Resolve 3 Forces & Find Resultant Force and Angle - Basic Question

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SUMMARY

The discussion focuses on resolving three forces: a 600N force at 50 degrees, a 300N force at 85 degrees, and a 700N force at 50 degrees from the negative x-axis. The participant calculated the components as Fx = -38.13 and Fy = 1294.7, leading to an initial angle of -88 degrees. However, this angle indicates a position in Quadrant 4, which contradicts the expected result in Quadrant 2. The correct angle is determined to be approximately 92 degrees counterclockwise from the positive x-axis, emphasizing the importance of accurately interpreting the signs of the components and visualizing the forces.

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Homework Statement



Force Resolution Question-

3 forces act on something,
-a 600N force which is 50 degrees from the +x axis
-a 300N force which is 85 degrees from the +x axis
-a 700N force which is 50 degrees from the -x axis

Resultant Force and angle

The Attempt at a Solution



Yes, very basic question, not hard to get, but i had a question about the angle I am getting.

This is the set up to a much larger more complicated problem, but i figured id post it here because this part of the question is basic.

I resolved the forces and got Fx=-38.13 and Fy=1294.7, now my question comes in here. For angle i did theta= arctan(1294.7/-38.13) which is equal to -88 degrees... That doesn't seem right to me and if i go with the wrong angle for this, ill bomb the whole question.

If i plot Fx and Fy on an xy plane its clear that it is slightly past 90 degrees.An angle of -88 degrees would hint that the resultant is in Quadrant 4, when it is in fact in quadrant 2. I noticed if i add 180 to it, i get roughly 91 degrees, which sounds like it would be along the same line of action as the actual resultant. Can someone tell me where I am going wrong here? I am either missing something very obvious or what, I am not sure
 
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It is always good to draw a sketch of the problem; otherwise, the plus and minus signs will drive you crazy. The resultant is in Quad 2, 88 degrees clockwise from the negative x axis, or 92 degrees counterclockwise from the positive x axis.
 
PhanthomJay said:
It is always good to draw a sketch of the problem; otherwise, the plus and minus signs will drive you crazy. The resultant is in Quad 2, 88 degrees clockwise from the negative x axis, or 92 degrees counterclockwise from the positive x axis.

Thanks jay,

Thats what i figured, I had the picture all drawn out and that's what I was getting. It was just odd to me that I wasnt getting theta= 92 degrees. I guess it just gave me the reaction force for some reason. My teacher is very nit picky so he would probably screw me for the whole problem if i just left the angle as -88 without explaining why.
 

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