# Feynman's (err, Gottlieb's) infinite pulley problem

1. Jul 27, 2016

### John1951

1. The problem statement, all variables and given/known data

2. Relevant equations

The right-hand mass is 1/(1-t). The sum of the left-hand masses (an infinite series) is also 1/(1-t).
3. The attempt at a solution

Therefore, the weight on both sides of the top pulley is equal and a = 0.

This clearly is not true, so where did I go wrong?

Thanks!

2. Jul 27, 2016

### haruspex

Not everyone will be familiar with this problem. Please post either a full description of it or a link to such.

3. Jul 28, 2016

### John1951

4. Jul 28, 2016

### TSny

An infinite series of pulleys and masses is arranged as shown, with m0 = 1/(1–t), and mi = t(i–1) for i > 0, with 0 < t < 1. At the moment the pulleys are released from rest, what is the acceleration of mass m0?

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5. Jul 28, 2016

### TSny

Consider the simpler problem shown below with just two pulleys. The total mass on the left side of the top pulley is 10 kg which is the same as the mass on the right side. Assume that the acceleration of $m_0$ is zero and show that you get a contradiction. (Consider the tensions $T_0$ and $T_1$ and treat the pulleys as massless.)

For the original problem with the infinite set of pulleys, there is one value of the parameter $t$ that makes the acceleration of $m_0$ equal to zero. But in general it is not zero.

Last edited: Jul 28, 2016
6. Jul 29, 2016

### SammyS

Staff Emeritus
m0 = 1/(1-t), mi= t^(i-1) for i > 0.
What possible range of values for $\ t\$ make sense in regard to $\ m_0, m_1, m_2, \dots\$ being masses?

If $\ t>1\,,\$ then $\ m_0\$ would be negative. Can't have that.

If $\ t=1\,,\$ then $\ m_0\$ would be undefined. Can't have that.

If $\ t<0\,,\$ then $\ m_i\$ would be negative for even values of $\ i\,,\ i>0\$. Can't have that.

If $\ t=0\,,\$ then $\ m_i\$ would be zero for $\ i>0\$. So, clearly the statement is false in this case.

Therefore, you need to check this situation for $\ 0<t<1\ .\$

To see how this all works, try the case of t = 0.5 .