Finding X and Y Components of a Vector

  • Thread starter Thread starter PonyGirl
  • Start date Start date
  • Tags Tags
    Components Vector
AI Thread Summary
To find the x and y components of vectors with magnitudes of 318 at 260 degrees and 140 at 180 degrees, it's essential to recognize the quadrant in which the angles lie. The angle of 260 degrees is in the third quadrant, which affects the signs of the components. Using trigonometric functions, the correct approach involves applying the cosine and sine functions with the appropriate angle adjustments for the quadrant. The confusion arose from not considering the angle's position relative to the standard axes. Understanding these concepts simplifies the process of calculating vector components accurately.
PonyGirl
Messages
3
Reaction score
0
Find the x and y components for the given vectors below.

Magnitude: 318, Angle: 260 degrees

Magnitude: 140, Angle 180 degrees

No units were specified.

I tried using trigonometric equations to figure them out, but it isn't working so far. What am I doing wrong? For instance, I used cos(260)*318 to get the x component, but Webassign said I was incorrect. Please help me!
 
Physics news on Phys.org
PonyGirl said:
Find the x and y components for the given vectors below.

Magnitude: 318, Angle: 260 degrees

Magnitude: 140, Angle 180 degrees

No units were specified.

I tried using trigonometric equations to figure them out, but it isn't working so far. What am I doing wrong? For instance, I used cos(260)*318 to get the x component, but Webassign said I was incorrect. Please help me!

Which quadrant is the angle 260 degrees in? Why does that matter?
 
Yes, I'm sure it does matter, but this is why I'm confused. We've never dealt with angles larger than ninety degrees in class and yet, we are expected to know how to do this. I started by subtracting the 260 from 360 and I got 100 degrees. So, I used that angle and divided the resulting triangle in half so I could have two right triangles.

Each triangle had a ninety degree angle, a fifty degree angle (100/2), and a forty degree angle (180-90-50). Since the hypotenuse/magnitude was 318, I divided that by two to get the x components of both triangles... And that's where I got confused. I'm completely lost. Sorry I can't explain myself very well. There wasn't much instruction other than what I gave beforehand.
 
Here's a simpler way to go about it. We know the angle is 260 right? It would fall into the third quadrant. 90 degrees in each quadrant. There is 180 in the first 2, and 260 -180 = the angle into the third quadrant.
 
Ohhhh! Okay! That helps a lot; I've never heard this sort of problem explained like that, but it made it much easier to see the problem visually. Thanks bunches!
 
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