# Motionless chain in a tube

Good evening, I come to this site with doubts on some very basic concepts.

I have started reading through Morin's Classical Mechanics. Among the problems on Chapter 2, Problem 3 says:

A frictionless tube lies in the vertical plane and is in the shape of a function that has its endpoints at the same height but is otherwise arbitrary. A chain with uniform mass per unit length lies in the tube from end to end, as shown in Fig. 2.9. Show, by considering the net force of gravity along the curve, that the chain doesn’t move.

I understand the solution provided by the textbook. However, when I try to think about the problem in terms of external forces, I cannot help but reach a contradiction:

Let the function describing the chain be $$f$$, $$f(a)=f(b)=0$$. Consider an element $$dx$$ of the chain, having mass $$dm = dx\sqrt{1+f\prime^2(x)}$$. The four forces acting on the element are gravity $$\vec{g}dm$$, the normal force $$\vec{N}$$ perpendicular to the line tangent locally to the function and the two tensions $$\vec{T_1},\vec{T_2}$$ that, in the limit where $$dx$$ is small, we can consider parallel to this same tangent. Since this means $$\vec{N}$$ and $$\vec{T_1}+\vec{T_2}$$ are perpendicular, the only values possible are $$N = g\cos\theta$$ and $$T_1+T_2=g\sin\theta$$, where $$\theta=\arctan\frac{f\prime(x)}{\sqrt{1+f\prime^2(x)}}$$. In particular the normal force is always smaller, in modulus, than the force of gravity. But the sum of the external forces is in this configuration equal to $$\int(d\vec{N}+\vec{g}dm)$$. Since the gravity vectors are all parallel, we have $$|\int{d\vec{N}}| < \int|d\vec{N}| < \int|\vec{g}dm| = |\int\vec{g}dm|$$. So we do not have equilibrium?

I do not understand why this happens. Is there an additional external force? Does the normal force not point perpendicularly to the tube, or the tension not point parallelly?

• The textbook question appears to assume that the chain is incompressible, whereas you have omitted any non-tension forces between chain links, so your model is for a limitlessly compressible chain. Consider a perfectly vertical pipe: no tension forces act on the chain, which is in free fall where it is not touching the ground, but if the chain is compressible, it obviously moves. (Or, since the question specifies the pipe ends are the same height, two vertical pipes side by side connected with a U joint at the bottom.)
– g s
Apr 30, 2022 at 18:08
• When you say that "the chain is incompressible", you mean that then there isn't tension forces?
– Jon
Apr 20, 2023 at 16:20

If you restrict $$\vec{T_1}$$ and $$\vec{T_2}$$ to being collinear, then the only admissible configuration according to the original question (which specifies matching endpoint heights) is a horizontal tube and chain. In this case, the normal force isn't smaller than the gravitational force—they're equal and opposite.

For a more general solution, you may wish to consider the possibility of $$\vec{T_1}$$ and $$\vec{T_2}$$ not being collinear.