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Resultant math help

  1. Jan 4, 2008 #1
    1. The problem statement, all variables and given/known data

    How am i going to compute for the 3rd force? the given are 163g 54degress S of E, 401g 12degrees E of N and the resultant 915g 180degrees or West.

    3. The attempt at a solution

    I computed it like the way a resultant is missing. I got 780.52g at 19.48degrees N of W. They say it is wrong. They say it should be computed differently.
  2. jcsd
  3. Jan 4, 2008 #2
    What was asked was to find a force, wich added to the first two forces in the problem, will give the third force. If the forces from the problem are A, B and C, you want to solve

    A+B+F = C for your unknown force F. (where A,B,C and F are 2-dimensional vectors)
  4. Jan 4, 2008 #3
    It might be easier to represent those vectors with cartesian coordinates (find their vertical and horizontal components) and solve for the missing vector algebraically.

    Also, of the two given vectors, the second one (401g, 12degrees E of N) is a lot stronger, and since it's going at a northward angle, I'd expect the missing vector to be southward.
  5. Jan 4, 2008 #4
    I would suggest drawing the two given vectors with their tails eminating from the origin, and then connect their heads with the vector you're trying to find.
  6. Jan 4, 2008 #5
    Wouldn't that just give you the resultant of those two vectors? I don't see how that resultant could equal the missing vector.
  7. Jan 4, 2008 #6


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    You're given vectors A, B, and R. You need to find vector C.

    [tex] \vec{A} + \vec{B} + \vec{C} = \vec{R} [/tex]


    [tex] \vec{C} = \vec{R} - \vec{A} - \vec{B} [/tex]

    Now break it down into componets

    [tex] C_x = R_x - A_x - B_x [/tex]


    [tex] C_y = R_y - A_y - B_y [/tex]

    The magnitude is then found by

    [tex] C = \sqrt{C_x + C_y} [/tex]

    and the angle by

    [tex] \theta = \tan^{-1}{\frac{C_y}{C_x}} [/tex]

  8. Jan 4, 2008 #7


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    One caveat with this that I forgot to mention: Be aware of what quadrant you are in because calculators will often give an apparently wrong angle from the arctan function.
  9. Jan 4, 2008 #8
    My mistake, I misread the problem. Since you have >2 vectors that make up the resultant, the component method (which stewartcs explained) would be easier. But the head-to-tail method would still work. If you compute the resultant of the two given vectors, and put that vector eminating from the same point as the resultant, the missing side of the triangle would be the vector you're looking for.
  10. Jan 5, 2008 #9
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