# Mathematical problem of balance **URL:** \ **Category:** Factual Questions\ **Created:** [October 16, 2014, 10:31am UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477 "2014-10-16T10:31:57Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [October 16, 2014, 10:31am UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/1 "2014-10-16T10:31:57Z")
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. ---
**Author:** ![kk\_fusion](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/kk_fusion/32/15276_2.png) [@kk\_fusion](https://boards.straightdope.com/u/kk_fusion)\ **Post date:** [October 16, 2014, 12:41pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/2 "2014-10-16T12:41:55Z")
Illustration [here](http://jwilson.coe.uga.edu/emt668/emat6680.2001/mealor/writeup4/image1.gif). 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. ---
**Author:** ![MikeS](https://avatars.discourse-cdn.com/v4/letter/m/919ad9/32.png) [@MikeS](https://boards.straightdope.com/u/MikeS)\ **Post date:** [October 16, 2014, 1:14pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/3 "2014-10-16T13:14:08Z")
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: -A/2 \* (cos(theta[sub]A[/sub]), sin(theta[sub]A[/sub])); B/2 \* (-cos(theta[sub]B[/sub]), sin(theta[sub]A[/sub])); and (C/2, 0). 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 cos(theta[sub]A[/sub]) = (A[sup]4[/sup] - B[sup]4[/sup] + C[sup]4[/sup])/(2 C[sup]2[/sup] A[sup]2[/sup]) and cos(theta[sub]B[/sub]) = (B[sup]4[/sup] - A[sup]4[/sup] + C[sup]4[/sup])/(2 C[sup]2[/sup] B[sup]2[/sup]). 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. ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [October 16, 2014, 1:39pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/4 "2014-10-16T13:39:58Z")
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: ) ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [October 16, 2014, 2:03pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/5 "2014-10-16T14:03:58Z")
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. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [October 16, 2014, 2:05pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/6 "2014-10-16T14:05:37Z")
Of course, if I hadn’t the answer would have been a trivial 120 degrees and simple symmetry. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [October 16, 2014, 2:18pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/7 "2014-10-16T14:18:15Z")
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. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [October 16, 2014, 2:27pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/8 "2014-10-16T14:27:40Z")
> [@MikeS](#): > > 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: > > -A/2 \* (cos(theta[sub]A[/sub]), sin(theta[sub]A[/sub])); B/2 \* (-cos(theta[sub]B[/sub]), sin(theta[sub]A[/sub])); and (C/2, 0). > > 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 > > cos(theta[sub]A[/sub]) = (A[sup]4[/sup] - B[sup]4[/sup] + C[sup]4[/sup])/(2 C[sup]2[/sup] A[sup]2[/sup]) > > and > > cos(theta[sub]B[/sub]) = (B[sup]4[/sup] - A[sup]4[/sup] + C[sup]4[/sup])/(2 C[sup]2[/sup] B[sup]2[/sup]). > > 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. I get a theta[sub]A[/sub] of zero, so something is off. I don’t know what, since you didn’t show your work. 😉 ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [October 16, 2014, 2:37pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/9 "2014-10-16T14:37:57Z")
> [@naita](#): > > Of course, if I hadn’t the answer would have been a trivial 120 degrees and simple symmetry. > [@Chronos](#): > > 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. You’re right. And I’m wrong again. It looks like it leads to even more horrible math. ---
**Author:** ![Boogly](https://avatars.discourse-cdn.com/v4/letter/b/f4b2a3/32.png) [@Boogly](https://boards.straightdope.com/u/Boogly)\ **Post date:** [October 16, 2014, 3:12pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/10 "2014-10-16T15:12:24Z")
Pardon me I am not a mathematician. There are 3 lines, Line A, line B and line C. Not using right angles. 1/2 of Line A is centered at 0,0 1/2 of Line B is centered at 0.0 1/2 of Line C is centered at 0.0 0,0 is equal to center mass. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [October 16, 2014, 3:15pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/11 "2014-10-16T15:15:59Z")
> [@Boogly](#): > > Pardon me I am not a mathematician. > There are 3 lines, Line A, line B and line C. > Not using right angles. > > 1/2 of Line A is centered at 0,0 > 1/2 of Line B is centered at 0.0 > 1/2 of Line C is centered at 0.0 > 0,0 is equal to center mass. That doesn’t give a Y shape. I intended to require the shared point to be an endpoint of A, B and C. ---
**Author:** ![Boogly](https://avatars.discourse-cdn.com/v4/letter/b/f4b2a3/32.png) [@Boogly](https://boards.straightdope.com/u/Boogly)\ **Post date:** [October 16, 2014, 3:19pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/12 "2014-10-16T15:19:09Z")
nm ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [October 16, 2014, 3:27pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/13 "2014-10-16T15:27:43Z")
> [@septimus](#): > > 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: ) 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. ---
**Author:** ![Boogly](https://avatars.discourse-cdn.com/v4/letter/b/f4b2a3/32.png) [@Boogly](https://boards.straightdope.com/u/Boogly)\ **Post date:** [October 16, 2014, 3:29pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/14 "2014-10-16T15:29:09Z")
Damn you **Naita** I spent the last couple hours trying to think of how to get this done on a 2D plane. I was sure I was on the cusp of figuring it out! And damn me for nothing recognizing the skewed Y. You want me to think in 3D? Without glasses? ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [October 16, 2014, 4:11pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/15 "2014-10-16T16:11:17Z")
> [@Boogly](#): > > Damn you **Naita** > I spent the last couple hours trying to think of how to get this done on a 2D plane. I was sure I was on the cusp of figuring it out! > > And damn me for nothing recognizing the skewed Y. You want me to think in 3D? Without glasses? Y is a 2D shape. We appear to be talking past each other here. 😃 ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [October 16, 2014, 4:12pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/16 "2014-10-16T16:12:55Z")
> [@Indistinguishable](#): > > 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. Oooh. That simplifies that case a lot. ---
**Author:** ![Boogly](https://avatars.discourse-cdn.com/v4/letter/b/f4b2a3/32.png) [@Boogly](https://boards.straightdope.com/u/Boogly)\ **Post date:** [October 16, 2014, 4:44pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/17 "2014-10-16T16:44:35Z")
> [@naita](#): > > Y is a 2D shape. We appear to be talking past each other here. 😃 I doubt anyone here is talking past me…rather over me. Again forgiveness if I am out of my league. If a triangle is not a right triangle could the lengths still be squared with weight as a measure? ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [October 16, 2014, 5:51pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/18 "2014-10-16T17:51:41Z")
Sure. Why couldn’t they be? ---
**Author:** ![md2000](https://avatars.discourse-cdn.com/v4/letter/m/73ab20/32.png) [@md2000](https://boards.straightdope.com/u/md2000)\ **Post date:** [October 16, 2014, 6:09pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/19 "2014-10-16T18:09:14Z")
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 D_C/2 = F_X - same moment. The variable needing adjustment is the angle between A and B such that these moment balance, D_C/2 = F_X Presumably we’re given the lengths and weights of A, B, C ? Is this what we are trying to solve? Interesting puzzle. ---
**Author:** ![Boogly](https://avatars.discourse-cdn.com/v4/letter/b/f4b2a3/32.png) [@Boogly](https://boards.straightdope.com/u/Boogly)\ **Post date:** [October 16, 2014, 6:35pm UTC](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477/20 "2014-10-16T18:35:37Z")
Wouldn’t this be a simple leverage formula? As long as the weight X length are equal on both sides it should equal right? Or (weight A x length) + (weight B x length) = weight C x length Assuming the weights of A,B,C are the same per length. Full disclosure…alcohol is involved. I beg forgiveness in advance. [Next page](https://boards.straightdope.com/t/mathematical-problem-of-balance/701477.md?page=2)