Vector forces and Trigonometry help!!! Please!!!
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Vector forces and Trigonometry help!!! Please!!!

[From: ] [author: ] [Date: 11-06-14] [Hit: ]
Then add their y components to find Y.I got M = 43.59 N @ 263.46° required for equilibrium.......
Five forces act on an object: (1) 60.0 N at 90°, (2) 40 N at 0°, (3) 80 N at 270°, (4) 40N at 180°, (5) 50 N at 60°. What are the magnitude and direction of a sixth force that would produce equilibrium?

I've worked out this far...2 and 4 cancel each other out, 1 and 3 cancel to 20 N at 270° and 50 N at 60°....I have no idea where to go from here??? How do I set up an equation from this? What is the answer??? I am so lost!

Thank you in advance!

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Take the remaining vectors (20 N @ 270° & 50 N @ 60°) and add their x components to find X. Then add their y components to find Y.

The resultant vector magnitude M = √[X²+Y²] at Θ = arctan[Y/X]

The answer will be a vector with the same magnitude and angle ß = Θ ± 180°

I got M = 43.59 N @ 263.46° required for equilibrium.
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