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Consider the double Atwood machine below:

enter image description here

I understand the classic approach to a solution for the acceleration of each mass involves setting the tension in the upper string equal to twice that of the lower string to achieve equilibrium in the lower pulley. The common reasoning is that "since the pulley is massless, it must be in equilibrium otherwise it would have infinite acceleration".

While I understand this reasoning must be correct, surely if the lower pulley is in equilibrium there would be no acceleration of m1 at all, which can't be correct? Therefore, under what conditions are the subsequent accelerations derived physically useful - are they just in the limit as the mass of real massive pulleys tend to zero?

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    $\begingroup$ By equilibrium they meant net force =0 remember f=ma so either a=0 or m=0 for f=0 $\endgroup$ Commented Dec 27, 2020 at 12:54
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    $\begingroup$ Found similar answer here (duplicate): Can a massless rope accelerate? $\endgroup$
    – user246795
    Commented Dec 27, 2020 at 19:45

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The lower pulley is holding to two masses, which account for the acceleration.

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A way to avoid confusion in analyzing motions of pulleys is to write the dynamical equations of motion sans the assumptions of low inertia, which may apply for some of the pulleys, and then neglect the masses of the pulleys. On doing so, it is clear that terms such as $m^{pulley}_i a^{pulley}_i$ and $m^{pulley}_i g$ which arise in the dynamical equations can be ignored if $m^{pulley}_i \rightarrow 0$. On noticing that applying this assumption required no additional assumption on the magnitude of $a^{pulley}_i$, it is evident that the pulley denoted by $i$ need not be in equilibrium in the analysis which includes the assumption of negligible mass. In general, such confusions are a result of conflating concepts of static analysis when conducting dynamical analysis and are readily avoided by doing the latter as a matter of policy towards as a safety measure (although it might make the analysis more time consuming than required when the system is indeed completely static, i.e. has no accelerating bodies).

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