- #1
almangar
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Homework Statement
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).
Homework Equations
a) Cx = Ax + Bx
b) Cy = Ay + By
c) Ax = ACos[tex]\Theta[/tex]
d) Ay = ASin[tex]\Theta[/tex]
e) Bx = BCos[tex]\Theta[/tex]
f) By = BSin[tex]\Theta[/tex]
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[tex]^{-1}[/tex](Cy/Cx) which gives me tan[tex]^{-1}[/tex](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!