Irodov 1.73: The Pulley That Rides Its Own Rope – Amit & Deepali Sinha Foundation

Problem Irodov 1.73

A block of mass $m_0$ rests on a frictionless horizontal table and is tied by a horizontal thread that runs to a fixed pulley mounted at the table’s edge and then straight down to the axle of a second, movable pulley hanging in the air. A separate thread passes around that movable pulley, carrying a block of mass $m_1$ on one end and a block of mass $m_2$ on the other. All pulleys and threads are massless, and there is no friction anywhere. Find the acceleration of $m_1$ , and examine the possible cases.

Block m_0 on a frictionless table, joined over a fixed pulley to a movable pulley; a second thread passing round the movable pulley carries m_1 on the left and m_2 on the right. — Block $m_0$ on a frictionless table, joined over a fixed pulley to a movable pulley; a second thread passing round the movable pulley carries $m_1$ on the left and $m_2$ on the right.

Given

$m_0$
$m_1$
$m_2$
frictionless table
massless pulleys and threads

Find

the acceleration of

m_1

, and the possible cases

Solution

Step 1Naming the two tensions

Call $T_1$ the tension in the thread that runs from $m_0$ over the fixed pulley down to the axle of the movable pulley, and $T_2$ the tension in the second thread, the one that passes round the movable pulley and carries $m_1$ on one end and $m_2$ on the other.

Step 2Force balance on the massless pulley

Because the movable pulley and both threads are massless, Newton’s second law applied to the pulley alone reads (mass) times (acceleration) with the mass equal to zero, so the net force on it is zero however fast it is accelerating.

The axle strand pulls the pulley up with $T_1$ ; the wrapped strand pulls it down with $T_2$ on each of its two sides, so

T_1 = 2T_2

(1)

Step 3Newton’s second law for each body

Let $w_0$ be the acceleration of $m_0$ along the table, taken positive toward the fixed pulley (the direction $T_1$ pulls it), and let $w_1$ , $w_2$ be the accelerations of $m_1$ , $m_2$ , both taken positive downward. The table is frictionless and horizontal, so the only horizontal force on $m_0$ is $T_1$ ; each hanging body carries its own weight and the single tension $T_2$ :

\begin{align*} T_1 &= m_0 w_0 \tag{2} \\ m_1 g - T_2 &= m_1 w_1 \tag{3} \\ m_2 g - T_2 &= m_2 w_2 \tag{4}\end{align*}

Step 4The two thread constraints

The thread over the fixed pulley has a fixed total length. Writing $x_0$ for $m_0$ ’s distance from the fixed pulley and $y_Q$ for the movable pulley’s depth below it, that length is $x_0+y_Q$ , so

x_0 + y_Q = \text{const}

Differentiating twice in time, and reading each acceleration in the positive sense already chosen ( $m_0$ toward the pulley, the pulley downward),

w_Q = w_0

(5)

where $w_Q$ is the movable pulley’s own downward acceleration. The second thread, wound round the movable pulley with $m_1$ on one end and $m_2$ on the other, also has fixed length: the sum of the two strand lengths, measured from the movable pulley down to each load, is constant,

(y_1 - y_Q) + (y_2 - y_Q) = \text{const}

Differentiating this twice in time as well,

\ddot y_1 + \ddot y_2 - 2\,\ddot y_Q = 0, \qquad w_1 + w_2 = 2w_Q

Substituting $w_Q = w_0$ from (5) gives the constraint

w_1 + w_2 = 2w_0

(6)

Step 5Solve for the acceleration of

m_0

Eliminating $T_1$ between (2) and the pulley balance (1) puts the wrapped-strand tension in terms of $w_0$ alone,

\begin{align*}T_2 &= \frac{T_1}{2}\\ &= \frac{m_0 w_0}{2}\end{align*}

Substituting that into (3) and (4) and solving each for its own acceleration,

w_1 = g - \frac{m_0 w_0}{2m_1}, \qquad w_2 = g - \frac{m_0 w_0}{2m_2}

Adding the two and using the constraint (6) on the left,

2g - \frac{m_0 w_0}{2}\left(\frac{1}{m_1}+\frac{1}{m_2}\right) = 2w_0

Multiplying through by $2m_1 m_2$ and collecting the $w_0$ terms,

\begin{align*}4m_1 m_2 g - m_0 w_0(m_1+m_2) &= 4m_1 m_2 w_0\\ 4m_1 m_2 g &= w_0\bigl[4m_1 m_2 + m_0(m_1+m_2)\bigr]\end{align*}

so that

w_0 = \frac{4m_1 m_2\, g}{4m_1 m_2 + m_0(m_1+m_2)} \tag{7}

Step 6Acceleration of

m_1

Substituting (7) back into Step 5’s $w_1 = g - m_0 w_0/(2m_1)$ ,

\begin{align*}w_1 &= g - \frac{m_0}{2m_1}\cdot\frac{4m_1 m_2\, g}{4m_1 m_2 + m_0(m_1+m_2)}\\ &= g - \frac{2m_0 m_2\, g}{4m_1 m_2 + m_0(m_1+m_2)}\\ &= g\left[\frac{4m_1 m_2 + m_0(m_1+m_2) - 2m_0 m_2} {4m_1 m_2 + m_0(m_1+m_2)}\right]\end{align*}

The numerator’s two $m_0$ terms collect,

m_0(m_1+m_2) - 2m_0 m_2 = m_0(m_1-m_2)

so that

w_1 = \dfrac{4m_1 m_2 + m_0(m_1-m_2)}{4m_1 m_2 + m_0(m_1+m_2)}\,g

(8)

Check

Let $m_0\to\infty$ : the formula reduces to $w_1 \to g(m_1-m_2)/(m_1+m_2)$ , the ordinary two-body Atwood result. That is the right limit, because an infinitely massive $m_0$ resists $T_1$ with infinite inertia, $w_0\to 0$ , the movable pulley stops descending, and $m_1$ , $m_2$ then hang from what is effectively one fixed pulley.

Remark

$w_1$ was defined positive downward, and the sign of its numerator, $4m_1m_2+m_0(m_1-m_2)$ , is not fixed: whenever $m_1\ge m_2$ every term is positive and $m_1$ accelerates down. But if $m_2$ is heavier and $m_0(m_2-m_1) > 4m_1m_2$ , the numerator turns negative and $m_1$ accelerates upward instead, dragged by $m_2$ through the movable pulley even though gravity alone pulls $m_1$ down. This sign change, not a separate machine, is what “look into possible cases” is asking for.