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Please need help

  1. Sep 26, 2007 #1
    please... need urgent help about vectors

    1. The problem statement, all variables and given/known data

    in the sum vector a + vector b = vector c, vector a has a magnitude of 12.0m and is angled 40.0 degrees counterclockwise from the +x direction, and vector c has a magnitude of 15.0m and is angled 20.0 degrees counterclockwise from the -x direction

    a)what is the magnitude
    b) what is the angle (relative to +x) of vector B.


    3. The attempt at a solution

    i first found the x and y component of each vector:

    so for vector a the x component would be 12 cos(40) and the y component would be 12 sin(40). for vector C the x component would be 15 cos (20) and the y component would be 15 sin (40). so do i just add the x components of the 2 vectors and the y component of the 2 vectors, do pythagorean theorem and find out what the magnitude is.
    so would i be (12cos(40)*15cos(20))^2+(12sin(40)+15sin(20))^2 and then square root?
     
    Last edited: Sep 26, 2007
  2. jcsd
  3. Sep 26, 2007 #2

    mgb_phys

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    That's the right idea.
    Draw a triangle, where the first side is 12m at 40deg above X, then from the end of that draw a line 15m at 20deg below X, then draw a closing line back to the start - this is the resulatant vector.
     
  4. Sep 26, 2007 #3

    Avodyne

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    Your problem said the second vector is at a 20 degree angle counterclockwise from the minus x direction. If that's not a typo, you need to think about whether the components of the second vector are positive or negative.
     
  5. Sep 26, 2007 #4
    yeah, its not a typo, so the second vector will have negative x and y right? does the problem change if vector b is the unknown not vector c.
    so what do you guys get, i am just checking my answers. i got 26.59 for the magnitude and 28.87 degrees. here's my work just to clarify my explanation
     
    Last edited: Sep 26, 2007
  6. Sep 26, 2007 #5
    bump...
     
  7. Sep 26, 2007 #6

    Avodyne

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    If b is unknown, then b = c-a, that is, you subtract the components of a from the components of c. So your magnitude above looks like it is correct.
     
  8. Sep 26, 2007 #7
    but i don't think i subtracted it though, i added Ax and Bx and Ay and By. so my answer in the pic is correct? and i think i get the same answer even if i subtracted
     
    Last edited: Sep 26, 2007
  9. Sep 27, 2007 #8

    Avodyne

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    Your Cx and Cy should both be negative, not positive (which should be clear from your picture). Then, since you are subtracting the positive A components from the negative C components, both terms in both Rx and Ry should be negative. Then, your magnitude of R is correct (I didn't actually check the numbers), but your angle is off by 180 degrees, since both Rx and Ry are negative.
     
  10. Sep 27, 2007 #9
    so i add 180 to my theta since both rx and ry are negative going to third quadrant. and the R would be the same because i square it later in the pythagorean theorem, making it positive right?
     
  11. Sep 27, 2007 #10
    bump....
     
  12. Sep 27, 2007 #11

    Avodyne

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    Right! (The magnitude of a vector is always positive.)
     
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