Finding the Components of a Vector in the Third Quadrant

[jpd5184]

Homework Statement

what are the x and y components of a vector a in the xy plane if its direction is 210(degrees) counterclockwise from the positive direction of the x-axis and its magnitude is 14 m.

Homework Equations

vector is in the third quadrant with an angle of 30(degrees)

The Attempt at a Solution

i got the angle to be 30(degrees) so i just did

14sin30 for the x component = 7.0
14cos30 for the y component = 12.1

should they both be negative since there in the third quadrant or do the negatives cancel to get a positive.

 

[Char. Limit]

Negatives cancel only when you multiply them. Since you aren't multiplying here, both parts are negative.

 

[jpd5184]

would my answers be correct then

x = -7.0

y= -12.1

 

[Char. Limit]

That seems right, yes.

Although, if significant figures were important, the y component might have to be rounded... I'm not sure how important they are here.

 

[mechmech]

14.cos 30=x . if it is 30 degree to x (210), x have to be bigger than y. (to correct)

 

[jpd5184]

well if its 210(degrees) counterclockwise then wouldn't the angle be 30(degrees) sin 210-180 = 30 degrees

 

[jpd5184]

14.cos 30=x . if it is 30 degree to x (210), x have to be bigger than y. (to correct)

why would x be cos and not y. if the vector is in the third quadrant and cos is adjacent/hypotenuse then adjacent would be x so the only side left would be y which you would be solving for.

 

[jpd5184]

anybody please

 

[Char. Limit]

why would x be cos and not y. if the vector is in the third quadrant and cos is adjacent/hypotenuse then adjacent would be x so the only side left would be y which you would be solving for.

I'm not sure what you mean...

If you mean that the x component should correspond to the cosine, this is because the cosine of the angle measures the length of the adjacent side to the angle, in this case, the x-axis.

 

[jpd5184]

ok i got it thanks.

14cos30 = -12.1 x-component
14sin30 = -7.0 y-component