- #1
singinglupine
- 15
- 0
The masses and coordinates of three particles are as follows: 20.0 kg, (0.50, 1.00) m; 309.0 kg, (-1.00, -1.00) m; 18.0 kg, (0.00, -0.50) m. What is the gravitational force on a 20.0 kg sphere located at the origin due to the other spheres, magnitude and direction? Give the direction as an angle in degrees counter clockwise with respect to the the + x-axis.
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?
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?