Friction Force Calculator

Enter the friction coefficient μ and the mass m in kilograms. The calculator finds the friction force T = μ·N from the Amontons–Coulomb law. On the flat the normal force is the weight, so μ = 0.4 and 10 kg give 39.24 N. The angle α is optional and lowers the normal force through cos α. The model does not solve for μ from a measured force and does not check whether the body is already sliding.

Inputs

Result

Enter the coefficient μ and the mass. The friction force, the normal force, and the chart of T against μ will appear here.

For the coefficient from the friction force and the normal force, use the μ = T/N calculator; for the acceleration of a body sliding down a slope, the incline with friction calculator.

How it works

How T is computed for μ = 0.4 and m = 10 kg on the flat
μ0.4
m·g98.10 N
T39.24 N
N T T = μ·N N = mg cosα
Friction force T = μ·N; on an incline the normal force is N = mg cos α.

Dry friction in the Amontons–Coulomb model is the product of the friction coefficient and the normal force: T = μ·N. On a flat surface the normal force equals the weight, N = m·g, with g = 9.81 m/s². For μ = 0.4 and a mass of 10 kg the calculator gives N = 98.10 N and T = 39.24 N. That is the textbook definition (Halliday, Resnick, Walker; Young and Freedman), written with the coefficient that Amontons and Coulomb put into the description of friction.

The calculator uses the numbers you enter: the dimensionless coefficient μ, the mass m in kilograms, and an optional angle α in degrees. A blank α means a flat surface, so cos α = 1. At 30°, cos 30° is 0.8660, and the same pair 0.4 and 10 kg gives N = 84.96 N and T = 33.98 N. The component of the weight parallel to the incline is then 49.05 N. The calculator does not subtract that component from T.

What the result panel shows. Besides the friction force, the calculator reports the normal force N, the weight m·g, cos α, the parallel component m·g·sin α, the share T/(m·g) = μ·cos α, and the angle it used (a blank field becomes 0°). For the reference values that is 39.24 N, 98.10 N, cos α = 1.0000, and a share of 0.4000. The bar chart shows how T grows with μ at the normal force from this calculation.

Limits of the model. There is one μ field: the calculator does not separate static friction from kinetic friction, and it does not decide whether the body starts to move. In this form Amontons’ law does not depend on the contact area or on the sliding speed. The calculator does not compute rolling resistance, lubrication, or air drag. An extra push into the surface, such as a hand pressed onto a table, has no field, so N is then not equal to m·g·cos α. The slide acceleration a = g(sin α − μ cos α) is covered by the incline-with-friction calculator, and the coefficient from measured T and N is covered by the friction-coefficient calculator.

The unit switch in the header changes the mass to pounds and the forces to pounds-force; the panel also keeps the value in newtons. The examples in the text stay in SI units as the reference. A comma and a period in the fields are equivalent. The mass must be greater than zero. An entered zero in the μ field gives T = 0. An angle outside 0° to 90° produces no result. At 90° the cosine is zero, so N and T are 0 N.

Examples on the page: 0.4 and 10 kg on the flat → 39.24 N; the same numbers at 30° → 33.98 N; 0.05 and 80 kg on ice → 39.24 N again; 0.7 and 1400 kg as the order of a dry tire → 9,613.80 N; 0.4 and 8 kg at 45° → 22.20 N. The button “Fill this example” enters those same numbers and refreshes the panel and the chart. The fields are empty when you open the page.

Formula

T = μ · N

N = m · g · cos α (a blank α means 0°, a flat surface)

g = 9.81 m/s²

Model: the Amontons–Coulomb law of dry friction. The calculator multiplies the mass by 9.81 and by the cosine of the angle in degrees, then by μ. The result is in newtons. The calculator does not compute μ = T/N or the acceleration a = g(sin α − μ cos α).

Theory, derivation, and domain

Definition of dry friction and the Amontons–Coulomb law

Dry friction is the resistance between two solid bodies whose surfaces touch. In the model used in class and in most textbook problems, that resistance is proportional to the force pressing the surfaces together: T = μ·N. The coefficient μ has no unit. It depends on the pair of materials and on the condition of the surfaces, and in this model it does not depend on the apparent contact area. The name “Amontons–Coulomb law” records the two authors who gave the proportion the form still used in class. The symbol μ itself is not named after one person.

The SI unit of force is the newton: 1 N = 1 kg·m·s⁻². The calculator reports T and N in newtons. After the header is switched to US units, the same quantity also appears in pounds-force, with 1 lbf = 4.4482216 N. The examples in the text stay in newtons, so they match the SI reading on the panel.

History: Amontons, Coulomb, and the coefficient μ

Guillaume Amontons, in a memoir presented to the Academy in 1699, described two regularities of dry friction: the resistance grows in proportion to the normal load and, to a first approximation, does not depend on the apparent area of contact. Charles-Augustin de Coulomb developed that account in Théorie des machines simples, presented to the Academy in 1781 and published in the Academy’s memoirs for 1785. He separated static friction, which holds a body up to a limit, from friction in motion, and he treated kinetic friction in the simple model as only weakly dependent on sliding speed.

The coefficient μ is a later, compact way to write the ratio T/N. Leonardo da Vinci recorded similar observations at the end of the fifteenth century, but the law entered textbooks through Amontons and Coulomb. This calculator computes the product μ·N. It does not reproduce Coulomb’s full account with separate thresholds for rest and for motion.

Derivation of N = mg cos α and T = μ·N

Step 1. On a flat surface the body does not accelerate perpendicular to the table. The weight m·g is balanced by the normal force N, so N = m·g. The calculator uses g = 9.81 m/s² and has no field for g. For 10 kg, N = 98.10 N. Step 2. In this model the friction force is the product T = μ·N. At μ = 0.4 the result is 39.24 N. The panel rounds forces to two decimal places; the multiplication uses the full value before that rounding is shown.

Step 3. On an incline of angle α, measured from the horizontal, the weight splits into a perpendicular component m·g·cos α and a parallel component m·g·sin α. If no extra force presses the body into the plane, the normal force is the perpendicular component: N = m·g·cos α. At 30°, with μ = 0.4 and a mass of 10 kg, cos 30° = 0.8660, N = 84.96 N, T = 33.98 N, and the parallel component is 49.05 N. Step 4. The calculator shows both components and the product μ·N. It does not compute the difference m·g·sin α − T, and it does not turn that difference into an acceleration.

Assumptions: one μ, a true normal force, no extra loads

The formula N = m·g·cos α holds when the only forces perpendicular to the surface are the weight and the normal force. A rope that pulls at an angle, a hand that presses a block down, or a cart with an extra load all change N. There is no field for that extra force: the mass you enter is the mass of the body, and the angle is the angle of the incline. If the problem already gives the normal force in newtons, this formula still starts from the mass and the angle. The coefficient is then computed with the friction-coefficient calculator, from μ = T/N.

The angle must lie between 0° and 90°, inclusive. A blank field is zero, which is a flat surface. At 90° the cosine vanishes, the normal force on a vertical “incline” is zero, and T = 0, unless something else presses the body against the wall. Gravity is fixed at 9.81 m/s². That is not 9.80665 m/s², and it is not a local g. On the Moon, or in an elevator, this result does not apply.

Static and kinetic friction, and the limit of motion

Static friction does not have a single value. It adjusts to the applied force, up to a limit of about μ_s·N. Only after that limit does the body start to move, and kinetic friction is then approximately μ_k·N, directed opposite the sliding velocity. Tables often list μ_k below μ_s. This calculator has one μ field and always returns the product of that coefficient and N. It does not know the pulling force, so it cannot say whether the body stays at rest.

On an incline the comparison is visible on the panel, and it is still not a verdict. The parallel component m·g·sin α and the force T sit side by side. If the μ you enter is a kinetic coefficient and the body is already sliding, the acceleration is a = g(sin α − μ cos α), and a separate calculator computes it. If μ is a static coefficient, the body starts down the incline only when m·g·sin α exceeds μ·N, which is when tan α exceeds μ. This calculator does not announce that inequality as a result.

Units, orders of magnitude, and friction nearby

Orders of magnitude from tables, not from a measurement on this page: ice is about 0.05, wood on wood is often 0.2 to 0.5, rubber on concrete is about 0.6, and dry asphalt under a tire is often quoted near 0.7. A real tire also depends on slip and on how the rubber deforms, so 0.7 × 1400 kg × 9.81 m/s² = 9,613.80 N is the order of the Coulomb model, not the specification of a particular tire. The same force, 39.24 N, comes from 0.4 and 10 kg and from 0.05 and 80 kg, because the product μ·m is the same.

Contact area does not enter this formula. A wide runner and a narrow runner with the same μ and the same normal force give the same T. That is Amontons’ assumption, not a law for every rubber. A soft tread and a lubricated slide depart from it. This page has no correction for those cases.

Sources

The definition, the split between static and kinetic friction, and the formula T = μ·N come from general-physics textbooks and from the classic memoirs: G. Amontons, memoir on resistance in machines, Histoire de l’Académie Royale des Sciences, 1699; C.-A. de Coulomb, Théorie des machines simples, Mémoires de Mathématique et de Physique de l’Académie Royale des Sciences, 1785 (text presented in 1781); D. Halliday, R. Resnick, and J. Walker, Fundamentals of Physics, the chapter on friction; H. D. Young and R. A. Freedman, University Physics, the chapter on friction; R. P. Feynman, R. B. Leighton, and M. Sands, The Feynman Lectures on Physics, Vol. I, Chapter 12. The newton: BIPM, The International System of Units (SI), 9th edition. This page is for educational purposes.

How to use

  1. Enter the coefficient μ, for example 0.4. That is a number with no unit.
  2. Enter the mass m in kilograms, for example 10.
  3. Leave α blank for a flat surface, or enter 30 if the body is on an incline.
  4. Click Calculate or wait for the live result. For 0.4 and 10 kg on the flat the calculator shows 39.24 N.
  5. Check the normal force, the weight, cos α, and the parallel component. The slide with friction, and the coefficient from T and N, are on separate pages.
  6. The clear button removes the entries and hides the panel. The unit switch in the header changes the mass to pounds and the forces to pounds-force.

Symbol glossary

T is the friction force, μ the coefficient, N the normal force, m the mass, and α the incline angle. For μ = 0.4 and 10 kg on the flat, T = 39.24 N.

T
Friction force in newtons. At μ = 0.4 and 10 kg on the flat it is 39.24 N, and at 30° it is 33.98 N.
μ
Dimensionless friction coefficient. One field does not separate μ_s from μ_k. Zero gives T = 0.
N
Normal force perpendicular to the surface. On the flat, N = m·g. On an incline, N = m·g·cos α. At 10 kg and 30° it is 84.96 N.
m
Mass in kilograms. It must be greater than zero. It is not a normal force already in newtons.
α
Incline angle in degrees, optional. A blank field leaves a flat surface. The range is 0°–90°.
g
Gravitational acceleration fixed in the code: 9.81 m/s². There is no field, and the header unit switch does not change it.

Real-life examples

Example 1

A crate on a warehouse floor: coefficient 0.4 and mass 10 kg. This is the reference value for this page. A blank angle means a flat surface.

T = 39.24 N · N = 98.10 N

Example 2

The same crate on a 30° ramp: the coefficient is still 0.4, and the mass is 10 kg. The cosine lowers the normal force, so the friction force is smaller than on the flat. The parallel component is 49.05 N, which is larger than T, and the slide acceleration is computed by a separate calculator.

T = 33.98 N · N = 84.96 N

Example 3

A boot on ice: a coefficient of about 0.05 and a mass of 80 kg. The friction force is 39.24 N again, because the smaller coefficient belongs to a larger mass.

T = 39.24 N · N = 784.80 N

Example 4

The order of magnitude of a dry tire on asphalt: coefficient 0.7 and mass 1,400 kg. A real tire also depends on slip. This result is only the product μ·m·g.

T = 9,613.80 N · N = 13,734.00 N

Example 5

A wooden crate on wood: coefficient 0.35 and mass 25 kg. The normal force is the weight, and the friction force is 0.35 of that weight.

T = 85.84 N · N = 245.25 N

Example 6

A carton on a 15° slope: coefficient 0.3 and mass 12 kg. Cos 15° is 0.9659, so the normal force is a little below the weight.

T = 34.11 N · N = 113.71 N

Example 7

A pallet on concrete: coefficient 0.25 and mass 200 kg. Friction is one quarter of the weight, if the μ you enter describes that pair of surfaces.

T = 490.50 N · N = 1,962.00 N

Example 8

A block on a board at 45°: coefficient 0.4 and mass 8 kg. At 45° the normal force and the parallel component are equal, 55.49 N each.

T = 22.20 N · N = 55.49 N

Example 9

A rubber sole on concrete: coefficient 0.6 and mass 50 kg. The share T/(m·g) is 0.6000, because the angle field is blank.

T = 294.30 N · N = 490.50 N

Example 10

A textbook block on a 30° incline: coefficient 0.2 and mass 5 kg. The angle matches the crate example, with a different coefficient and a different mass.

T = 8.50 N · N = 42.48 N

Ways to use this calculator

  • Find the force you have to overcome to start a crate across a flat floor when you know μ and the mass.
  • Compare that force on the flat and on an incline of a known angle, without computing the time of the slide.

Frequently asked questions

What does this calculator compute, and from which formula?

The calculator finds the dry-friction force from T = μ·N, the Amontons–Coulomb law. On the flat, N = m·g, and on an incline N = m·g·cos α, with g = 9.81 m/s². For μ = 0.4 and a mass of 10 kg on the flat, the result is 39.24 N at a normal force of 98.10 N.

Which units should I use for μ, mass, and the angle?

The coefficient μ has no unit. Enter the mass in kilograms, the angle in degrees, and read the result in newtons. After you switch the header to US units, the mass is in pounds and the forces are in pounds-force; the panel also shows the value in newtons. Gravity stays at 9.81 m/s² and has no field of its own.

What does a blank angle mean?

A blank α means a flat surface: cos α = 1 and N = m·g. An entered angle from 0° to 90° means an incline, and the normal force falls through cos α. At 30°, with μ = 0.4 and a mass of 10 kg, the normal force drops to 84.96 N and the friction force to 33.98 N.

Is this static friction or kinetic friction?

There is one μ field. The Amontons–Coulomb model used here does not separate μ_s from μ_k. If the problem gives one coefficient, enter that coefficient. The calculator does not compare an applied force with μ_s·N, and it does not decide whether the body starts to move.

How is this different from the coefficient page and from the incline page?

Here μ and the mass give the force T. On the friction-coefficient page, the force T and the normal force N give μ. The slide acceleration a = g(sin α − μ cos α) is computed by the incline-with-friction calculator. This page keeps the friction force and shows the parallel component, and it does not subtract that component from T.

Does the contact area change the result?

Not in this model. Amontons’ law takes dry friction to depend on the normal force and on the pair of surfaces, not on the apparent contact area. Soft rubber, a tire, and lubrication depart from that. The calculator does not apply those corrections.

Are a comma and a period equivalent?

Yes. The calculator reads 0,4 and 0.4 as the same coefficient. The fields are empty when you open the page. The hint in a field is not an entered value.

Why does μ = 0 produce a result, while a mass of zero does not?

An entered zero in the μ field means a model with no friction: T = 0 when the mass is positive. A mass of zero has no weight from which to compute a normal force, so the calculator shows no result. An angle outside 0° to 90° also produces no result. At 90° the cosine is zero, so N and T are 0 N.

Where is the acceleration a = T/m computed?

On the force page F = m·a. Here the result is T. On the flat, when friction is the only horizontal force, the acceleration is T/m. This calculator does not report that quotient as its own result.

Can gravity g be changed?

No. The calculator uses g = 9.81 m/s². That is not the standard 9.80665 m/s², and it is not a local g. At a different gravitational acceleration this result does not apply.

How do I calculate friction force with T = μ·N?

Multiply the coefficient μ by the normal force N. On the flat N = m·g: at μ = 0.4 and 10 kg you get N = 98.1 N and T = 39.24 N.

What is T at μ = 0.3 and N = 200 N?

60 N, because 0.3·200 = 60. You get the same result when a flat-surface mass satisfies m·g = 200 N.

How does the incline angle change the normal force?

N = m·g·cos α. At 30°, cos 30° ≈ 0.866, so normal force and friction drop about 13% versus the flat.

How is this different from the friction-coefficient calculator?

Here you enter μ and mass, and the result is force T. The coefficient page finds μ = T/N from a measured force and normal load.

Is N always equal to m·g?

On the flat with no extra vertical forces, yes. On an incline N = m·g·cos α. A downward push or an upward rope pull changes N outside this model.

The result is T = μ·N. On the flat, at μ = 0.4 and a mass of 10 kg, the friction force is 39.24 N. Slide acceleration is computed separately by the incline-with-friction calculator.

Knowledge sources

The definition, the history, and the derivation come from established physics and metrology sources.

Page updated November 3, 2026.