I’m playing silly games with creating an esthetically pleasing and mathematically interesting balance with three arms.
Given three lengths A, B, C, all of them different, A < B < C and A+B > C, it’s possible to connect them in a single point so that this point is also the center of mass. Think a skewed Y-shape.
Constructing this shape is easy, and I’ve figured out that given A, B and C that are also a pythagorean triple there’s a right angle between A and B. But trying to find functions for the other angles between the arms is proving too tricky for me. Since there’s only one Y-shape for every set of A, B and C, there should be such a function, but I’m thinking it might just be too complex to bother with, unless someone among the teeming millions finds this interesting and has an insight.
You have six unknowns, the angles at the center (or their complements) and the side lengths of the blue triangle spanned by the endpoints. You also have six cosine equations for the six small triangles meeting at the centroid G, so it’s solvable.
But yes, it gets ugly. I am also interested if there is a clever shortcut.
I used Cartesian geometry. Let the arm of length C lie along the positive x-axis, and let arm A make an angle theta[sub]A[/sub] below the negative x-axis and arm B make an angle theta[sub]B[/sub] above the negative x-axis. The coordinates of the centers of mass of the arms are then given by:
We then find the center of mass of these three objects (weighted by their masses, which are proportional to A, B, and C), demand that the coordinates of the center of mass are (0,0), and solve for the angles. If I’ve done my math right, these angles work out to be
I can’t yet prove, just by staring at this, that it works out as a right angle when A, B, and C are a Pythagorean triple; or that there’s always a solution if A + B > C. So I may be wrong, but hopefully my description above will let others check my work if that’s the case. If I have more time to think about this in the next couple of days, I’ll check back in.
Is the weight of an arm distributed along its length, or only at its distal end? I assume the latter.
In that case, are not the angles between the arms of a perfect balance each equal to 180° minus the corresponding angle of the triangle formed by the lengths? (Proof, if any, left as exercise. :smack: )
Dangnabbit. I appear to have calculated with arms of negligible mass and equal masses at each end of the balance. I may even made a conscious choice to do so, but forgotten it along the way.
If you assume mass distributed uniformly along the arms, and proportional to the length, then it’s again not trivial, but it also makes the statements in the OP untrue.
It doesn’t matter if the weights are distributed along the length or only at the distal end; the former puts the center of weight of each arm at half its length, which is simply a uniform scaling of the latter.
What does matter is whether each arm has the same total weight, or they have different total weights (e.g., weights proportional to their length).
Regardless, septimus gives the correct answer: take the arms’ lengths and scale them proportionally to their weights. We are looking to create vectors of corresponding lengths which sum to zero (putting their center of mass at the origin, a la MikeS’s approach) and then determine the angles between these vectors. This is as good as creating a triangle with these lengths and then determining the turning angles at each point, as septimus notes.
In particular, if all the arms have the same weight, regardless of length, then we find that Pythagorean triples induce a right angle between the lesser two lengths, exactly as the OP noted. We also have the existence of solutions precisely whenever the lesser two lengths sum to at least the largest length, again as the OP noted, by the standard theorem on existence of triangles with prescribed lengths.
If we take the arms to have weight proportional to their lengths, we can just square all the lengths and then proceed as in the previous case. But this doesn’t seem to be what the OP was getting at.
So if I understand the question properly -
The setup is 2-d (planar)
We have the long piece, C, with CoM at C/2 m and weight D kg.
So, we need to arrange A and B joined at one end of each at some angle so that -
the center of mass E is X m from the joint and the sum (A+B=F kg)
To put the CoM at the common joint, we then join C to AB at the common point such that C is pointing to/in line with E, the center of mass of A and B, and DC/2 = FX - same moment.
The variable needing adjustment is the angle between A and B such that these moment balance, DC/2 = FX
Presumably we’re given the lengths and weights of A, B, C ?
