Irodov 1.157: Falling Links Double the Push – Amit & Deepali Sinha Foundation

Problem Irodov 1.157

A uniform flexible chain hangs vertically from a thread, with its lower end just touching a horizontal table top. The thread is burned through, so the whole chain begins to fall freely; each link is undisturbed until it reaches the table, where it comes to rest and joins a growing pile. At some later moment a length $x$ of chain is piled at rest on the table while the rest is still falling. Show that at that moment the force the falling part exerts on the table (through the links arriving and stopping) is exactly twice the weight of the chain already piled on the table.

An instant after the thread is burned. The falling part has dropped x below its starting level (dashed) and reaches the table at speed v; the length x already down lies at rest on the table. The two dimensions marked x are equal. — An instant after the thread is burned. The falling part has dropped $x$ below its starting level (dashed) and reaches the table at speed $\vect v$ ; the length $x$ already down lies at rest on the table. The two dimensions marked $x$ are equal.

Given

uniform chain, linear density $\lambda$
released from rest touching the table

Find

show that the falling part’s push on the table is twice the resting part’s weight, at any

x

Solution

Step 1Each link falls freely until it lands

The top of the chain, and therefore every falling link, has dropped exactly the length $x$ already collected on the table, since the lower end started at the table surface. Falling freely from rest through that height, the standard constant-acceleration relation $v^2=v_0^2+2a\,\Delta y$ with $v_0=0$ and $a=g$ gives the arrival speed.

v^2 = 2gx \tag{1}

Step 2The resting part’s weight on the table

The length $x$ sitting motionless on the table has mass $\lambda x$ , so it presses on the table with its own weight alone.

N_{\text{rest}} = \lambda x g

(2)

Step 3The falling part’s push from arriving momentum

In a short time $dt$ the chain descends a distance $v\,dt$ , so a length $v\,dt$ of it crosses the table surface and stops. Its mass is that length times the linear density.

dm = \lambda v\,dt

That mass arrives at speed $v$ and is left at rest, so it gives up the whole of the momentum it carried. Writing $dp$ for the magnitude of that change, which is directed upward, against the motion,

dp = dm\,v - 0 = \lambda v^2\,dt.

The impulse-momentum theorem equates that change to the impulse of the net force on the arriving mass. Over the interval $dt$ its own weight contributes an impulse of order $dm\,g\,dt$ , second order in $dt$ and therefore negligible beside $dp$ , so the whole of the impulse comes from the extra force $F$ the table applies to stop it.

F\,dt = dp = \lambda v^2\,dt

Dividing through by $dt$ turns the impulse balance into a force, and by Newton’s third law the arriving chain pushes on the table with the same magnitude.

F = \lambda v^2 \tag{3}

Substituting the free-fall speed (1) into (3) turns this into a function of the collected length alone.

F = 2\lambda g x

(4)

Step 4Comparing the two contributions

Dividing (4) by (2) shows the ratio is fixed, independent of $x$ .

\frac{F}{N_{\text{rest}}} = \frac{2\lambda g x}{\lambda g x} = 2.

F = 2N_{\text{rest}}

(5)

Check

Dimensionally, $\lambda v^2$ is (mass/length)(length/time) $^2$ = mass $\cdot$ length/time $^2$ , a force, matching $\lambda x g$ . At $x=0$ (the instant the thread is burned) both $F$ and $N_{\text{rest}}$ vanish, and the total force on the table, $N_{\text{rest}}+F=3\lambda xg$ , grows linearly with the pile height throughout the fall precisely because $v^2=2gx$ is itself linear in $x$ : the momentum-flux term $F=\lambda v^2=2\lambda gx$ inherits that same linear dependence, not merely because a landing rate happens to exist.

Remark

The total table force, $3\lambda xg$ , is three times the weight of chain actually at rest on the table at that instant, not one time: a falling chain presses on what it lands on noticeably harder than its own resting weight would suggest, which is the same physics (momentum flux, not just weight) behind why a stream of water or sand striking a surface can outweigh its own static heap.