Calculate force in north/east directions given 2 components

[Rachael Hardman]

Hello,

My question is about using vector components to find a force in a particular direction.

I already have the answer, I just don't understand it and would like an explanation if anyone has one.

If I have two vectors with radial velocities (in this case it's measurements of ocean current speed and direction) of [itex] \nu_{R1} [/itex] and [itex] \nu_{R2} [/itex] whose directions, [itex] \theta_1 [/itex] and [itex] \theta_2 [/itex] respectively, correspond to the angle they make with the x-axis (anticlockwise i.e. 0 -> 360 degrees), how can I combine these to find the vector value in both the north and east directions (i.e. 90% and 0%)?

The answer I've found is:

[itex] u = \dfrac{\nu_{R1}\cos(\theta_2)-\nu_{R2}\cos(\theta_1)}{\sin(\theta_2-\theta_1)} \qquad v=\dfrac{\nu_{R2}\sin(\theta_1)-\nu_{R1}\sin(\theta_2)}{\sin(\theta_2-\theta_1)}, [/itex]

where [itex] u [/itex] and [itex] v [/itex] are the north and east components respectively.

I've checked this numerically and it works - a sound explanation would be great though!

Please let me know if any clarification is needed or any diagrams/etc are required.

Many thanks,
Rachael

 

[jtbell]

how can I combine these to find the vector value in both the north and east directions (i.e. 90% and 0%)?

Can you explain what you mean by "the vector value in both the north and east directions", preferably with a diagram? When I saw that, the first thing I thought of was the components of the vector sum of the two velocities (i.e. what is called the "resultant vector" in many textbooks). The two components in your notation would be simply be $$v = v_{R1} \cos \theta_1 + v_{R2} \cos \theta_2 \\ u = v_{R1} \sin \theta_1 + v_{R2} \sin \theta_2$$ See for example http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html