Re: Atwood's machine problem

[Bob LaMontagne <rlamont@POSTOFFICE.PROVIDENCE.EDU>]

Peter

I just assigned this as a homework problem! The CM argument doesn't
illuminate the solution here because the "system" includes the Earth. The
monkey and bananas can rise as long as the Earth moves in the opposite
direction.

The key is that the tension in the cord is exactly the same on both sides
of the pulley and the monkey and the bananas have identical weights. The
equation T - mg = ma applies to both, hence they have identical
accelerations at all times - i.e., identical motions.

Bob at PC

> Hello all,
>
> I just got a question I can't answer, and could use some help.
>
> A student got this problem out of a book (didn't tell which one) and
asked
> me, and I 'clutched':
>
> An Atwood's machine is in perfect balance, with equal weights on both
> sides.  One of the weights is a monkey, and the other is a stack of
> bananas.  The monkey begins to climb at a constant speed.  What happens
to
> the bananas; ie, do they rise, fall, or stay stationary.
>
> My first inclination was to use CM and say that if the monkey climbs the
> bananas must fall to keep the CM stationary.  However, the book he
Xeroxed
> this from, said the bananas rise with the monkey.
>
> Maybe it's just because it's Friday, but I don't see the 'mechanism' for
> that explanation.  Can somebody help?
>
> Thanks,
> Peter Schoch
> (sleep deprived with a 15 month at home)


---------------------------------------------
This message was sent using Endymion MailMan.
http://www.endymion.com/products/mailman/