Cramer%CSL60%ti-csl.csnet@CSNET-RELAY.ARPA (10/27/85)
From: Nichael <Cramer%CSL60%ti-csl.csnet@CSNET-RELAY.ARPA>
In an effort to: 1] divert attention from the Newman Machine discussion
[which has gone on far too long] 2] reintroduce some actual [albeit
low-level] physics back into this bboard and 3] to get a satisfactory answer
[which I've never found], I'd like to introduce an old chestnut from my
undergraduate mechanics course. [See diagram].
o
/ \
| |
| |
o/| /n\
()| |K|
\| |g|
| -
A [massless] Rope passes over a [frictionless, massless] Pulley. On one side
is a Weight [n Kg]. On the other side, level with the Weight, is a Monkey,
also weighing n Kg. Both the Monkey and the Weight are initially at rest.
The Monkey [as you might have guessed by now] starts climbing the Rope. What
is the motion of the Weight [or for that matter, that of the Monkey]?
What if the Pulley is not frictionless? What if the Rope is not massless?
Over the years I've seen at least FOUR different published solutions to this
problem [the monkey stays level as the weight rises, the weight stays level
as the monkey rises, the monkey and the weight rise at the same rate, and the
monkey rises and the weight LOWERS!!!]. Any takers??
No clever signoff,
Nichael
-------mwg@petrus.UUCP (Mark Garrett) (10/30/85)
++ > From: Nichael <Cramer%CSL60%ti-csl.csnet@CSNET-RELAY.ARPA> >A [massless] Rope passes over a [frictionless, massless] Pulley. On one side > is a Weight [n Kg]. On the other side, level with the Weight, is a Monkey, > also weighing n Kg. Both the Monkey and the Weight are initially at rest. >The Monkey [as you might have guessed by now] starts climbing the Rope. What > is the motion of the Weight [or for that matter, that of the Monkey]? OK, let's take a crack at it just for fun (it's been a long time). Assuming the monkey climbs so smoothly as to resemble a constant force on the rope (in addition to the force of his weight, which is balanced by the weight), then he pulls the rope down and the weight up without changing his height, because there is no fixed object which he could be pulling against to raise himself. How's that? If I'm wrong, I'll work out the Schroedinger equation for the monkey as penance. :-] I bet that if you were to hang from such a counterbalanced rope, with all the friction and non-linearities of real life, you could probably get the weight to move anywhere you wanted! -Mark
MJackson.Wbst@Xerox.ARPA (10/30/85)
Simple answers to these questions quickly get confounded in the ambiguity of "The Monkey. . .starts climbing the Rope." It is simplest if we take this to mean that the Monkey exerts a force Fm on the Rope which exceeds MG (where M is n Kg and G is the acceleration of gravity). Then by considering tension in the Rope it is clear that: If the Rope is massless and the Pulley is frictionless and massless then the Monkey and the Weight both accelerate upward at the same rate A = Fm/M - G. If the Pulley has friction (let us assume speed-independent friction Fp) then the Monkey accelerates upward faster than the Weight, which may not move at all. The Monkey's acceleration is as before and the Weight's acceleration is (Fm - Fp)/M - G [if Fm > MG + Fp], or zero [if Fm<= MG + Fp]. If the Rope has mass (let us assume mass per unit length R and total length L [Monkey and Weight are at opposite ends]) then the Monkey's acceleration is as before and the Weight's acceleration Aw is initially (Fm - MG)/(M + RL). However, as rope passes over the Pulley this increases the weight on the left (Monkey) side, which raises the rate of acceleration of Rope and Weight. If we denote the length of Rope on the left by X we have: (Fm - MG - LRG) + 2XRG = (M + RL)Aw and since Aw is also d(dX/dt)/dt the Rope acceleration increases exponentially until the Weight hits the Pulley. Note well the first paragraph. If one takes "The Monkey. . .starts climbing the Rope" to mean the Monkey begins moving with a fixed upward velocity with respect to the Rope then the answers may look somewhat different, and further specification of the "start-up conditions" (how did he accelerate to this velocity?) may be required. In particular, if the Rope has mass and is long enough the Monkey will eventually find himself *descending* under this interpretation. Loved your diagram, by the way. Mark
MJackson.Wbst@Xerox.ARPA (10/31/85)
From Rick McGeer: "Not quite correct. Since both the monkey and the rope weigh M, the total mass on the rope is 2M, hence Fm = 2M(G + A) hence A = Fm/2M - G. Picky, picky, picky." Also wrong, wrong, wrong. The Monkey exerts a force Fm on the Rope/Weight system, and the Rope/Weight system exerts a force Fm on the Monkey. A = Fm/M -G is correct. Consider the condition just before t = 0, when the Monkey is just holding himself up. Surely then A = 0 and Fm = MG. Mark
dgary@ecsvax.UUCP (D Gary Grady) (10/31/85)
> From: Nichael <Cramer%CSL60%ti-csl.csnet@CSNET-RELAY.ARPA> > The Monkey [as you might have guessed by now] starts climbing the Rope. What > is the motion of the Weight [or for that matter, that of the Monkey]? Can we bring an experimentalist in on this? -- D Gary Grady Duke U Comp Center, Durham, NC 27706 (919) 684-3695 USENET: {seismo,decvax,ihnp4,akgua,etc.}!mcnc!ecsvax!dgary
mcgeer@kim.berkeley.edu (10/31/85)
From: mcgeer@kim.berkeley.edu (Rick McGeer) Aieeee....this is what comes of doing mechanics at 1 am... Best, Rick.
bradford@AMSAA.ARPA (10/31/85)
From: Cymru am Byth! <bradford@AMSAA.ARPA>
<Simple answers to these questions quickly get confounded in the
<ambiguity of "The Monkey. . .starts climbing the Rope." It is simplest
<if we take this to mean that the Monkey exerts a force Fm on the Rope
<which exceeds MG (where M is n Kg and G is the acceleration of gravity).
<Then by considering tension in the Rope it is clear that:
Uh, yes, but HOW does he exert a force of greater than MG? Without first
exerting one of LESS than MG, etc, etc, etc...
PJB
"So you think being drunk feels good -- tell that to a glass of water!"wallace@ucbvax.BERKELEY.EDU (David E. Wallace) (10/31/85)
In article <661@petrus.UUCP> mwg@petrus.UUCP (Mark Garrett) writes: >Assuming the monkey climbs so smoothly as to resemble a constant force on >the rope (in addition to the force of his weight, which is balanced by the >weight), then he pulls the rope down and the weight up without changing >his height, because there is no fixed object which he could be pulling >against to raise himself. How's that? If I'm wrong, I'll work out the >Schroedinger equation for the monkey as penance. :-] Assuming the rocket engine fires so smoothly as to resemble a constant force, the spaceship doesn't move in outer space, because there is no fixed object which it could be pushing against, right? Now that we've proved that rockets don't work (:-), how about applying Newton's law to both. Since action = reaction, the rope exerts the same force on the monkey as the monkey exerts on the rope (which is transmitted to the weight, via the frictionless pulley and the massless rope). Hence both monkey and weight rise at the same rate. Let's see that Schroedinger equation (:-)! Dave Wallace (...!ucbvax!wallace wallace@ucbkim.berkeley.edu)
mcdaniel@uiucdcsb.CS.UIUC.EDU (11/03/85)
>> From: Nichael <Cramer%CSL60%ti-csl.csnet@CSNET-RELAY.ARPA> >> The Monkey [as you might have guessed by now] starts climbing the Rope. What >> is the motion of the Weight [or for that matter, that of the Monkey]? >Can we bring an experimentalist in on this? I'll provide the monkey if you'll provide the frictionless, massless pulley and the massless rope.
jordan@noscvax.UUCP (Martin C. Jordan) (11/06/85)
In article <667@ecsvax.UUCP> dgary@ecsvax.UUCP (D Gary Grady) writes: >> From: Nichael <Cramer%CSL60%ti-csl.csnet@CSNET-RELAY.ARPA> >> The Monkey [as you might have guessed by now] starts climbing the Rope. What >> is the motion of the Weight [or for that matter, that of the Monkey]? > >Can we bring an experimentalist in on this? >-- >D Gary Grady >Duke U Comp Center, Durham, NC 27706 Would a simulation do? If you can get me the massless rope and pulley (made of obscurium?), I can get a civil servant. The hard part will be training him to climb a rope! I posed this problem to a theoretical physicist here at NOSC. He mused over the problem yesterday. Today I received a memo from him. It started: "Assume a spherical monkey.........." "What would Atwood do?" Martin Jordan jordan@nosc.arpa
hes@ecsvax.UUCP (Henry Schaffer) (11/07/85)
> I posed this problem to a theoretical physicist ... > Today I received a memo from him. It started: > "Assume a spherical monkey.........." > > Martin Jordan jordan@nosc.arpa That reminds me of the study a local biomathematician did on milk production. It started out, "Assume a spherical cow radiating milk isotropically ..." (Being a biomathematician at times I can tell this story!) --henry schaffer (I feel sorry already.)