Addition of vectors and direction

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Homework Help Overview

The discussion revolves around the addition of vectors, specifically focusing on vector A with a magnitude of 56.0 m at an angle of 20.0° below the positive x-axis and vector B with a magnitude of 70.0 m at an angle of 45.0° above the positive x-axis. The objective is to determine the resultant vector C's magnitude and direction using the component method of vector addition.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the components of vectors A and B and subsequently vector C, raising questions about the direction of vector C and the interpretation of angles in different quadrants.

Discussion Status

Participants have provided feedback on the original poster's calculations, suggesting that the issue may lie in the syntax required by the automated system for inputting the direction. There is an acknowledgment of the original poster's frustration with the process and a resolution when the input was finally accepted.

Contextual Notes

The original poster mentions a lack of coverage on this topic by their professor and expresses uncertainty regarding the angle's interpretation due to the positioning of vector A in the fourth quadrant.

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\Theta
d) Ay = ASin\Theta
e) Bx = BCos\Theta
f) By = BSin\Theta

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!
 
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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).
 
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.
 
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.
 

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