View Full Version : Physics Questions About Vector
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.
Welcome to PF!
Post in some more detail why you don't get it.
Well I'm taking ap physics right now and I dont get it my teacher just puts two problems on the board every day and then goes on solving them. He doesnt teach, so far I dont 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 isnt the vector of A = 3?
since 3cos90 = 3
How would you solve this problem?
Use the paralelogram addition of vectors.
I have no clue what to do? I read the chapter three times and I just dont 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 wont get it. I have this text book
http://www.brookscole.com/cgi-wadsworth/course_products_wp.pl?fid=M20b&product_isbn_issn=003023798X&discipline_number=13
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?
yep, the coordinates of X would be (0,0) to (0,3)
the coordinates of Y would be (0,0) to (0,-4)
Good!
So the arrowhead of X lies at (0,3) and the arrowhead of Y at (0,-4)
Do you agree with this?
geometer
Oct3-04, 12:43 PM
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?
(3,0) to (3,-4) so it would look like
--->
....|
....|
....|
....|
....\/
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
Oct3-04, 12:50 PM
Looks like I was a day late and a dollar short here! Sorry for jumping in the middle of this!
(3,0) to (3,-4) so it would look like
--->
....|
....|
....|
....|
....\/
like that right? and then you draw a line? and use the pythagorean theorem
--->
\...|
.\..|
..\.|
...\|
....\/
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?
a squared + b squared = c squared?
3*3 + (-4)*(-4) = 25
square root of 25 = 5?
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)
45 degrees, going southeast, Oh when they ask what is A + B they are asking for the point at which it ends
45 degrees are incorrect; think again.
No, why did you think it was 45 in the first place?
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)
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?
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?
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))
..../\
../.|
./..|
/...|
--->
is it correct now?
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)
so its similar to my first graph for part A except you dont move it over?
--->
|.../
|../
|./
\/
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.
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)
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.
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
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).
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
/\--->
|..../
|.../
|../
|/
so magnitude and direction are still the same though right?
except its positive degrees
Yes, this is because X and Y are at right angles to each other.
(It is not true in general)
Note:
The "direction" is not the "same" since it uses positive degrees rather than negative degrees.
actually the angles different since
tan = opp/adj
so its tan^-^1 = 3/4
Nope, the angle between the vector (3,4) and the x axis has "4" as the opposite side, and "3" at the adjecent side
(3,4) is (x,y)
so X-Axis is 3
and Y-Axis is 4
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.
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
yep lol now theres this odd problem that I've been trying to solve but I cant get it. The answers in the back of the book but theres 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)____
Well, it's just the same procedures really; are you supposed to answer with a diagram or with coordinate values?
this one isnt hw, i just want to get a better understanding of how to do it. the answers in numbers and theres a degree angle too.
I know how to move it but then you cant do the theorem with this.
.......|
.......|
^.....|
|...../
|..../
|../
|/
----------- >
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
umm i did that... i get a different answer from the book
(2.6,1.5)
the other one is three but the same thing can be done to found it
3cos(90),3sin(90)
(0,3)
(2.6,1.5) looks right;
what does your calculator say when you type in cos(30)?
Multiply that number with 3.
i get .8660254038
*3 = 2.598076211
How would you get the answer after that?
Rx = Ax + Bx = 2.6 + 0
Ry = Ay + By = 1.5 + 3
Now I add Rx + Ry = 7.1
Answer in book is 5.2m at 60 above x-axis
HAVE YOU FORGOTTEN PYTHAGORAS??????
The resultant vector is:
(\frac{3\sqrt{3}}{2},\frac{9}{2})
The length is therefore:
\sqrt{\frac{27}{4}+\frac{81}{4}}=\frac{\sqrt{108}} {2}
how did u get the resultant vector?
\cos(30)=\frac{\sqrt{3}}{2},\sin(30)=\frac{1}{2}
(This is a rather well-known relation; you'll it later on)
Hence A has coordinates (\frac{3\sqrt{3}}{2},\frac{3}{2})
Summing A with B (coord. (0,3)) yields the resultant vector).
ahh indeed it does.. this makes sense now :D
yes I learned cos(30) = 3/2 in precalculus the 30-60-90 triangle
wohoo got B) correct how would you get the degree though?
ummm I know all the measurements of the sides yet no angles except that 30 degrees, but its on the outside
But you know its horizontal and vertical coordinates, right?
How can you calculate the tangent of the angle using that info?
the figure isnt a right triangle though
You are to find the resultant vectur's angle to the x-axis; how can you construct a triangle in such a manner that the coordinates you've been given will help you find that angle?
OOO I got it :D
extend the Line downward so the verticles of the triangle are
tan^-^1 = 4.5/2.6
(0,0) --> (2.6,0) --> (2.6,4.5)
You should get 60 degrees..
Yep I got 60 degrees 59.9 i think. Thanks for your help :D
nice i think i got section 3.2 and 3.3 of my text book down ahah... tommo I will probably ask about projectile motion, hopefully you'll help me again. The textbook leaves out alot of information. Like it leaves out little steps.
shweta tiwari
Mar9-11, 11:25 PM
its an emergency...i really need to know this.....
can two vectors representing two different physical quantities be equal????
can force and displacement be equal vectors if they are in same direction and have same magnitude?? does'nt representing different phy. quantities makes them different vectors???
Rayquesto
Mar9-11, 11:30 PM
I hate vectors! I'm in physics A and I don't get it, even though I understand all of the force, energy, and projectile motion stuff.
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.