Irodov 1.60: Which Thread Ties the Two Riders – Amit & Deepali Sinha Foundation

Problem Irodov 1.60

Three bodies of mass $m_0$ , $m_1$ and $m_2$ are connected as follows. Bodies $m_1$ and $m_2$ rest side by side on a horizontal table with friction coefficient $k$ against the table, and are joined to each other by a horizontal thread. A second thread runs from $m_1$ to a small pulley fixed at the table’s edge, and then straight down to the hanging body $m_0$ . The pulley and both threads are massless, and the pulley is frictionless. Starting from rest, find the acceleration $w$ with which $m_0$ moves, and the tension $T$ in the thread joining $m_1$ and $m_2$ , considering whatever cases the coefficient $k$ makes possible.

m_1 and m_2 sit on the table, tied to each other by a horizontal thread; a second thread runs from m_1 over the corner pulley and down to the hanging body m_0. — $m_1$ and $m_2$ sit on the table, tied to each other by a horizontal thread; a second thread runs from $m_1$ over the corner pulley and down to the hanging body $m_0$ .

Given

$m_0$ , $m_1$ , $m_2$
$k$

Find

w

;

T

(the thread between

m_1

and

m_2

)

Solution

Step 1Newton’s second law for each body

Let $T_0$ be the tension in the thread over the pulley and $T$ the tension in the thread joining $m_1$ and $m_2$ ; take $w$ positive in the sense $m_0$ descends. Each body gets one equation along its direction of motion.

\begin{align*} m_0 g - T_0 &= m_0 w \tag{1} \\ T_0 - T - k m_1 g &= m_1 w \tag{2} \\ T - k m_2 g &= m_2 w \tag{3}\end{align*}

Step 2Add the three equations to eliminate the internal tensions

Both $T_0$ and $T$ appear once with each sign across (1), (2) and (3), so summing the left and right sides cancels them and leaves a single equation in $w$ .

\begin{align*}(m_0 g - T_0) + (T_0 - T - k m_1 g) + (T - k m_2 g) &= m_0 w + m_1 w + m_2 w \\ m_0 g - k(m_1+m_2) g &= (m_0+m_1+m_2) w\end{align*}

w = \dfrac{m_0-k(m_1+m_2)}{m_0+m_1+m_2}\,g

(4)

Step 3Isolate

m_2

for the thread tension

Free body of m_2 alone: the thread to m_1, the normal, kinetic friction (both bodies move toward the pulley, so friction on m_2 resists that sense), and its weight. — Free body of $m_2$ alone: the thread to $m_1$ , the normal, kinetic friction (both bodies move toward the pulley, so friction on $m_2$ resists that sense), and its weight.

The tension asked for is the one binding $m_1$ and $m_2$ together, so only $m_2$ ’s own equation (3) is needed. Rearranging it for $T$ and then substituting (4) gives $T$ directly.

\begin{align*}T &= m_2 w + k m_2 g \\ &= m_2 g\left(\frac{m_0-k(m_1+m_2)}{m_0+m_1+m_2}+k\right) \\ &= m_2 g\,\frac{m_0-k(m_1+m_2)+k(m_0+m_1+m_2)}{m_0+m_1+m_2} \\ &= m_2 g\,\frac{m_0(1+k)}{m_0+m_1+m_2}\end{align*}

T = \dfrac{(1+k)m_0}{m_0+m_1+m_2}\,m_2 g

(5)

Step 4The possible cases

The system moves in the assumed sense only while $w>0$ in (4), and since its denominator is positive, the sign is decided by the numerator alone.

m_0 > k(m_1+m_2)

Both (4) and (5) are conditional on this case. In the opposite case the assembly stays at rest, and the three equations of Step 1 are rewritten with $w=0$ and each kinetic friction replaced by the static friction $f_1$ , $f_2$ the surface actually supplies, taken positive in the sense that resists the pull.

\begin{align*}T_0 &= m_0 g \\ T_0 - T - f_1 &= 0 \\ T - f_2 &= 0\end{align*}

Eliminating $T_0$ between the first two lines expresses both frictions through $T$ alone, and each is capped by its own static ceiling.

\begin{align*}f_2 &= T, & f_2 &\le k m_2 g \\ f_1 &= m_0 g - T, & f_1 &\le k m_1 g\end{align*}

The first cap is an upper bound on $T$ and the second a lower one, and a thread can only pull, so $T$ cannot be negative either.

\max\bigl(0,\ m_0 g - k m_1 g\bigr) \le T \le k m_2 g

The band is non-empty exactly when $m_0\le k(m_1+m_2)$ , which is the resting case itself; where inside it $T$ falls depends on how the static friction happens to split between $m_1$ and $m_2$ , which the given data does not fix. Exactly at $m_0=k(m_1+m_2)$ the band closes to the single point $T=k m_2 g$ and the system is on the verge of sliding.

Check

Setting $k=0$ recovers the frictionless three-body Atwood chain, $w=m_0 g/(m_0+m_1+m_2)$ and $T=m_0 m_2 g/(m_0+m_1+m_2)$ , the textbook result for a mass pulling two others in series. The threshold $m_0=k(m_1+m_2)$ found in Step 4 is exactly where the boxed $w$ vanishes, so the moving-case formula turns off precisely where it should.

Remark

The three individual equations in Step 1 never had to be solved separately for $T_0$ : adding them straight away is the free-body analogue of treating $m_1$ and $m_2$ as one lumped mass $m_1+m_2$ pulled by $m_0$ through the external thread tension $T_0$ . Step 2’s sum in fact cancels both tensions at once, $T_0$ (external to the $\{m_1,m_2\}$ pair, but internal once $m_0$ joins the sum) and $T$ (internal to the pair from the start), since each appears once with each sign across the three lines. That is exactly why $T$ cannot be read off the summed equation and Step 3 has to isolate $m_2$ alone instead.