Calculating the components of vectors

kgianqu2 · Sep 20, 2012

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?

Nessdude14 · Sep 20, 2012

Can you post your attempt at the problem? It makes it easier to see where you might have made a mistake.

kgianqu2 · Sep 20, 2012

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.

kgianqu2 · Sep 20, 2012

Correction: my first post should say, 41 degrees clockwise from the -y-axis)

Muphrid · Sep 20, 2012

If the tangent is greater than 1, how can the y component be smaller than the x component?

Nessdude14 · Sep 20, 2012

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 A_y.

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Calculating the components of vectors

kgianqu2

Nessdude14

kgianqu2

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Muphrid

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