I calculated the derivatives incorrectly, as the angle also changes. The correct approach is to either derive from ##(2r)^2 = c^2 + h^2## or use ##\delta \theta## to calculate ##\delta c## and ##\delta h##. This leads to ##F = -\frac{W}{\tan \theta}##, which resolves to 2 for the tension
The difference between this and calculating using projections is that using projections, ##F = \frac{W}{\tan\theta}##. I'm trying to understand what I did wrong with the virtual work principle
Here's a drawing. All the projections are done on the bottom plane. I calculated h as ##c \tan\theta##, which allowed me to calculate dh/dc above.
So, moving bottom balls inward slightly as a result of these tension forces will move the top ball slightly up, changing its potential energy as...
You're right, sorry. I tried being more rigorous, but it turned out to be more confusing.
For simplicity, we can omit the safety factor and just focus on ##F##. ##F## is the sum of all forces supporting the top ball in equilibrium. It acts through the centroid of the bottom triangle.
$$F =...
After applying a small displacement along the axis between the center of the bottom ball and the projected center of the fourth ball, I got 4 as the answer, which is twice the correct value. I’m assuming I didn’t account for the fact that the tension force acts on both sides, so I ended up...