Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding X-component of this force vecotr

  1. Aug 20, 2010 #1
    1. The problem statement, all variables and given/known data
    In big need of help trying to solve this equation. The newton part is throwing me off. I'm just wanting to ask what is the formula I would use and steps in trying to breaking it down to a final result.

    if a force vector has a magnitude of 2.47 newtons and points 76.2 degrees south of east, then what is its x-component?
  2. jcsd
  3. Aug 20, 2010 #2
    Re: X-components

    the first step is to define your reference. Though it's not stated, the problem probably assumes that the x-component is purely east and the y-component is purely north. However, without stating that in the problem, you can solve the problem an infinite many correct ways by assigning different references.

    Anyway, on with assuming 0 degrees is east. Your next step would be to convert the awkward angular presentation into a more meaningful one. 76.2 degrees south of east would be -76.2 degrees from our reference. You can see this by drawing a picture and by recognizing that going clockwise is a negative increment and going counterclockwise is a positive increment.

    Last, you need to find the x component by taking the cosine of your meaningful angle and multiplying it by the magnitude of your vector. The y-component would be the same except with a sine instead of a cosine.

    On a side note, I'm not sure what you mean by "the newton part is throwing me off" That's just a unit attached to the magnitude. It's like saying "13 inches" with 13 being the magnitude and inches being the units.

    Another approach students often use is to use the positive 76.2 degrees and modify the sign of the answer according to what is sensible. See, 76.2 degrees and -76.2 degrees give you the same x and y components except for a sign difference on the y-component. You can see all of this by just drawing the triangles and observing the trig.
    Last edited: Aug 20, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook