And if you isolate theta1 instead, I suppose you would have theta1 = arctan((1/3)*tan(theta2)), so in that case you'd have (X/2L) = cos(arctan((1/3)*tan(theta2)) + cos(theta2). Heh, I guess nobody said the answer had to be pretty.
The simplest I can see is X = 2*L*[cos(theta1) + cos(theta2)] So I suppose if you substitute theta2 = arctan(3*tan(theta1)) you'd have (X/2L) = cos(theta1) + cos(arctan(3*tan(theta1))... yeesh.
I got as far as writing out the force diagrams for each mass: Taking the left mass for instance, balancing the forces in the y direction gives T2*sin(theta2) = mg + T1*sin(theta1), and for the bottom mass, 2*T1*sin(theta1) = mg. Substituting gives 3*T1*sin(theta1) = T2*sin(theta2), and if you...
The way I see it, each string wants to reduce its tension by hanging straight down. The strings on top are pulled inward by the middle mass they have to support, and so they pull outward, while the strings on the bottom want to reduce their tension and pull inward.
Suppose you have 3 balls, all of equal mass, M. They are connected to each other by equal-length (each length L) strings of negligible mass, such that one ball is suspended in the middle and the two balls at the end are themselves suspended from the ceiling at two points, a distance X from one...