Irodov 1.159: Raft Recoils While Man Walks – Amit & Deepali Sinha Foundation

Problem Irodov 1.159

A raft of mass $M$ carrying a man of mass $m$ floats at rest on a lake, with water resistance neglected entirely. The man walks across the raft, moving a distance $l'$ relative to the raft with a known velocity $v'(t)$ relative to the raft, and then comes to rest relative to the raft again. Find:

the raft’s net displacement $\vect l$ relative to the shore;
the horizontal component of the force the man exerts on the raft, as a function of time, during the walk.

The raft of mass M floats at rest; the man of mass m walks across it with velocity v'(t) relative to the raft, covering relative distance l' before stopping. — The raft of mass $M$ floats at rest; the man of mass $m$ walks across it with velocity $\vect v'(t)$ relative to the raft, covering relative distance $l'$ before stopping.

Given

$M$
$m$
$l'$ (man’s net displacement relative to raft)
$v'(t)$ (man’s velocity relative to raft)

Find

(a) raft’s displacement

\vect l

relative to shore, (b) horizontal force

\vect F

on the raft from the man, as a function of time

Solution

Step 1The center of mass never moves

Let $\vect x_m(t)$ and $\vect x_r(t)$ be the man’s and the raft’s displacements, each measured from its own initial position, relative to the shore. Since the center of mass stays fixed at every instant, the mass-weighted sum of the two displacements vanishes at every instant.

m\vect x_m(t)+M\vect x_r(t) = \vect 0 \tag{1}

Step 2Relating the two displacements to the given relative motion

The man’s displacement relative to the raft at any time is $\vect x_m(t)-\vect x_r(t)$ ; at the end of the walk this equals the stated net relative displacement $\vect l'$ .

\vect x_m-\vect x_r = \vect l' \tag{2}

Substituting $\vect x_m=\vect x_r+\vect l'$ from (2) into (1) and solving for $\vect x_r$ ,

\begin{align*}m(\vect x_r+\vect l')+M\vect x_r &= \vect 0\\ (m+M)\vect x_r &= -m\vect l'.\end{align*}

\vect l = -\dfrac{m}{M+m}\vect l'

(3)

Step 3The raft’s instantaneous velocity from the man’s

Differentiating (1) and using $\vect v'(t)=\dot{\vect x}_m(t)-\dot{\vect x}_r(t)$ (the man’s velocity relative to the raft) to eliminate $\dot{\vect x}_m$ ,

\begin{align*}0 &= m\dot{\vect x}_m+M\dot{\vect x}_r\\ &= m\bigl(\dot{\vect x}_r+\vect v'\bigr)+M\dot{\vect x}_r\\ &= (M+m)\dot{\vect x}_r+m\vect v'.\end{align*}

\dot{\vect x}_r(t) = -\dfrac{m}{M+m}\vect v'(t)

(4)

Step 4The force on the raft from Newton’s second law

The only horizontal force on the raft (mass $M$ ) is the contact force from the man’s feet, so it equals $M$ times the raft’s own acceleration. Differentiating the raft velocity (4),

\vect F = M\ddot{\vect x}_r = M\frac{d}{dt}\left(-\dfrac{m}{M+m}\vect v'(t)\right).

\vect F = -\dfrac{mM}{M+m}\dfrac{d\vect v'}{dt}

(5)

Check

Integrating the force (5) over the whole walk, from $t=0$ to the moment the man stops (where $\vect v'=\vect 0$ at both ends), gives zero net impulse on the raft, consistent with the raft returning to a constant velocity (zero, since it started at rest) once the man is again at rest relative to it. And at any instant $\vect F=0$ exactly when $d\vect v'/dt=0$ , i.e. whenever the man walks at constant relative speed, matching the intuition that a steady walk needs no net horizontal push once under way.

Remark

The raft’s displacement (3) depends only on the man’s net relative displacement $\vect l'$ , never on the details of $\vect v'(t)$ : however he speeds up, slows down or reverses along the way, the raft ends up displaced by exactly $-m\vect l'/(M+m)$ once he stops. The force in part (b), by contrast, tracks $d\vect v'/dt$ at every instant, so two walks with the same net $l'$ but different speed profiles push the raft quite differently moment to moment even though they leave it in the same final place.