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Vector addition

  1. Sep 28, 2007 #1

    ~christina~

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    [SOLVED] Vector addition

    ]1. The problem statement, all variables and given/known data
    (a) Express the vectors A, B, and C in the figure below in terms of unit vectors.

    (b) Use unit vector notation to find the vectors R = A + B + C and S = C – A – B.

    (c) What are the magnitude and directions of vectors R and S?


    2. Relevant equations
    Ax= A cos (angle)
    Ay= A sin (angle)



    3. The attempt at a solution

    a) for a first of all how would I express the vectors in terms of "unit vectors"?
    would that include the [tex]\hat{} i [/tex] and the [tex]\hat{} j [/tex] ?

    b) for B I thought It would be okay unitll I saw the angles and now I'm not sure which to use like for example the [tex]\vec{}C[/tex] has a angle of 65 deg but is it's angle actually 245 deg ? The reason I'm asking is that 40 deg for [tex]\vec{} B [/tex] it would be Bx= Bcos (-40 deg) and since I know the value of B wouldn't I use the 15.0m for B ?

    c) magnitude and direction of vector R and S..wouldn't it be the resultant vectors??
    but I'll deal with this when I get past a, and b which has me confused.

    [​IMG][/URL][/IMG
     

    Attached Files:

    Last edited: Sep 29, 2007
  2. jcsd
  3. Sep 28, 2007 #2
    Hi

    How to get a unit vector of any given vector? Apply that formula and you wil get the unit vector of all the three given vectors.
     
  4. Sep 29, 2007 #3

    ~christina~

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    What about the i hat and j hat??

    Can anyone help me please??
     
    Last edited: Sep 29, 2007
  5. Sep 29, 2007 #4
    The unit vector will come in the terms of i j and k.
     
  6. Sep 29, 2007 #5

    ~christina~

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    That doesn't help me...just telling me the unit vector will be in terms of i, j and k....

    I need to know how that would be incorperated into the values I obtain if someone can tell me how to start this...

    but I think I'll have to wait untill the attatchment is approved first ...

    and I need to finish this by today...:cry:
     
    Last edited: Sep 29, 2007
  7. Sep 29, 2007 #6
    Ok ok dont panic

    [tex]u_{\vec{A}}[/tex] = [tex]\frac{\vec{A}}{A}[/tex]

    It means that unit vector in the direction of vector A is equal to the vector A divided by the magnitude of vector A.

    And attach the picture as a file so that we can see it.We are not able to see the picture which you have posted along.

    For example the unit vecor of 3i +5j+6k would be [tex]\frac{3i + 5j+6k}{\sqrt{70}}[/tex] =
     
    Last edited: Sep 29, 2007
  8. Sep 29, 2007 #7

    ~christina~

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    where did you get the 70?
     
  9. Sep 29, 2007 #8

    Kurdt

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    Well a unit vector is a vector that has a magnitude of one. Lets say we have any vector [itex]\mathbf{a}[/itex]. The unit vector in the direction of [itex]\mathbf{a}[/itex] is usually written as [itex]\mathbf{\hat{a}}[/itex] and of course has a magnitude of 1.

    the magnitude of a vector denoted by [itex]|\mathbf{a}|[/itex] is the length of the vector basically and is found using pythagoras' theorem.

    [tex]|\mathbf{a}| = \sqrt{a_1^2+a_2^2+a_3^2} [/tex]

    where the vector [itex]\mathbf{a}[/itex] is given by [itex]\mathbf{a}=(a_1,a_2,a_3)
    [/itex]

    Now to find the unit vector of any general vector [itex]\mathbf{a}[/itex] which is given by [itex]\mathbf{a}=(a_1,a_2,a_3)[/itex] we have to divide it by its magnitude. This ensure that the unit vector has a magnitude of 1.

    So in FedEx's example the square root of 70 comes from taking the magnitude of his vector defined by the position vector [itex]\mathbf{a} = (3,5,6) [/itex].

    For your question I believe they will want the vectors expressed in terms of the i, j and k unit vectors.

    P.S. I private messaged you about images if you need to put them up quickly.
     
    Last edited: Sep 29, 2007
  10. Sep 29, 2007 #9

    ~christina~

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    I uploaded the picture but I've also figured out how to get the unit vectors however I don't know how to convert back to magnitude/ angle notation.

    my answers for a and b
    a.)
    A = 12.00 (sin 37 i + cos 37 j) = 7.22 i + 9.58 j
    B = 15.00 (cos 40 i - sin 40 j) = 11.49 i - 9.64 j
    C = 5.00 (-cos 60 i -sin 60 j) = -2.50 i - 4.33 j


    b.)
    R= A+ B + C

    R= (7.22i + 9.58j) + (11.49i- 9.64j) + (-2.50i - 4.33j) = (16.21i - 4.39j)m

    S= C-A-B

    S= (-2.50i - 4.33j)- (7.22i + 9.58j) - (11.49i - 9.64j)= -2.50i - 4.33j- 7.22i - 9.58j -11.49i + 9.64j= (-21.21i- 4.27j)m

    c.) I'm not sure how to get the magnitude and direction for the angles..

    ~Thanks~
     
  11. Sep 29, 2007 #10

    Kurdt

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    Part a and part look ok to me. For part c, think of the vectors as the hypotenuse of a right angled triangle. How would you work out the length of the hypotenuse? Sticking with right angled triangles, how would you work out the angles? Its pretty much the reverse of part a.
     
  12. Sep 29, 2007 #11

    ~christina~

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    Um I was thinking about using tan theta but I'm not sure about that since the values I got were for the 4th quadrant and 3rd quadrant if I'm not wrong...

    I can't really figure out how to go backwards...it's kinda weird...

    actually thinking about it now to find the hypotenuse...wouldn't I use SOH CAH TOA...and I'm kind of lost after that...
     
  13. Sep 29, 2007 #12

    Kurdt

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    Ok so [itex]sin(\theta) =\frac{o}{h} [/itex] and [itex]cos(\theta)=\frac{a}{h}[/itex]. The opposite or adjacent will be either the x or y components. Its a bit strange because its standard to measure the angle anti-clockwise from the x-axis but in the diagram they don't seem to be bothered with that convention. So make sure you make it explicit as to where the angle is measured from (x or y axes) and in what direction (clockwise or anti-clockwise).
     
  14. Sep 29, 2007 #13

    ~christina~

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    Hm would it be like
    for R
    tan theta = -4.39/16.21= -.27 so finding theta
    isn't it that you always add 180 to a value of tan if it is a negative? Well assuming it is..
    tan^-1(-.27+180)= 89.68 deg


    tan theta= -4.27/-21.21= 0.201
    tan^-1(0.201)= 11.36 degrees

    for the magnitude I think I would use A^2= B^2 + C^2
    and use the values for the unit vectors that I obtained but I'm still not sure about that since the value of the vector units are for the x and y values...
     
  15. Sep 29, 2007 #14

    ~christina~

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    I guess what I did above was incorrect?

    well anyways..above post explains my confusion on the subject of what numbers to use
    seriosly confused as to what values to use since the numbers for the i and j are a point not a vector am I correct?
     
  16. Sep 29, 2007 #15

    Kurdt

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    Sorry. Yes you can use tan aswell. The angle calculated will be the angle clockwise from the positive x-axis for the vector R and the angle calculated for S will be anti clockwise from the negative x-axis.

    Heres a little summary of the quadrants. http://www.mathsrevision.net/alevel/pages.php?page=36

    For the magnitude you are correct in that you would use the pythagorean theroem. The reason this works is that you can consider the vectors as the sum of vectors along the coordinate axes. The magnitude of these vectors is given by the numbers that multiply the unit vectors along the axis. these act as the sides of a right angled triangle.

    I hope I explained that well enough I have a bit of trouble sometimes. :smile:
     
  17. Sep 29, 2007 #16

    ~christina~

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    I guess that it is fine but did I do the that correctly for tan?

    Well assuming I did that correctly...

    for finding the magnitude...if I do this correctly
    since R= (16.21i - 4.39j)m

    H^2 = (16.21)^2 + (-4.39)^2
    H= 16.79 ??

    That would be the magnitude?? somehow I'm thinking that looks funny
    continuing though...

    S=(-21.21i- 4.27j)m
    H^2 = (-21.21)^2 + (-4.27)^2
    H= 21.64 ??


    ~Thanks~

    P.S that site was helpful
     
  18. Sep 29, 2007 #17

    Kurdt

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    No I forgot to mention you shouldn't add anything on when working out the angle. You will get ~ -15 degrees which means as i said before ~ 15 degrees below the positive x-axis.

    You have worked the magnitudes out correctly.
     
  19. Sep 29, 2007 #18

    ~christina~

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    So in all the new magnitude and direction of R and of S would be

    R= 16.79m with a angle of -15 degrees with respect to the x axis

    and

    S= 21.64m with a angle of 11.36 degrees

    however with the second one I'm not sure how to describe the location but I think it's in the 3rd quadrant sinc that's the only place where tan is positive..


    ~Thanks Kurdt~
     
  20. Sep 29, 2007 #19

    Kurdt

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    The second one is corect you will have to work out the value of the first one again because I only gave an approximate value. But once you do that all is well. The third quadrant is anti-clockwise from the negative x axis.

    It would be easy to work out what they were anti-clockwise from the positive x-axis which is the convention but whoever set this question doesn't seem to be worried about that judging where the angles on the diagram are placed. So descriptions should be ok.

    If you want more information on vectors try chapter 9 of 'Guide to Mathematical Methods' by John Gilbert and Camilla Jordan. I like their approach very much, it might clearup a bit of the confusion better than I can.
     
    Last edited: Sep 29, 2007
  21. Sep 29, 2007 #20

    ~christina~

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    Thanks alot Kurdt!!

    I shall look up that book ..my text actually is a really bad book in my opinion.
     
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