What is the direction of the resultant force for two applied forces on a car?

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The discussion centers on determining the direction of the resultant force from two applied forces on a car, specifically focusing on the angle in relation to the horizontal. The calculated resultant force is 613.71 N, with an x component of approximately 613.41 N and a y component of about 19.38 N. The angle derived from these components is approximately 1.81 degrees, indicating it lies in the first quadrant above the positive x-axis. There is a debate regarding whether the angle should be negative, but participants agree it should be positive based on the quadrant location. The consensus confirms the resultant angle is correctly identified as positive.
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Homework Statement


Two forces, 422 N at 21◦ (above horizontal) and 256 N at 31◦ (below horizontal) are
applied to a car in an effort to accelerate it.

Find the direction of the resultant force
(in relation to forward, with counterclock-
wise considered positive, with −180◦ < <
+180◦).
Answer in units of ◦.

Homework Equations


The Attempt at a Solution


I found the resultant from previous question.
My resultant=613.711888 N
My x component of R=+613.405769
My y component of R=+19.3815275

Arctan (y/x)=arctan (19.3815275/613.405769)=1.809749

My teaching assistant told me that my angle should be negative, but both of my resultant vector components lie in quadrant one, so shouldn't my resulting angle be positive as well?
 
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Hi strag

I got the same result as you. :smile:
 
Yes, the resultant angle is 1.8 degrees above the positive x axis, in the first quadrant (ccw from the positive x axis). So that looks positive to me. (And to songoku:wink:).
 
Nice I got the problem right. Thanks for cross-checking!
 

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