Resolving Vector Along Non-Standard Axes

[srg]

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!

 

[CWatters]

As a check... R has no vertical component so do the vertical components of Ra and Rb sum to zero?

 

[HallsofIvy]

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.

 

[ehild]

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.