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Vector Component angles

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data

    A velocity of [tex]10ms^{-1}[/tex] is to be replaced by two components, [tex]7.0ms^{-1}[/tex] and [tex]5.0ms^{-1}[/tex]. What must be the angle between the two components?

    2. Relevant equations

    3. The attempt at a solution

    Now I think that the answer to the solution lies in using trig to work out the angles, and that solving this equation [tex](5 sin\Theta)^2 + (7 + 5cos\Theta)^2 = 10^2[/tex] should give me the respective answers. What I don't understand is WHY I am doing that. So if someone could be so kind as to tell me how I would reach the conclusion that I should do those steps I would be very grateful.
  2. jcsd
  3. Oct 4, 2009 #2
    Hmm, you should use the fact that

    Vy=Vsin(theta) and that Vx^2+Vy^2 = V^2
    where Theta is the angle between the components. I'm not sure why you have a (7+5cos(\theta))^2 there.
  4. Oct 4, 2009 #3
    Well in my book I am basically told to use this equations (Where Vr is the resultant):

    [tex]V_r^2 = (V_1 sin\Theta)^2 + (V_1 cos\Theta + V_2)^2[/tex] - I want to know why I would use this forumla...
  5. Oct 5, 2009 #4
    Have you learned something called the cosine rule before? The cosine rule says,
    that if I have 2 vectors a and b, |a+b|^2=a^2+b^2+2abcos(theta).

    Now, relate that to the formula written in the book.
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