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?

Answer 1

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.

Answer 2

yeah, i do.
I don't understand how to do it or i would practice.

Answer 3

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

Answer 4

tan theta = -23.3/-25

Answer 5

25 * ( sin 40 )