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Displacement vectors help

  1. Feb 14, 2008 #1
    1. The problem statement, all variables and given/known data
    Each of the displacement vectors A and B have a magnitude of 3.00 m. Vector B is vertical and is on the y axis, vector A forms a 30 degree angle to the positive x axis. Report all angles counterclockwise from the positve x axis. Find A+B, A-B, B-A, A-2B
    This is similar to what the diagram looks like:

    2. Relevant equations
    magnitude=square root(Ax+Bx)^2+(Ay+By)^2

    3. The attempt at a solution

    1. A+B
    magnitude=square root (3cos30+3cos90)^2+(3sin30+3sin90)^2


    2. A-B
    magnitude=square root (3cos30+3cos90)^2+-(3sin30+3sin90)^2=square root (-13.5) This then goes to 3.67, right?

    3. B-A

    4. A-2B
    magnitude=square root(3cos30+3cos90)^2+-(2(3sin30+3sin90)^2=5.8

    I would really appreciate if someone could help me with these 4 parts. I've been working on this for two days and I don't understand it. I don't see what I'm doing wrong.

    Thank you very much
  2. jcsd
  3. Feb 14, 2008 #2
    There are a couple of things going on here, one is I'm not quite sure how you're getting the answers you're getting.

    First let's simplify things a bit. Vector B is vertical and on the y axis, therefore there is no x component, so you should just use 0 when referring to the x component of B. Also, you can just use 1 as the sin90, so the y component of B is just 3.

    And you have A split correctly, the x = 3cos30, the y = 3sin30.

    So, for A+B you have:
    x component (3cos30 + 0) = 2.6
    y component (3sin30 + 3) = 1.5 +3 = 4.5

    So the magnitude of A+B =[tex]\sqrt{2.6^2 + 4.5^2} = \sqrt{27.01} = 5.2[/tex]

    Redo the problem and see if you get the same answers, then recalculate the angle.

    For A-B, you need to be careful what you subtract from what. It should be:

    x component (3cos30 - 0) = 2.6
    y component (3sin30 - 3) = 1.5 - 3 = -1.5

    From here you can recalculate the magnitude and angle. Once again, be careful where you put the angle as the x is positive, but the y is negative.
  4. Feb 14, 2008 #3
    Thank you very much

    Does this look correct? Aren't the sin angles 90°?

    direction of a+b=tan-1(2.6/4.5)=30.02
    180-30.02=150 Do I need to subtract it from 180 or 360?

    magnitude of a-b=square root of (2.6)^2+(-1.5)^2=3

    direction of a-b=tan-1(2.6/-1.5)=-60.02

    x component=0-3cos30=-2.6
    y component=3-3sin30=1.5

    magnitude=square root ((-2.6)^2+(1.5)^2)=9.01



    x component=3cos30-2(0)=2.6
    y component=3sin30-2(3)=-4.5

    magnitude=square root of (2.6^2+-4.5^2)=5.2

    Thank you
  5. Feb 14, 2008 #4
    Not sure what you mean by this. The sin 90° = 1, so 3sin90 = 3*1 = 3

    Be careful, the x value is 2.6, the y value is 4.5, and tan is y/x (not x/y as you have it).
    So, your angle isn't right. As to what to subtract it from, look at where it would be based on the signs of the x and y. Both are positive, so the angle is in the first quadrant, no need to subtract it from anything.

    The magnitude of a-b is correct, re-do the angle as above. As to where it's located, the x is positive, the y is negative, so where would the angle be?

    Check your math for the magnitude. Then recalculate your angle

    The magnitude looks good. Redo the angle.
    You seem to be understanding it pretty well, If I were making these mistakes, and believe me I have, it would be because I was going to fast, and not being deliberate with the calculations. Since you're just using two vectors to do all the calculations, you can do the original calculations at the top, and label them such as:
    Ax = 3cos30 = 2.6
    Ay = 3sin30 = 1.5

    etc., then just plug in the values you need to each equation.
  6. Feb 14, 2008 #5
    Thank you very much

    "As to where it's located, the x is positive, the y is negative, so where would the angle be?"

    It would be in the third quadrant, right? So, I just need to subtract it from 180, right? If it were in the third or fourth quadrant, I would need to subtract it from 360, right?

    thank you
    Last edited: Feb 14, 2008
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