Irodov 1.124: Integrating Friction as the Chain Leaves

Problem Irodov 1.124

A uniform chain of mass $m=0.80$ kg and length $l=1.5$ m lies on a rough horizontal table, with part of its length hanging vertically over the edge. The chain is on the verge of sliding off entirely by itself once the hanging fraction reaches $\eta=1/3$ of the total length. Taking that instant as the start of the motion, find the total work done by the friction forces on the chain by the moment the whole chain has left the table.

The chain at the threshold instant, an overhanging length η l pulling against a length (1-η)l still lying on the table; the table extends further left than the chain itself. — The chain at the threshold instant, an overhanging length $\eta l$ pulling against a length $(1-\eta)l$ still lying on the table; the table extends further left than the chain itself.

Given

$m=0.80$ kg
$l=1.5$ m
$\eta=1/3$

Find

A_{\mathrm{fric}}

Solution

Step 1Friction coefficient from the threshold overhang

At the threshold the weight of the hanging length $\eta l$ exactly equals the greatest friction the part still on the table, of length $(1-\eta)l$ , can supply; that balance fixes $k$ in terms of $\eta$ alone.

\eta mg = k(1-\eta)mg

k = \dfrac{\eta}{1-\eta}

(1)

Step 2Friction force at a general overhang

x

Once sliding starts, let $x\ge\eta l$ be the length already hanging. The remaining length $l-x$ still rests on the table and carries the whole normal force there, so kinetic friction on that stretch is

f(x) = k\,\frac{m}{l}(l-x)\,g.

Step 3Work of friction over the whole slide

Integrating $f(x)$ , always opposing the chain’s motion, from the threshold $x=\eta l$ to $x=l$ , when the chain leaves the table entirely, gives the total friction work.

\begin{align*}A_{\mathrm{fric}} &= -\int_{\eta l}^{l} f(x)\,dx \\ &= -\frac{kmg}{l}\int_{\eta l}^{l}(l-x)\,dx \\ &= -\frac{kmg}{l}\left[-\frac{(l-x)^2}{2}\right]_{\eta l}^{l}\end{align*}

A_{\mathrm{fric}} = -\dfrac{kmgl(1-\eta)^2}{2}

(2)

Step 4Eliminating

k

and evaluating

Substituting (1) for $k$ into (2) collapses the $(1-\eta)$ factors to a single power, and the given numbers evaluate the result.

A_{\mathrm{fric}} = -\frac{\eta}{1-\eta}\cdot\frac{mgl(1-\eta)^2}{2}

A_{\mathrm{fric}} = -\dfrac{(1-\eta)\eta\, mgl}{2} = -1.3\text{ J}

(3)

Check

Dimensionally $[m][g][l]=\mathrm{kg\cdot(m/s^2)\cdot m}=\mathrm J$ and $\eta(1-\eta)$ is dimensionless, so the boxed quantity is an energy, as required. Numerically $\eta(1-\eta)=(1/3)(2/3)=2/9$ and $mgl=0.80(9.8)(1.5)=11.76$ J, giving $A_{\mathrm{fric}}=-(2/9)(11.76)/2\approx-1.31$ J, matching the boxed value to the given precision.

Remark

The answer is symmetric under $\eta\leftrightarrow1-\eta$ even though the physical situation is not: a chain released with only a small fraction $\eta$ hanging dissipates the same energy as one released with almost the whole chain already hanging ( $\eta$ near 1). Both extremes in fact give $A_{\mathrm{fric}}\to0$ , but for different reasons at each end: as $\eta\to0$ the friction coefficient $k=\eta/(1-\eta)$ itself vanishes even though nearly the whole chain, and so nearly the greatest possible normal force, is still on the table; as $\eta\to1$ almost no chain is left on the table to rub at all, however large $k$ has grown. The dissipation is largest near the middle, at $\eta=1/2$ , where neither effect has yet taken hold.