Law of gravitation

1. Nov 19, 2006

huh

law of gravitation- is this better?

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?)

(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: Nov 19, 2006
2. Nov 19, 2006

Office_Shredder

Staff Emeritus
You have to break them up into x and y axis components, add those, then use Pythagoras to add them back up.

3. Nov 19, 2006

rsk

I'm not sure of your reasoning here.

I would calculate the new distance by pythagoras and use that.

You're right - the attaraction DOES half, but you've used that fact to half your distance, squared your distance and substitued it into the equation which has given you a GREATER force for that diagonal one... which can't be right!

4. Nov 19, 2006

huh

Okay let's try again:

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

use cos(0) and sin(0)
x-component: 1.23 x 10^-3 y-component: 0
use cos(90) and sin(90)
x-comp: 0 y-comp: 1.23 x 10^-3

(G*600*600)/[(.14^2)+(.14^2)]= 6.13 x 10^-4 for the mass diagonal from the one used
(I saw something like this denominator in an example I found)
use cosine and sine of 45 degrees
x and y-comp: 4.33 x 10^-4