That only gives you the correct answer if A and B are parallel. The challenge is in the angle between them, if C lies along the x-axis, what you get is A’s and B’s components along the y-axis have to cancel out, while the sum of them has to equal C.
When length and weight are the same, your version gives us A^2+B^2=C^2, which I just discovered myself when I tried to find a solution for A=3, B=4 and C=5 and it told me the “angle”* between A and B was zero.
*I’m not calculating the actual angle.
Using GeoGebra and a mess of geometric features I don’t necessarily need I can generate solutions for the case of lengths A, B, C with masses A, B, C. For instance this one, where all three arms are connected at (0, 0) A is 3, B is 4 and their other ends are at:
A (1.337437, -2.6667)
B (3.4641, 2)
C (-4.24023, 0)
Multiplying each coordinate by the length of the arm shows they cancel out.
The general formula is:
A (sqrt(A^4-x^2) / A, x/A)
B (sqrt(B^4-x^2) / B, x/B)
C (C, 0)
where x is the solution to the equation: sqrt(sqrt(A^4-x^2) + sqrt(3^4 - x^2)) = C
OK, so taking the model of m mass/unit length, length C on the X axis, length A above the X axis by angle α, and length B below the X axis by angle β we get:
m(C/2)*C = m(A/2)Acos(α) + m(B/2)Bcos(β)
and
m(A/2)Asin(α) = m(B/2)Bsin(β)
All the ms and 2s drop out and you get:
C^2 = A^2cos(α) + B^2cos(β)
and
A^2sin(α) = B^2sin(β)
OK, I move the A^2* cos(α) to the other side, square both sides of each equation, then add the equations together to get:
A^4 + C^4 – 2(AC)^2*cos(α) = B^4
Or
α = acos((A^4 + C^4 - B^4)/(2(AC)^2))
if you move the B^2*cos(β) over instead and repeat, you get:
β = acos((B^4 + C^4 - A^4)/(2(BC)^2))
or, just what MikeS said.
Square the Pythagorean theorem and substitute for the C term in the numerator of the alpha equation. This gives:
cos(α) = (2A^4 + 2(AB)^2)/(2(AC)^2)
or
cos(α) = (A^2 + B^2)/(C^2)
Using the Pythagorean theorem again that gives us 1/1 or α is zero. Same with the β equation.
So, if you have a Pythagorean triple, you need to the two smaller lengths together exactly opposite the big length.
ETA: Naita discovered that this holds for point masses when the “length and mass are the same” which is a bit specious, but the math then works out the same for distributed mass lengths.
I got the same thing with your numbers but I would assume that was because there are both Line A and B double up on a line point from the fulcrum where Line C only has a single line going out.
Couldn’t the angles of Line A and B be calculated using ‘Length’ from the fulcrum? If you know the length and distance from 0,0 you can figure the angle right?
And only now did I get that this is what MikeS formula showed.
The general solution is as already noted: take A, B, and C to be the products of lengths and weights. The angle between arms corresponding to A and B is given by the law of cosines: it has cosine equal to (C^2 - A^2 - B^2)/(2AB). And analogously for the other pairs of arms.
I was reading that the Pythagorean and Law of Cosines apply to right triangles only. I thought that the OP was about triangles other than 90 degrees.
The OP had two questions. The first was if there was a general equation for the angles. MikeS and I came up with the same answer for that. The second was what if the lengths form a Pythagorean triple (when A^2 + B^2 = C^2). We showed that for that special case the two small lengths have to be together and opposite the large one.
Therein lie my problems. What most people wrote looked like english, well it had english words, but it read like a foreign language.
I read Pythagoreans, and Cos and Sin all over and became befuddled and focused on other than right triangles.
Ignorance not only fought but pummeled. This has been frustratingly fun. Thank you all for putting up with me.
Sure, if weight is proportional to length; as noted, this corresponds to traversing a triangle with side lengths of A^2, B^2, and C^2. If the former two sum to the latter, this triangle is a flat line, with the traversal consisting of two hops in one direction and then a returning hop in the opposite direction.
More generally, with more arms, the question amounts to the relative orientations of the legs of a(n oriented) polygon of prescribed lengths.
(This was in response to two posts up)
The Pythagorean Theorem only applies to right triangles. The Law of Cosines applies to all triangles, and in fact the Pythagorean Theorem is just the special case of it where the angle is 90 degrees.