Illustrated Physics
physics, drawn
Series

Friction — the series

9 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. What friction returns, against what it is asked for. The friction force on a block under a 50 N normal load, against the force applied to it. Below 30.0 N — the static limit μs·N — friction returns exactly what is asked for and nothing moves, so the curve is the 45° line and the coefficient never appears. At that point the surface gives way and the force drops to μk·N = 22.5 N, where it stays however hard the block is pushed. The gap above the flat line is the surplus that accelerates it: 22.5 N at the right-hand edge of the axis.

    The force that takes what it needs

    Static friction has no value of its own. It supplies exactly what equilibrium demands and not a newton more, right up to the moment it cannot — which is the only instant in the whole business at which a coefficient of friction means anything at all.

    part 1 · mechanics
  2. The same steady pull, twice. Spring force divided by the normal load, against time, with the far end drawn away at 100 µm/s in both traces and every property of the contact the same. The soft holder at 0.4 k_c produces events every 3.75 s, each of which reaches 0.94 m/s — 9.4e+3 times the speed it is being pulled at. The stiff one at 1.6 k_c settles to a straight line and stays on it.

    The chatter a stiffer holder removes

    A brake squeals, a bow sounds a string, a fault slips in jerks. The usual explanation is that static friction exceeds kinetic friction — and that explanation, taken seriously, predicts the jerking would happen no matter how the thing were held. It does not, and what decides is a length nobody mentions.

    part 2 · mechanics
  3. Rough contact grows as W^1.000, and one smooth bump as W^0.667. Real contact area against load, both logarithmic, three ways. Plastic flow of the asperities gives A = W/H, a slope of exactly one, which is the account Bowden and Tabor's argument uses. A single elastic sphere gives Hertz's answer, A ∝ W^(2/3), and a friction coefficient falling as W^(−1/3) — Amontons' law would be false. A rough elastic surface, with many asperities spread exponentially in height and integrated here rather than quoted, gives a slope of 1.0000, because pressing harder mostly recruits new contacts rather than enlarging existing ones. The dotted curve repeats that integration with a Gaussian spread of heights and gives 0.9695 — the upper tail of a Gaussian is nearly exponential, so the answer is nearly and not exactly linear. Amontons' law does not need plasticity. It needs roughness, it is exact for one distribution and good to three per cent for another, and it survives whichever way the individual junctions deform.

    The grip that is not a coefficient

    A friction coefficient is independent of load and of area, and the reason is usually given as plastic flow of the asperities. It is not — a rough elastic surface gives the same law, integrated here to a slope of 1.0000, because pressing harder recruits new contacts rather than enlarging old ones. What breaks the law is a contact that is smooth, or a material that dissipates in its bulk — where μ passes one and stops being a coefficient at all.

    part 3 · mechanics
  4. How much a wrap holds, against how many turns it is. The ratio of the two tensions a rope can hold across, against the number of turns it is wrapped, for coefficients of 0.1, 0.25, 0.5. The axis is logarithmic because the law is exponential, so each line is straight and its slope is the coefficient. At µ = 0.1 one turn multiplies by 1.9, two turns by 4 and three by 7 — so a person pulling with the strength of one arm holds a load that a small crane would be needed to lift. The practical consequence is the one a sailor states as a rule: turns are cheap and each is worth as much as the one before it, which is a statement about a constant factor rather than a constant force.

    The part of the wrap that is actually gripping

    The capstan equation gives the largest tension ratio a wrap can hold, and almost nothing spends its life at that limit. Below it the wrap divides in two: an idle arc doing nothing at all and an active arc creeping and carrying the whole exponential — and the division explains a belt's speed loss and its squeal.

    part 4 · mechanics
  5. Force against slip, and the knee between them. The force a tyre delivers against how much faster its tread is going than the road, for a patch 120 mm long under 4000 N with a friction coefficient of 1. The curve is the integral over the bristles, and the dashed line is the cubic the brush model gives in closed form; they agree to 0.00 per cent of the sliding force. The first slope is 80 kN per unit slip, and it belongs entirely to the elasticity of the rubber — at vanishing slip nothing is sliding, so no friction coefficient can appear in it. Full sliding is reached at 15.0 per cent slip and not before. Everything a driver calls grip lives on the rising part of this curve, at a few per cent of slip, where the patch is partly stuck and partly sliding — and the quantity that decides handling in that region is the slope rather than the friction coefficient at the top.

    The grip that needs a little slipping

    A wheel that transmits any force at all is not rolling. Part of its contact patch is stuck to the road and part is already sliding, and the force it delivers is a measure of how much has given up. The useful part of the curve is a few per cent of slip, the peak is not the end of it, and everything past the peak is unstable.

    part 5 · mechanics
  6. The shear a pushed contact carries, and where it gives. Across a Hertzian contact of radius a, the normal pressure as a fraction of its peak, which is also the friction limit μp in units of μp₀ (dashed), and the shear traction for a sideways push of 0.3, 0.6, 0.9 times the load that would slide it, μP. Where the shear reaches the dashed ceiling the surfaces slip; inside, they stick. At 0.3 μP the stuck disc has a radius of 0.89a, so 21 per cent of the contact area is slipping; at 0.6 μP the stuck disc has a radius of 0.74a, so 46 per cent of the contact area is slipping; at 0.9 μP the stuck disc has a radius of 0.46a, so 78 per cent of the contact area is slipping. The shear is highest just inside the slipping ring and, in an elastic contact that simply stuck everywhere, would be infinite at the rim: slip at the edge is what removes that singularity.

    The contact that slips before it slides

    Static friction is supposed to hold perfectly until the load reaches μ times the normal force and then give way. Two elastic bodies pressed together do something else. Push a ball sideways on a flat with any force at all and a ring at the edge of their contact is already slipping, while the centre stays stuck. As the push grows the ring spreads inward, and gross sliding is only the moment the stuck centre shrinks to nothing. Shake the push back and forth and the ring slips each way on every cycle. The contact then loses energy though it never slides, as the cube of how hard it is shaken.

    part 6 · mechanics
  7. The pressure under a wheel rolling on a soft surface. The pressure under a rigid wheel of radius 30 cm carrying 3 kN per metre of width as it rolls over a rubber-like bed, against position along the contact (ahead of the axle to the right), scaled to the largest pressure and the widest contact, at three speeds. The bed relaxes with times of 2 and 20 ms. At 1 mm/s the pressure is symmetric and its resultant passes through the axle. At 1.0 m/s the material ahead is being squeezed faster than it can flow and pushes back harder than the recovering material behind, which lags and lets go early: the resultant sits 6.39 mm ahead of the axle. At 1000 m/s the bed has no time to relax at all and behaves as a stiffer elastic solid, symmetric again. A wheel is held back by where its support pushes, not by any sliding.

    The push a millimetre ahead of the axle

    A wheel rolling freely on a flat road slows down, and nothing in it slides. What holds it back is where the road pushes on it: not directly under the axle but a little ahead, because the material in front of the contact is being squeezed and pushes back harder than the material behind, which is recovering and lags. A car tyre's push is about three millimetres ahead of its axle, a train wheel's about two-thirds of a millimetre, and that offset, divided by the radius, is the whole of rolling resistance. It peaks at the speed where the time under the wheel matches the time the material takes to relax, and the same loss, at a different frequency, is what gives a tyre its grip.

    part 7 · mechanics
  8. Two ways for a swing to die. A mass on a spring released 10.6 dead-band widths from rest, against time in units of 1/ω. Red: sliding on a dry surface, whose friction is a constant force; each swing is exactly a cosine, and the turning points fall in a straight line by two dead-band widths per half-cycle until the mass stops, 2.5 cycles after release, at −0.60. Blue: the same mass with viscous damping (damping ratio 0.05), whose force is proportional to speed; the swings shrink by the same fraction each cycle, along an exponential, and never stop. The period is the same in both and the same as with no damping at all to within the viscous case's 0.13 per cent.

    The swing that dies in a straight line

    A pendulum in air loses the same fraction of its swing every cycle, so its amplitude falls along an exponential and, in principle, never quite reaches zero. A mass on a spring sliding over a dry surface does something else. Its friction is a fixed force, so it loses a fixed amount of amplitude every half-swing — the swings shrink in a straight line — and it stops completely, after a countable number of swings, somewhere inside a band around its natural position whose width is the friction divided by the stiffness. Where in that band it stops depends on where it started, which is why a barometer with friction in it is tapped before it is read.

    part 8 · mechanics
  9. A floor that has to hit a rod nobody dropped. The normal force on the lower end of a uniform rod at 63° with μ = 1.6, sliding outward at √(gl) and not yet turning, against time on a logarithmic axis, for a floor whose contact stiffness is 10⁴, 10⁵ and 10⁶ times mg/l. The rigid equations have no solution here, and the compliant floor shows why: the end digs in, the deeper it digs the harder friction twists it further down, and the force climbs until the sliding stops. The stiffer the floor, the higher and briefer the spike — 20, 56, 163 times the rod's weight, ending after 41, 18, 7 thousandths of √(l/g) — a collision with no approach.

    The rod that friction cannot decide about

    Newton's laws, a rigid rod and Coulomb's law of friction are the first things a mechanics course teaches, and in 1895 Paul Painlevé showed that together they can fail to give any answer at all. A rod sliding on a rough enough floor, at the right angle, has no motion consistent with the three — or has two. The way out is not a new law but a forgotten fact: real contacts give a little. With that, the rod jams, and the floor delivers a sudden blow to a rod that nothing struck — an impact without a collision, which is how a stick of chalk comes to hop.

    part 9 · mechanics

All series