Coefficient of Friction Table: Values and Ranges

There is no standard coefficient of friction. Common surface pairs run from roughly 0.03 for oiled steel to above 1 for soft rubber on dry road, and any single pair can sit outside its usual band. Coefficients are measured, not looked up, and an AP question states the value it wants you to use.

AP Physics: Unit 2 (topics 2.7 Kinetic and Static Friction). Coefficients of friction belong to Topic 2.7 of AP Physics 1 Unit 2, Force and Translational Dynamics, weighted at 18 to 23 percent of the multiple-choice section and about 22 to 27 class periods. No coefficient values are required content. EK 2.7.A.2.i states only that the coefficient of kinetic friction depends on the material properties of the surfaces in contact, and EK 2.7.B.3 that the static coefficient is typically the greater of the two. The values on this page are approximate bands for orientation, not sourced measurements, and no AP question expects them to be recalled.

Typical coefficient of friction values

Read every cell below as a band, not a value. A coefficient is a measured property of two specific surfaces in a specific condition, so "wood on wood" is a family of numbers rather than one. Use this to calibrate your intuition and to sanity check an answer. Do not memorize it, and do not carry a number from it into a problem that gave you its own.

Surface pair	Static, $\mu_s$	Kinetic, $\mu_k$	What moves the number most
Rubber on dry asphalt or concrete	roughly 0.7 to 1.1	roughly 0.6 to 0.9	rubber compound, road texture, temperature
Rubber on wet asphalt	roughly 0.3 to 0.6	roughly 0.2 to 0.5	water depth and tread; standing water drops it much further
Shoe sole on a wooden floor	roughly 0.3 to 0.6	roughly 0.2 to 0.5	sole material, polish, dust
Wood on wood	roughly 0.25 to 0.5	roughly 0.2 to 0.4	grain direction, finish, moisture
Steel on steel, dry	roughly 0.5 to 0.8	roughly 0.4 to 0.6	oxide layer, polish, cleanliness
Steel on steel, oiled	roughly 0.05 to 0.15	roughly 0.03 to 0.1	the lubricant, not the metal
Glass on glass	roughly 0.4 to 1.0	roughly 0.3 to 0.6	cleanliness swamps everything else
Steel on ice	roughly 0.02 to 0.1	roughly 0.02 to 0.05	temperature; near melting a water film forms
PTFE (Teflon) on itself	roughly 0.05 or below	roughly 0.05 or below	close to the low end for unlubricated solids

Those bands are deliberately coarse. They are not measurements of any particular sample, and no number on this page should be treated as authoritative. If your work needs a coefficient for a real pair of surfaces, measure it. If your work is an AP problem, the problem will give it to you.

The whole table spans more than a factor of thirty, from oiled steel near 0.03 to soft rubber above 1. That spread, not any individual entry, is the useful thing to carry around.

Why every entry has to be a range

Naming two materials does not describe two surfaces. The AP Physics 1 CED makes this point itself, in the appendix discussion of what an object model does and does not capture: two wooden blocks will slide across each other differently if they are covered with sandpaper than if they are covered in grease. Same two materials by name, different surfaces, different coefficient.

The same appendix goes further. In its list of the subtleties that an introductory treatment deliberately sets aside, it asks whether the coefficient of friction decreases slightly as the abrasion between a block and an incline smooths the surfaces. In other words the coefficient can drift during a single slide, which is a strong hint that a two decimal figure was never on offer.

So the things that actually set the number include:

Surface finish. Polish, grit, machining marks, and how worn the surfaces are.
Contamination. Water, oil, dust, oxide layers, and skin oils from handling. For glass on glass and for metal on metal, this usually matters more than the material.
Temperature and humidity. Most visible with ice and with polymers.
History. Whether the surfaces have been sitting in contact, and whether they have already been slid over each other.

None of that is in a materials name, which is why two published tables can disagree by a factor of two for the same row and both be reporting honest measurements. Treat a published coefficient as somebody else's experimental result on somebody else's sample.

What AP Physics actually does with coefficients

No coefficient value is on the equation sheet, and none is required course content. Here is precisely what the exam materials do provide.

The AP Physics 1 equation sheet prints one relation involving friction:

\left|\vec{F}_f\right| \leq \left|\mu \vec{F}_N\right|

Counting the Mechanics and Fluids equations on that sheet, it is the only one with a $\mu$ in it, and the sheet's symbol list defines $\mu$ as simply the coefficient of friction, with no subscript and no value. The Table of Information alongside it prints constants such as $g$ and $G$ , prefixes, unit symbols, and a short table of trigonometric values. There is no friction table.

The CED does not print coefficient values either, except inside sample questions, where they are handed to the student:

The CED's block-on-a-ramp case study, used to show how one scenario supports several different skills, gives $\mu_k = 0.1$ in the stem of its calculation question.
An AP Physics C: Mechanics sample multiple-choice question prints a small data table listing each block's mass, contact area, and coefficient of static friction (0.20 and 0.40), then asks which block slides first when the board is tilted. The point of the question is that only the coefficient decides it.

That second example is worth sitting with. The question supplies the coefficients rather than expecting them, and two of its four options offer mass and contact area as the deciding factor. The keyed answer is the one that picks the block with the smaller coefficient of static friction, which is the whole point: nothing about the block itself decides it.

Turn any coefficient into an angle to judge it

A bare number like 0.55 is hard to have an opinion about. The slip angle is not. Tilt a surface until an object just begins to slide, and

\mu_s = \tan\theta_c

so any coefficient converts straight into a picture of a ramp. That gives you a plausibility check that needs no table at all.

Coefficient $\mu$	Just slips at about	What that looks like
0.05	3 degrees	ice; almost any noticeable slope starts it moving
0.1	6 degrees	a lubricated or very smooth pair
0.2	11 degrees	slippery, like a polished floor in socks
0.3	17 degrees	a shallow ramp holds it, a moderate one does not
0.5	27 degrees	an ordinary dry pair of everyday materials
0.75	37 degrees	grippy; the sheet's trig table prints $\tan 37^\circ = 3/4$
1.0	45 degrees	the surface has to reach 45 degrees before it lets go
1.5	56 degrees	rare; soft rubber on a rough dry surface

The angles are rounded to the nearest degree. Two of them come free on the exam: the AP Physics 1 Table of Information prints $\tan 37^\circ = 3/4$ and $\tan 45^\circ = 1$ , so those conversions need no calculator.

This also settles the question that gets asked about any coefficient near 1. A coefficient of 1 is not a ceiling; it is simply the value at which the slip angle reaches 45 degrees. Values above 1 mean the friction force exceeds the normal force, which grippy pairs manage routinely. The derivation of the tangent rule, and the reason mass never appears in it, is in static vs kinetic friction.

Units: the force has them, the coefficient does not

Two different questions get asked with the same words here, so answer them separately.

The friction force is a force, so its SI unit is the newton (N), the same as weight, tension, and the normal force. In base units that is $\text{kg} \cdot \text{m}/\text{s}^2$ .

The coefficient of friction has no units at all. It is defined as a ratio of two forces,

\mu = \frac{F_f}{F_N}

and newtons divide out. That is why you never write "0.4 N" for a coefficient, and why a coefficient means the same thing to somebody working in pounds. It is also a fast error check: if the number you calculated for $\mu$ came out carrying newtons or kilograms, you divided the wrong two quantities.

A useful consequence of being dimensionless: the coefficient does not change if you scale the whole problem. Double the mass on a level surface and both $F_f$ and $F_N$ double, so $\mu$ holds still. That is the same cancellation that makes the slip angle independent of mass.

How a coefficient is actually produced

Every number in any friction table started as somebody pulling something across something else. The CED's own suggested activity for this material is exactly that: have students determine the coefficient of kinetic friction between their shoe and the surface of their desk by pulling the shoe across the surface with a spring scale at constant speed, then compare the coefficients they get for athletic shoes, slippers, and sandals, and discuss the relationship between the type of shoe and the coefficient.

Notice what that activity is teaching. Not a value. A method, and the fact that the value depends on which shoe.

Three standard routines, all of which come down to measuring $F_f$ and $F_N$ and dividing:

Constant speed pull. Zero acceleration, so the pull equals kinetic friction. Divide by the normal force.
Deceleration. Let the object skid to a stop with friction as the only horizontal force, so $\mu_k = a/g$ .
Tilt to slipping. Raise one end until it just goes, and take the tangent of the angle for $\mu_s$ . No force sensor needed.

The step-by-step versions are in how to find the coefficient of friction, and you can run the third one on a virtual ramp in the inclined plane simulator before you try it on a real one.

Whichever route you take, repeat it. Real friction data scatters by a few percent between trials, which is the honest reason a table entry is two significant figures at best.

Sanity checks on a coefficient you calculated

When a problem asks you to find $\mu$ rather than giving it to you, these catch most of the ways it goes wrong.

No units. A coefficient carrying newtons means the division was wrong.
Convert to an angle. Take $\arctan\mu$ . If the answer says the surfaces would hold at 70 degrees, the number is almost certainly too big.
Suspect the normal force first. An implausibly large $\mu$ nearly always means $F_N$ came out too small. The usual causes are using $mg\cos\theta$ on a surface that was actually horizontal, or forgetting that a downward-angled push adds to $F_N$ . Work through how to find normal force if it is not obvious which applies.
Check the pair, not the object. If two blocks of different mass on the same surface give you different coefficients, that is measurement error, not new physics.
Expect $\mu_s \geq \mu_k$ . The CED says the static coefficient is typically the greater one for a given pair, so data that reverses it points at the analysis.
Watch the significant figures. Coefficients are ratios of measured forces, so two figures is usually the honest limit.

Nothing in that list requires a table, which is the point. A coefficient is checked against physics and against your own measurement, not against a published number.

Where the number goes once you have it

A coefficient is never the answer to an AP question; it is an input to one. Once you have it:

Multiply by the normal force to get the kinetic friction force, or the static ceiling. The friction calculator runs either direction and shows the steps.
Decide static or kinetic first, because the same $F_N$ gives two different friction forces. That decision is the whole of static vs kinetic friction.
Feed the friction force into Newton's second law, which is where most Unit 2 problems actually live. See how to find net force.
On a ramp, remember that $F_N = mg\cos\theta$ before the coefficient does anything. Inclined plane problems covers the geometry.
In energy problems, kinetic friction is the term that removes mechanical energy, which is how friction shows up again in Unit 3.

The CED framing for all of this is Topic 2.7, inside Unit 2.

What a range does to your answer

A crate is sliding across a level floor at $6.0\ \text{m/s}$ when the push is removed. A table says the coefficient of kinetic friction for this pair of surfaces is somewhere in the range 0.30 to 0.60. How far does the crate slide before stopping?

Set the convention: positive $x$ is the direction of motion. Kinetic friction is the only horizontal force, so $\mu_k m g = m a$ and the mass cancels: the deceleration is $a = \mu_k g$ .
Use the kinematic relation with final speed zero: $0 = v_0^2 - 2 a d$ , so $d = \dfrac{v_0^2}{2 \mu_k g}$ .
Lower end of the band, $\mu_k = 0.30$ : $d = \dfrac{(6.0)^2}{2(0.30)(9.8)} = \dfrac{36}{5.88} = 6.12\ \text{m}$ .
Upper end, $\mu_k = 0.60$ : $d = \dfrac{(6.0)^2}{2(0.60)(9.8)} = \dfrac{36}{11.76} = 3.06\ \text{m}$ .
Note the structure: $d$ is inversely proportional to $\mu_k$ , so a factor of two in the coefficient is exactly a factor of two in the distance. No amount of care with the arithmetic narrows that.

Between about 3.1 m and 6.1 m. The honest answer to a problem built on a table lookup is a range, and here it is a factor of two wide. This is why an AP question hands you a single coefficient: it wants one answer, so it removes the ambiguity itself rather than leaving you to pick a row.

Judging whether a quoted coefficient is plausible

A lab group reports $\mu_s = 0.85$ for a wooden block on a wooden board. Without looking anything up, decide whether that is believable, and say what you would check.

Convert the coefficient into the thing you can picture. At the verge of slipping, $\mu_s = \tan\theta_c$ , so $\theta_c = \arctan 0.85 = 40.4^\circ$ .
Interpret it. The claim is that a wooden block will sit still on a wooden board tilted to just over 40 degrees, and only let go past that. That is a steep board, well beyond the shallow tilt that usually starts wood sliding on wood.
Compare with the band in the table above, roughly 0.25 to 0.5 for wood on wood, which corresponds to slip angles of about 14 to 27 degrees. The reported value sits well outside that band, so it needs an explanation rather than a rejection.
List the explanations, in order of likelihood. The normal force was underestimated, which inflates $\mu = F_f / F_N$ directly. Or the block was pushed rather than allowed to just slip, so the measured force exceeded the true maximum. Or the surfaces are genuinely unusual: unfinished, rough-sawn, or slightly damp wood really can grip much harder than a planed board.
Design the follow-up. Tilt the board and measure the angle at which the block first moves. If it goes at about 40 degrees, the value stands and the surfaces are the story. If it goes near 20 degrees, $\tan 20^\circ = 0.36$ and the original force measurement was the problem.

It is possible but suspicious. 0.85 means a slip angle of about $40^\circ$ , roughly double the tilt that a planed wooden board usually needs, so the value should be re-measured by the tilt method before it is used. Notice that the check never required a published table, only the tangent relation and a ramp.

Producing a coefficient from measured data

A student drags a block at constant speed across a bench with a spring scale, loading the block with extra masses between runs. They record normal force and pulling force: 9.8 N and 3.2 N, 19.6 N and 6.3 N, 29.4 N and 9.7 N. Find the coefficient of kinetic friction and say how many figures it deserves.

Constant speed means zero acceleration, so the pulling force equals the friction force in every run. The three friction forces are 3.2 N, 6.3 N, and 9.7 N.
Take the ratio $\mu_k = F_f / F_N$ for each run separately: $3.2/9.8 = 0.327$ , $6.3/19.6 = 0.321$ , $9.7/29.4 = 0.330$ .
Each run sits within about 1.5 percent of the mean of the three. That scatter is normal for friction measurements and it is the real limit on the answer.
Treat the data as a graph of $F_f$ against $F_N$ , which the model says is a straight line through the origin with slope $\mu_k$ . A least-squares slope through the origin gives $\mu_k = 0.327$ , essentially the same as the mean of the ratios, 0.326.
Round to the precision the data supports. The inputs carry two significant figures, and the run-to-run spread is already in the third, so report two: $\mu_k = 0.33$ .

$\mu_k = 0.33$ , with the three runs agreeing to within about 1.5 percent. Reporting 0.327 would claim a precision the scatter does not support. This is what every entry in every friction table is underneath: a slope through a handful of noisy points, taken on one particular pair of surfaces on one particular day.

Frequently asked questions

What is the SI unit of friction?

Friction is a force, so the SI unit of the friction force is the newton (N), which is kg m/s^2 in base units. The coefficient of friction is a different quantity and has no units at all, because it is defined as the friction force divided by the normal force and the newtons cancel. So a friction force of 12 N can be correct, but a coefficient of 0.4 N never is.

What is a standard coefficient of friction?

There is not one. A coefficient is a measured property of two specific surfaces in a specific condition, not a constant of a material, so there is no standard value the way there is a standard value for the speed of light. Published tables give typical values for common pairs, and they disagree with each other because they are reporting different samples. For AP Physics, this does not cost you anything: the exam states the coefficient inside the question whenever a number is needed.

What is the range of coefficient of friction values?

Most everyday pairs fall between about 0.03 and 1.1. The low end is lubricated or very smooth contact, such as oiled steel, steel on ice, or PTFE, where values below 0.1 are normal. The middle, roughly 0.2 to 0.6, covers most dry ordinary materials such as wood on wood or a shoe on a floor. The high end, 0.7 and above, is grippy contact such as rubber on dry road. Values above 1 are perfectly possible and just mean the friction force can exceed the normal force.

Why do different tables give different coefficients for the same materials?

Because naming two materials does not describe two surfaces. Finish, wear, dust, moisture, oxide layers, and temperature all change the number, often more than the material choice does. The AP Physics 1 CED makes the same point when it notes that two wooden blocks slide across each other differently when covered with sandpaper than when covered in grease. Two tables reporting 0.3 and 0.6 for wood on wood can both be honest measurements of different samples.

Will the AP exam give me the coefficient of friction?

Yes, whenever a numerical value is needed. No coefficient values appear on the AP Physics 1 equation sheet or in the Table of Information, and no coefficients are listed as required content in the CED. Where the CED's own sample questions need one, they state it in the problem: the block-on-a-ramp case study gives the coefficient of kinetic friction as 0.1, and an AP Physics C: Mechanics sample question prints a data table of static coefficients for the blocks involved. Memorizing values is not part of the course.

Does the coefficient of friction change when a surface is wet or cold?

Yes, and often by more than switching materials would. Water on a road can roughly halve the coefficient between tire and asphalt, and standing water can cut it much further. Ice gets more slippery close to its melting point, because a thin liquid film forms at the contact. Oil or grease can drop a metal pair by a factor of ten. This is why a friction table always has to specify the condition, and why a value copied without its condition is close to meaningless.

Is the coefficient of friction on the AP Physics 1 equation sheet?

The symbol is, the values are not. The sheet prints one friction relation, the magnitude of the friction force is less than or equal to the magnitude of mu times the normal force, and its symbol list defines mu as the coefficient of friction with no subscript. You supply the subscript yourself: static or kinetic, decided by whether the surfaces are sliding. The maximum static friction, mu_s times the normal force, is listed in the CED as a derived equation and is not printed on the sheet.