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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