Net Gravitational Force

  • #1
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?
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
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Welcome to PF!

Hi singinglupine! Welcome to PF! :smile:

(have a square-root: √ :wink:)
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))

Sorry, but I don't understand what formula you've applied here. :redface:

Use mMr/r3 :smile:
 
  • #3


Hi singinglupine! Welcome to PF! :smile:

(have a square-root: √ :wink:)


Sorry, but I don't understand what formula you've applied here. :redface:

Use mMr/r3 :smile:

Thanks for the welcome :) I used the law of universal gravitation where
F = GmM/r2 and G is the gravitational force constant. The parts with something else multiplied by the force are the x, cos(θ) and y, sin(θ) components. I used pythagorean theorem to figure out the length of the hypotenuse, so that cos would be adjacent/hypotenuse and sin opposite/hypotenuse.
 

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