Re: Double Atwood

[John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU>]

On Sat, 16 Oct 1999, Ed Schweber wrote:

> Hi;
>
>    David Abineri posed a question about inertial and non-inertial reference
> frames in solving a double Atwood problem.
>
>    I have a solution below that explitly uses only inertial reference
> frames. However, I do not get the same answer that Bob Sciamanda says he got
> in his posting.

And I will contribute a solution that makes use both of "noninertial"
frames and of what Ed calls "more intuitive approaches to pulley
problems."  The answer I get--24/7 kg-- agrees with Eds.  (BTW, for those
who might wonder, no, I didn't have enough confidence in this solution to
post it until I had verified it with a more standard approach.)

Notation: (see David's message for original problem statement)
   m1 = 3 kg, m2 = 2 kg, other notation should be obvious

Kinematic relations:
   At all times the speed of the string over the lower pulley must be
   equal to the speed of the other string over the upper pulley in order
   for a2 = 0.  Thus, a1 = 2A (where A is the acceleration of M).

For the top pulley we have:

   Force down on more massive side - force down on less massive side
            = sum of (masses * respective accelerations)
giving

   (m1 + m2)*g - Mg = m1*(2A) + M*A

For the bottom pulley in its own "noninertial" frame in which the
"acceleration due to gravity" is g-a, the *same* equation gives

   m1*(g-a) - m2*(g-a) = (m1 + m2)*A

Eliminating A from the two equations and solving for M gives

   M = 4*m1*m2/(3*m1-m2) = 24/7 kg

John Mallinckrodt      mailto:ajm@csupomona.edu
Cal Poly Pomona          http://www.csupomona.edu/~ajm