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Need help finding the difference of two forces

  1. Sep 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Find P-Q (P=11.5 kN, Q=12.5kN)

    2. Relevant equations
    x-component = cos(angle)*force
    y-component = sin(angle)*force
    Pythagorean Theorem (to find resultant magnitude): c2 = a2 + b2
    arctan (y-component / x-component) : to find angle of resultant

    3. The attempt at a solution
    The answers for this problem are given to me in my book: magnitude = 14.63 kN @ 160.6°
    However, I am having trouble getting the direction.

    I first solved for the x and y components of P:
    x-component = cos(75)*11.5 = 2.97642
    y-component = sin(75)*11.5 = 11.1081

    Then I replaced Q with -Q and solved for the x and y components:
    x-component = cos(-30)*12.5 = 10.8253
    y-component = sin(-30)*12.5 = -6.25

    I then solved for the magnitude by adding the x-components and y -components, and then used Pythagorean theorem:
    x-component = 10.8253+2.97642 = 13.8017
    y-component =-6.25 + 11.1081 = 4.8581

    √(13.8017)2 + (4.8581)2 ≈ 14.63 kN (correct)

    Then I solve for the arctan of the components:

    arctan (4.8581 / 13.8017) ≈ 19.3917

    My confusion arises here, since the book states that the angle is 160.6° which places the resultant in the second quadrant...however how can this be if the x and y components of my resultant are both positive, which suggests that it is in the first quadrant?

    If I subtract this from 180, I do get 160.608 which is the answer, but it doesn't make sense to me why I would do that?
  2. jcsd
  3. Sep 28, 2015 #2


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    Gold Member

    Uh ... how did you get a positive X component for -Q ?
  4. Sep 28, 2015 #3
    When I typed in cos(-30)*12.5 into my calculator it gave me +10.8253
  5. Sep 28, 2015 #4
    You actually made me rethink and realize what I did. I was supposed to literally move Q in the completely opposite direction. What I did was only reverse it in the y-direction. Now my math makes sense, I have negative x-component and y-component and my answers make sense. Thank you very much. I know it seems like I didn't think this through first, but I was actually stumped for a while because my book only gave a me one example of this type of problem :/
  6. Sep 28, 2015 #5


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    Homework Helper

    You can't ignore the reference points for the angles which define the vectors P and Q. Take the origin of the coordinate system at the point where the tail ends of both vectors coincide. The y-components of P and Q are both positive, but the x-components should be of opposite signs.
  7. Sep 28, 2015 #6


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    Another way to do this: The vector P- Q is the vector from the tip of vector Q to vector P. Ignoring, temporarily, the direction of that vector, it is the third side of a triangle with two sides of length 11.5 and 12.5 with angle 180- 75- 30= 180- 105= 75. By the cosine law, the length of the third side, and so the length of the difference vector, is where [itex]c^2= (11.5)^2+ (12.5)^2- 2(11.5)(12.5) cos(75)[/itex]. The angle
  8. Sep 28, 2015 #7


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    We all make this kind of obvious error from time to time (well, I do anyway :smile:)
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