Re: Chain Problem.

["Richard W. Tarara" <rtarara@saintmarys.edu>]

In the idealized problem the rope/chain falls straight down, but if it is
moving horizontally just before leaving the table, then there must be a
horizontal force that opposes that motion just as the chain leaves the
table.  That force must increase with time since the horizontal momentum of
each chain segment increases with time (or as we get near the end of the
chain) and the time period in which the chain segment must change its
motion from horizontal to vertical is decreasing.  This is certainly a
problem that one does not want to look at too closely--at least not until
graduate school!

Rick
----------
> From: Leigh Palmer <palmer@sfu.ca>


> To see that the horizontal component is present (Donald knows
> this already, which is why he suggested the frictionless tube
> in the first place) one need merely consider what the rope is
> doing at the midpoint of its slide off the table. The portion
> of the rope which is still on the tabletop will be moving
> horizontally, thust the rope has a horizontal component of
> momentum. It is easy to see, then, that there must have been
> a horizontal force acting on the rope.