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Re: Double Atwood



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:
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----- 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.