At a container port such as Nhava Sheva near Mumbai, a crane lifts a loaded container off a truck. For a moment after the container leaves the truck it bobs up and down a little on the cable, and then it settles. The cable looks rigid, yet the container bounces as if it hung from a spring.
Hooke's law in spring form
Rearrange the definition of Young's modulus, \(Y = FL/(A\,\Delta L)\), to give the force: \(F = (YA/L)\,\Delta L\). The force is proportional to the extension. That is exactly the law of a spring from the chapter on Forces and Newton's Laws, \(F = kx\). So a wire within its elastic limit is a spring, with a spring constant, or stiffness, fixed by its material and its shape.
A steel wire \(2\,\text{m}\) long and \(1\,\text{mm}^2\) in section has \(k = 2 \times 10^{11} \times 10^{-6}/2 = 10^5\,\text{N/m}\). A load of \(100\,\text{N}\), about ten kilograms, stretches it by \(100/10^5 = 10^{-3}\,\text{m}\), one millimetre. A coiled spring of the same steel would stretch by centimetres. The straight wire is a spring thousands of times stiffer.
The formula also tells what makes a wire stiff. The material enters through \(Y\). A larger area gives a proportionally stiffer wire, since there is more metal side by side to share the pull. A longer wire is softer, since there are more lengths of metal one after another, each stretching its share.
A copper wire is \(1\,\text{m}\) long and \(2\,\text{mm}^2\) in section. Take \(Y = 1 \times 10^{11}\,\text{Pa}\) for copper. Find its spring constant, in N/m.
Cutting, doubling and joining
Everything you learnt about springs in the chapter on Potential Energy and Conservation of Energy now applies to wires. Cut a wire in half and each half has twice the stiffness, \(2k\), because \(k \propto 1/L\). A wire of twice the diameter has four times the area and four times the stiffness.
Hang two wires side by side from one support, joined at the bottom to one bar. They are springs in parallel. The bar stays level, so both stretch by the same amount, and their pulls add: \(k = k_1 + k_2\). Join two wires end to end instead. They are springs in series: one tension runs through both, their extensions add, and \(1/k = 1/k_1 + 1/k_2\).
Try the chips of Fig 3.1 in order. One wire \(2\,\text{m}\) long stretches \(1\,\text{mm}\). Half of it stretches \(0.5\,\text{mm}\). The two halves side by side stretch \(0.25\,\text{mm}\), a quarter: twice the area and half the length. A wire twice as long stretches \(2\,\text{mm}\). A metre of steel followed by a metre of copper stretches \(0.5 + 1 = 1.5\,\text{mm}\).
A steel wire \(1\,\text{m}\) long is joined end to end to a copper wire \(1\,\text{m}\) long; both are \(1\,\text{mm}^2\) in section. The pair hangs from a ceiling and carries \(100\,\text{N}\). Take \(Y = 2 \times 10^{11}\,\text{Pa}\) for steel and \(1 \times 10^{11}\,\text{Pa}\) for copper. Find the extension of each wire and the stiffness of the pair.
The steel has \(k_1 = 2 \times 10^{11} \times 10^{-6}/1 = 2 \times 10^5\,\text{N/m}\), and the copper \(k_2 = 10^5\,\text{N/m}\).
One tension, \(100\,\text{N}\), runs through both. The steel stretches \(100/(2 \times 10^5) = 0.5\,\text{mm}\) and the copper \(100/10^5 = 1\,\text{mm}\).
The pair stretches \(1.5\,\text{mm}\), so \(k = 100/(1.5 \times 10^{-3}) = 6.7 \times 10^4\,\text{N/m}\). Check: \(1/k = 1/(2 \times 10^5) + 1/10^5\) gives the same.
Sharing a load side by side
Now hang a light bar from a steel wire and a copper wire side by side, each \(1\,\text{m}\) long and \(1\,\text{mm}^2\) in section. The steel is twice as stiff as the copper. If the bar is to stay level, both wires must stretch by the same amount. Then the steel pulls up with twice the force of the copper.
The load of Fig 3.2 is \(300\,\text{N}\). Equal stretches \(x\) need \(2 \times 10^5 x + 10^5 x = 300\), so \(x = 1\,\text{mm}\), with \(200\,\text{N}\) in the steel and \(100\,\text{N}\) in the copper. The bar also must not turn. Taking moments about the load point, \(200 \times d_1 = 100 \times d_2\), so the load hangs twice as far from the copper as from the steel.
A light bar hangs level from two wires of the same length side by side. One is steel of \(1\,\text{mm}^2\) (\(Y = 2 \times 10^{11}\,\text{Pa}\)), the other copper of \(2\,\text{mm}^2\) (\(Y = 1 \times 10^{11}\,\text{Pa}\)). A \(300\,\text{N}\) load keeps the bar level. How is the load shared?
A rod has stiffness \(k\). Which statements are true?
A load that bobs
Hang a load of mass \(m\) on a wire and pull it down a little. Released, it bobs up and down about its resting position, exactly as a load on a spring does. The chapter on SHM derives the period of a load on a spring, \(T = 2\pi\sqrt{m/k}\). Here it is quoted as a result. For a wire, \(k = YA/L\).
A \(1\,\text{kg}\) block hangs from a steel wire \(2\,\text{m}\) long and \(1\,\text{mm}^2\) in section, with \(Y = 2 \times 10^{11}\,\text{Pa}\). It is pulled down slightly and released. Find the period and the frequency of its vertical oscillation. Take \(\pi^2 = 10\).
The wire has \(k = YA/L = 10^5\,\text{N/m}\).
\(T = 2\pi\sqrt{1/10^5} = 2\sqrt{\pi^2 \times 10^{-5}} = 2\sqrt{10^{-4}} = 0.02\,\text{s}\).
The frequency is \(1/T = 50\,\text{Hz}\). The bob is far too fast to follow by eye; it is heard as a hum more than seen.
The last form of the formula is worth a second look. The resting stretch is \(\Delta L_0 = mg/k\), so \(m/k = \Delta L_0/g\). A load that stretches its wire very little bobs very fast. The block above stretches the wire only \(0.1\,\text{mm}\), and it bobs fifty times a second.
The \(1\,\text{kg}\) block of the example is replaced by a \(4\,\text{kg}\) block on the same wire. Find the new period of vertical oscillation, in seconds. Take \(\pi^2 = 10\).
Treat a port crane's cable as a steel rod \(50\,\text{m}\) long with \(10\,\text{cm}^2\) of metal, a simplification of a rope of many strands. Its stiffness is \(k = 2 \times 10^{11} \times 10^{-3}/50 = 4 \times 10^6\,\text{N/m}\). A \(10\)-tonne container stretches it by \(10^5/(4 \times 10^6) = 2.5\,\text{cm}\). So the container bobs with \(T = 2\pi\sqrt{0.025/10} \approx 0.31\,\text{s}\), about three times a second, until friction in the rope damps it out.
Return to the container at the port. The steel cable is a spring with \(k\) in the millions of newtons per metre. When the container leaves the truck, the cable takes the load, stretches a couple of centimetres and overshoots a little. The container then bobs about three times a second on that stiff spring before the bobbing dies away.
- Taking a wire of twice the diameter to be twice as stiff.The area goes as the diameter squared, so the stiffness is four times as large.
- Adding the stiffnesses of wires joined end to end.End to end the extensions add, so the compliances add: \(1/k = 1/k_1 + 1/k_2\).
- Thinking a shorter piece of wire is softer.\(k = YA/L\): half the length is twice as stiff.
- Using the natural length for \(\Delta L_0\) in \(T = 2\pi\sqrt{\Delta L_0/g}\).\(\Delta L_0\) is the extension produced by the load at rest, \(mg/k\).