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Two Vector Problem

  1. Sep 20, 2015 #1
    1. The problem statement, all variables and given/known data

    The diagram below shows two vectors, A and B, and their angles relative to the coordinate axes as indicated.

    vector.png

    DATA: α=45.7°; β=55.6°; |A|=3.50 cm. The vector AB is parallel to the −x axis. Calculate the y-component of vector B.


    2. Relevant equations

    rsinθ=y

    3. The attempt at a solution

    A-B is parallel to x so A_{y} must equal B_{y} ? Therefore, 3.5sin45.7 = A_{y} = B_{y} = 2.50 cm ; which is wrong. I'm a bit confused here. It tells me "Since the sum vector (AB) has no y-component, vector A must have the same y-component as vector B. As shown, 'east' is for +x, and 'north' for +y, thus the answer can be negative. " Did I mess up the calculation?
     
  2. jcsd
  3. Sep 20, 2015 #2

    andrewkirk

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    Why do you say it's wrong? They haven't given you a magnitude for B so you just need to choose a magnitude for B that, given its angle, makes its vertical component equal to 2.50.
     
  4. Sep 20, 2015 #3
    It's part of an online assignment, to which when I enter 2.50 cm it tells me it's incorrect.
     
  5. Sep 20, 2015 #4

    andrewkirk

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    Perhaps they really meant to ask you the magnitude of B, because if they didn't want that they wouldn't have needed to give you the angle ##\beta##. Try calculating and entering that and see if it accepts it.

    If that doesn't work, the next thing I'd try is the x component of B.
     
  6. Sep 20, 2015 #5
    Sorry, I should of specified. I supposed they give you β because there's other parts to the question:

    Calculate the x-component of the vector AB.

    Calculate the magnitude of the vector A+B.

    I do think it's being explicit though, in wanting the y-component rather than the vector. I could try but I have four more tries left.
     
  7. Sep 20, 2015 #6

    andrewkirk

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    Perhaps they want a minus sign on the 2.5. The word 'component' can refer either to an unsigned vector or to a signed scalar. If they mean the latter, the answer would be -2.50 since the vertical component is downwards and the positive y direction is up..
     
  8. Sep 20, 2015 #7
    Wow...

    wow....png

    I still don't understand though. So is it that if the vector is 'moving downwards' in a Cartesian plane, the y-component, regardless of the quadrant it resides in, will be negative? Vice versa for 'moving upwards'?
     
  9. Sep 20, 2015 #8

    andrewkirk

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    Yes that's right, although it would be more accurate to say pointing downwards rather than moving.

    Vectors don't have a location. A vector is just a direction and a magnitude. In fact it's a little odd that the two vectors are drawn the way they are, rather than head-to-tail as is the usual convention. I guess that in this case they didn't want to draw them head to tail because they want you to calc things about both A - B and A + B and those two give different head to tail diagrams.
     
  10. Sep 20, 2015 #9
    The y component of A is -2.5. The y component of A - B is yA - yB=0. Therefore, yB=yA= -2.5
     
  11. Sep 22, 2015 #10
    Having trouble with "Calculate the magnitude of the vector A+B. " My friends and I are stuck and what seems like a easy problem.
     
  12. Sep 22, 2015 #11
    What are the x- and y components of the vectors A and B?
     
  13. Sep 22, 2015 #12
    A_{y} = 2.5
    B_{y} = -2.5
    A_{x} = 2.44
    B_{x} = 3.65 (-3.65? it shouldn't matter since Pythagoras)

    We did:

    [tex] \sqrt{(2.44+3.65)^{2} + (2.5+2.5)^{2} } = 7.88 cm [/tex]

    Which it tells me is wrong.
     
  14. Sep 22, 2015 #13
    Would A+B be parallel to the y-axis?
     
  15. Sep 22, 2015 #14
    No. In the figure, is the x component of A positive or negative? In the figure, is the x component of B positive or negative?

    Chet
     
  16. Sep 22, 2015 #15

    I got it now, thanks. Answer was 5.14 cm
     
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