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2D Vectors - Addition and Subtraction of Successive Displacement Vectors

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bearjew11
#1
Jan24-13, 05:28 PM
P: 4
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 Thumbnails
vector addition problem.jpg  
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Andrew Mason
#2
Jan24-13, 11:46 PM
Sci Advisor
HW Helper
P: 6,672
Quote Quote by bearjew11 View Post
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. (?)
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
tms
#3
Jan25-13, 12:07 AM
P: 616
Quote Quote by bearjew11 View Post
So, my understanding is that the net displacement of the successive displacement vectors is the overall sum of the successive displacement vectors. (?)
Correct.
ΔA
Hypotenuse = 18km
Height = 18sin60
Base = 18cos60
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).

tms
#4
Jan25-13, 12:10 AM
P: 616
2D Vectors - Addition and Subtraction of Successive Displacement Vectors

Quote Quote by Andrew Mason View Post
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').
The resultant is also the sum of the three given vectors. The OP is correct on this point.
Andrew Mason
#5
Jan25-13, 09:16 AM
Sci Advisor
HW Helper
P: 6,672
Quote Quote by tms View Post
The resultant is also the sum of the three given vectors. The OP is correct on this point.
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
tms
#6
Jan25-13, 09:41 AM
P: 616
Quote Quote by Andrew Mason View Post
I responded before the drawing was available, thinking P and P' were displacement vectors from a common origin.
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.
bearjew11
#7
Jan25-13, 11:35 AM
P: 4
Quote Quote by Andrew Mason View Post
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.).

AM
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.

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).
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.
tms
#8
Jan25-13, 12:52 PM
P: 616
Quote Quote by bearjew11 View Post
Can you elaborate on this a bit more, please?
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].
Andrew Mason
#9
Jan25-13, 02:24 PM
Sci Advisor
HW Helper
P: 6,672
Quote Quote by bearjew11 View Post
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.
Right.


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.
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
physicsmybane
#10
Oct31-13, 06:46 PM
P: 8
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.
physicsmybane
#11
Oct31-13, 06:50 PM
P: 8
Please note that in the figure in post #1, all the 60 degree angles are supposed to 30 degrees.
tms
#12
Oct31-13, 07:15 PM
P: 616
Quote Quote by physicsmybane View Post
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.
Show what you've done so far.
physicsmybane
#13
Oct31-13, 08:02 PM
P: 8
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 Thumbnails
Untitled.jpg  
physicsmybane
#14
Oct31-13, 08:06 PM
P: 8
My final answer is PP' is 11.2 km and 2.1 degrees North of East.
tms
#15
Oct31-13, 09:58 PM
P: 616
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.

Quote Quote by physicsmybane View Post
Angle wBP' is 60 degree and BP' is 12 km.
wP' = (sin 60) 12 = 10.39km.
wB = (cos 60) 12 = 6km.
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]
physicsmybane
#16
Oct31-13, 10:19 PM
P: 8
What would the magnitude of P' be?
physicsmybane
#17
Oct31-13, 10:34 PM
P: 8
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.
tms
#18
Oct31-13, 10:45 PM
P: 616
Quote Quote by physicsmybane View Post
The co-ordinates of A is 9,15.6.
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.

B is 17.2, 10.85 and P' is 11.2 and 0.41.
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.

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.
No, you add all three vectors. If you calculate the components properly, some will be negative.


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