Atwood Machine, two masses + two pulleys (w/ mass)

AI Thread Summary
The discussion centers on solving an Atwood machine problem involving two pulleys and four masses. The setup includes two wheels with different masses and radii, along with two hanging masses. The equations of motion are established using torque and tension, leading to the need to eliminate tension for a solution. The final expression for acceleration is derived as a = (g(m3 - m4)) / (m1 + m2 + m3 + m4). This approach effectively simplifies the problem and provides a clear solution for the acceleration of the system.
destrow111
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Homework Statement


An Atwood machine is constructed using two
wheels (with the masses concentrated at the
rims). The left wheel (m1) has a mass of 2.1 kg and
radius 21.44 cm. The right wheel (m2) has a mass
of 2.8 kg and radius 33.94 cm. The hanging
mass on the left (m3) is 1.53 kg and on the right (m4)
1.13 kg.



Homework Equations


Sum of Torque1= (m3)g(r1) - T(r1) = (m1)(r1)a
Sum of Torque2= T(r2) - (m4)g(r2) = (m2)(r2)a


The Attempt at a Solution


used subsitution to solve for a...
a = ((m3)g - (m4)g) / ((m1) + (m2))
 
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welcome to pf!

hi destrow111! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)
destrow111 said:
Sum of Torque1= (m3)g(r1) - T(r1) = (m1)(r1)a
Sum of Torque2= T(r2) - (m4)g(r2) = (m2)(r2)a

i'm not sure what the set-up is, but looking at those two equations, if T is an unknown, then you need to eliminate T …

how would you do that? :smile:
 


I actually figured it out!

there were actually three different tensions so i added sum of forces equations for each of the masses then subsituted a couple times until acceleration was the only unknown!

ended up being a = (g(m3 - m4)) / (m1 + m2 + m3 + m4)
 
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