Sum & Angle of Three Vectors (College Physics)

AI Thread Summary
The discussion revolves around calculating the resultant vector from three given vectors with specified lengths and angles. The user initially calculated the x and y components using the formulas Rx = Ax - Bx + Cx and Ry = Ay - By + Cy, but faced discrepancies with the homework site's answers. Participants suggested that the subtraction of vector B might be incorrect, as it may not represent an opposing force based on the angle convention. Additionally, there were concerns about the precision of significant figures and the calculation of the resultant angle using the tangent function. The user is seeking clarification on their calculations and the correct approach to vector addition.
bebe087
Messages
7
Reaction score
0
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 homework is due Saturday.
 
Physics news on Phys.org
bebe087 said:
(a) Rx = Ax-Bx+Cx Ry = Ay-By+Cy

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 :smile:

Furby said:
Your formula appears to be prematurely subtracting a force that may not necessarily be an opposing force.

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 homework 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.
 
bebe087 said:
We're subtracting because that's the homework question as it was written by the instructor.
Okay, I missed that when I first read the problem.

I added this course late and do not know how to add vectors. I've ordered the textbook, but it hasn't arrived yet.
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)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top