Irodov 1.295: Compression That Tapers to Nothing – Amit & Deepali Sinha Foundation

Problem Irodov 1.295

A uniform elastic plank slides over a frictionless horizontal plane, driven by a constant force $F_0$ applied uniformly over one of its end faces, of area $S$ . The material’s Young’s modulus is $E$ . Find the plank’s overall compressive strain in the direction the force acts.

The plank, of length l and cross section S, sliding on a smooth horizontal plane and driven by F_0 applied uniformly over its left end face. — The plank, of length $l$ and cross section $S$ , sliding on a smooth horizontal plane and driven by $\vect F_0$ applied uniformly over its left end face.

Given

$F_0$
$S$
$E$

Find

\varepsilon

, the compressive strain along the direction of

\vect F_0

Solution

Step 1The internal force needed to accelerate the rest of the plank

The same plank with a highlighted slice dx a distance x from the driven end. Everything beyond the section at x must be pushed by the force carried across it. — The same plank with a highlighted slice $dx$ a distance $x$ from the driven end. Everything beyond the section at $x$ must be pushed by the force carried across it.

Let the plank have length $l$ and density $\rho$ , so its mass is $M=\rho S l$ . Since the plane is smooth, $F_0$ alone accelerates the whole plank, $a = F_0/M$ . At a cross section a distance $x$ from the driven end, the material lying beyond it, of mass $\rho S(l-x)$ , must itself be pushed at that same acceleration; the internal force $N(x)$ carried across that section supplies it.

\begin{aligned} N(x) &= \rho S(l-x)\,a\\ &= \rho S(l-x)\cdot\frac{F_0}{\rho S l} \end{aligned}

N(x) = F_0\left(1-\frac{x}{l}\right)

(1)

Step 2The plank’s overall shortening

The local compressive strain at $x$ is $N(x)/(SE)$ by Hooke’s law. The plank’s total change in length is that local strain integrated over its whole length.

\Delta l = \int_0^l\frac{N(x)}{SE}\,dx = \frac{F_0}{SE}\int_0^l\left(1-\frac{x}{l}\right)dx = \frac{F_0}{SE}\left[x-\frac{x^2}{2l}\right]_0^l

\Delta l = \frac{F_0 l}{2SE}

(2)

Step 3The overall compressive strain

Dividing the shortening (2) by the plank’s own length $l$ gives the strain asked for; $l$ cancels, leaving a result independent of the plank’s size.

\varepsilon = \frac{\Delta l}{l} = \frac{F_0 l}{2SEl}

\varepsilon = \frac{1}{2}\frac{F_0}{ES}

(3)

Check

$[F_0]/([E][S]) = \text{N}/(\text{N/m}^2\cdot\text{m}^2) = \text{N/N}$ , dimensionless, correct for a strain. Had the plank instead been in static equilibrium, held by equal and opposite forces $F_0$ at both ends, every cross section would carry the full $F_0$ and the strain would be $F_0/(ES)$ , twice the answer above; the factor $1/2$ is exactly the signature of the internal force tapering from $F_0$ down to zero rather than staying uniform.

Remark

This result generalizes beyond a plank: for any uniform slender body of constant cross section, pushed from one end while sliding freely on a frictionless surface, Newton’s second law forces the same triangular internal-force profile, full at the driven end and zero at the free end, so the average strain always comes out to exactly half the value naively computed by pretending the whole body carries a uniform stress $F_0/S$ . A train of freight cars coupled and pushed from the locomotive end carries the same triangular profile, its buffer forces greatest at the locomotive and tapering to nothing at the last car; that taper is the cause and the factor of one half is its consequence, not the other way round. Real cars also drag against rolling resistance, so the profile of an actual train is only approximately triangular.