- #1

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I have split the forces into x and y components using the F = (Gm1m2)/r^2 equation.

So for my x-forces I get:

F = (309)(20)G/2 * (1/sqrt(2)) ----> this is a negative force pulling down

F = (20)(20)G/1.25 * (.5/sqrt(1.25))

y-forces:

F = (309)(20)G/2 * (1/sqrt(2)) ----> this is a negative force pulling down

F = (18)(20)G/.25 ----> this is a negative force pulling down

F = (20)(20)G/1.25 * (1/sqrt(1.25))

Then I took the sum of the Xs squared and the sum of the Ys squared and then the square root to find the magnitude of the resultant vector. I got 2.612E-7 N.

Then for the degrees I took the inverse tan ( sum of the Ys/sum of the Xs) +180 deg to get it in the third quadrant for a result of 282.2 deg. Somewhere I've done something wrong because I'm getting this incorrect, where did I make a mistake?