Physics

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?

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  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.

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

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  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

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  4. tan theta = -23.3/-25

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  5. 25 * ( sin 40 )

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