# Addition of vectors and direction

1. Sep 10, 2009

### almangar

1. The problem statement, all variables and given/known data

A vector A has a magnitude of 56.0 m and points in a direction 20.0° below the positive x axis. A second vector, B, has a magnitude of 70.0 m and points in a direction 45.0° above the positive x axis.

Using the component method of vector addition, find the magnitude and direction of the vector C (counterclockwise from the positive x-axis).

2. Relevant equations

a) Cx = Ax + Bx
b) Cy = Ay + By
c) Ax = ACos$$\Theta$$
d) Ay = ASin$$\Theta$$
e) Bx = BCos$$\Theta$$
f) By = BSin$$\Theta$$

3. The attempt at a solution

I have calculated out all the relevant equations:
a) Cx = 52.6 + 49.5 = 102.1m
b) Cy = 49.5 - 19.2 = 30.3m
c) Ax = 56Cos20 = 52.6m
d) Ay = 56Sin20 = -19.2m (negative due to y in the negative direction per my written graph)
e) Bx = 70Cos45 = 49.5m
f) By = 70Sin45 = 49.5m

I have come up with vector C to be 106.5m using the Pythagorean theorem, but I cannot seem to get the correct direction of vector C. I have tried using the equation tan$$^{-1}$$(Cy/Cx) which gives me tan$$^{-1}$$(30.3/102.1) = 16.5, but when I plug it into the answer it tells me to check my syntax. By my graph, Vector C is in Q1 so I would assume from x-axis counter clockwise the answer should be 16.5deg. What I am unsure of is the fact that Vector A is in QIV and therefore the angle is intersected by the x-axis, so maybe that is where I am having my problem. My professor didn't really cover anything to this degree, so I have been relying on the book and it hasn't been much help either. Your help is greatly appreciated.

As this is my first post to this forum, I was trying to follow the rules to the letter, but if I don't need to type everything I did, would someone please let me know. Thank you very much.

Ahhh....The joy of returning to school after a 10 year hiatus!!

2. Sep 10, 2009

### gabbagabbahey

Your answer looks fine to me, so it must be that the automated system you are plugging it into requires it to be in a certain form (syntax).

3. Sep 10, 2009

### almangar

What other form could there be. It is asking for the direction counterclockwise from the x-axis and is asking for the answer in degrees.

4. Sep 10, 2009

### almangar

Nevermind, I was racking my brain to try and figure out what the problem was, so I just decided to forget it and try re-inputting my answer again after about 4 different times of working out the problem and it finally took my answer @ 16.5 deg which is what I had put into the automated system originally. Thanks for your help.