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Physics: VECTOR help

  1. Sep 12, 2004 #1
    Can someone please help me with this?
    1) A resultant vector is 5 units long and makes an angle of 23 degrees measured counter-clockwise with respect to the positive x-axis. What are the magnitude and angle (measured counter-clockwise with respect to the positive x-axis) of the equilibrant vector?

    2) Find the x- and y-components of the vector whose magnitude is 8.73 units making an angle of 155 degrees measured counter-clockwise with respect to the positive x-axis.
     
  2. jcsd
  3. Sep 12, 2004 #2
    These kinds of questions have already been answered, look at some other posts.

    Think of the fact that the x-component of a vector of magnitude F and angle x with the x-axis is
    equal to F*cos (x). The y-component is F*sin(x)

    Once the components are given you can work the other way around and determin the magnitude of the vector as F = sqrt(x² + y²)

    regards
    marlon
     
  4. Sep 12, 2004 #3
    But what does the counter-clockwise mean? Does it mean that the angle is negative? I'm confused.
     
    Last edited: Sep 12, 2004
  5. Sep 12, 2004 #4

    Pyrrhus

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    Homework Helper

    Clockwise means a negative angle, and counter-clockwise means a positive angle.
     
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