Net force due to gravity (sort of conceptual questions)

[LogicX]

Homework Statement

http://edugen.wiley.com/edugen/courses/crs4957/art/qb/qu/c13/fig13_37.gif

Masses m are the same on the bottom, and the net gravitational force is 0 on m4 which is located at the center of an equilateral triangle. What is the mass of M?

Homework Equations

Fg=GmM/r^2

The Attempt at a Solution

Ok, obviously the answer is that M=m because its an equilateral triangle so the masses on all three sides must be the same. But I was thinking about it conceptually a different way and I'm not sure why it doesn't work.

Lets say you find the center of mass of the bottom two. It is located halfway between them both. So you can consider both masses of m as if they were in that point right? So the center of mass (with mass 2m) is in line with M and the same distance away from m4 as M, because m4 is in the center of the triangle. So in order for M to offset 2m to 0, M must equal 2m. What am I doing wrong here?

Here is another one.

http://edugen.wiley.com/edugen/courses/crs4957/art/qb/qu/c13/qu_1.8.gif

What is the net force on the center mass, in component form? (and you are given the different masses of all the particles, and a side length of the square)

I got this one correct using the regular method of getting each force from each mass on the center mass (with r being half of the diagonal through the square) to get a vector and then using sin45 and cos45 to divide the vector into component form. But then I got it wrong by doing it this way:

Instead of getting the magnitude of the total vector I tried to solve for each component separately. So taking an example force, say the y component of the force of mass 2 on mass 5. I tried to use an r value of side length/2 because that is how far away it is in the y direction. Shouldn't this have worked?

Thanks.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Doc Al]

Lets say you find the center of mass of the bottom two. It is located halfway between them both. So you can consider both masses of m as if they were in that point right?

No. If the two masses were in a uniform gravitational field and you wanted to find the net force acting on them, then you can use that approach. But that's not what's going on here. Here you are trying to find the field created by those masses.

Actually do the calculation of the field created by the two masses at point m4. Then compare it to the field from a single mass of 2m directly below m4.

 

[Doc Al]

Instead of getting the magnitude of the total vector I tried to solve for each component separately. So taking an example force, say the y component of the force of mass 2 on mass 5. I tried to use an r value of side length/2 because that is how far away it is in the y direction. Shouldn't this have worked?

No. Take this example. Mass m is at the origin. The point P is at (1 m, 10000000 m). Clearly the force at P is small because it's over 10000000 m away from mass m. If you computed the x-component based on a distance of only 1 m, you'd be wildly wrong.