Materials Science · Contact, friction, hardness and wear

Contact mechanics and friction

Connect local surface contacts, adhesion and deformation to conditional friction models. Distinguish a static limit from an actual force, calculate the normal reaction before using a coefficient, and compare contacting pairs with their surface history and test evidence.

Useful background: Elasticity, plasticity and tensile strength; Processing, structure and useful properties. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Distinguish nominal contact area and pressure from the smaller load-bearing contacts of a rough interface.
Explain junction shear, ploughing and deformation loss as possible sources of tangential resistance.
Derive a bounded real-contact-area model and identify why it can predict approximate nominal-area independence.
Use static-friction inequalities and sliding-force balance without assigning the limiting force prematurely.
Interpret a pair-and-condition comparison while separating friction force, dissipated power and wear.

Surfaces touch through local regions

Even apparently smooth surfaces contact through smaller asperities and deformation zones. Adhesion, ploughing and interfacial films can contribute to tangential resistance. Nominal contact area does not by itself describe the true contact structure.

An asperity is a local high region of a rough surface. Under normal loading, some high regions touch, deform and carry load before the surrounding valleys do. The nominal area A₀ is the macroscopic footprint used to describe the geometry. The real contact area Aᵣ is the sum of the actual load-bearing contact patches in the chosen description. A smooth-looking face can still be rough at a smaller observation scale.

If a normal force of magnitude N passes through a nominal area A₀, the nominal mean pressure is p₀ = N/A₀. If only Aᵣ carries that load, the mean pressure over those contacts is N/Aᵣ, which can be much greater. Neither mean gives the full local stress distribution: edge shape, curvature, elasticity and plastic deformation determine how pressure varies. A sharp particle, a rounded asperity and a broad flat contact do not produce identical stress fields.

Increasing load may enlarge existing contacts and bring additional high regions into contact. The response can be elastic, plastic or a mixture. This is the bridge to hardness: confined local deformation matters, but the reading from one indentation test is not automatically the actual pressure at every rough contact.

What consumes energy during sliding?

At a contact junction, interactions between the two surfaces or their films can resist relative tangential motion. Shearing those junctions costs work. Adhesion here means attractive interfacial interactions; it need not mean a visible glue layer or a weld between the whole nominal faces. Surface chemistry, adsorbed species and transfer films can change the resistance of a junction.

Ploughing is different: a harder protrusion or trapped particle pushes through a softer region, displacing material into a groove and its sides. The forward motion must continually deform material. Some displaced material may stay beside the track, so a groove does not by itself measure how much material has left the system. A friction force can include both junction shear and deformation contributions, which can interact rather than behave as independent fixed constants.

In a viscoelastic solid such as a rubber, deformation can lag behind the changing load. Recovering its shape does not necessarily return all the supplied mechanical energy; that lag can dissipate energy as surface features pass. Thus a surface’s elasticity, time response and contact geometry matter alongside junction shear. The dominant contribution is a question for the particular pair and conditions.

Roughness alone is not a complete dissipation mechanism: an ideal reversible elastic climb and descent can return stored energy. Irreversible deformation, interfacial rearrangement or another loss process is needed to sustain a net sliding resistance over a repeated route. This also explains why polishing does not guarantee zero friction. Removing peaks may reduce some deformation while changing the real contact and interfacial interactions.

A simple model links true contact area to force

Build a limited model for a rough interface whose separated microcontacts deform plastically. Suppose their mean load-bearing pressure is an approximately constant p*, and their total projected area is Aᵣ. Ignore an additional normal attraction between the bodies, bulk shape change and fluid load support. Normal force balance then gives N = p* × Aᵣ, so Aᵣ = N/p*. The asterisk marks an assumed effective contact pressure, not a universal constant or an instruction to substitute a traditional HV number.

Now suppose the mean shear resistance per unit real contact area is a constant τⱼ. The junction contribution to the sliding resistance is Fⱼ ≈ τⱼAᵣ. Substitution gives Fⱼ ≈ (τⱼ/p*)N and μⱼ = Fⱼ/N ≈ τⱼ/p*. Both τⱼ and p* have pressure units, so their ratio is dimensionless. This is one illustrative route to a friction force proportional to normal load; it is not a derivation of every observed friction law.

A₀ does not appear in this final junction expression. If the same N, p* and τⱼ remain applicable, doubling the nominal footprint halves p₀ without necessarily changing Aᵣ or Fⱼ. The result comes from the model’s contact response, not from asserting that geometry never matters. Its validity requires Aᵣ to remain a small part of A₀ and the assumed deformation regime to survive the change.

Do not extrapolate this plastic-contact model to every elastic, very smooth, soft adhesive or fully conforming interface. Contacts may interact, change pressure or approach full contact; junction shear resistance can also change with pressure, speed and chemistry. Adding a separate ploughing contribution can be a useful bookkeeping approximation, but those mechanisms need not be independent. The following original model case uses stipulated p* and τⱼ rather than fitted material measurements.

The original selected model has normal force 200 N and a square nominal footprint 10 mm on each side, area 100 mm². At 22 pixels per millimetre the left square is 220 by 220 pixels. Three equal circular contact patches have total area 0.200 mm², a fraction 0.200%; their radius is about 0.14567 mm. They illustrate stipulated contact area, not measured asperities. Mean nominal pressure is 2.00 MPa and assumed mean pressure over true contacts is 1000 MPa. With mean junction shear resistance 100 MPa, the model gives junction friction 20.0 N and coefficient 0.100. A separate enlarged schematic section shows finite contacting high regions and separated valleys with normal loading and relative sliding directions; it has no quantitative magnification or pressure field.

A coefficient is conditional

During sliding, tangential resistance divided by normal force defines a sliding-friction coefficient for the stated test. A limiting static coefficient instead uses the force at the onset of sliding; static friction below that limit adjusts to the imposed tangential load. These values depend on the pair, surface state, speed, load, environment and motion history. It is not an immutable property of one isolated material.

In a simple rigid-body force model, write the magnitude of static friction as |Fₛ| ≤ μₛN. The maximum μₛN is the limiting value, not the friction force at every smaller push. First calculate the tangential force required to prevent relative sliding; then check whether the interface can supply it. Static friction opposes the impending relative slip at the contact, which need not be the same direction as the body’s overall motion.

Once the two surfaces slide relative to each other, use Fₖ ≈ μₖN only in a regime where an approximately constant sliding coefficient describes the measurements. Fₖ here is a positive magnitude, with its force direction opposite the relative sliding. It does not automatically equal the applied driving force: equality requires the appropriate tangential acceleration to be zero. Many pairs have a lower sliding resistance than their initial static limit, but this is not a universal inequality for all materials and histories.

Always find N from force balance rather than replacing it automatically by weight. On a horizontal support with no vertical acceleration, a pull P inclined upward by θ has N + P sinθ − W = 0, hence N = W − P sinθ, where W is the weight force. Its tangential component is P cosθ. These equations assume continued contact, N ≥ 0, and no other vertical forces. A downward push changes the sign of the vertical component.

This threshold model suppresses local partial slip, contact ageing, real force fluctuations and apparatus dynamics. A block may show stick–slip when the contacting pair and loading system interact; a difference between two constants alone does not predict its full time history. Report how the interface was loaded and observed when assigning a coefficient.

The original ideal force model uses N = 100 N, static coefficient 0.50 and sliding coefficient 0.30. Starting at rest and increasing a positive horizontal push P, static friction matches P up to the 50 N limit. The separate after-onset sliding branch has friction magnitude 30 N. A dashed connector marks the model change without predicting its actual time history; an open circle distinguishes the lower branch at the threshold. The model does not determine the force from P alone for an arbitrary previous motion. A separate block force diagram shows normal reaction N up, weight W down, friction left and a 30 degree upward pull P. With no vertical acceleration and maintained contact, N = W − P sinθ. Arrow lengths are schematic, not scaled force magnitudes.

Force, sliding work and surface history are different records

For a constant resisting sliding force Fₖ over relative sliding distance s, the mechanical energy supplied against that resistance is D = Fₖs. At relative speed v, the corresponding power is Ḋ = Fₖv. Work has units N·m = J and power has units J/s = W. If Fₖ varies, adding the force multiplied by each small distance interval approximates the total instead of using one unrepresentative peak.

That power is not a temperature. Heat partition between the bodies, contact size, thermal properties and heat removal affect the local temperature. Some mechanical work can also go into irreversible surface change. Likewise, friction force is not a wear volume: the next lesson treats how material moves, accumulates or leaves a contact.

During running-in, initially contacting regions can deform, smooth, accumulate debris or acquire transferred material. A transfer film contains material deposited from the other body or generated at the interface; later sliding can therefore occur against a changed surface rather than the one first prepared. A coefficient at the start and one after many passes need not describe the same state. Distance, dwell before motion and repeat-cycle history can matter as well as speed.

Compare the interaction, not a material label

A metal against rubber and the same metal against a ceramic are different contacting systems. The counterface can change deformation, chemistry, debris and the route that dissipates energy. A different coefficient is not automatically a faulty result. To test a particular explanation, compare matched geometry, preparation, normal loading, speed, environment and sliding history, then inspect the contacting surfaces as well as the force record.

For a concrete research example, Jia and colleagues studied diamond-like carbon (DLC) films sliding against steel, silicon nitride and DLC in dry and 40% relative-humidity air. The primary abstract reports coefficients near 0.05 across the studied cases while describing different transfer-layer and wear responses. Similar friction coefficients therefore need not imply identical damage. This is evidence about those prepared films and tests, not a constant for every carbon coating.

The selected NIST database transcription of that study lists the values below and a normal load of 9.8 N with sliding speed 1.6 cm/s. It also notes a coefficient that changes with repeated passes before reaching a more nearly constant regime. The table reports selected conditioned values, not a universal ranking of steel and silicon nitride. No uncertainty intervals are supplied here, so differences in its rounded entries do not alone establish statistical significance or a unique causal explanation.

Keep measurement and interpretation separate. A force ratio establishes the coefficient for that reading or averaging interval. A film mechanism needs surface evidence; a claim of lower wear needs a wear measurement; a service-life claim needs still more. This makes the pair, evolving surface state and specified test conditions part of the result.

Selected friction-coefficient entries from NIST SCD record Z00374 for the Jia et al. (1995) DLC study. These are source-reported test values, not the invented coefficients used in the force-model exercises.
Counterface against DLC	Relative humidity (%)	Reported sliding coefficient
Silicon nitride	0	0.04
Silicon nitride	40	0.05
Steel	0	0.04
Steel	40	0.06

Worked examples

A small true area can carry a large local pressure

Use an original rough-plastic-contact model case: N = 200 N, nominal footprint A₀ = 100 mm², effective contact pressure p* = 1.00 GPa and junction shear resistance τⱼ = 100 MPa. Find Aᵣ, its fraction of A₀, nominal pressure and junction friction. What changes if only A₀ doubles while the same model assumptions remain valid?

Convert p* to 1000 MPa = 1000 N/mm². Then Aᵣ = N/p* = 200/1000 = 0.200 mm², only 0.200/100 = 0.00200 = 0.200% of A₀.
The nominal pressure is p₀ = 200/100 = 2.00 MPa. The stipulated mean pressure over the real contacts is 1000 MPa, five hundred times larger; neither number is a local peak stress.
Fⱼ ≈ τⱼAᵣ = (100 N/mm²)(0.200 mm²) = 20.0 N. Thus μⱼ ≈ 20/200 = 0.100, equal to τⱼ/p*.
At A₀ = 200 mm², p₀ becomes 1.00 MPa. If p*, τⱼ and the separated plastic-contact regime truly remain unchanged, Aᵣ stays 0.200 mm² and Fⱼ stays 20.0 N. A changed real contact regime would invalidate that prediction.
If an independently stipulated ploughing contribution were another 20 N and the two terms were approximately additive here, total resistance would be 40 N and total μ would be 0.200. Junction shear alone would no longer describe the whole result.

The model gives Aᵣ = 0.200 mm², a 0.200% fraction, nominal pressure 2.00 MPa and junction resistance 20.0 N. Doubling the nominal area alone halves nominal pressure; the other predictions are conditional on the unchanged microcontact model.

Check the required force before using the limit

An original horizontal-block case has weight W = 100 N, no other vertical force, μₛ = 0.50 and μₖ = 0.30. It starts at rest. First apply a horizontal force P = 35 N; in a separate loading from rest apply P = 60 N. Find the friction and horizontal net force in each case. During subsequent rightward sliding at v = 0.20 m/s, find the frictional power.

Vertical balance gives N = W = 100 N. The static limit is μₛN = 50 N.
At P = 35 N, the required opposing static force is 35 N, below the limit. Thus friction is 35 N left and the horizontal net force is zero. Assigning 50 N would violate equilibrium.
At P = 60 N, remaining at rest would require 60 N of friction, greater than the 50 N limit. After rightward sliding begins, the model gives Fₖ = 0.30 × 100 = 30 N left. The net horizontal force is 60 − 30 = 30 N right, so constant speed would require a different driving force.
At the specified instantaneous sliding speed, Ḋ = Fₖv = 30 × 0.20 = 6.0 W. This is mechanical power against resistance, not an inferred temperature or wear rate.

At 35 N from rest: 35 N static friction and zero net force. After sliding starts under 60 N: 30 N opposing friction and 30 N net force to the right. The stated sliding power is 6.0 W.

A pull changes the normal reaction

A block of weight W = 100 N rests on a horizontal surface. Pull it at θ = 30° above horizontal, with μₛ = 0.40. Assume vertical equilibrium, maintained contact and no other forces. Derive the limiting pull P at impending rightward sliding and compare it with the 40 N limit for a horizontal pull.

Vertical balance gives N = W − P sinθ. Tangential equilibrium at the positive static limit gives P cosθ = μₛN.
Substitute N: P cosθ = μₛ(W − P sinθ). Collect terms to obtain P(cosθ + μₛ sinθ) = μₛW.
Therefore P = μₛW/(cosθ + μₛ sinθ) = 40/(0.866025 + 0.40 × 0.5) = 37.5226 N.
At that limit N = 100 − 37.5226 × 0.5 = 81.2387 N, still positive. The tangential component is 32.4955 N, equal to 0.40 × N. The resultant pull is smaller than 40 N because its upward component reduces N; its horizontal component is smaller than its full magnitude.

The limiting inclined pull is about 37.52 N, with N ≈ 81.24 N. Using N = 100 N throughout would miss the coupling between pull direction and friction limit.

Read a pair-and-condition comparison

A metal has different friction against rubber and against a ceramic. Is one result invalid? Then use the source table’s steel-counterface DLC entries, 0.04 in dry air and 0.06 at 40% relative humidity, to infer the sliding-force magnitudes at the reported N = 9.8 N. What can their difference establish?

Changing the counterface changes the system. Deformation and interfacial processes can change; compare the pair and conditions rather than assigning a coefficient to the metal alone.
For the selected steel-counterface readings, Fdry = 0.04 × 9.8 = 0.392 N and Fhumid = 0.06 × 9.8 = 0.588 N.
Their difference is 0.196 N. The coefficient changes by 0.02 in absolute value, or (0.06 − 0.04)/0.04 = 50% relative to the dry entry. A relative percentage does not turn the small rounded entries into a high-precision causal law.
These are arithmetic implications of the reported conditioned values. Without uncertainty and fuller experimental comparison, they do not establish a universal humidity multiplier, a statistically resolved effect or the corresponding wear. The actual study’s similar friction and differing wear responses warn against equating the two.

Different counterfaces can legitimately give different results. The selected DLC entries imply 0.392 N and 0.588 N, but those forces remain tied to the stated test values and do not supply a general material or wear law.

Try the reasoning

In the stipulated rough-plastic-contact model, double A₀ at fixed N, p* and τⱼ. Does friction double? A second contact is nearly fully conforming and elastically deformed. Can the same nominal-area conclusion be copied to it?
Show the answer and reasoning
The junction-model friction stays unchanged if all its assumptions hold; nominal pressure halves. The nearly full elastic contact is outside those assumptions, so the same conclusion needs a different justified model or measurement.
A₀ disappears only after using the particular Aᵣ = N/p* response. The result is not a general claim that area and geometry never matter.
Why report sliding speed when two tests give the same coefficient? Suppose both have N = 80 N and μₖ = 0.25, but speeds are 0.10 and 0.50 m/s. Calculate the resisting forces and mechanical powers. Do equal coefficients establish equal temperatures or wear?
Show the answer and reasoning
Both resisting forces are 20 N. The powers are 2 W and 10 W. Equal coefficients establish neither equal temperatures nor equal wear.
Power uses Fₖv and changes with speed even when the force ratio is unchanged. Speed can also alter deformation, film response and heating, so a constant coefficient over one range still requires evidence.
In the 100 N normal-load case, μₛ = 0.50 and μₖ = 0.30. A block is already sliding right when the rightward pull is reduced to 20 N. What friction and net horizontal force does the model predict while it still slides? Once it stops, can it stay at rest under the same pull?
Show the answer and reasoning
While sliding right, friction is 30 N left and the net force is 10 N left, slowing it. Once it stops, a 20 N static force can balance the pull because that is below the 50 N limit.
Motion history selects which model applies. Friction during sliding does not become 20 N merely to match the driving force; nor does static friction jump automatically to 50 N after stopping.
For another stipulated plastic-contact case, N = 120 N, p* = 600 MPa, τⱼ = 90 MPa and A₀ = 50 mm². Find Aᵣ, its nominal-area fraction, Fⱼ and μⱼ. Could substituting a hardness number written simply as “600 HV” for p* in MPa justify the same answer?
Show the answer and reasoning
Aᵣ = 0.200 mm², Aᵣ/A₀ = 0.004 = 0.4%, Fⱼ = 18 N and μⱼ = 0.15. An unconverted traditional HV number is not the same quantity or unit convention as the stipulated effective contact pressure.
Use N/mm² for MPa and keep the contact-model assumption distinct from a hardness measurement. Even conversion of units would not alone validate a particular test hardness as p*.
Return to the 100 N weight and μₛ = 0.40 case, but push at 30° below horizontal. At impending rightward sliding, derive P from N = W + P sin30° and P cos30° = μₛN. Compare the required push with the upward-pull result.
Show the answer and reasoning
P = μₛW/(cos30° − μₛ sin30°) = 40/(0.866025 − 0.20) ≈ 60.06 N. It exceeds the upward-pull result of 37.52 N because it increases the normal reaction.
The vertical sign changes the normal force and therefore the limit. The denominator is positive for this chosen case; the formula should not be treated as a solution for every angle or loss-of-contact condition.
A constant measured resistance of 12 N acts over 0.75 m of relative sliding. Find the work against resistance. A second contact produces the same work but visible grooves rather than an intact transfer film. Does the energy equality establish the same wear volume or mechanism?
Show the answer and reasoning
D = 12 × 0.75 = 9.0 J. Equal work does not establish equal wear volume or the same mechanism.
Junction shear, plastic displacement and surface changes can consume energy differently. Wear needs its own measurements; the next lesson develops that distinction.
A specimen remains at rest under a measured tangential force of 24 N and normal force of 100 N. A report assigns μₛ = 0.24 as the limiting static coefficient. What does the single observation actually establish within the simple model?
Show the answer and reasoning
It establishes that the limiting μₛ is at least 0.24 for those conditions, not that the limit was reached.
Static balance requires 24 N, while |Fₛ| ≤ μₛN. The inequality yields μₛ ≥ 24/100. A test of onset is needed to estimate the limiting value.
A new counterface is tested only after a long running-in period, while the old counterface is represented only by its first few passes. The new record has lower friction. What prevents attribution of the difference solely to counterface composition, and what comparison would help?
Show the answer and reasoning
Counterface identity and surface history changed together. Compare both counterfaces at matched preparation, geometry, load, speed, environment and corresponding sliding histories, retaining the force evolution and surface evidence.
A transfer layer or altered roughness can change what is actually sliding. The lower measured ratio can be real while its proposed single-variable explanation remains unproved.

What to carry forward

A nominal footprint can contain a much smaller set of load-bearing contacts; mean pressure depends on the area being used.
Junction shear, ploughing and time-dependent deformation can contribute differently to resistance.
Static friction supplies the required force up to a limit; sliding friction need not balance the driving force.
Find the normal reaction from the actual force directions before using a coefficient.
Force, mechanical power and wear are different quantities, and the contacting pair evolves with its history.
Carry the pair, preparation, geometry, conditions, history and measurement limits with every friction comparison.

References

Teaching resources and research supporting this lesson’s explanations and examples.