h4tt3n Posted June 13, 2009 Posted June 13, 2009 If two stiff or rigid objects are connected with a stretched spring, it will apply equal opposite forces on the objects and pull them together (let's assume a frictionless environment). Since this is a closed system of free moving bodies, the net force on the system will remain constant (let's assume zero for cimplicity). If instead two chains are connected with a stretched spring, it will apply opposite equal forces on the two links it is attached to, which again applies opposite equal forces on the next two links and so on, until the whole chain has been influenced by the spring force. First I assumed that the forces simply propagate unchanged from link to link, but this can't be true, since we'll end up with a change in net force with chains of different number of links. What is the right way to calculate force propagation from link to link in this case? Cheers, Mike
swansont Posted June 13, 2009 Posted June 13, 2009 If the chain is vertical, you have to worry about the weight. But it sounds like we're looking at a horizontal chain. What's the net force on any link in the chain? Zero. There's a force to the left, and one to the right, and they add to zero. It doesn't matter how many links there are.
h4tt3n Posted June 13, 2009 Author Posted June 13, 2009 (edited) If the chain is vertical, you have to worry about the weight. But it sounds like we're looking at a horizontal chain. For cimplicity, let's just assume the chains are floating in a gravity-less, frictionless space, and that they are stretched out so every link is alrady in physical contact with its neighbors. What's the net force on any link in the chain? Zero. There's a force to the left, and one to the right, and they add to zero. It doesn't matter how many links there are. But if there's no net force on any particular link in the chain, then they won't move, right? I realise that the number of chain links doesn't matter in real life, I just can't see how to express this in terms of force propagation from link to link, especially if the chains have a different number of links: force <- force <- spring -> force -> force -> force -> force (For all it matters, this could be rephrased into a question about how to calculate the force propagating from atom to atom in a seemingly rigid object which is beeing influenced by an external contact force.) Cheers, Mike Edited June 13, 2009 by h4tt3n
swansont Posted June 13, 2009 Posted June 13, 2009 But if there's no net force on any particular link in the chain, then they won't move, right? I realise that the number of chain links doesn't matter in real life, I just can't see how to express this in terms of force propagation from link to link, especially if the chains have a different number of links: force <- force <- spring -> force -> force -> force -> force (For all it matters, this could be rephrased into a question about how to calculate the force propagating from atom to atom in a seemingly rigid object which is beeing influenced by an external contact force.) Cheers, Mike They won't accelerate; we can then assume we are in the chain's frame of reference and all will be happy. I've modified your diagram a little: force <-> force <-> spring <-> force <-> force <-> force Any link exerts a force on whatever is on either side of it, i.e it exerts two forces, and as a result has two forces exerted on it, and they add to zero. Regardless of the number of links.
h4tt3n Posted June 13, 2009 Author Posted June 13, 2009 Ok, I don't get what you're trying to say here... In a Newtonian / classical context an object accelerates only when influenced by a force, right? When the extended spring between the two sections of chain contracts, it applies a force on them and makes them accelerate, right? So, looking at a single one of those chain links, the only way it can accellerate - which it does in a stationary frame of reference - is because it was influenced by a force "passed over to it" from one or both of its neighboring links. How can you physically describe or calculate this phenomenon, either in terms of force, acceleration, impulse or other? Cheers, Mike
swansont Posted June 13, 2009 Posted June 13, 2009 Ok, I don't get what you're trying to say here... In a Newtonian / classical context an object accelerates only when influenced by a force, right? When the extended spring between the two sections of chain contracts, it applies a force on them and makes them accelerate, right? So, looking at a single one of those chain links, the only way it can accellerate - which it does in a stationary frame of reference - is because it was influenced by a force "passed over to it" from one or both of its neighboring links. How can you physically describe or calculate this phenomenon, either in terms of force, acceleration, impulse or other? Cheers, Mike I thought we were looking at a static case. What is accelerating?
J.C.MacSwell Posted June 14, 2009 Posted June 14, 2009 (edited) If two stiff or rigid objects are connected with a stretched spring, it will apply equal opposite forces on the objects and pull them together (let's assume a frictionless environment). Since this is a closed system of free moving bodies, the net force on the system will remain constant (let's assume zero for cimplicity). If instead two chains are connected with a stretched spring, it will apply opposite equal forces on the two links it is attached to, which again applies opposite equal forces on the next two links and so on, until the whole chain has been influenced by the spring force. First I assumed that the forces simply propagate unchanged from link to link, but this can't be true, since we'll end up with a change in net force with chains of different number of links. What is the right way to calculate force propagation from link to link in this case? Cheers, Mike It is not true. If a chain or rod is being pulled along it's length unconstrained it will accelerate. There will be a force gradient along that length, greatest at the pulled end and zero at the other depending on the distribution of mass. The force at any point will simply be proportional to the amount of mass behind that point. If the links are all the same, the force will be proportional to the number of links remaining behind the point in question. The force is reduced between each successive link, due to the acceleration/reaction of the preceding link. If you drew a free body diagram of any link, or set of links, the acceleration would be proportional to the net force, and inversely proportional to the mass of the link or links. So if, say, you had a 10 lb force pulling a chain of 10 links of mass 1 slug each the acceleration would be 1 foot per second per second. The force on the last 1 slug link would be 1 lb and it would nicely accelerate at 1 foot per second per second along with the rest of the chain. Or if, say, you had a 10 newton force pulling a chain of 10 links of mass 1 kilogram each the acceleration would be 1 meter per second per second. The force on the last 1 kilogram link would be 1 newton and it would nicely accelerate at 1 meter per second per second along with the rest of the chain. Edited June 14, 2009 by J.C.MacSwell 1
h4tt3n Posted June 14, 2009 Author Posted June 14, 2009 (edited) @ MacSwell Thank you for the comprehensive answer, now I believe I understand how it works: Every link in the chain is beeing influenced by the same net force, which is the external force / number of links (assuming each link has the same mass). Let's use your example with a chain of ten links, each weighing 1 kg, beeing pulled with a force of 10 N: The first link is pulled with 10 N but pulls the remaining 9 links with 9 N, giving a net force of 1 N. The second link is pulled with 9 N but pulls the remaining links with 8 N and so on right down to the last link, which is beeing pulled by 1 N. Boiled down, each link is influenced by a force of 1 N or total force / number of links. If there is a chain of just 5 links weighing 1 kg each in the other end of the spring, then each link is influenced by 10 N / 5 links = 2 N per link, accelerating each link twice as fast as in the longer chain - just as we would expect in real life. Thanks, Mike Edited June 14, 2009 by h4tt3n
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now