Resolving Vector Along Non-Standard Axes

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Homework Help Overview

The discussion revolves around resolving a vector \(\vec{R}\) along two non-standard axes, labeled as axes a and b, which do not intersect at a right angle. The original poster seeks to verify their approach to this problem, which involves using the law of sines and graphical reasoning to determine the components of the vector.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster describes their method of creating a triangle to resolve the vector and calculates the components using the law of sines. Some participants question the vertical components of the resolved vectors and their sum. Others suggest projecting the vector onto the axes and using trigonometric functions to find the components.

Discussion Status

Participants are exploring different methods to resolve the vector, with some providing alternative approaches and clarifications. There is no explicit consensus on the best method, but various interpretations and techniques are being discussed.

Contextual Notes

The original poster mentions limited examples in their textbook, which may affect their confidence in the method used. There is also a consideration of the angles involved in the resolution of the vector.

srg
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Hi guys,

I have a problem in which I have to resolve \vec{R} 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

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

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

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

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

Thank you PF!
 
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As a check... R has no vertical component so do the vertical components of Ra and Rb sum to zero?
 
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). \vec{R}= (800 cos(30))\vec{a}- (800 cos(20))\vec{b} where \vec{a} and \vec{b} are unit vectors in the directions of lines a and b, respectively.
 
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 R= R_a \hat a + R_b \hat b, sum of two vectors, parallel to a and b like in the attached picture.
 
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