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Vector Addition

  1. Sep 5, 2006 #1
    Hey guys!

    I feel like I should know how to do this problem from Trig, but I am totally blanking out right now. We have to find the resultant from the vectors that he has given us. One of the vectors is lying on the negative x-axis. How do I find cosine and sine when the line is on the x-axis. Thanks!
     
  2. jcsd
  3. Sep 5, 2006 #2

    Hootenanny

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    What do you mean, is the vector parallel to the x-axis? If that is the case, then there is no y-component so the vector has a sin or cosine of zero.
     
  4. Sep 5, 2006 #3
    the vector is on the x-axis as in horizontal. Does that mean that there is no x component?
     
  5. Sep 5, 2006 #4
    No; it means that the vector has no y component.
    What are you trying to find the sine and cosine of? If you are using it to find the components of the vector, the angle is measured from the +x axis to the vector counter-clockwise. In this case, since the vector is on the -x axis, the angle is 180 degrees.

    Split each vector into its x and y components, and then add the components. Then you don't have to think about trigonometry except in finding the components. In this case, since the vector is lying on the x-axis, you do not need trigonometry to figure out the x and y components.
     
  6. Sep 5, 2006 #5
    Hmm okay, so if my teacher says that the hypot. is equal 60, that means that I use 180 as my angle because the line is on the negative x-axis? Therefore, I don't need to worry about the y component and then add the other components from the other vectors that I have together. Ah! I see now. So if the line was on the y-axis would the angle be 90 degrees? Thanks!
     
  7. Sep 5, 2006 #6
    ooo, wait, you said that I don't need to use trig. So I don't have to find cosine and sine then? Hmmm, I think I am confused. How do I use the 60 that my teacher gave me with the problem?

    Thanks!
     
  8. Sep 5, 2006 #7
    I think you could still use sine and cosine. Did your teacher give you the magnitude of anything if he did then you would take the magnitude times cosine of zero for X since your vector lies on the X-axis. If 60 is the verticle direction's (Y) magnitude then you would take that times sine of whatever your angle is. then after you get these numbers you can use pythag. to get the magnitude of the resultant. Next you will need to use inverse tangent to get the angle of the resultant. Hopefully this will help you.
     
  9. Sep 5, 2006 #8
    I understand the pythag and the inverse tangent part. He didn't give us the angles for the vector that is lying on the negative x-axis. That's where I'm confused. What would that angle be? He said that when we formed the right triangle, the hypothenuse would be 60. I think that is what the magnitude is supposed to be then. Since it's also on an axis I didn't know if I should use cosine or sine with the angle. Thanks everyone! You have all been helpful so far.
     
  10. Sep 5, 2006 #9
    I think you should use 0 as your angle if that doesn't work then use 180. The reason I say 0 is becasue vectors can be moved around the X/Y coordinate system. You always use cos to find the X component and sin to find the Y component.
     
  11. Sep 6, 2006 #10
    *sigh* I think I am doing something wrong because I'm not getting the right answer. Since it's on the negative x-axis, it's going to be negative 60sine/cos 0 right? My prof told me in class today that the x component is not always cos and the y component is sine. He did an example on the board. But either way I tried it I got the wrong answer. I think I'm doing something wrong with the calculator. I was pretty sure that it was 60*cos/sin 0. Something's wrong! LOL
     
  12. Sep 6, 2006 #11
    You can always solve it graphically.
     
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