Solving Vector Sum A+B=C: Graphically

[bilalbajwa]

Help Please: MY effort is here!

Homework Statement

In the vector sum A + B = C , vector A has a magnitude of 11.3 m and is angled 49° counterclockwise from the +x direction, and vector C has a magnitude of 15.0 m and is angled 20.0° counterclockwise from the -x direction. What are
(a) the magnitude and
(b) the direction (relative to +x
of vector B?
Warning! Make a graphical solution first so you will know what quadrant B lies in.

Please show me the graphical representation/soloution of the vactors
I can solve it theoretically but people please tell me where this B vector will be. How we find out that in which quad this vector is, what sign "-, +" it bears and also what "relative to +x or -x means"?

 

[dontdisturbmycircles]

If we do your homework for you, you gain nothing! Show some effort.

So, did you draw the vectors? I'll give you a hint here... If vector A added to vector B yields vector C, what would the vector C - A look like?

+x means on the positive x-axis of the X-Y plane,

look familiar? :p

Here is an example of what they mean from direction from +x.. just replace the word 'east' with +x

http://www.glenbrook.k12.il.us/gbssci/Phys/Class/vectors/u3l1a4.gif

(I notice you added this part to your post : " How we find out that in which quad this vector is, what sign "-, +" it bears and also what "relative to +x or -x means"?" , I would have answered it but didn't notice it until later ;-))

 

[husky88]

To find out where the B vector is, you can try to draw a diagram to scale. Measure the angles and the distances to scale. This will provide an accurate picture. To find out the sign of the vector is easy. Just look if it is on the negative or the positive side.
Relative to -x means the opposite direction of x. Imagine x pointing the other way. In the question, relative to +x means the angle between x and B.

 

[bilalbajwa]

Here is what i am thinking

Please tell me am i thinking in a right way? When i looked up for answers in the book why B's direction is -147.61?

 

[husky88]

(I don't want to be mean, but have you ever heard of the words "Thank you"?
Not for me, because I didn't help that much, but other answers you receive are quite comprehensive and you don't even bother to post a thank you. And I am referring to all the threads that you started.)

 

[bilalbajwa]

Thank you every body for helping me.
But please tell me am i thinking in a right way in my last thread before this?

 

[husky88]

Yep that is correct.
You said you got the numerical value, but I don't understand why B = -147.61 either. Normally they would ask for an angle, but you could also say that B is in the direction of -x, although not really laying on -x. So I guess relative to +x, B is in the opposite direction.

 

[dontdisturbmycircles]

So you now have done the graphical solution. So now you must do the algebraic one.

You will want to use vector decomposition to solve for the magnitude and direction of vector B algebraically.

I will decompose vector A for you and then you can decompose vector C and do the math and find out the magnitude and direction of vector B.

If vector A has a magnitude of 11.3m and is angled 49° from the positive x axis, then you can decompose into its horizontal/vertical components as follows.

A_{x}=11.3m(cos49^{\circ})=7.4135m

A_{y}=11.3m(sin49^{\circ}) = 8.5282m

Now decompose vector C into its horizontal and vertical components and then subtract the vertical component of A from the vertical component from C and then record that number as B_{y}. Do the same for the horizontal components of C and A. Then use pythagoras's theorem to find the magnitude of B and use trigonometry to find the direction of B.

I suggest you draw lots of diagrams while you do this so that you fully understand what you are doing. Good luck. ;)

 

[bilalbajwa]

Thanks for Reply

I have done all the calculations. Thats why i am coming back to you guys again and again. If my visual graphical represenation (that i posted earlier) is correct then why my vector B direction is -147

My calculations for vector B=Bx+By
B=-21.508i-13.658j
B(magnitude)=24.47m
Direction=Bx/(B magnitude)*cos inv=147.58
But the answer in the book is -147.58

The sign with 147.58 is really confusing me.!

How you guys post equations and pics here. tell me the process. Picture resolution?

 

[dontdisturbmycircles]

Okay I am having a bit of trouble finding out where you are getting your answer of +147.58° from, but I understand what the book means so let me explain it.

When they say the answer for the direction of B is -147.58, negative angles are always measured CLOCKWISE FROM THE POSITIVE x axis. So -90° is the same as +270°.

You are semi-correct on the direction(it is -147.58° or 212.42°) but your answer for the magnitude is incorrect. (maybe it was a typo?) It is 25.479

 

[bilalbajwa]

Thanks a lot for helping me out and clarifying the concept!
Opps sorry i missed type the magnitude!
It means whenever the book says that right your answer relative to +x; i have to go clockwise from +x axis to the vector!?
They must have to tell us its either clockwise or counter clockwise; they do gave the direction for Vector A and C but for Vector B they said "relative to +x of vector B".
How we know from where to measure it?

 

[dontdisturbmycircles]

If you said it was 212.42° from +x, there is no question about it that is a correct and valid answer, and I would say that it makes much MORE sense than measuring it clockwise and stating the angle as -147.58° but the fact is that they mean the exact same thing.

Note that your original answer of +147.58° would not be considered correct though because that would be in quadrant 2.

I always measure the angle counter-clockwise from the positive x-axis as that is the traditional way to do it, but again, if your textbook measures it clockwise and states it as a negative measurement, they mean the same thing. So no worries.

 

[bilalbajwa]

Q1 ...Sorry, the magnitude is 25.479
So it means both answers are correct
212.42
-147.58

Q2. How u guys put degree symbol, pictures and equations in your message?

Thanks a lot. Now my concept is polished.

 

[dontdisturbmycircles]

Yes both answers are correct.

q2. You put the degree symbol in your first post and I just copy/pasted. ;)

For the top two pictures I just googled "cartesian coordinate system" and "angle measurement" or something and right clicked on the pictures -> properties -> used tags to put the images into the posts.

You can press the "quote" button on any of my posts to see how to use the [PLAIN] tags.

Here is an introduction on how to use latex to write out your equations :

[url]https://www.physicsforums.com/misc/howtolatex.pdf[/url]

No problem. ;)