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

[John1951]

Homework Statement

OK, I've worked up my nerve to ask a stupid question about this problem. I've read the various discussions of it, but I'm clearly missing something.2. Homework Equations [/B]

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

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!

 

[haruspex]

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

 

[John1951]

Homework Statement

OK, I've worked up my nerve to ask a stupid question about this problem. I've read the various discussions of it, but I'm clearly missing something.

The problem is as follows:

A pulley is fixed to the ceiling. On one side hangs a mass m0. On the other side hangs a second pulley.

On one side of the second pulley hangs a mass m1, on the other hangs a third pulley, and the series of pullies and weights continues as infinitum.

M0 = 1/(1-t), mi = t**(i-1) for i > 0.

To find: the acceleration of m0 at the moment the weights are released.
2. Homework Equations [/B]

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

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!

 

[TSny]

An infinite series of pulleys and masses is arranged as shown, with m₀ = 1/(1–t), and m_i = 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?

 

[TSny]

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

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.

 

[SammyS]

Homework Statement

OK, I've worked up my nerve to ask a stupid question about this problem. I've read the various discussions of it, but I'm clearly missing something.[/B]

m₀ = 1/(1-t), m_i= t^(i-1) for i > 0.

Homework Equations

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

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!

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 .

Added in Edit:
Actually, t = 0.5 is not a representative case. It's a special case. Try it anyway.

 

[John1951]

Thanks very much for all the help on this. It led to a lot of thought and finally to understanding of this problem.