1. You are given three vectors of lengths A=65.7, B=38.4, and C=43.7. The angles are theta a=29.1 degree and theta b=57.2 degree, and C points along the negative y-axis. (a) Determine the length of the vector A-B+C. (b) Calculate the angle of this vector 2. (a) Rx = Ax-Bx+Cx Ry = Ay-By+Cy (b) R=square root of Rx^2+Ry^2 3. Rx=(65.7cos29.1)-(38.4cos57.2)+(43.7cos270) Rx=57.407-20.802+0 Rx=36.605 Ry=(65.7sin29.1)-(38.4sin57.2)+(43.7sin270) Ry=31.95-32.28-43.7 Ry= -44.03 R=square root of 36.605^2+(-44.03)^2 R=57.26 The HW site says my answers are wrong. Please help me and let me know what I did wrong. My hw is due Saturday.
If you draw your vectors on a scratch of paper coming from the same point (just a quick sketch, nothing too precise), and draw their X and Y components, solving the problem should be much clearer and obvious. Components in the same direction should work together, combining. Your formula appears to be prematurely subtracting a force that may not necessarily be an opposing force.
Welcome to PF, bebe087 and Furby Yes, why are Bx and By being subtracted here? By convention, the description that "theta b=57.2 degree" usually means from the +x direction going counter-clockwise. Is there reason to think otherwise?
We're subtracting because that's the hw question as it was written by the instructor. I added this course late and do not know how to add vectors. I've ordered the textbook, but it hasn't arrived yet.
Okay, I missed that when I first read the problem. Looks to me like you have correctly added the vectors, by adding/subtracting x and y components. It may be that you gave an answer with too many significant figures (the original vector lengths were known to the nearest 0.1). Also, what did you get for the vector's angle?
I entered 57.3 for part a and the answer was still wrong. I got -50.3°. Perhaps my calculations are wrong...do the equations look correct to you? This is how I calculated the angle: tanθ=Ry/Rx tan−1(inverse tan)(Ry/Rx)