Re: [Phys-l] October Physics Challenge

[brian whatcott <betwys1@sbcglobal.net>]

On 11/6/2010 9:20 AM, Carl Mungan wrote:
> > It might be helpful to know if the ratio of masses is greater or less than the ratio of static and kinetic coefficients of friction.
> Ho ho, you're very insightful! That key piece of info does indeed
> arise. We are told that M>m (and physics requires mu_k<= mu_s). But
> we also told to consider ALL cases. So please go ahead and suppose
> that mu_k/mu_s<= m/M, which is indeed required if the heavy block is
> to slip on the ribbon while the light block does not. -Carl

Quoting your challenge:

"How many of you look at the Physics Challenges in TPT? Well, anyhow,
go look at the October challenge and its solution at:

http://scitation.aip.org/journals/doc/PHTEAH-home/challenges/oct2010.pdf

"I find myself in a bit of disagreement over the parenthetical note at the beginning of Case 2 which is supposed to explain that the heavy block can never slip on the ribbon.

"I agree that both blocks cannot continuously slip (unless we allow the ribbon to bunch up near the top of the prism!). But I think that for steep angles, either block could slip. For the sake of argument,
suppose I initially force the light block to move with the ribbon (by
some unspecified coupler) but allow the heavy block to slip on it. If I now release the light block coupling, there's no reason it'll suddenly start slipping. Once started with the heavy block slipping and the light block moving with the ribbon, that configuration is stable."

*************************************************
If I consider the limits of the given assumption:
that mu_k/mu_s<= m/M,

then I must consider the case where mu_k  goes to zero.

1) Neither block slips on shallow inclines.

a = (T – mg sin q)/m = (Mg sin q – T)/M.
T = 2g sin q[Mm/(M + m)].
a = {[2g sin q][Mm/(M + m)] – mg sin q}/m
= (g sin q)[M – m)/(M + m)]   ...as given.

2) Small block begins slipping on steeper inclines:

2.g sin(theta).[M.m/(M + m)] = mu_s.m.g cos theta →
tan theta = (mu_s/2)(1 + m/M).
When the smaller block is sliding, the ribbon tension is
equal to the sliding friction between the smaller block and
the ribbon: T = mu_k.m.g cos theta. For mu_k = 0, T = 0
...as given.

3) The setup is manipulated to start the large block slipping first
by holding the small mass and ribbon in place at a high inclination:
The ribbon and small mass descend uninhibited at m.g. sin(theta)
the ribbon tension is zero, the large block descends at M.g.sin(theta)

This result is unsurprizing, and so it is not the case that you find interesting, which must occur when mu_k/mu_s = m/M, at the largest mu_k for consideration.
But I will pause at this point, in case I am missing the crux.

Brian W