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Calculating Resultant Vectors Using Trig

  1. Jul 3, 2012 #1
    Given Info: 4 m/s [N30°E] and 5 m/s [S17°E]. FIND: The Resultant Vector And it's angle.


    So i'm having a big problem with figuring the angles between the two vectors... I feel like i'm doing it correctly, but at the same time i'm not sure...


    Here's my diagram. http://i47.tinypic.com/x38osl.jpg

    So the angle between them, from what i see, is 30 degrees + 17 degrees = 47 degrees..

    The reason i'm not comfortable with this... is because when i place my drawing on the same coordinate system, I will have 60 degrees between the X axis and vector a, and 73 degrees between the X axis and vector b... Therefore the angle between them should be 133 degrees...

    So i'm not sure now which angle is correct.. this trigonometry method seems a lot tougher than the component method.


    .... But actually... I see something... 47 degrees is = 180 - 133

    And the 133 degrees is when they are cnonected head to head.. and the 47 degrees is connected from head to tail..

    So the angle should be 47 degrees bewteen them? is this right? and if so, does this mean the angle between two vectors is when they are connected head to head, or head to tail?
     
    Last edited: Jul 3, 2012
  2. jcsd
  3. Jul 3, 2012 #2
    First break the vectors up into their components

    a=Acosθi+Asinθj

    then do the same for your second vector, except remember you're now moving in the -y direction, so you're j-hat component of the second vector will be negative.

    Also look at the axis you placed at the head/tail of the vectors. the angle is a -θ there. So do 90-17 to find that angle.

    After finding all 4 components, add your i-hats and your j-hats together.

    Then take the inverse tangent of the y-component divided by your x-component.
     
    Last edited: Jul 3, 2012
  4. Jul 3, 2012 #3
    Thank you for your help with component method. but i understand how to do this method already, i'm just trying to learn how to do it using trigonometry so i have more than 1 tool.

    I need to know if the steps i'm taking in the trigonometric process for determining angles is correct.
     
  5. Jul 3, 2012 #4
    The angle between two vectors is the angle made when the vectors are connected tail-to-tail, or head-to-head, as both situations will give the same angle from the parallelogram law.

    I suppose one way to apply trigonometry is taking the magnitudes of each vector, and the 47 degree angle between the head-to-tail connection, and plugging them into the law of cosine to find the magnitude of the resultant vector. Then you can use the law of sines to find the lower left angle of the triangle. This angle, when added to 33 degrees, will give you the angle of the resultant vector East of North.
     
  6. Jul 3, 2012 #5
    c.c here calculations for the rest of the question, i'm just stuck on this stupid angle problem i'm having...

    So if my angle of 47 is the angle between the two vectors, that means i can use cosine law.


    |R| = sqrtall |a|^2 + |b|^2 - 2|b||a|cos47
    |R| = 3.70

    SinA/|a| = Sin47/|R|

    A = 52.25 degrees

    therefore 17 + 52.25 = 69.25

    Ans: R = 3. 70 M/S [S69.25W]
     
  7. Jul 3, 2012 #6

    I don't know that this is a true statement... Because in a question i did


    The angle BETWEEN the HEAD HEAD connection, allowed you to solve for the angle between a head to tail connection..

    Ie:http://postimage.org/image/sj72x4nv9/

    For the Pink angle, lets say it is 50 degrees..

    This is the Head To Head connection.


    Now you can solve for the Red angle (Head To Tail Connection0

    By doing 180 - (PINK ANGLE)

    180 - 50 = 130 degrees


    Thus the angles are not the same?)

    And sorry i just read the comment said HEAD TO HEAD or TAIL TO TAIL.

    My initial question was in regards to Head To Tail, Or Head To Head.



    **** Last thing, so that angle 47 degrees could work as the angle between a head to tail connection of these vectors?****
     
    Last edited: Jul 3, 2012
  8. Jul 3, 2012 #7
    Looking at the given data: v1 = 4 m/s [N30°E] and v2 = 5 m/s [S17°E], the resultant vector should point in the South East direction, because v2 is larger in magnitude and is deflected less away from the vertical axis.

    Using the law of sines to solve for the lower left angle of the triangle,
    SinA/|v2| = sin(47)/3.70m/s
    A = arcsine[5m/s * sin(47)/3.70m/s] = 81.2 degrees

    To find the angle east of south, let's call it B, we can say that
    B + 81.2 + 30.0 = 180.0
    B = 68.8

    I'm not sure why you chose to solve for the lower right angle and why you added 17, but the results saying it is pointing in the south west direction is wrong.

    47 is the angle between the head-tail connections, unless you also consider the larger angle, 360 - 47.
     
    Last edited: Jul 3, 2012
  9. Jul 3, 2012 #8

    yea it's not even possible for it to point soutwest logically, all of the horizontal components point east lollololol im a silly goose.

    so if i do

    Sin47/|R| = SinB/|b|

    b = 81.23 degrees


    adding it with 30 i get 111.23 degrees

    so 111-90= 21.23degreees

    so if this is the deviation north of east... it then would go into the 4th quadrant.. making it a West Of South?

    so 111.23 - 90 = 21.23

    3.7 m/s [W21.23S]


    is this how i approach the end, because i think initially i tried to do this angle and add it to 30 but i saw the problem of an angle greater than 90 degrees.. and it was confusing me..
     
    Last edited: Jul 3, 2012
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