Morin's mechanics problem 2.3 (motionless chain)

[Adgorn]

Homework Statement

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 does not move.

Relevant Equations

F=ma

In his solution, Morin solves the problem as the hint suggests: cutting the chain into small pieces, taking the component of the external forces along the curve (which is just the component of gravity here) and summing up an in integral, obtaining 0. He then claims that because the "total external force along the curve" is 0, that the chain will not move (accelerate) along the curve. The problem is I have no idea why this statement is justifiable.

Newton's laws (which are just barely introduced in this first chapter, so it seems that's all I am expected to work with here) say that if the net force on a point particle is 0, the particle will not accelerate. Here Morin seems to take a chain, which is basically a bunch of connected particles, and say that if the sum of the components of the forces on all particles along the curve vanishes, then the entire chain will not move along the curve, a statement which requires so many leaps of logic that I'm not sure how to begin deconstructing it. He summed up a bunch of different forces in a bunch of different directions and treated them as forces in a single direction "of the curve", treating the chain like a rigid object in 1-D space. I don't know how to start justifying this treatment without literally modeling the chain as a bunch of point particles connected by rods or springs and analyzing the internal forces to see how the force acting on a single point affects the entire chain.

I tried solving this problem without using this treatment but its proving difficult. It suffices to show the total force on the chain is 0, so that the center of mass does not move, meaning the chain does not move. This requires finding the normal force on each small piece of chain, which depends on the change in the direction and magnitude of the tension along the small piece, which in turn requires 2nd order approximation of the curve. This becomes rather convoluted, especially when I don't know a priori whether the chain actually moves or not.

At any rate, I was caught off guard by this nonchalant leap from forces on point particles to forces on pliable objects constrained to manifolds, so I would love some clarification here...

 

[TSny]

I don't know how to start justifying this treatment without literally modeling the chain as a bunch of point particles connected by rods or springs and analyzing the internal forces to see how the force acting on a single point affects the entire chain.

Consider a small segment of the chain. Let $s$ denote arc length along the chain, let $T(s)$ denote the tension in the chain as a function of arc length, let $dN$ denote the normal force that the tube exerts on the segment, and let $dW$ be the weight of the segment.

Show that Newton’s second law for the direction tangent to the segment may be written $$dT - \lambda g ds \sin \theta = (\lambda ds)a_t$$Here, $dT = T(s+ds) – T(s)$. $\lambda$ is the linear mass density and $a_t$ is the tangential acceleration of the segment.

$a_t$ is the same for all segments since we assume the chain cannot stretch.

You want to show that $a_t$ is zero if the two ends of the chain are at the same vertical height $y$.

Try integrating the equation above from the left end of the chain to the right end of the chain and note that you can express $ds \sin \theta$ as $dy$. What can you say about the tension at each end of the chain?

 

[Adgorn]

Try integrating the equation above from the left end of the chain to the right end of the chain and note that you can express $ds \sin \theta$ as $dy$. What can you say about the tension at each end of the chain?

That cleared it up. Thank you for the clear and illustrated answer!