Irodov 1.160: Climbing Man Shifts the Whole System – Amit & Deepali Sinha Foundation

Problem Irodov 1.160

A rope passes over a fixed, frictionless pulley (massless rope, negligible axle friction). One end carries a ladder with a man of mass $m$ standing on it; the other end carries a counterweight of mass $M$ . Before the man moves, the assembly hangs motionless. The man then climbs a net distance $l'$ up the ladder, measured relative to the ladder itself, and comes to rest again relative to the ladder. Find the net displacement $l$ of the center of inertia of the whole system (ladder, man and counterweight together).

The ladder, carrying the man of mass m, hangs from one strand of the rope, the counterweight M from the other, over a fixed pulley. The arrow on the left is the man's climb l' relative to the ladder. — The ladder, carrying the man of mass $m$ , hangs from one strand of the rope, the counterweight $M$ from the other, over a fixed pulley. The arrow on the left is the man’s climb $l'$ relative to the ladder.

Given

$m$ (man)
$M$ (counterweight)
$l'$ (man’s net climb relative to the ladder)

Find

displacement

l

of the center of inertia of the whole system

Solution

Step 1Static start fixes the ladder’s own mass

Writing $M_L$ for the ladder’s own mass, the tension supporting the ladder side (ladder plus man) and the tension supporting the counterweight side must separately equal the weight each side carries at rest.

(M_L+m)g = T = Mg.

M_L = M-m

(1)

Step 2The rope ties the ladder’s and the counterweight’s motion together

Taking the upward direction as positive and measuring every displacement from the instant before the man starts climbing, the ladder’s displacement $x_L(t)$ and the counterweight’s displacement $x_C(t)$ satisfy

x_C(t) = -x_L(t). \tag{2}

Step 3The center of inertia moves with the man’s climb alone

The system’s total mass is $M_L+m+M=2M$ by (1). Writing the center of inertia’s displacement as the mass-weighted average of the three bodies’ own displacements, with $x_m(t)$ the man’s displacement,

\begin{align*}2M\,x_{\text{cm}}(t) &= M_L\,x_L(t)+m\,x_m(t)+M\,x_C(t)\\ &= (M-m)x_L(t)+m\,x_m(t)-M\,x_L(t)\end{align*}

The two $x_L$ terms cancel, leaving the man’s displacement relative to the ladder alone.

2M\,x_{\text{cm}}(t) = m\bigl[x_m(t)-x_L(t)\bigr]

(3)

Step 4Evaluate once the man has stopped

Once the man has climbed and stopped, $x_m-x_L$ has reached its final value, the given net climb $l'$ (climbing up, the positive sense chosen in Step 2). Substituting into (3) at that instant,

2M\,l = m\,l'.

l = \dfrac{ml'}{2M}

(4)

Check

Dimensionally, $ml'/(2M)$ is a mass ratio times a length, a length, as required. For a check that leans on the dynamics rather than on (3), integrate Newton’s second law for the Atwood pair, $2M\ddot x_L+m\ddot x_{\text{rel}}=0$ , from rest, which fixes the ladder’s own displacement at $x_L=-ml'/(2M)$ once the man has stopped. Now take the extreme case $m=M$ , where (1) makes the ladder itself massless and the only two massive bodies left are the man and the counterweight. There $x_L=-l'/2$ , so the counterweight rises by $l'/2$ , and the man, climbing $l'$ along a ladder that has meanwhile dropped $l'/2$ , rises by $l'/2$ as well. Two equal masses each displaced $l'/2$ upward carry their center of inertia $l'/2$ upward, which is exactly what $ml'/(2M)$ returns when the two masses are equal.

Remark

The relation (3) holds at every instant, not only once the man has stopped: $x_{\text{cm}}(t)=m[x_m(t)-x_L(t)]/(2M)$ tracks the man’s climb relative to the ladder in real time, so the center of inertia does drift while he climbs and only settles at $l=ml'/2M$ once he stops. Nothing in the argument needed the man’s speed profile on the ladder, only his net relative displacement, exactly as Irodov 1.159’s raft problem turned on the net relative walk and not on how it was carried out.