Static vs Kinetic Friction: Which One Do You Use?

Ask one question: are the two surfaces sliding across each other right now? If they are, use kinetic friction, fixed at the coefficient of kinetic friction times the normal force. If they are not, use static friction, which takes whatever value prevents sliding, up to a maximum.

AP Physics: Unit 2 (topics 2.7 Kinetic and Static Friction). The static and kinetic split is Topic 2.7 of AP Physics 1 Unit 2, Force and Translational Dynamics, which the CED weights at 18 to 23 percent of the multiple-choice section and about 22 to 27 class periods. LO 2.7.A covers kinetic friction and gives its magnitude as an equality; LO 2.7.B covers static friction and gives it as an inequality, with the maximum listed separately as a derived equation. EK 2.7.B.3 states that the coefficient of static friction is typically greater than the coefficient of kinetic friction for a given pair of surfaces. AP Physics C: Mechanics covers the same material in its own Unit 2.

Which one applies? One question decides it

Before any algebra, ask whether the two surfaces are sliding across each other at the instant the problem describes.

They are sliding. Use kinetic friction. Its magnitude is settled: $\left|\vec{F}_{f,k}\right| = \left|\mu_k \vec{F}_N\right|$ . Nothing in the problem can move that number except the normal force.
They are not sliding. Use static friction. Its magnitude is not settled: $\left|\vec{F}_{f,s}\right| \leq \left|\mu_s \vec{F}_N\right|$ . You have to work out what value it is actually taking.

Two details in that question do real work. At the instant the problem describes, because a block can be static at the start of a problem and kinetic two seconds later. And across each other, because the AP Physics 1 CED defines the kinetic direction rule as opposite to the motion of each surface relative to the other surface (EK 2.7.A.1.i), not relative to the ground. A crate sitting on a conveyor belt that is pulling ahead of it feels friction pushing it forward, in the belt's direction of travel.

That is the whole decision. Everything below is what to do once you have made it, and how the two cases meet at the moment of slipping. If you want the CED framing of the same distinction, that is Topic 2.7; if you want the number, the friction calculator will run either case.

Static vs kinetic friction, side by side

	Static friction	Kinetic friction
When it acts	The surfaces are in contact and not sliding on each other	The surfaces are sliding on each other
Magnitude	Anything from zero up to $\mu_s F_N$	Fixed at $\mu_k F_N$
Relation	$\lvert F_{f,s} \rvert \leq \lvert \mu_s F_N \rvert$	$\lvert F_{f,k} \rvert = \lvert \mu_k F_N \rvert$
Direction	Whatever direction prevents slipping, forward just as readily as backward	Opposite the motion of each surface relative to the other
Push harder along the surface	The friction force rises to match, until it cannot	Nothing changes; the push is not in the formula
Coefficient	$\mu_s$ , typically the larger of the two	$\mu_k$
Mechanical energy	No sliding at the contact, so nothing is dissipated there	Dissipated as thermal energy while sliding continues
How it gets measured	Tilt until it just slips, or push until it breaks loose	Slide at constant speed, or work back from the deceleration

Read the second and third rows together and the rest of the page follows. Kinetic friction is a value. Static friction is a permission slip with a limit printed on it.

Static friction is an inequality, and that is where points go missing

The quantity $\mu_s F_N$ is a ceiling, not an answer. Static friction supplies exactly as much force as the object needs to not slide, and no more. So finding it is a two step job, and the coefficient is only involved in step two.

Find the requirement. Apply Newton's second law along the surface and ask how much friction would have to act for the object to keep whatever acceleration it actually has. On a level floor with a horizontal push and no motion, that is just the push. On a ramp with the block at rest, it is $mg\sin\theta$ . Inside an accelerating vehicle, it is $ma$ .
Compare with the ceiling. If the requirement is at or below $\mu_s F_N$ , static friction supplies the requirement, full stop. If it exceeds the ceiling, the surfaces slide and you switch to kinetic friction at its own fixed value.

Three consequences worth having reflexes for:

A book lying on a level table with nothing pushing it sideways has zero friction acting on it, not $\mu_s mg$ . Nothing needs canceling.
A block resting on a ramp has friction $mg\sin\theta$ up the slope, not $\mu_s mg\cos\theta$ . The second expression is only the largest value it could have supplied. The inclined plane guide runs that comparison in full.
Static friction reaches the ceiling at exactly one moment: the verge of slipping. That is why every measurement of $\mu_s$ is a measurement of the breaking point.

Kinetic friction is one number, and here is what it ignores

Once sliding starts, kinetic friction is $\mu_k F_N$ and stays there. Read the formula for what is missing from it:

No applied force. Push a sliding crate twice as hard and kinetic friction does not budge. The extra push goes into acceleration.
No speed. In the AP model, a block sliding at $0.5\ \text{m/s}$ and the same block at $5\ \text{m/s}$ feel the same friction force.
No contact area. EK 2.7.A.1.ii states that the force of friction between two surfaces does not depend on the size of the surface area of contact. Stand a brick on its narrow edge and the friction force is unchanged.
No mass, directly. Mass enters only through the normal force. That matters because the ratio $F_f / F_N$ is what the coefficient is, so mass cancels out of it.

What it does depend on is the pair of materials in contact, per EK 2.7.A.2.i, and the normal force. The normal force is where sliding problems actually go wrong, because $F_N = mg$ only holds on a level surface with no other vertical forces. Angle the push downward and $F_N$ goes up, so friction goes up. Tilt the surface and $F_N = mg\cos\theta$ . Get $F_N$ first, every time, using the normal force guide if the geometry is not obvious.

Why the static coefficient is the larger one

For a given pair of surfaces you expect $\mu_s \geq \mu_k$ . The CED states it as EK 2.7.B.3: the coefficient of static friction is typically greater than the coefficient of kinetic friction for a given pair of surfaces.

Keep the word typically. It means two things for you. First, $\mu_s \geq \mu_k$ is the expected relationship rather than a law you can lean on to reject a problem. If a question hands you two coefficients that happen to be equal, that is not a misprint. Second, if your own lab data comes out the other way round, the physics is not new; the force analysis or the constant-speed condition is off somewhere.

The CED does not explain why, and the exam will not ask you to. The usual explanation is that surfaces at rest have had time to settle into each other: contact happens at a small number of microscopic high points, and while nothing is moving those junctions have time to develop. Once the surfaces are sliding, no junction lasts long, so it takes less force to keep going than it took to start.

You feel the difference constantly. A stuck drawer that will not budge and then flies open is $\mu_s > \mu_k$ in one motion: the force that finally broke it loose is more than the sliding friction that follows, and the surplus goes straight into acceleration.

The instant of slipping: what jumps and what does not

Raise the applied force on a stationary block steadily. Static friction climbs to match it, one for one, until it reaches $\mu_s F_N$ . At the next fraction of a newton the block breaks loose and friction falls immediately to $\mu_k F_N$ .

The applied force did not change. The friction force did: it dropped. So on a level surface the acceleration jumps from zero to

a = \frac{\mu_s F_N - \mu_k F_N}{m} = g(\mu_s - \mu_k)

with the mass canceling, which makes it a clean result to check yourself against. The block does not ease into motion; it starts with that acceleration already.

Repeat the cycle fast enough and you get stick-slip: the surfaces grab, break loose, slow down, grab again. A chair leg juddering across a floor, a squealing brake, and a bow drawing a note out of a violin string are all the same $\mu_s$ to $\mu_k$ drop happening many times a second.

If you want the picture rather than the algebra, sketch the friction force against a steadily rising applied force. Topic 2.7 covers that graph and what each part of it means; the inclined plane simulator shows the same transition on a ramp, where you raise the angle instead of the push and watch the block let go.

The angle of friction: mu is the tangent of the slip angle

Put an object on a surface and tilt the surface slowly. At some angle $\theta_c$ the object just begins to slide. At that instant the gravity component along the surface has reached the static ceiling:

mg\sin\theta_c = \mu_s\, mg\cos\theta_c

Both $m$ and $g$ cancel, which leaves the result that gets searched for by name:

\mu_s = \tan\theta_c

That angle has a name. It is the angle of friction, sometimes written $\phi$ , and it is the coefficient expressed as an angle instead of a ratio. Here is why the two are the same thing. The surface exerts two forces on the object, a normal force perpendicular to it and a friction force along it. Add them as vectors and you get a single contact force tilted away from the normal by an angle $\phi$ with

\tan\phi = \frac{F_f}{F_N}

At the verge of slipping $F_f$ has reached $\mu_s F_N$ , so $\tan\phi = \mu_s$ . Tilting the surface to $\theta_c$ tilts the normal direction by the same amount, which is why the critical ramp angle and the friction angle are one number. For a heap of loose material the closely related angle is called the angle of repose.

Three things this buys you:

A measurement with no force sensor. An angle and a tangent give you $\mu_s$ . The routine is in how to find the coefficient of friction.
Mass independence you can quote. A heavier block slips at the same angle. Nothing about the object survives the cancellation, only the pair of surfaces.
A kinetic version. If the object slides down at constant speed at angle $\theta$ , the same cancellation gives $\mu_k = \tan\theta$ . One surface pair therefore has two angles: it starts sliding at $\arctan\mu_s$ and coasts at constant speed at the smaller $\arctan\mu_k$ .

The AP Physics 1 equation sheet helps here more than students expect. Its table of trigonometric values for common angles prints $\tan 37^\circ = 3/4$ and $\tan 45^\circ = 1$ , so a block that lets go at $37^\circ$ has $\mu_s = 0.75$ and one that holds until $45^\circ$ has $\mu_s = 1$ , with no calculator involved.

Where students pick the wrong one

Every item here is the same mistake: reading the object's motion instead of the relative motion at the contact.

A wheel rolling without slipping. The contact patch is momentarily at rest against the road, so the friction there is static, even though the car is moving at 30 m/s. Skid the tires and it becomes kinetic.
A car turning on a flat curve. The tires are not sliding sideways, so the centripetal force is supplied by static friction, and the inequality is what sets the fastest safe speed. That is Topic 2.9.
Walking. Your shoe does not slide, so the friction on it is static, and it points forward. Friction opposing motion is a rule about sliding surfaces, not about objects.
A block held on a ramp by friction alone. Static, and the value is $mg\sin\theta$ , not the ceiling.
A box in an accelerating truck bed. Static, and it points in the direction the truck accelerates, because that friction is the only thing making the box speed up.
A crate on a moving conveyor. Compare the crate's velocity with the belt's, not with the ground's.

The reflex to build: name the two surfaces, then ask whether they are moving with respect to each other. A free-body diagram makes it hard to skip, because you have to commit to a direction for the friction arrow before you can write anything down.

What the equation sheet gives you, and what it leaves out

The AP Physics 1 equation sheet compresses both regimes into one line:

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

Counting the sheet's Mechanics and Fluids equations, that is the only one with a $\mu$ in it, and its symbol list defines $\mu$ as simply the coefficient of friction, with no subscript. So the sheet gives you the inequality and nothing else. You supply:

the subscript, from the sliding test at the top of this page,
the equality for the kinetic case, since the printed relation is written as an inequality covering both,
the maximum, $F_{f,s,\text{max}} = \mu_s F_n$ , which the CED lists under EK 2.7.B.2.ii as a derived equation. The CED says plainly that not all equations in the course framework appear on the exam's equation sheet, and this is one of them, so write it out yourself when a problem needs the ceiling.

None of that is a problem once you know it. The failure mode is treating the printed inequality as the static formula and forgetting that kinetic friction sits at the equality. Both cases live on that one line, and the sliding test is what tells them apart. For the values themselves, see the coefficient of friction table: they are not on the sheet, and an AP problem will state the one it wants.

Static friction as the force that speeds something up

An 8.0 kg toolbox sits on the flat bed of a pickup truck. The coefficient of static friction between the box and the bed is $\mu_s = 0.40$ . The truck accelerates forward at $2.5\ \text{m/s}^2$ and the box rides along without sliding. Find the friction force on the box, then find the largest acceleration the truck could have without the box slipping.

Set the convention: positive $x$ is the direction the truck accelerates, positive $y$ is up. The box does not slide on the bed, so this is a static friction problem, and the box's acceleration equals the truck's.
Find the requirement, not the ceiling. Friction is the only horizontal force on the box, so it must be the entire net force: $F_{f,s} = ma = (8.0\ \text{kg})(2.5\ \text{m/s}^2) = 20\ \text{N}$ , pointing forward, in the $+x$ direction.
Now check it against the ceiling. The bed is level and nothing else pushes vertically, so $F_N = mg = (8.0\ \text{kg})(9.8\ \text{m/s}^2) = 78.4\ \text{N}$ , and $F_{f,s,\text{max}} = \mu_s F_N = (0.40)(78.4\ \text{N}) = 31.36\ \text{N}$ .
Since $20\ \text{N} < 31.36\ \text{N}$ , static friction can supply the requirement, so the box rides along and the answer to the first part is 20 N, not 31 N.
For the largest acceleration, set the requirement equal to the ceiling: $ma_{\max} = \mu_s mg$ , so $a_{\max} = \mu_s g = (0.40)(9.8\ \text{m/s}^2) = 3.92\ \text{m/s}^2$ . The mass cancels, so a heavier toolbox slips at the same acceleration.

Friction on the box is 20 N forward, in the direction of the acceleration, and the truck can accelerate at up to $3.9\ \text{m/s}^2$ before the box slides. Both parts break the habit of reading friction as a backward force: here friction is the only thing moving the box at all.

One surface pair, two critical angles

A block on a plank has $\mu_s = 0.62$ and $\mu_k = 0.45$ . The plank is tilted slowly from horizontal. At what angle does a resting block first start to slide, and at what angle does an already sliding block travel at constant speed? What happens between the two angles?

Resting block. It lets go when the gravity component along the plank reaches the static ceiling: $mg\sin\theta = \mu_s mg\cos\theta$ , so $\tan\theta_s = \mu_s$ and $\theta_s = \arctan 0.62 = 31.8^\circ$ .
Sliding block at constant speed. Zero acceleration means the gravity component along the plank exactly equals kinetic friction: $mg\sin\theta = \mu_k mg\cos\theta$ , so $\tan\theta_k = \mu_k$ and $\theta_k = \arctan 0.45 = 24.2^\circ$ .
Note that $\theta_k < \theta_s$ , which has to be true whenever $\mu_k < \mu_s$ , since the tangent increases with angle over this range.
Read the three zones. Below $24.2^\circ$ : a resting block stays, and a sliding block decelerates to a stop. Above $31.8^\circ$ : a resting block cannot stay, and a sliding block speeds up. Between $24.2^\circ$ and $31.8^\circ$ : the outcome depends on what the block was already doing.
Check the middle zone at $28^\circ$ . The gravity component needs $\tan 28^\circ = 0.532$ of the normal force to be balanced. That is below $\mu_s = 0.62$ , so a block at rest holds. It is above $\mu_k = 0.45$ , so a block already sliding keeps accelerating.

A resting block first slides at about $31.8^\circ$ , and a sliding block moves at constant speed at about $24.2^\circ$ . Between roughly $24^\circ$ and $32^\circ$ the same plank at the same angle either holds the block or lets it accelerate, depending only on whether it was already moving. Nothing about the block's mass appears anywhere in this.

The jump at the moment it breaks loose

A 15 kg crate sits on a level floor with $\mu_s = 0.50$ and $\mu_k = 0.35$ . A horizontal push is increased very slowly from zero. Find the push at which the crate breaks loose, and the crate's acceleration immediately afterward, while the push is still at that same value.

Convention: positive $x$ is the direction of the push. The floor is level and the push is horizontal, so $F_N = mg = (15\ \text{kg})(9.8\ \text{m/s}^2) = 147\ \text{N}$ throughout.
Breakaway push. Up to this point static friction has matched the push exactly, so the crate lets go when the push reaches the ceiling: $F = \mu_s F_N = (0.50)(147\ \text{N}) = 73.5\ \text{N}$ .
The instant it slides, kinetic friction takes over at its own fixed value: $f_k = \mu_k F_N = (0.35)(147\ \text{N}) = 51.45\ \text{N}$ .
Net force with the push unchanged at 73.5 N: $F_{net,x} = 73.5\ \text{N} - 51.45\ \text{N} = 22.05\ \text{N}$ , so $a_x = 22.05\ \text{N} / 15\ \text{kg} = 1.47\ \text{m/s}^2$ .
Confirm it against the general result. Substituting $F = \mu_s mg$ gives $a = g(\mu_s - \mu_k) = (9.8\ \text{m/s}^2)(0.50 - 0.35) = 1.47\ \text{m/s}^2$ , with the mass gone. Any crate on this floor, pushed to its own breakaway force, starts with the same acceleration.

The crate breaks loose at 73.5 N and immediately accelerates at $1.47\ \text{m/s}^2$ , with the push held constant. The acceleration is discontinuous because the friction force drops by 22 N the moment sliding begins. Both coefficients are given to two significant figures, so report $1.5\ \text{m/s}^2$ .

Frequently asked questions

What is the difference between static and kinetic friction?

Static friction acts between surfaces that are not sliding on each other, and it takes whatever value and direction is needed to prevent slipping, up to a maximum of mu_s times the normal force. Kinetic friction acts between surfaces that are sliding, and its magnitude is fixed at mu_k times the normal force, directed opposite the motion of each surface relative to the other. Static friction is a range, kinetic friction is a value, and for a given pair of surfaces mu_s is typically the larger coefficient.

What is the angle of friction, and why does tan phi equal the coefficient of friction?

The angle of friction is the coefficient written as an angle instead of a ratio. A surface exerts a normal force perpendicular to itself and a friction force along itself; added as vectors, they make one contact force tilted away from the normal by an angle phi, with tan phi equal to the friction force divided by the normal force. At the verge of slipping the friction force has reached mu_s times the normal force, so tan phi equals mu_s. This is why tilting a surface until an object just slides gives mu_s = tan of that critical angle, and why the answer does not depend on mass.

How do you find the static friction force?

Not by multiplying mu_s by the normal force, unless the object is exactly on the verge of slipping. Apply Newton's second law along the surface and work out how much friction is needed to give the object the acceleration it actually has: on a level floor with a horizontal push and no motion that is the push itself, on a ramp at rest it is mg sin theta, and inside an accelerating vehicle it is ma. Then compare that requirement with mu_s times the normal force. If the requirement is smaller, the requirement is the answer. If it is larger, the object slides and kinetic friction takes over.

Is friction static or kinetic when a wheel rolls without slipping?

Static. Rolling without slipping means the contact patch is momentarily at rest against the road, so the two surfaces are not sliding on each other, however fast the vehicle is traveling. That is why a car's grip in cornering and braking is governed by the static coefficient and by the inequality, not by a fixed force. Lock the brakes so the tires skid and the contact starts sliding, which switches the friction to kinetic at mu_k times the normal force, and mu_k is typically smaller.

Can friction make an object speed up?

Yes, and static friction routinely does. A toolbox in an accelerating truck bed is accelerated by friction alone, pointing in the direction the truck accelerates. Your shoe does not slide on the ground when you walk, so the friction on it is static and points forward. A rolling wheel is driven by static friction at the contact patch. The rule that friction opposes motion is really a rule about sliding surfaces: kinetic friction opposes the relative sliding, while static friction points whichever way prevents it.

What happens to the friction force the instant an object starts to slide?

It drops. Static friction has been climbing to match the applied force, and it peaks at mu_s times the normal force. The moment the surfaces slide, friction falls to mu_k times the normal force, which is typically smaller, and it stays there no matter how much harder you push. Since the applied force has not changed, the acceleration jumps discontinuously from zero to g times the difference of the two coefficients on a level surface. Repeated quickly, that same drop is what makes brakes squeal and chair legs judder.

Does static friction always equal mu_s times the normal force?

No, and this is the most common error on the topic. That product is the maximum static friction can reach, not its value. A book lying on a level table with nothing pushing it sideways has zero friction on it, because nothing needs canceling. A block at rest on a ramp has friction mg sin theta up the slope, which is smaller than mu_s mg cos theta whenever the block is not on the verge of slipping. Static friction equals the maximum at exactly one moment: the instant slipping begins.