Find Magnitude and Angle of Vector B

[kevinf]

please... need urgent help about vectors

Homework Statement

in the sum vector a + vector b = vector c, vector a has a magnitude of 12.0m and is angled 40.0 degrees counterclockwise from the +x direction, and vector c has a magnitude of 15.0m and is angled 20.0 degrees counterclockwise from the -x direction

a)what is the magnitude
b) what is the angle (relative to +x) of vector B.

The Attempt at a Solution

i first found the x and y component of each vector:

so for vector a the x component would be 12 cos(40) and the y component would be 12 sin(40). for vector C the x component would be 15 cos (20) and the y component would be 15 sin (40). so do i just add the x components of the 2 vectors and the y component of the 2 vectors, do pythagorean theorem and find out what the magnitude is.
so would i be (12cos(40)*15cos(20))^2+(12sin(40)+15sin(20))^2 and then square root?

 

[mgb_phys]

That's the right idea.
Draw a triangle, where the first side is 12m at 40deg above X, then from the end of that draw a line 15m at 20deg below X, then draw a closing line back to the start - this is the resulatant vector.

 

[Avodyne]

Your problem said the second vector is at a 20 degree angle counterclockwise from the minus x direction. If that's not a typo, you need to think about whether the components of the second vector are positive or negative.

 

[kevinf]

yeah, its not a typo, so the second vector will have negative x and y right? does the problem change if vector b is the unknown not vector c.
so what do you guys get, i am just checking my answers. i got 26.59 for the magnitude and 28.87 degrees. http://img178.imageshack.us/img178/1294/image0027nb2.jpg

 

[kevinf]

bump...

 

[Avodyne]

If b is unknown, then b = c-a, that is, you subtract the components of a from the components of c. So your magnitude above looks like it is correct.

 

[kevinf]

but i don't think i subtracted it though, i added Ax and Bx and Ay and By. so my answer in the pic is correct? and i think i get the same answer even if i subtracted

 

[Avodyne]

Your C_x and C_y should both be negative, not positive (which should be clear from your picture). Then, since you are subtracting the positive A components from the negative C components, both terms in both R_x and R_y should be negative. Then, your magnitude of R is correct (I didn't actually check the numbers), but your angle is off by 180 degrees, since both R_x and R_y are negative.

 

[kevinf]

so i add 180 to my theta since both rx and ry are negative going to third quadrant. and the R would be the same because i square it later in the pythagorean theorem, making it positive right?

 

[kevinf]

bump...

 

[Avodyne]

Right! (The magnitude of a vector is always positive.)