Morin's mechanics problem 2.3 (motionless chain)

In summary, Morin solves the problem by 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.
  • #1
Adgorn
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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...
 
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  • #2
Adgorn said:
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.

1642013042462.png


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?
 
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  • #3
TSny said:
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!
 
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Related to Morin's mechanics problem 2.3 (motionless chain)

1. What is Morin's mechanics problem 2.3?

Morin's mechanics problem 2.3 is a physics problem that involves a chain suspended between two points and its motion when released from rest.

2. What is the purpose of Morin's mechanics problem 2.3?

The purpose of this problem is to apply principles of mechanics, such as Newton's laws of motion and conservation of energy, to analyze the motion of a chain in a real-world scenario.

3. What are the assumptions made in Morin's mechanics problem 2.3?

The problem assumes that the chain is massless, inextensible, and has no friction. It also assumes that the chain is released from rest and that the only forces acting on it are gravity and tension.

4. How do you solve Morin's mechanics problem 2.3?

To solve this problem, you can use the principles of mechanics to set up equations for the forces acting on the chain and solve for the unknown variables, such as the acceleration and tension in the chain. You can also use conservation of energy to determine the velocity of the chain at different points in its motion.

5. What are some real-life applications of Morin's mechanics problem 2.3?

This problem can be applied to real-life scenarios involving objects suspended by chains, such as a hanging lamp or a bridge with suspended cables. It can also be used to understand the motion of objects released from rest and the forces involved in their motion.

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