Ed Schweber (edschweb@ix.netcom.com)
Physics Teacher at The Solomon Schechter Day School, West Orange, NJ
To obtain free resources for creative physics teachers visit: http://www.physicsweb.com
----- Original Message -----
From: Bob Sciamanda <trebor@VELOCITY.NET>
To: <PHYS-L@lists.nau.edu>
Sent: Saturday, October 16, 1999 12:19 AM
Subject: Re: Double Atwood
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.
I use the following conventions:
Mass M as an upward acceleration a(1) with respect to an
inertial reference frame
The tension in the connecting string connecting mass M to the
moving pulley is T(1)
The 3-kg mass has a downward acceleration of a(2) with
respect to the non-inertial reference frame of the moving
pulley
The 2 kg mass has an upward acceleration of a(2) with
respect to the non-inertial reference frame of the movable
pulley
The acceleration due to gravity is 10 m/s/s.
Applying the 2nd Law to mass M yields
T(1) - 10M = Ma(1) eq. 1
Applying the 2nd law to the (assumed massless) movable
pulley gives
T(1) = 2T(2) eq. 2
The downward acceleration of the 3 kg mass in an inertial
reference frame is a(1) + a(2). The 2nd law then yields
T(2) - 30 = -3(a(1) + a(2)) eq 3
The upward acceleration of the 2 kg mass in an inertial
frame is a(2) - a(1)
For the 2 kg block to have no acceleration in an inertial
frame implies
(a1) = a(2)
eq 4
That is, the downward acceleration of the movable pulley
in the inertial frame just counterbalances the upward
acceleration of the 2 kg mass in the reference frame
of the pulley
Applying the second law to the 2 kg mass yields
T(2) - 20 = 2(a(2) - a(1)) eq (5)
Using eq(4) in eq(5) gives
T(2) = 20 N eq
(6)
This can also be seen less formally. If the 2 kg mass is
not accelerating, it is in equilibrium with its weight
balancing the tension
Eq (2) then immediately implies that
T(1) = 40 N eq
(7)
Using eq(4) and eq (6) in eq(3) gives
a(1) = 1.67 m/s/s
eq(8)
Using eq(8) and eq (7) in eq 1 gives
M = 3.43 kg
Ed Schweber
BTW: In a previous posting on another topic I did speak of the
advantages of using more intuitive approaches to pulley problems. Obviously,
there comes a time where you have to use a formal free body diagram
approach.