Irodov 1.294: The Sag That Cubes the Load – Amit & Deepali Sinha Foundation

Problem Irodov 1.294

A steel wire of diameter $d = 1.0$ mm is stretched taut between two fixed clamps a distance $l = 2.0$ m apart. A small weight of mass $m = 0.25$ kg is hung from the wire’s midpoint, and the wire sags a little to accommodate it. Find how far the midpoint descends, in centimeters.

The wire, originally the straight line AB, sags to the point O once the weight is hung. Each half makes a small angle θ with the original line, and x is O's descent. — The wire, originally the straight line $AB$ , sags to the point $O$ once the weight is hung. Each half makes a small angle $\theta$ with the original line, and $x$ is $O$ ’s descent.

Given

$d = 1.0$ mm
$l = 2.0$ m
$m = 0.25$ kg

Find

x

, the descent of

O

, in cm

Solution

Step 1The strain each half acquires

Before loading, each half of the wire runs straight from a clamp to $O$ , length $l/2$ . Once $O$ drops by $x$ , that half instead runs the hypotenuse of a right triangle with legs $l/2$ and $x$ , length $\sqrt{(l/2)^2+x^2}$ . Since $x \ll l$ , expand the square root to first order in $(2x/l)^2$ :

\sqrt{\left(\frac{l}{2}\right)^2+x^2} = \frac{l}{2}\sqrt{1+\left(\frac{2x}{l}\right)^2} \approx \frac{l}{2}\left(1+\frac{2x^2}{l^2}\right) = \frac{l}{2}+\frac{x^2}{l}.

So each half stretches by $x^2/l$ , and its strain is

\varepsilon = \frac{x^2/l}{l/2} = \frac{2x^2}{l^2}. \tag{1}

Step 2The tension this strain produces

By Hooke’s law the tension in the (now stretched) half wire is $T = EA\varepsilon$ , with cross-sectional area $A = \pi d^2/4$ . Substituting $A$ and the strain (1) gives $T$ as a function of the sag.

T = EA\varepsilon = \frac{\pi d^2 E}{4}\cdot\frac{2x^2}{l^2}

T = \frac{\pi d^2 E x^2}{2l^2}

(2)

Step 3Vertical equilibrium at

O

Each half pulls back up along its own length, at angle $\theta$ to the horizontal. For small $x$ the angle itself is small, so $\sin\theta$ is well approximated by $\tan\theta$ , the ratio of the drop $x$ to the horizontal half-span $l/2$ .

\sin\theta \approx \tan\theta = \frac{x}{l/2} = \frac{2x}{l}. \tag{3}

The two vertical components of the tension support the weight.

2T\sin\theta = mg. \tag{4}

Substituting (2) and (3) into (4):

\begin{aligned} 2\cdot\frac{\pi d^2 E x^2}{2l^2}\cdot\frac{2x}{l} &= mg\\ \frac{2\pi d^2 E x^3}{l^3} &= mg\\ x^3 &= \frac{mgl^3}{2\pi d^2 E} \end{aligned} \tag{5}

Step 4Numbers

Steel: $E = 2.0\times10^{11}$ Pa. With $m=0.25$ kg, $g=9.8$ m/s $^2$ , $d=1.0\times10^{-3}$ m, $l=2.0$ m, evaluate the ratio under the cube root, take the cube root, then scale by $l$ .

\frac{mg}{2\pi d^2 E} = \frac{0.25\times9.8}{2\pi\times(10^{-3})^2\times2.0\times10^{11}} = \frac{2.45}{1.257\times10^6} = 1.95\times10^{-6},

\sqrt[3]{1.95\times10^{-6}} = 1.25\times10^{-2}.

x = l\times1.25\times10^{-2} = 2.0\times1.25\times10^{-2} = 2.5\times10^{-2}\ \text{m} = 2.5\ \text{cm}.

x = l\sqrt[3]{\frac{mg}{2\pi d^2 E}} \approx 2.5\ \text{cm}

(6)

Check

$x/l \approx 0.0125$ : the sag is only about one percent of the span, which is what justified the small angle expansion behind (1). Had $x/l$ come out order unity, the whole derivation would need revisiting.

Remark

Because $x$ scales as $E^{-1/3}$ , the sag is remarkably insensitive to the exact value of Young’s modulus: doubling $E$ only shrinks $x$ by a factor of $2^{-1/3}\approx0.79$ , about twenty percent. A taut, initially straight wire is therefore a poor way to measure $E$ precisely, even though the same cubic softness is exactly why it makes such a sensitive scale for small weights.