Irodov 1.78: A Peg Anchored Thread Drags the Block – Amit & Deepali Sinha Foundation

Problem Irodov 1.78

A wedge of mass $M$ stands on a frictionless horizontal floor and is free to slide on it. Body $m$ rides against the wedge’s vertical face. A thread tied to $m$ runs up over a pulley fixed at the top of that face, down over a second pulley fixed at the bottom of the same face, and then horizontally to a peg fixed in the floor, independent of the wedge. The pulleys and thread are massless and frictionless; the coefficient of friction between $m$ and the wedge’s face is $k$ , and this is the only friction in the system. Find the magnitude of body $m$ ’s acceleration relative to the floor.

Body m against wedge M's vertical face; the wedge slides freely on the hatched floor. The thread runs from m over the apex pulley, down to a second pulley at the face's foot, then horizontally to a floor peg. — Body $m$ against wedge $M$ ’s vertical face; the wedge slides freely on the hatched floor. The thread runs from $m$ over the apex pulley, down to a second pulley at the face’s foot, then horizontally to a floor peg.

Given

Find

the acceleration of

m

relative to the floor

Solution

Step 1One horizontal slide couples to one vertical slide

Since body $m$ is pressed against the wedge’s face, it shares the wedge’s own horizontal acceleration $w$ , taken as positive toward the peg. Write $X$ for the wedge’s horizontal position in that same sense, $y$ for body $m$ ’s height above the floor, $x_{\text{peg}}$ for the peg’s fixed position, and $h$ for the apex pulley’s fixed height. The thread is the strand from $m$ up to the apex pulley, the pulley to pulley strand, and the horizontal strand out to the peg, so its total length is

L = (h-y) + c + (x_{\text{peg}} - X - d)

with $c$ and $d$ constants fixed by the wedge’s own geometry. Everything in that line except $y$ and $X$ is constant, so their sum is constant too,

y + X = \text{const}

Differentiating twice with respect to time,

\ddot y + \ddot X = 0

and taking magnitudes, body $m$ slides down the face at exactly the rate the wedge advances toward the peg,

w_{m,\text{vert}} = w

(1)

Body $m$ ’s acceleration relative to the floor therefore has a horizontal component $w$ (moving with the wedge) and a vertical component of the same magnitude $w$ (sliding along the wedge’s face), at right angles to each other,

w_{\text{rel}} = \sqrt{w^2+w^2} = w\sqrt2 \tag{2}

Step 2Newton’s second law for body

m

Taking the wedge’s acceleration $w$ toward the peg as positive, body $m$ shares that horizontal acceleration and, by (1), slides downward along the face at the same rate $w$ , with friction $f=kN$ opposing that downward slide.

\begin{align*}mw &= N \\ mw &= mg - T - kN\end{align*}

Substituting $N=mw$ from the first line into the second and solving for $T$ ,

mw = mg - T - kmw

T = mg - mw(1+k)

(3)

Step 3Newton’s second law for the wedge

The only horizontal forces on the frictionless wedge are the reaction $-N$ from body $m$ and the thread’s pull $T$ toward the peg, transmitted through the two pulleys mounted on it.

Mw = T - N

Substituting $N=mw$ and $T$ from (3),

\begin{align*}Mw &= mg - mw(1+k) - mw \\ Mw + mw(1+k) + mw &= mg \\ w\bigl(M+2m+km\bigr) &= mg\end{align*}

w = \dfrac{mg}{M+2m+km}

(4)

Step 4Combine with Step 1

Substituting $w$ from (4) into the relative acceleration (2),

w_{\text{rel}} = \sqrt2\cdot\dfrac{mg}{M+2m+km}

w_{\text{rel}} = \dfrac{g\sqrt2}{2+k+M/m}

(5)

Check

Dimensionally $g$ divided by a pure number is an acceleration, as required. As $M\to\infty$ , (4) gives $w\to0$ and hence $w_{\text{rel}}\to0$ : an immovably heavy wedge cannot be pushed, so body $m$ , pinned against its face, cannot move relative to the floor either.

Remark

The factor $\sqrt2$ comes entirely from the rope constraint (1), which forces $m$ ’s vertical speed relative to the floor to equal the wedge’s horizontal speed; no incline angle enters the problem at all, since $m$ rides the wedge’s flat vertical face throughout, not a slope. It is the fixed pulley to pulley segment, both ends riding on the wedge, that makes this particular two pulley arrangement transmit length changes one for one between the peg strand and the $m$ strand.