posted by Jaime .
Four identical masses of 800kg each are placed at the corners of a square whose side length is 10.0cm. What is the net gravitational force (magnitude and direction) on one of the masses, due to the other three?
I think I have to separate the four masses but I'm not sure how to do that??? Thanks
No, don't separate them. Use arguement of symettry.
Pick a corner. The force on it is from across the scquare (the diagnol ), so the force is along the diagonal. Call this force 1.
The other forces (adjacent corners) are along the line connecting them, but resolve that into two components: along the diagonal, and perpendicular to the diagonal. But if you examine the perpendicular force, each of the adjacent corners are in opposite directions, and cancel. So all that is left is the force component of each along the direction of the diagonal. Call this force2
Force net: Force1 + 2*force2 (both adjecent corners)
= GMM/(1.41s)^2 + 2GMMcos45/s^2 where s is side length.
For the radius it is supposed to be r^2. Would I use 10.0cm^2 or 5cm^2? and would I convert that value to meters? Also where did the 1.41s come from? Thank you
Huh? Radius on a square? The sides are 10cm center to center. You have to convert to meters. the 1.41s? The distance across the square diagonal...
thank you for that answer you are amazing!
hey bob its frank here, your younger brother.. well done answering this question you always understood physics better than i did back in high school that's for sure!!
Uhm, to get the length of the diagonal, you should compute it using pythagorean theorem sqrt(.1^2+.1^2)=0.141