines.

    Please study the material I linked to previously. If you have any questions about that, I will be happy to answer them.
     
  7. Dec 24, 2015 #6

    gneill

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    As was mentioned already, The u and v axes are not orthogonal (do not meet at a right angle). So you can't treat them as if they do.

    If the using the laws of sines or cosines is throwing you off, you could approach the problem as two equations in two unknowns.

    Fig1.png

    Suppose that Fu and Fv are two vectors with the given directions and you need to find Fu and Fv such that they sum to a 600 N horizontal resultant. Impose axes x-y on the image such that the x-axis coincides with the horizontal 600 N force. Then write equations summing the x and y components of Fu and Fv accordingly.
     
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