# co.combinatorics – Can a generic rope exert the same force as a rope with loops?

This is a spin-off of my previous question. I will take a moment to reintroduce the notation. The problem concerns elastic “ropes” in $$mathbb{R}^3$$, which are modeled as sequences of points $$gamma=(x_1,x_2,dots,x_m)$$. A rope is supported on the union of line segments
$$S(gamma) := bigcup_{j=1}^{m-1} overline{x_jx_{j+1}} subsetmathbb{R}^3.$$
The $$j$$-th segment of the rope has direction
$$tau_j := frac{x_{j+1} – x_j}{|x_{j+1}-x_j|},$$
and the force $$F_gamma$$ associated to the rope $$gamma$$ is described by the vector-valued measure
$$F_gamma := sum_{j=1}^{m-1} tau_j (delta_{x_{j+1}} – delta_{x_j}).$$

This time I am curious about “generic” ropes $$gamma$$. We say that $$gamma$$ is generic if the points $${x_1,x_2,dots,x_m}$$ are in general position. That is, the points are all distinct, no three are collinear, and no four are coplanar. In particular, no two of the line segments $$overline{x_jx_{j+1}}$$ and $$overline{x_kx_{k+1}}$$ intersect except when $$j=k+1$$ or vice versa (so there are no loops in $$S_gamma$$).

Question: Suppose that $$gamma$$ and $$gamma’$$ are two ropes with the same force, so
that $$F_gamma = F_{gamma’}$$. If $$gamma$$ is generic, does it follow that
$$gamma’$$ is also generic?

What I believe I can show is that, if $$gamma’$$ is not generic, it must have at least three loops. I would also be curious if there was a reasonable strengthening of the “generic rope” condition that satisfied this conjecture.