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Homework Help: Resolving Vector Along Non-Standard Axes

  1. Sep 19, 2014 #1


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    Hi guys,

    I have a problem in which I have to resolve [itex]\vec{R}[/itex] along two axes, a and b. However, those axes don't have a right angle between them (hence, non-standard). See the image below.

    http://srg.sdf.org/images/PF/VectorHW.png [Broken]

    I believe I'm doing this correctly, however my textbook has very limited examples and I'd like to verify my work.

    My method for solving this is to create a triangle by "moving" the axes around and then solving for the two components of the vector.

    http://srg.sdf.org/images/PF/VectorHW2.png [Broken]

    In which case, using the law of sines, I get:
    [tex]\frac{R_a}{\sin{110}} = \frac{800}{\sin{40}} \therefore R_a=1169.5[/tex]
    [tex]\frac{R_b}{\sin{30}} = \frac{800}{\sin{40}} \therefore R_b=622.3[/tex]

    Thinking about the results logically/graphically, it seems to make sense that [itex]R_a[/itex] has a higher magnitude than [itex]R_b[/itex] and that the two components make up the proper angle for [itex]\vec{R}[/itex].

    Again, I believe this to be correct, however my textbook as limited examples and I'd like to confirm.

    Thank you PF!
    Last edited by a moderator: May 6, 2017
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  3. Sep 20, 2014 #2


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    As a check... R has no vertical component so do the vertical components of Ra and Rb sum to zero?
  4. Sep 20, 2014 #3


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    The fact that there are two lines is not really important at first- just find the projection of the given vector on each separately. To find the projection of the vector onto a imagine dropping a perpendicular from the tip of the given vector to a. That gives a right triangle with angle 30 degrees and hypotenuse of length 800N. Its projection onto a is the "near side", 800 cos(30). Similarly, the projection of the give vector on b is 800 cos(-110)= 800 cos(110)= -800 cos(20). [itex]\vec{R}= (800 cos(30))\vec{a}- (800 cos(20))\vec{b}[/itex] where [itex]\vec{a}[/itex] and [itex]\vec{b}[/itex] are unit vectors in the directions of lines a and b, respectively.
  5. Sep 20, 2014 #4


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    You solved the problem correctly, but I would show a different picture. Resolving the vector R means that you write it as the linear combination [itex]R= R_a \hat a + R_b \hat b [/itex], sum of two vectors, parallel to a and b like in the attached picture.
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