# Recent content by akorn3000

1. ### 3 balls connected by massless strings

Thank you for all your help, I appreciate it.
2. ### 3 balls connected by massless strings

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.
3. ### 3 balls connected by massless strings

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.
4. ### 3 balls connected by massless strings

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...
5. ### 3 balls connected by massless strings

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.
6. ### 3 balls connected by massless strings

They must be balanced, so T1*cos(theta1) = T2*cos(theta2)
7. ### 3 balls connected by massless strings

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...