Irodov 1.93: Why the Two Tensions Differ – Amit & Deepali Sinha Foundation

Problem Irodov 1.93

A weightless thread passes over a fixed pulley whose rim is rough. It carries a mass $m_1$ at one end and a mass $m_2$ at the other, and the friction is such that the thread starts slipping over the pulley when the ratio $m_2/m_1$ reaches $\eta_0$ . Find:

the coefficient of friction between the thread and the pulley;
the acceleration of the two masses once $m_2/m_1 = \eta$ exceeds $\eta_0$ and the thread slides over the pulley.

The two masses hang from the two ends of one weightless thread passing over a single fixed pulley, whose rim is rough rather than smooth.

Given

$m_2/m_1=\eta_0$ (onset of slipping)
$m_2/m_1=\eta>\eta_0$ (after slipping)

Find

(a) the friction coefficient

k

; (b) the acceleration

w

of the two masses

Solution

Step 1A differential balance on the thread

Consider a short element of the thread lying on the pulley’s rim, spanning the small angle $d\alpha$ (figure below). At its two ends the thread carries tension $T$ and $T+dT$ , both directed tangentially and pulling the element away from itself, exactly as a taut string does across any small piece of itself. Because the thread is weightless, the element itself carries no mass: Newton’s second law for it reads $m_{\text{elt}}\vect a=\vect 0$ for $m_{\text{elt}}=0$ , so the element is in equilibrium at every instant, whether the two hanging masses are at rest or already accelerating. That single fact is what lets one differential relation serve both parts of this problem.

A small element of the thread on the pulley's rim, subtending dα, with the tension at each end and the pulley's normal and frictional reaction on it. — A small element of the thread on the pulley’s rim, subtending $d\alpha$ , with the tension at each end and the pulley’s normal and frictional reaction on it.

Resolve the two tension forces along the radius through the element. Being tangential at each end and separated by the small angle $d\alpha$ , their radial components very nearly cancel except for a small net component directed into the pulley; to first order in $d\alpha$ (the correction from $dT$ is second order) that net component has magnitude $T\,d\alpha$ , and it is balanced by the pulley’s normal reaction $dN$ pushing outward on the element:

dN = T\,d\alpha.

Resolve along the tangent instead. The far end pulls with $T+dT$ and the near end with $T$ , in opposite tangential senses, so the net tangential force on the element is $dT$ . Equilibrium requires the pulley’s frictional reaction to supply exactly this, and since the thread is on the point of slipping, or already slipping, everywhere along the wrap, that reaction takes the full Coulomb value $k\,dN$ :

dT = dF_{fr} = k\,dN.

Combining the two relations above eliminates $dN$ :

\frac{dT}{T} = k\,d\alpha.

Step 2Integrating over the half turn

The thread leaves the pulley toward $m_1$ at $\alpha=0$ , where its tension is $T_1$ , and toward $m_2$ at $\alpha=\pi$ , a half turn later, where its tension is $T_2$ . The relation just found holds at every point of contact, so it integrates straight across the whole wrap:

\int_{T_1}^{T_2}\frac{dT}{T} = \int_0^{\pi} k\,d\alpha \quad\Longrightarrow\quad \ln\frac{T_2}{T_1} = k\pi.

T_2 = T_1\,e^{k\pi}

(1)

Step 3Part (a): the friction coefficient at the point of slipping

By definition, $\eta_0$ is the largest ratio $m_2/m_1$ the friction can still hold without the thread sliding: at that ratio the two masses are still in static equilibrium, not yet accelerating, so $T_1=m_1g$ and $T_2=m_2g$ . Their ratio is exactly $\eta_0$ , and substituting into (1) gives

\eta_0 = e^{k\pi}

(2)

Solving for $k$ :

k = \dfrac{\ln\eta_0}{\pi}

(3)

Step 4Part (b): Newton’s second law once the thread slips

For $\eta>\eta_0$ the friction can no longer hold the thread, so it slides over the pulley and both masses accelerate: $m_2$ downward and $m_1$ upward, by the same magnitude $w$ , since the thread is inextensible and its two strands lengthen and shorten at the same rate. Newton’s second law along the vertical for each mass, then solving each for its own tension, gives

\begin{aligned} T_1 - m_1 g &= m_1 w, & T_1 &= m_1(g+w),\\ m_2 g - T_2 &= m_2 w, & T_2 &= m_2(g-w). \end{aligned}

The differential balance of Step 1 never assumed the system was at rest, only that the thread is weightless, so the same relation (1) still connects the two tensions while the thread slides, with $k$ now fixed by part (a): $e^{k\pi}=\eta_0$ , from (2). So

T_2 = \eta_0\,T_1.

Step 5Solving for the acceleration

Substitute the two tensions found above and $m_2=\eta m_1$ , then collect the terms in $w$ :

\begin{aligned} \eta m_1(g-w) &= \eta_0\,m_1(g+w)\\ \eta(g-w) &= \eta_0(g+w)\\ \eta g - \eta_0 g &= \eta w + \eta_0 w \end{aligned}

w = \dfrac{g(\eta-\eta_0)}{\eta+\eta_0}

(4)

Check

At $\eta=\eta_0$ exactly the formula gives $w=0$ , recovering the static threshold of part (a). As $\eta\to\infty$ (an overwhelmingly heavy $m_2$ ) it gives $w\to g$ , the free-fall limit, and it never exceeds $g$ for any finite $\eta$ , since the lighter mass and the pulley’s friction can only resist $m_2$ ’s fall, never accelerate it past free fall.

Remark

The part (a) formula carries its own sanity check: $\eta_0=1$ , equal masses on the verge of slipping, means no friction is needed at all, and indeed $k=\ln 1/\pi = 0$ .