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Help with vectors (using force tables)?
Alright, so I'm doing a lab and cannot figure out this one part.
On a force table, I've got three vectors: Vector A is at an angle of 0 degrees, and has a force of .6 N. Vector B is at an angle of 90 degrees, and has a force of .8 N. Vector C has a force of 1.0 N (in the opposite direction.) What angle does Vector C need to be at in order to balance A and B (aka, be at equilibrium)?
I'm assuming it has something to do with the formulas y=rsin(theta), x=rcos(theta) and r=sqrt of (x squared + y squared.) But I might be wrong.
Any help would be VERY much appreciated!! I am confused out of my mind.
Thank you!
Yep. It's at the point on the force table marked 0 degrees. The force table is just a circle of 360 degrees. (So, technically, it's also at 360 degrees.)
2 Answers
- Andrew SmithLv 79 years agoFavorite Answer
This problem has been made particularly simply because the vectors are perpendicular.
A and B form two sides of a triangle.the OPPOSITE side is at 90 degrees and has a value of 0.8
The ADJACENT side to the angle you seek is 0.6
and using trigonometric ratios opposite / adjacent = tan( theta)
so theta = atan(0.8 / 0.6) = 53 degrees.
This is the resultant of A and B
Now the balancing force must lie along the same line but be in the opposite direction.
Technically we would specify this by adding 180 degrees to the angle
ie it has a magnitude of 1.0 N in the direction 233 degrees. ( which is the opposite direction to 53 degrees)
