Is it possible to simplify this Atwood machine problem?

[linusjak]

Homework Statement

So this might be completly wrong but i'm thinking since we only care about pulley A and everything moves in a vertical line i should just be able to simplify the entire second pulley to a weight with mass 3m, is this possible to do? why or why not?

Relevant Equations

Lagrange Equations

 

[kuruman]

You could do that if $m_B=m_C$, i.e. there is no relative acceleration of the two masses. The relative acceleration changes the tension in the upper string from what it would be if the masses were equal. Remember that in a simple Atwood machine with one pulley and two masses the tension in the string connecting the two masses is$$T=\frac{2m_1m_2}{m_1+m_2}g$$ in which case the tension in the string that connects the pulley to the ceiling is twice that and equal to the sum of the masses only when they are equal.

On edit
The answer to your title question is yes, you can replace the Atwood machine with a single equivalent mass but not the way you propose. How you do that is hinted in my answer above. It is also Rule I in this article. However, the problem is asking you specifically to use Lagrange's equations and that you must do by considering the Atwood machine masses as separate. Perhaps you could check your answer by using the equivalent mass shortcut as well.

 

[linusjak]

You could do that if $m_B=m_C$, i.e. there is no relative acceleration of the two masses. The relative acceleration changes the tension in the upper string from what it would be if the masses were equal. Remember that in a simple Atwood machine with one pulley and two masses the tension in the string connecting the two masses is$$T=\frac{2m_1m_2}{m_1+m_2}g$$ in which case the tension in the string that connects the pulley to the ceiling is twice that and equal to the sum of the masses only when they are equal.

Thank you that helps a lot, makes sense now!!