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Vector components

  1. Oct 18, 2005 #1
    I have a problem that includes an acceleration vector 'a' located in the 3 quadrant and it makes an angle (theta) 30 degrees to the Negative Y axis.
    The solution manual shows the the vector resolved like
    a = (12.0 sin 30 m/s^2) i - (12.0 cos 30 m/s^2) j
    Why take the sin of 30 for the x direction and the cos of 30 for y? I've always done it cos of x and sin of y.
  2. jcsd
  3. Oct 18, 2005 #2


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    Because the angle is here defined differently than usual...
  4. Oct 18, 2005 #3
    The reason you flip them is because the reference angle is made with the y axis instead of the x. Usually we measure the angle from the x axis.. The reason it works when you flip then is due to the two properties of sin and cos:

    [itex] \sin(x-90) = \cos(x) [/itex]

    [itex] \cos(x-90) = \sin(x) [/itex]
  5. Oct 18, 2005 #4
    Thanks!! That leands me to another question...
    In this problem, the angle is 30 degrees with the negative axis so shouldn't cos 240 be equal to sin 30? They come to -0.5 and 0.5 respectively.
  6. Oct 18, 2005 #5


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    You're turning in the wrong direction: 30 degrees starting from the negative Y axis (=270 degrees) gives you 300 degrees, not 240...
  7. Oct 18, 2005 #6

    the vector is in the 3rd quadrant so isn't it 240?
  8. Oct 18, 2005 #7
    [itex] \cos(240)[/itex] measures the (60 deg) reference angle from the negative x axis in the third quadrant, where cosine is always negative. However if you are going to refer to the angle from hte positive x axis, then you are also taking [itex] \sin(240) [/itex] and not [itex] \sin(30) [/itex]. Those relations are technically supposed to be used in the first quadrant, or just as a relative measure.
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