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Law of sine/cosines to find resultant force

  1. Dec 18, 2007 #1
    1. The problem statement, all variables and given/known data
    So I am using Law of sine/cosines to find resultant force R and its direction.

    [​IMG]

    My teacher gave me a hint to decompose the 600 and 800 into x and y components...but I have done this and cannot see what it helps me to derive? Anyone else see it?

    Casey

    Also, I have drawn parellogram law
     
  2. jcsd
  3. Dec 18, 2007 #2
    I just don't see the relationship here. It looks like the y components might add up to the y component of R...but I am not sure how to prove it or if that can even help me here.
     
  4. Dec 18, 2007 #3
    I'm going postal as we speak....I just thought you should know.
     
  5. Dec 18, 2007 #4

    stewartcs

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  6. Dec 18, 2007 #5
    So if A+B=R then [itex]A_x+B_x=R_x[/itex] and [itex]A_y+B_y=R_y[/itex] and [tex]R=\sqrt{(R_x^2+R_y^2)}[/tex]

    Is this what I just read?! If so I did this earlier and got the wrong answer...but most likly because of a stupid mistake.

    Is this correct though?
     
  7. Dec 18, 2007 #6

    stewartcs

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    When you add vectors graphically, which way must the be aligned? Tail to Tip. Otherwise, you will have a sign problem and add when you should subtract.

    So,

    Rx = Ax + Bx, but A (the 600 N vector) is tail to tail with the B vector, so what does this tell you?
     
  8. Dec 18, 2007 #7
    So, since the x components are in opposite directions, I need to take one as negative....thanks stewartcs! I knew I was overlooking the obvious!

    Casey
     
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