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posted by  gathan on 9/2/2008 10:33:08 PM  |  status: Live  

Vector Subtraction - Life Saver

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Determine the magnitude of the vector difference V'=V2 - V1 and the angle θx which V' makes with the positive x-axis.

V2 is 21 units at and angle of 126.87o from the positive x-axis.
V1 is 27 units at an angle of 30o from the positive x-axis.

I think the answers are V'=36.1 units and θx=174.8o, but I need confirmation
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posted by palmreader on 9/2/2008 11:51:00 PM  |  status: Live
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gathan's comment:
"Great, Thanks!"
Response Details:
recall that x = rcosθ
               y = rsinθ

for V1, x1 = 27cos30 = 23.4
          y1 = 27sin30 = 13.5

for V2, x2 = 21cos126.87 = -12.6
          y2 = 21sin126.87 = 16.8

V' = √[(-12.6-23.4)2+(16.8-13.5)2]
    = 36.13 units


to find the angle,
   θ = arctan(y/x) = arctan[(16.8-13.5)/(-12.6-23.4)]
                         = -5.24deg

since V'y = 16.8-13.5 = 3.3
        V'x = -12.6-23.4 = -36.0
then we know that V' lies in the second quadrant.
∴ θx = 180deg + -5.24deg = 174.8deg

You're right!
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