I have a question regarding Atwood's machines. Consider the following machine and text accompanying it:
The tension in the short string connected to $m_1$ is $2T$ because there must be zero net force on the massless left pulley, because otherwise it would have infinite acceleration.
My problem is that I don't see why the net force on the left pulley must be zero, since it means that the acceleration of the left pulley must be zero, although it clearly is moving upwards (and thus accelerating) when the mass $m_2$ is accelerating downwards.
Best Answer
The "massless pulley" approximation is to make the system easier to understand without having to get into junk about rotational inertia.
Suppose that the pulley isn't massless but has some little mass $\delta m \ll m_1$, and that $m_1$ is hanging from it by another rope. That other rope will have an extra tension for us to keep track of. We'll still ignore any rotational inertia in the pulley. Here's the new part of the setup:
In that case, Newton's second law for each of the two masses gives
\begin{align} \delta m\, a_\delta &= 2T - T_\text{extra} - \delta m\, g \\ m_1 \, a_1 &= T_\text{extra} - m_1\,g \end{align}
Your author seems to be taking a counterfactual approach. The acceleration of the pulley is,
$$ a_\delta = \frac{2T - T_\text{extra}}{\delta m} - g $$
and in the limit $\delta m \to 0$, this acceleration becomes very large unless $T_\text{extra} \approx 2T$; your author states that this is bad and therefore can't be happening.
My instinct is to assume the accelerations are the same and to eliminate the bookkeeping tension in the line connecting your mass to the massless pulley:
\begin{align} T_\text{extra} &= 2T - \delta m(a_\delta + g) \approx 2T \\ m_1\,a_1 &= \left( 2T - \delta m\,(a_\delta+g) \right) - m_1\,g \\ \delta m\,a_\delta + m_1 a_1 &= 2T - (\delta m + m_1)g \\ \widetilde m_1\,a &= 2T - \widetilde m_1\,g \tag1 \end{align}
Since $\delta m \ll m_1$, there's not very much difference between $m_1$ and $\widetilde m_1 = m_1 + \delta m$, and the upward force on the latter is $2T$. The final line (1) encodes the statement that your author makes, that the upward force on $m_1$ is $2T$, and that approximation gets better in the limit that $\delta m$ is very small.