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Vector Subtraction- Magnitude

  1. Jan 30, 2012 #1

    T54

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    1. The problem statement, all variables and given/known data
    https://smart.physics.illinois.edu/images/phys101/Homework/02/vector.gif
    Two vectors A and B are shown in the x-y plane. Vector A has a magnitude of 5 m and makes an angle of 55° with the positive x-axis vector B has a magnitude of 2 m and makes an angle of 33° with the negative x-axis. (See the figure above.) Vector C (not shown in the diagram) is the difference of A and B (C = A - B).
    What is the magnitude of C?
    2. Relevant equations
    [v] = sqrt(vx2-vy2)


    3. The attempt at a solution
    A=5cos55deg+5sin55deg
    B=-2cos33deg+2sin33deg
    So I plugged it in to the equation above and got 6.391
    But apparently my answer is not correct. I am wondering where I went wrong
     
  2. jcsd
  3. Jan 30, 2012 #2
    Welcome to PF!

    Hmm, I get an answer a little lower than yours for the magnitude of C. Maybe you made a calculator error? Show us what you've done.
     
  4. Jan 30, 2012 #3
    In what quadrant is -B?
     
  5. Jan 30, 2012 #4

    T54

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    sqrt(2.868+4.096)2-(-1.677+1.089)2)
    sqrt(48.497-.346)
    = 6.989
    This time I got a little higher than my previous answer.
     
  6. Jan 30, 2012 #5

    T54

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    Qaud II
     
  7. Jan 30, 2012 #6

    T54

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    I think I was making it to complicated.
    The angle between vector A and vector B is 90°, so if I just the pythagorean theorem
    I would get magnitude of vector C = sqrt(5² + 2²) = sqrt29
    which is 5.385
     
  8. Jan 30, 2012 #7
    There is part of your problem. B is in Quad II. When you take the inverse of a vector, like -B, you extend the line right out of his butt in the opposite direction and of the same magnitude. -B is in Quad IV.

    EDIT: I want to add something to make sure this is clear. A - B = A + (-B), just as if these were regular variables.

    If you draw that line of B backwards into Quad IV, you have -B and the angle is -33° (or 327°, if you prefer)...and keeping that 33° reference angle.

    When you are dealing with vectors, there is two methods to deal with signs. Either you reduce everything to reference angles and then tack on the signs for the appropriate quadrant or you ignore the signs entirely and let the angles take care of it.

    In Quad I: cos 33° is positive and sin 33° is positive
    In Quad IV: cos 33° is negative and sin 33° is negative, so you either need to tack on the negatives yourself or use the -33° or 327° angle measures.

    Once you understand A-B the rest is pretty easy peasy.

    The magnitude of the combined vector will be the square root of the squares of the sums of the X and Y components. Which is a fancy way of saying that:

    (A-B)=√( (Ʃx)^2 + (Ʃy)^2)

    Try recalculating with the new information and give a shout if something is still unclear.
     
    Last edited: Jan 30, 2012
  9. Jan 30, 2012 #8
    The angle between A and B is NOT 90°. Add up the angles you were given (33° + 55° = 88°). We know that the total angle measures from 0 to ∏ is 180°. 180°-88°=92° between the two angles.

    I know it's very tempting, but you need to avoid slamming numbers into equations. You can bang your head against the wall for quite a while that way. Draw your sketch and label your vectors and then create your equations based upon the geometry, not by the utility of any particular equation.
     
  10. Jan 30, 2012 #9

    PeterO

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    Homework Helper

    Note: if C = A - B, it follows that C + B = A

    What vector must you add to B, in order to get A, for that is indeed C.
     
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