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What is the effective mass of a an accelerating pulley block system?

Consider the setup shown above.

Setup

I was trying to find the mass $m$, which would produce the same acceleration of the moving pulley system (as a whole) i.e. $a_1$ when the moving pulley system(along with the 2 blocks) is replaced by a block of mass $m$. Using Newton's second law and the constraint relations, I got the following system of equations.Let $2T$ be the tension in the topmost string(The one in the diagram where $a_1$ is writen). Here, $g$ is the acceleration due to gravity. $$mg-2T=ma_1\\Mg-T=Ma_2\\2Mg-T=2Ma_3\\a_2+a_3=2a_1$$ On solving the system of equations, I got $$m=\frac{8M}{3}$$

But, since we are taking the movable pulley along with the 2 blocks as our system, shouldn't the total mass of the system be the sum of the masses of the individual masses of the system(in this case, the $m$ should be $3M$)?

MAIN QUESTION:
Is it that the total mass of a system is not in general the sum of masses of its individual components?In what cases is it valid? Where am I wrong?

Thanks for any answers!

EDIT:I have read the answer on this link, but I still didn't understand why we cannot consider certain systems as "systems" in classical mechanics

Why can't I choose blocks attached with pulley B as a system?

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    $\begingroup$ I think you need an equation for the upper mas including a force from the spring. Also i think last equation relating accelerations is wrong. $\endgroup$
    – R.W. Bird
    Commented Jun 23, 2020 at 17:52

3 Answers 3

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A system in classical translational mechanics is assumed to be a rigid system. An assumption in the system is that the acceleration of every particle in the system is the same, which is why we can replace it with a large particle of the same acceleration. Here there are elements with various accelerations in the 'system' you want to consider. What then, will you assume to be the net acceleration of the system?

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Taking the two hanging masses and pulley as a system, the center of mass is accelerating downward. That means that the downward force of gravity must be greater than the upward force provided by the cord supporting the pulley. I'll assume the the cords and pulleys are mass-less and friction-less, and that the hanging masses have enough cord to move for some period of time. Let's start by assuming the sliding mass is not moving. Then for the hanging masses and cord: a = (2Mg – Mg)/(2M + M) = g/3. For the left side mass, T – Mg = M(g/3) and T = 4Mg/3 For the upper cord the tension is 8Mg/3 (as you found) which is balanced by the force from the spring. If the upper mass is accelerating to the left, that puts the hanging masses in an accelerating frame, effectively increasing g. For the upper mass, kx – 8M(g+a)/3 = 2Ma and kx – 8Mg/3 = (2M+8M/3)a. As before, the 8Mg/3 determines the equilibrium position and the 8M/3 increases the mass which determines the angular frequency.

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You can consider the whole pulley system as a single body of an equivalent mass m for convenience. By taking simple FORCE EQUATIONS of the two bodies attached to pulley of masses m1 and m2, you can find that:

$m=(4*m_1*m_2)/(m_1+m_2)$

For example, in the above given scenario, $m_1=M$ and $m_2=2M$ so that makes:

$m=8M/3$

as observed.

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