Thread: Find y-component of vector A

Vector A is in the direction 34.0 degrees clockwise from the -y axis. The x-component of vector A is Ax = - 16.0m

What is the y-component of vector A?

Im not sure how to solve it since the magnitude is not known. CI drew a picture of how i think its suppose to be.
Can someone help me set it up and how they got that answer?

I think since the hypotenuse is not known i use tan somehow.

Thank you

Is A supposed to be counter-clockwise from the -y-axis? This is the way you have drawn it and is the only way the x-component can be negative.

Otherwise, your hunch about using the tangent function is a good one...

Originally Posted by MarkFL2

Is A supposed to be counter-clockwise from the -y-axis?

No, Vector A is clockwise from the -y axis.

D'oh...it seems I forgot how clocks move...

Originally Posted by icelated

Vector A is in the direction 34.0 degrees clockwise from the -y axis. The x-component of vector A is Ax = - 16.0m
What is the y-component of vector A?
Im not sure how to solve it since the magnitude is not known.

. From that you can find .

90 - 34 ?

Originally Posted by icelated

90 - 34 ?

I have no clue what you mean by this. You have a right triangle with angle 34 degrees and "opposite side" of length 16. Since you mention the tangent function, yes, if you call the length of the near side "y" then you have [itex]tan(34)= 16/y[/itex]. You can solve that for the absolute value of y (since lengths are always positive).