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Calculating the components of vectors

  • Thread starter kgianqu2
  • Start date
  • #1
16
0
Vector A is in the direction 41.0 degrees clockwise from the y-axis. The x component of A is = -15.0 .

A)What is the y component of vector A?

B)What is the magnitude of vector A?

I got -13.0m for part a, and 19.8m for part b, but mastering physics says they are wrong. Any ideas?
 

Answers and Replies

  • #2
172
2
Can you post your attempt at the problem? It makes it easier to see where you might have made a mistake.
 
  • #3
16
0
I kind of worked backwards since I knew that the angle is 229 degrees (41 degrees clockwise from the -y axis). And I know that the x=-15. I took tan(229) which = 1.15, which should be the fraction of the y length over the x length. Since I know the x=-15 I just solved for y to get -13.0. And then I calculated the magnitude from that, which give me 19.8.

I thought this method would work, and I am not sure where I went wrong.
 
  • #4
16
0
Correction: my first post should say, 41 degrees clockwise from the -y-axis)
 
  • #5
834
2
If the tangent is greater than 1, how can the y component be smaller than the x component?
 
  • #6
172
2
So you did:
[itex]\displaystyle tan(229°)=\frac{A_y}{A_x}[/itex]

[itex]\displaystyle A_y=A_x tan(229°)[/itex]

and you're claiming that:
[itex]\displaystyle A_y=-15 tan(229°)=-13.0[/itex]

right? however, I get:
[itex]\displaystyle A_y=A_x tan(229°)=-17.25[/itex]

You should try re-calculating your value for Ay.
 

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