- #1

huh

- 15

- 0

**law of gravitation- is this better?**

Could someone check this please?

Four identical masses of mass 600 kg each are placed at the corners of a square whose side lengths are 14.0cm (.14m).

What is the magnitude of the net gravitational force on one of the masses, due to the other three?

It's asking for the absolute value of the force.

Ok so the equation to use is (GMm)/r^2

And superposition of forces should be used, correct?

This is what I am trying: for the mass in the upper left hand corner

G= 6.67 x 10^-11

(G*600*600)/(.14^2)= 1.23 x 10^-3 for the masses beside and below

(G*600*600)/(.07^2)= 4.90 x 10^-3 for the mass diagonal from the one used

(.07 because the attraction would decrease by half, right?)

So adding these together:

(2)(1.23 x 10^-3) + 4.90 x 10^-3= 7.35 x 10^-3

or

(1.23 x 10^-3)+(-1.23 x 10^-3)+ 4.90 x 10^-3 = 4.90 x 10^-3

Last edited: