1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Resolving Vector Along Non-Standard Axes

  1. Sep 19, 2014 #1

    srg

    User Avatar
    Gold Member

    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
  2. jcsd
  3. Sep 20, 2014 #2

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    ehild

    User Avatar
    Homework Helper
    Gold Member

    resvec.JPG


    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Resolving Vector Along Non-Standard Axes
  1. Resolving vectors (Replies: 2)

  2. Resolving A Vector (Replies: 3)

Loading...