Posted by Kiley on Sunday, November 2, 2008 at 12:03pm.
Five forces act simultaneously on Point A: the first 60N at 90 degrees, the second 40N at 0 degrees, the third 80N at 270 degrees, the fourth 40N at 180 degrees, and the fifth 50N at 60 degrees. What is the magnitude and direction of a sixth force that produces the equilibriant at Point A?

Physics  drwls, Sunday, November 2, 2008 at 1:35pm
The sixth force is the opposite (negative) of the sum of the first five vectors. Add them up.
You obviously need some practice in vector addition.

Physics  Damon, Sunday, November 2, 2008 at 2:24pm
I assume you are measuring angle from the +x axis (not North)
Add up the five X components
60 * cos 90 = 0
40 * cos 0 = 40
80 * cos 270 = 0
40 * cos 180 = 40
50 * cos 60 = 25
sum = 25 N in x direction so the sixth has 25 N in x direction
Then y components
60 * sin 90 = 60
40 * sin 0 = 0
80 * sin 270 = 80
40 * sin 180 = 0
50 * sin 60 = 43.3
sum = 23.3 so force six has y component = 23.3
magnitude of F6 = sqrt (25^2 + 23.3^2)
negative x and negative y means quadrant 3
tan theta = 23.3/25
theta = 43 degrees below x so
180+43 = 223 degrees

typo  Damon, Sunday, November 2, 2008 at 2:25pm
tan theta = 23.3/25

Physics  Omar, Wednesday, November 16, 2011 at 8:50pm
25 * ( sin 40 )
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