Angle of vevtor in component form

AI Thread Summary
The discussion focuses on determining the angle of a vector in component form, specifically for the vector A = 3.50i - 7.70j. The correct angle below the positive x-axis is calculated as 65.56 degrees, which is derived from the arctangent of the ratio of the y-component to the x-component. While some participants suggested alternative angles like -65 degrees or 294 degrees, the question specifically asks for the angle measured clockwise from the x-axis, confirming that 65 degrees is the appropriate answer. The confusion arises from differing conventions in angle measurement, but clarity is provided that the specified direction dictates the final answer. Understanding these conventions is crucial for accurately interpreting vector angles.
arukia
Messages
3
Reaction score
0
Angle of vector in component form

Find the vector's magnitude and direction. Use right as the +x direction and up as the +y direction.

\vec{A} = 3.50 \hat{i} - 7.70 \hat{j}

Find \theta_A degrees below the positive x-axis.

\theta_A = arctan\left ( \frac{7.70}{3.50} \right )

=65.55604522^\circ

360^\circ - 65.55604522^\circ = 294.4439548^\circ

But this is wrong. I also tried: -65^\circ and 336^\circ
 
Last edited:
Physics news on Phys.org
Usually the angle is measured anticlockwise from the x axis, in which case your answer of 294.44 degrees would be correct. When you found the angle below the positive axis, that was an intermediate step? Or is that what the question asked you for? If the question asked for that, then the answer would be 65.56 degrees.
 
No my 294^\circ was the final answer and what was being asked.
Yes, 65^\circ turned out to be correct but I don't understand why.
Since the question asked: Find \theta_A degrees below the positive x-axis. I thought it would be -65^\circ or 294^\circ
By the way are links to a non-copyrighted PDF allowed in the forums ?
 
I'm not sure about the PDF's... I would suppose so. About the problem, they ask for degrees below the positive x-axis, so this means the angle as measure clockwise from the x-axis, which is why the answer is +65 degrees. In the usual convention of measuring anti-clockwise, the answer would have been 294 degrees, but they specify to not use the convention.
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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