Adding vectors using geometry

In summary, vector addition using geometry is accomplished by attaching the tail of one vector (in this case B) on the tip of the other (A) and using the laws of plane geometry to find the length C, and angle phi, of the resultant (or sum) vector, C=A+B.
  • #1
sami23
76
1

Homework Statement


The vectors A and B have lengths A and B, respectively, and B makes an angle [tex]\theta[/tex] from the direction of A.
Vector addition using geometryVector addition using geometry is accomplished by putting the tail of one vector (in this case B) on the tip of the other (A) (attached fig.) and using the laws of plane geometry to find the length C, and angle phi, of the resultant (or sum) vector, C=A+B

1. C=[tex]\sqrt{A^2 +B^2 -2ABcos(c)}[/tex]
2. [tex]\phi[/tex] = sin-1([Bsin(c)]/C)

Vector addition using components

Vector addition using components requires the choice of a coordinate system. In this problem, the x-axis is chosen along the direction of A (attached fig.) Then the x and y components of B are B and B respectively. This means that the x and y components of C are given by

3. Cx = A + Bcos([tex]\theta[/tex]),
4. Cy = Bsin([tex]\theta[/tex]).

Which of the following sets of conditions, if true, would show that the expressions 1 and 2 above define the same vector C as expressions 3 and 4?

Pick all that apply.
a. The two pairs of expressions give the same length and direction for C.
b. The two pairs of expressions give the same length and x component for C.
c. The two pairs of expressions give the same direction and x component for C.
d. The two pairs of expressions give the same length and y component for C.
e. The two pairs of expressions give the same direction and y component for C.
f. The two pairs of expressions give the same x and y components for C.

Homework Equations



1. C=[tex]\sqrt{A^2 +B^2 -2ABcos(c)}[/tex]
2. [tex]\phi[/tex] = sin-1([Bsin(c)]/C)
3. Cx = A + Bcos([tex]\theta[/tex])
4. Cy = Bsin([tex]\theta[/tex])

The Attempt at a Solution



Attached diagram drawn.
sinC = B/y
C = sin-1(B/y) = sin-1(B/(BsinC))

I believe the vector A needs to be on the x axis
a. The two pairs of expressions give the same length and direction for C.
c. the 2 pairs of expressions give the same direction and x components for C.
 

Attachments

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  • #2
1. Use equations 3 and 4 to calculate the magnitude C.
2. Find a relation between cos(c) and cos(θ)
3. Compare C that you get in step 2 with equation 1 in view of equation 2.

This should get you going. I am curious, though, what is φ supposed to be? There is no such thing in the drawing, so its relevance to the problem is unknown.
 
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  • #3
using eqn 3 and 4: C = [tex]\sqrt{A^2 + (Bcos\Theta)^2 + (Bsin\Theta)^2}[/tex]
simplify with trig indentity: sin2(theta) + cos2(theta) = 1
C = [tex]\sqrt{A^2 + 2B^2}[/tex]

This would be the length of C.
 
  • #4
Did you ever get an answer to this? This is the same question I'm looking for help with. There seems to never have been any consensus on this particular problem.
 
  • #5
Sami23 gave you the answer in post #3... [edit] but he got it wrong!
The thread looks incomplete because it is a homework question - if we provided a complete answer right off the bat, that would be the same as doing the homework - which is against the rules.

It is up usually to the OP to complete the thread.

For this sort of problem, you add the vectors head-to-tail: physically sketch them out on paper.
It helps to see what is happening if you use some axis - draw A along the x-axis so A=Ai and B=iBcosθ + jBsinθ

This means that A + B = i(A + Bcosθ) + jBsinθ = C

The magnitude of C is the sum of the squares of it's components and the angle C makes to A is the inverse-tangent of the ratio of the j to i components.

From that you can see where sami made a mistake.
 
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  • #6
But the correct answer should actually be [itex]\sqrt{A^{2}+B^{2}+2ABCos\vartheta}[/itex]

And

[tex]\sqrt{A^2 + 2B^2}[/tex] ≠ [itex]\sqrt{A^{2}+B^{2}+2ABCos\vartheta}[/itex]

So post # 3 can not be correct.

Further, suppose that even was the correct answer in # 3 (which it's not), how does [tex]\sqrt{A^2 + (Bcos\Theta)^2 + (Bsin\Theta)^2}[/tex] even reduce to [tex]\sqrt{A^2 + 2B^2}[/tex]? Where does that extra 2 in front of the B come from?

And while i's and j's are not the language included in this poster's question, you did get the correct components, but if you were to sum the squares as you say, it would produce a very different result than post # 3.

This is a MasteringPhysics online H/W and is the same H/W I have right now. In fact, only one angle symbol in this thread poster's diagram differs from the one in mine. And the language of i's and j's are not included in this course, although I kind of have an idea of them. My professor explained unit vectors and immediately after that brief overview on the first day of class, said he would not be using them for the remainder of the course. But sure i'll be using unit vectors in future courses.

Anyway, back to the problem in this thread, this is a tricky problem for me at my current level so please forgive my anxiousness to understand this clearly and fully. Thanks. And I undoubtedly appreciate help. It's just that this particular problem is a buildup to more problems similar to it, but more difficult. I already posted the problem I'm currently trying to understand. Anyway, thanks again.
 
  • #7
Yeah - sami made a mistake.
Looks like you were writing your answer while I was making that explicit.

sami did the right method but blew the algebra.

What I got will derive what you expect.
You do have to do some work yourself :)
 
  • #8
Simon Bridge said:
Yeah - sami made a mistake.
Looks like you were writing your answer while I was making that explicit.

sami did the right method but blew the algebra.

What I got will derive what you expect.
You do have to do some work yourself :)

Yes sir :)

BTW, I had already done some work on my to come up with the same components that you came up with. I should have provided it with my above comment. I think a lot of the confusion is coming from this MasteringChemistry HW.

Basically, you found C the same way me or anyone would, by finding the x and y components, then squaring their sum and taking their root. That is the way to do it.

However...this MasteringChemistry is also asking students to find these vectors a completely different way as well. It's asking us to find the vectors, "geometrically." In otherwords, using the Law of Cosin and Law of Sin formula's from Trigonometry.

I was able to find C using Law of Cosines. However, the next question in the H/W asks me to find Cy, geometrically. And I couldn't figure out how to use Law of Cosines to find Cy. But now, from the thread that I created regarding these questions, I like Serena points out that I can use Law of Sin, which I had not even considered because I was so stuck on Law of Cosine.
 
  • #9
However...this MasteringChemistry is also asking students to find these vectors a completely different way as well. It's asking us to find the vectors, "geometrically." In otherwords, using the Law of Cosin and Law of Sin formula's from Trigonometry.
Yeah - I'm in your other thread too.

I think we should continue this discussion in the this other thread or maybe this one?

But you seem to be multiplying threads on this HW problem ... naughty.
 
  • #10
Simon Bridge said:
Yeah - I'm in your other thread too.

I think we should continue this discussion in the this other thread or maybe this one?

But you seem to be multiplying threads on this HW problem ... naughty.

Yes, I was having some trouble seeking help on this problem. I searched this problem on PF and found that this question had been asked previously so wanted to see if this person had ever gotten an answer to this question. I didn't realize it would eventually get responses on multiple threads. One topic, one thread definately. Sorry about that. I would prefer to continue the conversation on the thread that I like Serena responded on.
 
  • #11
LearninDaMath said:
Yes, I was having some trouble seeking help on this problem. I searched this problem on PF and found that this question had been asked previously so wanted to see if this person had ever gotten an answer to this question. I didn't realize it would eventually get responses on multiple threads. One topic, one thread definitely. Sorry about that. I would prefer to continue the conversation on the thread that I like Serena responded on.
Other thread (singular) ?

I believe you have two threads on this topic.

The one with the screen shots from 'Mastering Physics' or whatever, actually gives enough background for us to understand what it is that you're asking.
 
  • #12
SammyS said:
Other thread (singular) ?

I believe you have two threads on this topic.

The one with the screen shots from 'Mastering Physics' or whatever, actually gives enough background for us to understand what it is that you're asking.



I didn't mean to cause any discontinuity with the thread flow. I apologize.
 
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What is the concept of adding vectors using geometry?

The concept of adding vectors using geometry is a method of combining two or more vectors to determine their resultant vector. This is done by using geometric techniques such as vector addition, subtraction, and scalar multiplication.

How do you add vectors using the geometric method?

To add vectors using the geometric method, you first need to determine the direction and magnitude of each vector. Then, using the head-to-tail method, you draw each vector starting from the tail of the previous one. The resultant vector is drawn from the tail of the first vector to the head of the last vector, and its direction and magnitude can be calculated using trigonometric functions.

What is the difference between adding vectors algebraically and geometrically?

The main difference between adding vectors algebraically and geometrically is the method used. In algebraic addition, vectors are represented using their components, and the addition is done using the rules of vector addition. In geometric addition, vectors are represented using arrows and added using geometric techniques such as the head-to-tail method.

Can vectors be added in any order using the geometric method?

Yes, vectors can be added in any order using the geometric method. This is because the geometric method is based on the properties of vectors, and the resultant vector will be the same regardless of the order in which the vectors are added.

What is the purpose of adding vectors using the geometric method?

The purpose of adding vectors using the geometric method is to determine the resultant vector, which represents the combined effect of the individual vectors. This is useful in many scientific and mathematical applications, such as calculating displacement, velocity, and force.

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