# infinite atwood machine problem

## infinite atwood machine problem

A peculiar Atwood’s machine

The Atwood’s machine shown in the figure consists of masses m,m/2,m/4,........,m/(2^(N-1)). All the pulleys and springs are mass less, as usual

a) Put a mass m/(2^(N-1)) at the free end of the bottom spring. What are the acceleration s of all the masses

Remove the mass m= (m/(2^(N-1)) [which was arbitrarily small, for very large

N] that was attached in part (a). What are the accelerations of all the

Masses, now that you’ve removed this infinitesimal piece?

The Atwood’s machine shown in the figure consists of masses m,m/2,m/4,........,m/(2^(N-1)). All the pulleys and springs are mass less, as usual

a) Put a mass m/(2^(N-1)) at the free end of the bottom spring. What are the acceleration s of all the masses

Remove the mass m= (m/(2^(N-1)) [which was arbitrarily small, for very large

N] that was attached in part (a). What are the accelerations of all the

Masses, now that you’ve removed this infinitesimal piece?

**Pinnaka**- Admin
- Posts : 36

Join date : 2008-05-20

Age : 31

Location : India

## Re: infinite atwood machine problem

n part (a), there is perfect balance at all points, so each mass will have acceleration = 0.

Proof: At the last pulley, there are equal masses on each side, so it will not turn. Now on the next pulley above it, the mass on the left equals the sum of the masses on the right, so it will not turn. This continues all the way to the first pulley.

In part (b), all the string tensions are zero, so each mass will accelerate downward with acceleration = g.

Proof: There is a mass on one side of the last pulley, but no mass on the other side. Therefore, the string around the last pully has no tension, so the last mass falls at acceleration g. Since there is no tension in the last string, there is no downward force on the last pulley, so there is no force in the string supporting it. Therefore, the mass on the other end of that string falls freely with acceleration g. This continues all the way to the first pulley.

It is surprising that an arbitrarily small mass can make such a large difference, isn't it?

Hope this helps!

Proof: At the last pulley, there are equal masses on each side, so it will not turn. Now on the next pulley above it, the mass on the left equals the sum of the masses on the right, so it will not turn. This continues all the way to the first pulley.

In part (b), all the string tensions are zero, so each mass will accelerate downward with acceleration = g.

Proof: There is a mass on one side of the last pulley, but no mass on the other side. Therefore, the string around the last pully has no tension, so the last mass falls at acceleration g. Since there is no tension in the last string, there is no downward force on the last pulley, so there is no force in the string supporting it. Therefore, the mass on the other end of that string falls freely with acceleration g. This continues all the way to the first pulley.

It is surprising that an arbitrarily small mass can make such a large difference, isn't it?

Hope this helps!

**Pinnaka**- Admin
- Posts : 36

Join date : 2008-05-20

Age : 31

Location : India

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