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Vector question

  1. Dec 14, 2005 #1
    For the vectors in the picture, we're supposed to break down each vector into its x and y components. I dont understand why the x component is given by cos(theta). It seems like it should be sin(theta) to me

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  3. Dec 14, 2005 #2


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    Staff: Mentor

    While waiting for approval, let's assume a Cartesian (x,y) coordinate system with x-axis horizontal and y-axis vertical with positive coordinate in upper right quadrant.

    Take F to be in the right half, either above or below. If the angle [itex]\theta[/itex] between F and the x-axis, then the component Fx would be given by F cos [itex]\theta[/itex]. If however, the angle was taken from the y-axis, then Fx would be given by sin [itex]\theta[/itex].

    With respect to F, Fx, Fy, think of F as the hypotenuse of a triangle and Fx and Fy as the legs, and then apply the Pythagorean theorem, i.e. appropriate trigonometric relationship.
  4. Dec 14, 2005 #3


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    The x components are gven by the sine of the respective angles in the diagram you showed. Who said otherwise?
  5. Dec 14, 2005 #4


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    The x component of either vector cannot be [itex]sin(\theta)[/itex] or [itex]cos(\theta)[/itex]. There is no [itex]\theta[/itex] in the picture!

    If, as is often done- but not in this picture, [itex]\theta[/itex] is measured from the positive x-axis, then the x component of the vector would be given by the length of the vector times [itex]cos(\theta)[/itex].
  6. Dec 14, 2005 #5


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    Well, v1x, would be given by v1 sin [itex]\theta_1[/itex], and v2x, would be given by v2 sin [itex]\theta_2[/itex], that is assuming the vertical axis is the y-axis, and x-axis is horizontal. These give the magnitudes, but realize that v2x is in the +x direction, and v1x is in the -x direction.

    For future reference - http://en.wikipedia.org/wiki/Trigonometric_function
    Last edited: Dec 14, 2005
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