The discussion revolves around subtracting vectors, specifically the problem of calculating the resultant of 3 m/s [E] and 5 m/s [N]. Participants are exploring the methods of vector subtraction and the implications of directionality in vector components.
Some participants have offered guidance on the graphical interpretation of vectors and the concept of flipping vectors for subtraction. There is an ongoing exploration of how to find the magnitude and direction of resultant vectors, with various interpretations being considered.
Participants mention the potential complexity introduced by angles and the need for trigonometric functions when vectors do not form right triangles. There is also a reference to homework constraints that may limit the methods available for solving the problem.
Is this a graphical or algebraic method? I don't remember how to add vectors, either.RJLiberator said:Based on your question you have 3m/s E. This leads me to believe that the vector is due east. Likewise with the 5m/s north.
Do you recall how to add vectors?
Subtracting is merely A-B = A+(-B).
RJLiberator said:If you look at it graphically, we have 3 m/s due east, meaning right on the x-axis. x=3
We have 5 m/s due north, meaning y = 5
A+(-B) would mean you flip B, and then place the tail of B to the head of A. Now you have a right triangle, what can do you with a right triangle to find the missing values?
Thank you!RJLiberator said:In this case you could break up the actual components of each vector.
If vector a was 5 m/s and at 45 degrees north of east, then you have an angle and the hypotenuse.
You should know/recall from trig that cos(angle) = a/h so h*cos(angle) = a
Now you have the velocity of that vector in the due east direction. You can do the same with sin to find the velocity in the due north direction.
Once you break up the components of the vectors, you can add/subtract the components as necessary and then pull everything back together.