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Adding vectors using geometry

  1. Aug 30, 2010 #1
    1. The problem statement, all variables and given/known data
    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 geometry


    Vector 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.


    2. Relevant 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])

    3. 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.
     

    Attached Files:

    Last edited: Aug 30, 2010
  2. jcsd
  3. Aug 30, 2010 #2

    kuruman

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    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.
     
    Last edited: Aug 30, 2010
  4. Aug 31, 2010 #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.
     
  5. Jan 16, 2012 #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 concensus on this particular problem.
     
  6. Jan 16, 2012 #5

    Simon Bridge

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    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.
     
    Last edited: Jan 16, 2012
  7. Jan 16, 2012 #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.
     
  8. Jan 16, 2012 #7

    Simon Bridge

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    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 :)
     
  9. Jan 16, 2012 #8
    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.
     
  10. Jan 16, 2012 #9

    Simon Bridge

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    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.
     
  11. Jan 16, 2012 #10
    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.
     
  12. Jan 16, 2012 #11

    SammyS

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    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.
     
  13. Jan 17, 2012 #12


    I didn't mean to cause any discontinuity with the thread flow. I apologize.
     
    Last edited: Jan 17, 2012
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