The gold mines of Kolar in Karnataka were among the deepest in the world, with shafts about three kilometres deep. Steel cables lowered the cages of miners down those shafts. Near the bottom a cable held up only the cage. At the top it held up the cage and every metre of cable below as well.
The tension depends on where you cut
So far a wire under a load has had one tension along its whole length. Now let a uniform rod hang from a ceiling with nothing at its lower end. The rod has a length \(L\), an area of cross-section \(A\), a density \(\rho\) and Young's modulus \(Y\). Imagine a cut across it at a height \(x\) above the lower end.
The part below the cut hangs at rest. Two forces act on it: its weight pulling down, and the pull of the rod above the cut, the tension \(T\), pulling up. The part below has a volume \(Ax\), a mass \(\rho A x\) and a weight \(\rho g A x\). So the tension at the cut is \(T(x) = \rho g A x\).
Move the cut in the figure above. At the bottom the tension is zero, because nothing hangs below. At the top it is the whole weight, \(80\,\text{N}\). In between it grows in a straight line, \(8\,\text{N}\) for every metre. With the \(80\,\text{N}\) load added, every tension rises by \(80\,\text{N}\), and the line starts at \(80\,\text{N}\) instead of zero.
Adding up the stretch
The stress changes along the rod, so the rod does not stretch evenly. Its upper part, under more tension, stretches more than its lower part. Take a thin slice of length \(dx\) at the height \(x\). The tension across it is \(T(x)\), which hardly changes over so short a length. So the slice stretches by \(T\,dx/(AY)\), and the rod stretches by the sum over all slices.
\[ T(x) = \rho g A x \]
Why. the tension at a height \(x\) holds up the weight of the rod below it.
\[ d(\Delta L) = \dfrac{T\,dx}{AY} = \dfrac{\rho g x\,dx}{Y} \]
Why. a slice of length \(dx\) obeys \(Y = (T/A)/(d(\Delta L)/dx)\), so it stretches by \(T\,dx/(AY)\).
\[ \Delta L = \displaystyle\int_0^L \dfrac{\rho g x}{Y}\,dx = \dfrac{\rho g L^2}{2Y} \]
Why. add the stretches of all the slices from the bottom, \(x = 0\), to the top, \(x = L\).
\[ \Delta L = \dfrac{WL}{2AY} \]
Read it. the weight of the rod is \(W = \rho g A L\), so \(\rho g = W/(AL)\).
Compare \(WL/(2AY)\) with the stretch of a light wire carrying the same weight \(W\) at its end, \(WL/(AY)\). The rod stretches half as much. The tension in the rod climbs evenly from \(0\) to \(W\), so on average it is \(W/2\). A rod under its own weight behaves as if half its weight hung at its lower end.
A steel rod \(10\,\text{m}\) long hangs from a ceiling. For steel \(\rho = 8000\,\text{kg/m}^3\) and \(Y = 2 \times 10^{11}\,\text{Pa}\). Take \(g = 10\,\text{m/s}^2\). Find its elongation under its own weight. Compare it with the stretch of a light wire of the same size carrying the same weight.
\(\Delta L = \dfrac{\rho g L^2}{2Y} = \dfrac{8000 \times 10 \times 100}{2 \times 2 \times 10^{11}} = 2 \times 10^{-5}\,\text{m} = 0.02\,\text{mm}\).
The area has cancelled: a thick rod and a thin rod of the same steel stretch equally under their own weight.
With the same weight at the end of a light wire of the same size, the stretch doubles to \(0.04\,\text{mm}\).
A steel rod \(20\,\text{m}\) long hangs from a ceiling. Take \(\rho = 8000\,\text{kg/m}^3\), \(Y = 2 \times 10^{11}\,\text{Pa}\) and \(g = 10\,\text{m/s}^2\). Find its elongation under its own weight, in mm.
How deep a cable can hang
The stress is greatest at the top, where the whole weight acts: \(\rho g A L / A = \rho g L\). The area has cancelled again. A thicker cable carries more weight, but it has a larger area to carry it, in exactly the same ratio. So the stress at the top depends only on the density and the length.
A cable breaks when the stress at the top reaches the breaking stress of its material. So there is a longest length that a cable can hang, set by its material alone. For a steel of breaking stress \(4 \times 10^8\,\text{Pa}\), that length is \(4 \times 10^8/(8000 \times 10) = 5000\,\text{m}\). A deep mine hoist must therefore use a strong steel and a large safety margin, and the deepest shafts are worked in stages.
A steel cable hanging freely would break under its own weight at a length of \(5\,\text{km}\). What is the breaking length of a steel cable of twice the diameter?
A rod hanging in a lift that accelerates upward at \(a\) needs more tension. Each part below a cut must be pulled up hard enough to hold it and also to accelerate it. The tension becomes \(\rho A x (g + a)\), and every result above holds with \(g\) replaced by \(g + a\). The \(10\,\text{m}\) steel rod in a lift accelerating upward at \(2\,\text{m/s}^2\) stretches \(0.024\,\text{mm}\).
In his Two New Sciences of 1638, Galileo asked how long a rope or a metal rod could hang. At some length it must break under its own weight. He argued that making it thicker cannot help. A thicker rod is stronger, but it is heavier in exactly the same proportion. Only a stronger or lighter material hangs further. It was one of the first arguments in which the size of a thing, and not only its shape, decided whether it could stand.
A rod pulled on a smooth table
Lay a uniform rod of mass \(m\) and length \(L\) on a smooth table and pull its right end with a force \(F\). Nothing pulls the left end, so the rod accelerates at \(a = F/m\). Every part of the rod has this same acceleration, and each part needs a net force to give it that acceleration.
Cut the rod at a distance \(x\) from the free end. The piece to the left of the cut has mass \(mx/L\), and only the tension at the cut pulls it. So \(T(x) = (mx/L)\,a = Fx/L\). The tension is zero at the free end, \(F/2\) at the middle and \(F\) at the pulled end.
The stretch again adds up slice by slice, and the tension grows evenly from \(0\) to \(F\). So the rod stretches as if its average tension, \(F/2\), acted along the whole length: \(\Delta L = FL/(2AY)\). That is half the stretch of the same rod pulled by \(F\) at both ends and held at rest.
Now pull the two ends in opposite directions with unequal forces, \(F_1\) at one end and \(F_2\) at the other, \(F_1 > F_2\). The rod accelerates towards the larger force. The tension grows evenly from \(F_2\) at one end to \(F_1\) at the other, as the last chip of Fig 2.1 shows. The average tension is \((F_1 + F_2)/2\).
A rod \(1\,\text{m}\) long with area \(1\,\text{mm}^2\) and \(Y = 2 \times 10^{11}\,\text{Pa}\) lies on a smooth table. Forces of \(30\,\text{N}\) and \(10\,\text{N}\) pull its two ends in opposite directions. Find the tension at the middle and the elongation.
The tension grows evenly from \(10\,\text{N}\) to \(30\,\text{N}\), so at the middle it is \(10 + 20/2 = 20\,\text{N}\).
\(\Delta L = \dfrac{(30 + 10) \times 1}{2 \times 10^{-6} \times 2 \times 10^{11}} = 10^{-4}\,\text{m} = 0.1\,\text{mm}\).
This is the same extension as \(20\,\text{N}\) at both ends: only the average tension matters.
A uniform rod on a smooth table is pulled by a force of \(60\,\text{N}\) at one end only. Find the tension at a point one third of the way along the rod from the pulled end, in N.
A rod that turns
A uniform rod of mass \(m\) and length \(L\) turns on a smooth table about a fixed end O at angular velocity \(\omega\). Every point moves in a circle, and a point at a distance \(r\) from O needs a centripetal acceleration \(\omega^2 r\) towards O. Only the rod itself can supply the force. So the rod is stretched, and the tension changes along it.
The tension at \(x\) must supply the centripetal force for everything beyond \(x\). A slice of length \(dr\) at \(r\) has mass \((m/L)\,dr\) and needs a force \((m/L)\,\omega^2 r\,dr\). Adding these from \(r = x\) to \(r = L\) gives the tension at the cut.
\[ dF = \dfrac{m}{L}\,dr\;\omega^2 r \]
Why. a slice of length \(dr\) at \(r\) has mass \((m/L)\,dr\) and needs a centripetal force of mass times \(\omega^2 r\).
\[ T(x) = \displaystyle\int_x^L \dfrac{m\omega^2}{L}\,r\,dr \]
Why. the tension at the cut is the only inward force on the whole outer part, so it equals the sum of the slices' needs.
\[ T(x) = \dfrac{m\omega^2(L^2 - x^2)}{2L} \]
Read it. the integral of \(r\) from \(x\) to \(L\) is \((L^2 - x^2)/2\).
The tension is greatest at the axis, \(m\omega^2 L/2\), where the whole rod must be held in. It falls to zero at the free end, where nothing lies beyond. In a frame turning with the rod, the same tension balances the centrifugal force on the outer part. The chapter on Dynamics of Circular Motion used that frame.
A steel rod \(1\,\text{m}\) long and \(1\,\text{cm}^2\) in section (mass \(0.8\,\text{kg}\)) turns at \(50\,\text{rad/s}\) about one end on a smooth table. Take \(Y = 2 \times 10^{11}\,\text{Pa}\). Find the tension at the axis and at the middle, and the elongation.
At the axis \(T = m\omega^2 L/2 = 0.8 \times 2500 \times 1/2 = 1000\,\text{N}\). The stress there is \(1000/10^{-4} = 10^7\,\text{Pa}\).
At the middle \(T = 0.8 \times 2500 \times (1 - 0.25)/2 = 750\,\text{N}\), not half of \(1000\,\text{N}\): the tension falls slowly near the axis and fast near the tip.
\(\Delta L = m\omega^2 L^2/(3AY) = 2000/(3 \times 10^{-4} \times 2 \times 10^{11}) = 3.3 \times 10^{-5}\,\text{m}\), about \(0.03\,\text{mm}\).
A uniform rod turns about one end on a smooth table. Where is the tension in it greatest?
A rod hanging from the roof of a lift stretches \(0.020\,\text{mm}\) under its own weight when the lift is at rest. How much does it stretch while the lift accelerates upward at \(2\,\text{m/s}^2\)? Take \(g = 10\,\text{m/s}^2\).
Return to the mine cable at Kolar. Its tension rose from the weight of the cage at the bottom to the cage and three kilometres of steel at the top. The top of such a cable is at a stress near \(\rho g L\) before the cage adds anything. A thicker cable cannot help; only a stronger steel can.
- Using the full weight \(W\) in \(WL/(AY)\) for a rod under its own weight.The average tension is \(W/2\), so \(\Delta L = WL/(2AY)\).
- Measuring \(x\) from the pulled end in \(T = Fx/L\).\(x\) is measured from the free end, where the tension is zero.
- Taking the tension in a turning rod to fall evenly from the axis.\(T \propto L^2 - x^2\): at the middle it is three quarters of its value at the axis.
- Thinking a thicker cable can hang longer.The stress at the top is \(\rho g L\), independent of the area.