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Calculating Vector Components

  1. Sep 18, 2011 #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.

    [URL]http://loncapa.physics.mun.ca/res/mun/PHYSICS/msuphysicslib/Graphics/Gtype07/prob01a_vectors2.gif[/URL]

    DATA: α=43.7°; β=53.4°; |A|=4.30 cm. The vector A−B is parallel to the −x axis. Calculate the y-component of vector B.

    2. Relevant equations

    [itex]A_x = |A| cos \theta, A_y = |A| sin \theta[/itex]

    3. The attempt at a solution

    I easily found that [itex]A_x = 4.30 cm cos(43.7) = 3.10 cm[/itex] and [itex]A_y = 4.30 cm sin(43.7) = 2.97 cm[/itex].

    The problem I'm having is figuring out how to calculate [itex]B_x[/itex] with only the information I have. If I had [itex]|B|[/itex] then I could easily solve for [itex]B_x[/itex]. The problem states that [itex]\vec{A-B} || -x-axis[/itex], but what role does this play in solving the problem?
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Sep 18, 2011 #2
    Try drawing a picture. It can be very difficult to sort this sort of thing without a good reference.

    The key to this problem is the statement that [itex]\vec{A} - \vec{B}[/itex] is // to the -x axis. What does it mean if a vector is parallel to an axis? Also, how do you subtract two vectors? If you can answer these two questions, you should be able to figure out By.
     
  4. Sep 18, 2011 #3
    A vector being parallel to an axis can tell you something about it's direction. In this case, the vector is parallel to the -x axis. Is the fact that it's parallel to the negative x axis (as opposed to the positive x axis) important here? If so then that means that [itex]\theta_{A-B} = 180 deg[/itex].

    When you subtract a vector [itex]\vec{A}[/itex] from another vector [itex]\vec{B}[/itex], you are essentially adding the negative of [itex]\vec{B}[/itex] to [itex]\vec{A}[/itex].

    So, from this I gather that [itex]\theta_{A-B} = 180 deg[/itex], but I'm still having trouble seeing how that helps me find [itex]B_y[/itex]. To find [itex]B_y[/itex], I need to find the magnitude of [itex]B[/itex], don't I?
     
  5. Sep 18, 2011 #4
    A vector being parallel to an axis can tell you something about it's direction. In this case, the vector is parallel to the -x axis. Is the fact that it's parallel to the negative x axis (as opposed to the positive x axis) important here? If so then that means that [itex]\theta_{A-B} = 180 deg[/itex].

    When you subtract a vector [itex]\vec{A}[/itex] from another vector [itex]\vec{B}[/itex], you are essentially adding the negative of [itex]\vec{B}[/itex] to [itex]\vec{A}[/itex].

    So, from this I gather that [itex]\theta_{A-B} = 180 deg[/itex], but I'm still having trouble seeing how that helps me find [itex]B_y[/itex]. To find [itex]B_y[/itex], I need to find the magnitude of [itex]B[/itex], don't I?
     
  6. Sep 18, 2011 #5
    Technically, the positive x-axis and neg x-axis would both be parallel. The fact that the problem specifies the negative is suggestive, but I would ask the professor if I were you. There's a more important point here though, and it has to do with how you subtract vectors.

    Remember that you have to perform algebraic operations on the components of vectors. [itex]\vec{A} - \vec{B}[/itex] is not the same as [itex]\vec{A-B}[/itex]. So, if [itex]\vec{A} - \vec{B}[/itex] is parallel to an axis, what does that tell you about the components? If your having a problem with this, I suggest drawing a picture or going over vector calculus (search the web, look to your textbooks or ask your professor).
     
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