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2D Vectors - Addition and Subtraction of Successive Displacement Vectors

  1. Jan 24, 2013 #1
    The figure shows the successive displacements of an aircraft flying a search pattern. The initial position of the aircraft is P and the final position is P'. What is the net displacement (magnitude and direction) between P and P'?


    My attempt:
    So, my understanding is that the net displacement of the successive displacement vectors is the overall sum of the successive displacement vectors. (?)
    Also, referring to the rules of alternate interior angles, ∠A, ∠B, and ∠C = 60° (?)

    ΔA
    Hypotenuse = 18km
    Height = 18sin60°
    Base = 18cos60°

    ΔB
    Hypotenuse = 9.5km
    Height = 9.5sin60°
    Base = 9.5cos60°

    ΔC
    Hypotenuse = 12km
    Height = 12sin60°
    Base = 12cos60°

    My attempt to find the base of the triangle formed by the point P and P' and the x-axis was to add the bases of ΔA and ΔB and subtract the base of ΔC from the sum.

    With my calculations (using rounded decimals and plugging in the exact values):
    I got ~13km for →P (incorrect because my book reads 11.2km)
    and for ∠θ I got 57.1° (incorrect, because the remaining angle should be 27.7° and is 32.9° with my calculations)

    I am doing something wrong. HELP
     

    Attached Files:

  2. jcsd
  3. Jan 24, 2013 #2

    Andrew Mason

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    This is where you are going wrong. The displacement between P and P' is the vector going from the head of P to the head of P' (i.e. P' - P or the displacement vector that, when added to P, results in P').

    AM
     
  4. Jan 25, 2013 #3

    tms

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    Correct.
    It's best not to think in terms of triangles, hypotenuses, bases, and heights, but to think in terms of magnitudes and directions and components.

    In this particular case, you're using the wrong angle (assuming 60° is the angle between the [itex]y[/itex]-axis and the first vector).
     
  5. Jan 25, 2013 #4

    tms

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    The resultant is also the sum of the three given vectors. The OP is correct on this point.
     
  6. Jan 25, 2013 #5

    Andrew Mason

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    Of course you are right. I responded before the drawing was available, thinking P and P' were displacement vectors from a common origin. Sorry about the confusion.

    Part of the problem here is that the diagram is misleading. The vector lengths and angles do not fit. For example, the heights of ΔA and ΔB are the same. Using the figures given, however, the height of ΔA is 18sin(30) =9 km and the height of ΔB 9.5cos(30) = 8.22 km. (In other words, arctan 9.5/18 = 27.8 deg ≠ 30 deg.).

    I would suggest that you add the vector components using the angles given and disregard the triangles, as tms has suggested. The answer given is correct (11.2).

    AM
     
  7. Jan 25, 2013 #6

    tms

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    P and P' can be thought of as displacement vectors from the origin (or some other point); you were right on that point. But P' is also the sum of the given vectors; the OP was right about that. You were both right.
     
  8. Jan 25, 2013 #7
    I see. So, since this is how I interpreted the diagram, then I was visualizing it incorrectly. You're saying, because of the angles and magnitudes given, the head of →B is not at on a point on the x-axis.

    Can you elaborate on this a bit more, please?
    I will upload the original diagram from my book. Maybe I will be able to better visualize it rather than using my quick sketch of the diagram drawn in MS paint.

    Thanks.
     
  9. Jan 25, 2013 #8

    tms

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    A vector [itex](r, \vartheta)[/itex] can be decomposed into components using [itex]x = r\;cos\; \vartheta[/itex] and [itex]y = r\; sin\; \vartheta[/itex].
     
  10. Jan 25, 2013 #9

    Andrew Mason

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    Right.


    The resultant sum of the two vectors (x1, y1) and (x2, y2) would be a vector (x1 + x2, y1 + y2). The angle of the resultant vector to the +x axis would be arctan((y1+y2)/(x1+x2)).

    AM
     
  11. Oct 31, 2013 #10
    I had to solve the same problem. I finally solved it but not completely. I am not able to determine the angle that PP' makes.
    I would appreciate some help. I am not able to figure this part out. Thanks.
     
  12. Oct 31, 2013 #11
    Please note that in the figure in post #1, all the 60 degree angles are supposed to 30 degrees.
     
  13. Oct 31, 2013 #12

    tms

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    Show what you've done so far.
     
  14. Oct 31, 2013 #13
    I had to redraw the figure and this took me a while. Alright, here is my work.

    Angle APz is 60 degrees and AP is 18km.
    Py = 18 cos 60 = 9km.
    This makes Ay = 15.6km.

    Angle BAx is 60 degrees and AB is 9.5km.
    Ax = (cos 60) 9.5 = 4.75 km.
    This makes Bx = 8.22km.

    xy = Ay-Ax = 10.85km.

    Angle wBP' is 60 degree and BP' is 12 km.
    wP' = (sin 60) 12 = 10.39km.
    wB = (cos 60) 12 = 6km.

    The x - co-ordinate for B is Py+Bx = 9+8.22 = 17.22km.

    This makes Pz 17.22-wB = 17.22-6 = 11.2km.

    zP' = wP'-wz = wP'-xy = 10.39-10.85 = - 0.41km

    This means that P' co-ordinates are 11.2km, 0.41km.

    This makes PP' square root of (11.2)^2 - (0.41)^2
    = square root of 125 = 11.2km.
     

    Attached Files:

  15. Oct 31, 2013 #14
    My final answer is PP' is 11.2 km and 2.1 degrees North of East.
     
  16. Oct 31, 2013 #15

    tms

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    A couple of pieces of advice: First, don't use x, y, and z to label point; those letters are too easily confused with the coordinates. For instance, your Ay is really the x coordinate of point A.

    Second, in this problem, you are given three vectors in polar coordinates (length and direction), which you must add together. The easiest way is to convert each vector to rectangular coordinates (x, y) and then simply add them. This will eliminate all the extra points you created, and if you use the right angles, will take care of the signs automaticaly.

    You seem to have mixed up the vertical and horizontal here. 30 degrees S of W is the same as 210 degrees (mathematically speaking, with 0 being the positive x direction). Therefore,
    [tex]
    \begin{align}
    BP' &= (r, \vartheta) = (12, 210^\circ) \\
    &= (r \cos \vartheta, r \sin \vartheta) \\
    &= (-10.4, -6)
    \end{align}[/tex]
     
  17. Oct 31, 2013 #16
    What would the magnitude of P' be?
     
  18. Oct 31, 2013 #17
    The co-ordinates of A is 9,15.6.
    B is 17.2, 10.85 and P' is 11.2 and 0.41.

    The object first moves to A then B and then P'.

    So, do I have to subtract A-B? If I do that the co-ordinates of C and the final answer is very different.
     
  19. Oct 31, 2013 #18

    tms

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    Right. But you should consider the vector from P to A, not the point A by itself, in which case the components of the vector are as you state.

    Again, look at the components of the vectors, not the coordinates of the points. It is much easier that way, and easier to visualize.

    For instance, the vector from A to B has a length of 9.5 and a direction of -30 degrees (in polar coordinates), which leads to x and y components of (8.22, -4.75). Do that for all three vectors, and then add them.

    No, you add all three vectors. If you calculate the components properly, some will be negative.
     
  20. Oct 31, 2013 #19
    Thank you very much. I will follow what you said.
     
  21. Oct 31, 2013 #20
    So, following the method you mentioned, x and y components of P' will be 6,10.39 or will it be -6 and -10.39?

    I just want to clarify addition of vectors, A+B+C = (Ax+Bx+Cx)+(Ay+By+Cy)?
     
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