Expression for force of gravity on an arbitrary curve with endpoints fixed

Question

Consider the following problem:

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. Show, by considering the net force of gravity along the curve, that the chain doesn’t move.

The given solution is as follows:

Let the curve be described by the function $f(x)$ and let it run from $x=a$ to $x=b$. Consider a tiny segment of the chain between $x$ and $x+dx$. The mass of this piece is $\rho\sqrt{1+f'^2}dx$ where $\rho$ is the mass per unit length. The component of gravitational accelaration along the curve is $-gsin(\theta) = \frac {-gf'}{\sqrt{1+f'^2}}$. The total force is \begin{align} F = \int_a^b-gsin(\theta)dm &= \int_a^b \frac {-gf'}{\sqrt{1+f'^2}}.\rho\sqrt{1+f'^2}dx\\ &=-g\rho\int_a^bf'dx\\ &=-g\rho (f(b) - f(a)) = 0 \end{align} However, since the force of gravity along the curve points along different directions for each differential element, isn't this solution incorrect? If we write the differential force as $$ d\vec F = \frac {-gf'}{\sqrt{1+f'^2}}.\rho\sqrt{1+f'^2}dx\hat t$$ where $\hat t = cos(\theta)\hat x + sin(\theta) \hat y$ is the tangential unit vector, and then integrate, we end up with a totally different expression. Which of these methods is incorrect?

Answer 1

The given solution integrates forces in the direction of the chain such that constraint forces from the frictionless tube can be ignored.

Your solution does give the force due to gravity, but lacks the normal force from the frictionless tube, so it gives the incorrect net force.

Answer 2

If the chain is flexible, but in-compressible, and is fully supported by the tube, then it will act like a fluid. The compression force at each point will depend on the height of the chain above that point. The rest of the weight will be supported by the normal forces from the tube. Your first integral [which reduces to σg(Δy)] is correct. In your last expression, be careful that you have not used the sin(θ) twice.