If the object was symmetrical, I would be able to figure it out. With a non-symmetrical object, I honestly have no idea where to start. If anyone could tell me the relevant general equations I would really appreciate it. I know you guys don't generally give answers, but I'm kinda hoping that since I'm not asking for answers, just the general equations (with variables only) you all might help me out

A rod that is 2 meters long and weighs 2935 grams has weights attached to the end that weigh 405.1 grams (each weight is approximately 100g - the total of all weights added is 405.1g).

The weights have a diameter of approximately 3cm, and have been taped to the end of the rod in such a way that the end of the masses are at the end of the rod.similar to this:
=======OO (the rod extends under the masses, and ends at the same point as the masses end.)

There are two of the weights on the front and two of the weights on the back of the bar, so the cross section of the end with the weights looks like this:

0||0
0||0

The pivot point is .7 meters away from the end of the rod that has no additional weights.

Hi and welcome to the forum!
I don't really get the picture of the system. If you're looking for a formal definiton of center of mass, check out this page : Center of mass - Wikipedia, the free encyclopedia , which involves a triple integral.
So if you haven't taken calculus III yet, I guess your problem can be simplified. But as I said, I'm not able to visualize the system.
Let's wait another helper help you. Good luck.

__________________
Isaac
If the problem is too hard just let the Universe solve it.

Hmmm. Well, imagine a two by four, with four of these http://bodybuilderfitness.com/librar...se_weights.JPG taped to one end of the two by four. Two of the weights on the front (the wide end that is on top in this picture http://upload.wikimedia.org/wikipedi...ted_timber.JPG) and two of the weights on the back - imagine holding the two by four upright, with the weights facing you. The back would be directly behind the wood on the other side of the weights.

Here's a very badly drawn MS Paint diagram

It will be approx:

(1/6) x mass x (width (0.1 m)^2 + length from axis (1.3m)^2)

The pivot point is .7 meters away from the end of the rod that has no additional weights

I dont understand this bit. If the pivot point is as stated, the system wont balance as there is clearly a larger torque on the side of the weights.

If you wish to find the point at which it remains horizontal, you could try the following.

The mass / unit length would be 2935/2000 = 1.4675 .

Let the balance point be at a distance d from the end with the weights.
which are assumed to be on the right side.

The counterclockwise torque provided by the free portion of the rod will be

1.4675 (2000 -d) g x (2000 -d) / 2.
weight x distance



Since the addititional weight is 405.1 gm it should be mass and not weight. We can take each mass to be thus 405.1/4 = 101.275.

The clockwise torque will be provided by 3 forces the weight of the rod to the right of the pivot, the pair of masses closer to the end, and the pair of masses closer to the balance point which are

1.4675 d (d/2)

2 x 101.275 ( d -3/2)

2 x 101.275 [ d - (3 + 3/2)].

Thus we can write

1.4675(2000 -d)g x (2000 -d)/2 = 1.4675 d (d/2) + 2 x 101.275 ( d -3/2)

+ 2 x 101.275 [ d - (3 + 3/2)].

Solve for d.