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.

 

[HalfLostAndSearching]

Hi there,

Your method for solving this problem is correct. By using the law of sines, you have correctly determined the components of \vec{R} along the non-standard axes a and b. Your graphical representation also helps to visualize the problem and confirm your solution. Well done!

It's always good to double check your work and seek confirmation when needed. Keep up the good work in your studies and don't hesitate to ask for help when needed. Good luck with your future scientific endeavors!