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Find the acceleration of any mass for this type of system. At the lowest pulley there are two masses connected, $m_n$ and $m_{n'}$. I having a hard time tackling the finite problem but the solution as $n->\infty$ should be easier. For $m$ is a constant. Need to solve for $T_1$

$$a(P_1) = 0$$ $$T_r = 2T_{r+1}$$ $$a(P_r) = 2\cdot{a(P_{r-1})} - a(m_{r-1})$$ $$a(m_n) = a(P_n)$$

$$a(m_{r}) = g - \frac{T_r}{m} = g - \frac{T_1}{m\cdot{2^{r-1}}}$$

For $n->\infty$,

$$a(m_n) = g$$

So substituting recursively into equation number 3, using the help of equation 1, 4, 6:

$$g + 2^{n-2}\cdot{a(m_{1})} + 2^{n-3}\cdot{a(m_{2})} + ... 2^{0}\cdot{a(m_{n-1})} = 0$$

Please confirm the steps above

jg mr chapb
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  • FWIW this same question was posted then deleted from a different account. – dxiv Aug 29 '17 at 21:20
  • It would be helpful to post the formula if you just have $m_1,m_2,P_1,P_2$. For those of use who don't remember college/HS physics this will save a lookup. Also if you did so, please don't post the same question from multiple accounts. – Χpẘ Aug 29 '17 at 21:27

1 Answers1

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This is the so-called infinite Atwood machine. Solutions can be obtained based on Newton's laws of motion (with an equivalent mass / tension approach). See for instance

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