How Do You Calculate Vector Sums and Differences Graphically?

[soul814]

Vector A is 3.00 units in length and points along the positive x-axis. Vector B is 4.00 units in length and points along the negative y-axis. Use graphical methods to find the magnitude and direction of the vectors.
Find
A) A+B
B) A-B

I don't get it.

 

[arildno]

Welcome to PF!
Post in some more detail why you don't get it.

 

[soul814]

Well I'm taking ap physics right now and I don't get it my teacher just puts two problems on the board every day and then goes on solving them. He doesn't teach, so far I don't get any thing in physics.

Thats a question from my textbook. Is A) 3+-4? That seems wrong. Since R = Vector of (A) + Vector of (B)

but isn't the vector of A = 3?
since 3cos90 = 3

How would you solve this problem?

 

[Pyrrhus]

Use the paralelogram addition of vectors.

 

[soul814]

I have no clue what to do? I read the chapter three times and I just don't get it. I know what hte parallelogram addition of vectors is.

Its when you make a diagonal line between the two constants.

Can someone solve it with an explanation for me? Since even if you say all these rules I still won't get it. I have this textbook

http://www.brookscole.com/cgi-wadsw...uct_isbn_issn=003023798X&discipline_number=13

 

[arildno]

All right, let's take it easy!
1. Suppose you draw an arrow on a piece of paper.
That arrow has two features:
a) It has some length
b) It has a direction
You get that?

 

[soul814]

yep, the coordinates of X would be (0,0) to (0,3)
the coordinates of Y would be (0,0) to (0,-4)

 

[arildno]

Good!
So the arrowhead of X lies at (0,3) and the arrowhead of Y at (0,-4)
Do you agree with this?

 

[soul814]

yes i got that much

 

[geometer]

A useful property of vectors is that they remain the same vector if you move them as long as you don't change their magnitude or direction. So, you can move vector B out along the positive x-axis until its "foot" is at the "head" of vector A, and it's still the same vector. Now, what would the vector be that you could draw from the origin to the tip of vector B in it's new position?

 

[soul814]

(3,0) to (3,-4) so it would look like

--->
...|
...|
...|
...|
...\/

 

[arildno]

Allright!
At this position, both X and Y has the "foot" at the origin.
You are asked to ADD Y to X.
That means:
"Slide" the foot of Y along X (without rotating Y!), such that when the foot of Y coincides with the arrowhead of X, you've got a line segment parallell to Y attached to the arrowhead of X (that line segment (the "translated Y") is of equal length as Y).
Draw this.

We say that the SUM of X and Y is the vector which has its foot in the origin, and its arrowhead coincident with the arrowhead of "translated Y"

Get that?

 

[geometer]

Looks like I was a day late and a dollar short here! Sorry for jumping in the middle of this!

 

[soul814]

(3,0) to (3,-4) so it would look like

--->
...|
...|
...|
...|
...\/

like that right? and then you draw a line? and use the pythagorean theorem

--->
\...|
.\..|
..\.|
...\|
...\/

 

[arildno]

Precisely!
This is the graphical way in summing two vectors together; how can you do this arithmetically if you're given the vector's components?

 

[soul814]

a squared + b squared = c squared?
3*3 + (-4)*(-4) = 25

square root of 25 = 5?

 

[arildno]

The Sum of (3,0) and (0,-4) is given by:
(3,0)+(0,-4)=(3+0,0-4)=(3,-4)
(It seems you did this somewhere..)
The magnitude (length) is, as you said, 5.
What is the vector's direction (given, for example, as the angle the vector makes with the x-axis)

 

[soul814]

45 degrees, going southeast, Oh when they ask what is A + B they are asking for the point at which it ends

 

[arildno]

45 degrees are incorrect; think again.

 

[soul814]

negative 45?

 

[arildno]

No, why did you think it was 45 in the first place?

 

[soul814]

3-4-5 triangle?

 

[arildno]

That's definitely correct, but there's no 45 degrees angle in that triangle.
Lets look at the tangent value in a 3-4-5 angle:
You have:
tan(\theta)=\frac{oppositeside}{adjacentside}
In your case (let's just look at "4" rather than "-4") you have:
tan(\theta)=\frac{4}{3}
Agreed?
now, what angle \theta satisfies that equation?
(You'll need to use your calculator)

 

[soul814]

ummm about 53 degrees

 

[arildno]

That's right, so you'll agree the angle the vector (X+Y) is about -53 degrees respective to the x-axis?

Now, for your second problem, how would you approach this one?

 

[soul814]

A - B

Umm my guess is
A (0,0) to (3,0)
B (0,0) to (0,4) Since its Negative

^
|.\
|..\
|...\
|--->

Same magnitude, and this time its positive 53?

 

[arildno]

Your drawing is incorrect:
Let C=X-Y.
Hence, we have:
Y+C=X
(that is C is the vector added to Y so that the result is X)
Draw this.
(By the way, the coordinates of the arrowhead of C when its foot is at the origin is (3,4))

 

[soul814]

.../\
../.|
./..|
/...|
--->

is it correct now?

 

[arildno]

You've certainly plotted the vector (3,4) correctly, however:
1. Plot Y, that is, the line segment from the origin down to (0,-4)
2. Plot X, that is, the line segment from the origin to (3,0)
3. Draw the line segment from the arrowhead of Y to the arrowhead of X.
4. Verify that that line segment is parallell to, and of equal length to the vector going from the origin to (3,4)

 

[soul814]

so its similar to my first graph for part A except you don't move it over?

--->
|.../
|../
|./
\/

 

[arildno]

Note that when you draw (X+Y) and (X-Y) in this manner, you draw up the diagonals of the parallellogram (in this case, a rectangle!) given by X and Y.

 

[arildno]

So, just to finish this:
The drawn line segment is the "translated (3,4)", where by (3,4) I mean the vector with its foot in the origin, and its arrowhead in (3,4).
(You've translated it by dragging it down along Y)

 

[soul814]

I'm confused by what you said. The resultant (R) of Vector A and Vector B is a straight line from the starting point to the ending point. If its two points its usually a diagonal.

 

[soul814]

Ohh a negative would flip the direction of the Y axis right and since vectors are allowed to move anywhere as long as its direction and magnitude is the same I can move it upward to (0,0) and (0,4) then shift it over to (3,0) and (3,4)

therefore the vector will be starting at (0,0) to (3,4)

right? the magnitude would still be 5 and the angle would still be 53 but it would be positive this time

 

[arildno]

Now look:
1. If you add Y and (X-Y) together, then your resultant is X, the diagonal in a parallellogram determined by Y and (X-Y).
(This is what I asked you to draw)
(X-Y) itself is the vector stretching from the origin up to (3,4)
(The leg adjoining Y in the origin in the parallellogram determined by them.)
2. You could also look at X-Y as X+(-Y).
Adding the vector (-Y) to X is what you did in your next to previous post; your resulting diagonal (in the parallellogram determined by X and (-Y)) is then (X-Y).

 

[soul814]

ehh did a seach on google, since at first I didnt get what you said, now after looking at some pictures I think I get what you mean.

First picture (A + B)
|--->
|
|
\/

Second Picture (A + (-B))
/\
|
|
|
|--->

Third Picture Shift (A) Upwards
/\--->
|
|
|
|

Now Draw the Line
/\--->
|.../
|.../
|../
|/

 

[arildno]

Precisely!

 

[soul814]

so magnitude and direction are still the same though right?

except its positive degrees

 

[arildno]

Yes, this is because X and Y are at right angles to each other.
(It is not true in general)

 

[arildno]

Note:
The "direction" is not the "same" since it uses positive degrees rather than negative degrees.

 

[soul814]

actually the angles different since

tan = opp/adj

so its tan^-^1 = 3/4

 

[arildno]

Nope, the angle between the vector (3,4) and the x-axis has "4" as the opposite side, and "3" at the adjecent side

 

[soul814]

(3,4) is (x,y)

so X-Axis is 3
and Y-Axis is 4

 

[arildno]

Definitely!
But its angle with respect to the x-axis, is to regard the triangle with hypotenuse up to (3,4), the vertical line segment down to (3,0) (and then along the horizontal to the origin).
What you're looking at is the vector's angle with respect to the y-axis, not with respect to the x-axis.

 

[soul814]

Ahh I see you calculate this

/\--->
|.../
|../
|./
|/X

and to calculate that you would have to shift the vectors
.../\
.../.|
../..|
./...|
/X..|

therefore its still tan = 4/3

 

[arildno]

Then we agree?

 

[soul814]

yep lol now there's this odd problem that I've been trying to solve but I can't get it. The answers in the back of the book but there's no explanation.

Each of the displacement vectors A and B shown in Figure P3.3 has a magnitude of 3. Graphically find (a) A + B (b) A - B (c) blah blah (d) blah blah

I think If I got one of it I would get the rest.

A is the diagonal line
B is Bolded the vertical line

/\
|.../
|.../
|.../
|../
|./
|/30(degrees)____

 

[arildno]

Well, it's just the same procedures really; are you supposed to answer with a diagram or with coordinate values?

 

[soul814]

this one isn't hw, i just want to get a better understanding of how to do it. the answers in numbers and there's a degree angle too.

I know how to move it but then you can't do the theorem with this.

...|
...|
^...|
|.../
|.../
|../
|/
----------- >

 

[arildno]

True enough, but let's find the coordinates of the skew line (A).
We know that its length is 3, and the angle to the x-axis is 30.
This means that its coordinates is:
(3\cos(30),3\sin(30))
You are now able to find the coordinates to for example, the sum of A+B