# Simple Net Gravitation problem

1. Nov 26, 2009

### godtripp

Here's the question.

"Four identical masses of 800 kg each are placed at the corners of a square whose side length is 10.0 cm. What is the net gravitational force on one of the masses due to the other three?"

for convenience, I replaced mass with "M" and the distance with "d"

So, I figure that Net Force is equal to the sum of the gravitational forces between the three masses.

So if we call the force between the vertical mass and horizontal mass F1 and F2 respectively then F1=F2

Calculating out F1 gives me F= (GM^2)/(d^2)

And by Pythagorean the distance to the horizontal mass is d$$\sqrt{2}$$

so F3= (GM^2)/(2d^2)

So net force FN = 2F1+F3

or (5GM^2)/(2d^2).

However this is not the proper answer....

Proper answer is 8.2 x 10^-3 N

Can someone help me with this please? Thank you!

2. Nov 26, 2009

### alphysicist

Hi godtripp,

Remember that here the net force is a vector sum, so it will be the vector sum of the three individual forces.

Their magnitudes are equal, but their directions are different.

When you perform the vector sum, there will be some cancellation occuring between F1 and F2 (so you cannot simply add the magnitudes together). Do you see what to do?

3. Nov 26, 2009

### godtripp

Thank you so much for your reply. I totally neglected that the net force would have superposition.

Thanks again

4. Nov 26, 2009