AP Physics 1 · Topic 2.7

Topic 2.7: Kinetic and Static Friction

Unit 2: Force and Translational Dynamics18-23% of the multiple-choice section

Topic 2.7 splits friction in two. Kinetic friction acts while two surfaces slide across each other, and its magnitude is the coefficient of kinetic friction times the normal force. Static friction acts when they do not slide, and it takes whatever value prevents slipping, up to a maximum.

AP Physics: Unit 2 (topics 2.7 Kinetic and Static Friction). Topic 2.7 carries two CED learning objectives: 2.7.A, describe kinetic friction between two surfaces, and 2.7.B, describe static friction between two surfaces. The CED gives kinetic friction as an equality and static friction as an inequality, with a separate derived equation for the static maximum. The CED's suggested skills for this topic are 1.C, 2.B, 2.C, and 3.B. Unit 2 is weighted at 18 to 23 percent of the multiple-choice section, tied with Unit 3 for the largest share, and about 22 to 27 class periods.

What Topic 2.7 requires

Two learning objectives, and the split between them is the whole topic. LO 2.7.A asks you to describe kinetic friction between two surfaces. LO 2.7.B asks you to describe static friction between two surfaces. Kinetic friction gets an equality. Static friction gets an inequality. Almost every mistake on this topic comes from blurring those two.

Under 2.7.A the CED lists five essential knowledge statements covering when kinetic friction occurs, which way it points, why contact area does not enter, how its magnitude is set, and what the normal force is. Under 2.7.B it lists five more covering when static friction occurs, how it takes on a value, what slipping means, that a maximum exists, and how the two coefficients compare.

The CED lists four suggested skills for this topic: 1.C (create qualitative sketches of graphs), 2.B (calculate or estimate an unknown quantity with units), 2.C (compare physical quantities between two or more scenarios or at different times), and 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim). Those are skills sitting beneath the three Science Practices, not practices in their own right.

Topic 2.7 sits in Unit 2, which the CED weights at 18 to 23 percent of the multiple-choice section and about 22 to 27 class periods.

Kinetic friction: one value, set by the surfaces

Kinetic friction occurs when two surfaces in contact move relative to each other. Its direction, per EK 2.7.A.1.i, is opposite to the motion of each surface relative to the other surface.

Read that direction rule slowly, because it says relative to the other surface, not relative to the ground. A crate sliding backward along the belt of a forward-moving conveyor has friction pushing it forward, in the direction the belt moves. The ground frame never enters the rule.

The magnitude is the product of the normal force and the coefficient of kinetic friction:

|\vec{F}_{f,k}| = |\mu_k \vec{F}_n|

Once the surfaces are sliding, that number is locked. Push twice as hard along the surface and kinetic friction does not budge, because the push appears nowhere in the equation. Only two things can change it: the normal force, and the pair of materials in contact.

That fixed value is what makes sliding problems tractable. Find the normal force once, multiply by $\mu_k$ , and friction becomes a known constant you can carry through Newton's second law and then into the kinematic equations.

Static friction adopts whatever value is required

Static friction may occur between the contacting surfaces of two objects that are not moving relative to each other, and EK 2.7.B.2 states the key property: static friction adopts the value and direction required to prevent an object from slipping or sliding on a surface. The CED writes it as an inequality:

|\vec{F}_{f,s}| \leq |\mu_s \vec{F}_n|

There is a maximum value for which static friction will prevent slipping (EK 2.7.B.2.ii), and the CED gives it as a derived equation:

F_{f,s,\text{max}} = \mu_s F_n

So $\mu_s F_n$ is a ceiling, not an answer. Every static problem runs in the same order:

Work out how much force static friction would have to supply to keep the object from sliding.
Compare that requirement with the ceiling $\mu_s F_n$ .
If the requirement is at or below the ceiling, static friction supplies exactly the requirement and nothing accelerates. If it exceeds the ceiling, the surfaces slide and kinetic friction takes over at its own fixed value.

The clearest consequence: a block resting on a level floor with nothing pushing it sideways has zero friction acting on it, not $\mu_s mg$ . Nothing needs canceling, so static friction supplies nothing.

What the equation sheet prints, and what it leaves to you

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

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

No subscript on $\mu$ , no separate kinetic equation, no maximum. You supply all three. Read the line as the static case when the surfaces are not sliding, with $\mu_s$ , and as the kinetic case sitting at the equality with $\mu_k$ when they are. The derived form $F_{f,s,\text{max}} = \mu_s F_n$ is listed in the CED but is not printed on the sheet, so write it out yourself when a problem needs the ceiling.

The sheet also cannot tell you which coefficient a problem wants. That decision comes from one question, asked at the instant the problem describes: are these two surfaces sliding across each other right now? Sliding means kinetic. Not sliding means static, including a wheel rolling without slipping, where the contact patch is momentarily at rest against the road.

One more piece of CED wording worth keeping exact. EK 2.7.B.3 says the coefficient of static friction is typically greater than the coefficient of kinetic friction for a given pair of surfaces. Typically is the CED's word, so treat $\mu_s \geq \mu_k$ as the expected relationship rather than a law.

The normal force is the input, and it is rarely just mg

Both friction expressions take the normal force as their input, so a wrong normal force ruins everything downstream. EK 2.7.A.2.ii defines it precisely: normal force is the perpendicular component of the force exerted on an object by the surface with which it is in contact, and it is directed away from the surface.

Perpendicular to the surface, not vertical. On a level floor with no other vertical forces those happen to coincide and $F_N = mg$ . Tilt the surface, angle the applied force, or accelerate the floor and they part company:

Ramp at angle $\theta$ : $F_N = mg\cos\theta$ , worked through in inclined plane problems.
Push angled downward: the vertical component of the push adds to $F_N$ , so friction goes up.
Pull angled upward: the vertical component of the pull subtracts, so friction goes down.
Elevator accelerating: $F_N$ is no longer $mg$ at all.

The normal force guide covers each case. Get $F_N$ first, then friction, never the other way around.

Worked answers on this page use $g = 9.8 \, \mathrm{m/s^2}$ , the value printed in the CED's Table of Information. The CED's Topic 1.3 boundary statement says the exam will use $g \approx 10 \, \mathrm{m/s^2}$ wherever a numerical value of $g$ is required, and that students are not penalized for correctly using the more precise commonly accepted values of $9.81 \, \mathrm{m/s^2}$ or $9.8 \, \mathrm{m/s^2}$ .

What the coefficient depends on, and what it ignores

EK 2.7.A.1.ii is blunt: the force of friction between two surfaces does not depend on the size of the surface area of contact. Lay a brick on its narrow edge instead of its broad face and, in this model, the friction force is unchanged. The statement is written about friction generally, so it covers both regimes.

EK 2.7.A.2.i says the coefficient of kinetic friction depends on the material properties of the surfaces that are in contact. Rubber on concrete and steel on ice are different numbers because the materials differ, not because the objects differ.

Mass does affect the friction force, but only through the normal force. Double the mass of a block on a level floor and $F_N$ doubles, so both the kinetic force and the static ceiling double. The coefficient itself does not move: it is a ratio of two forces, so it carries no units and no dependence on how heavy the object is. That is why tilting a surface until an object first slips measures $\mu_s$ without you ever needing the mass, the method laid out in how to find the coefficient of friction.

Treat all of this as the model the CED asks for, not as a claim about every real surface. The friction calculator applies the same model if you want to check a result.

The graph that holds the whole topic

Skill 1.C for this topic is creating qualitative sketches of graphs, and there is one sketch worth being able to draw from memory: the friction force on an object plotted against a steadily increasing applied force, with the object starting at rest on a level surface.

It has four features, in order:

A rising straight segment. While the object stays put, static friction matches the applied force exactly, so the graph climbs along the line of slope one.
A peak. The rise stops at $\mu_s F_N$ , the largest value static friction can reach.
A drop. The instant the object breaks loose, friction falls to $\mu_k F_N$ , lower because $\mu_k$ is typically the smaller coefficient.
A flat line. Kinetic friction then stays constant no matter how much harder you push.

Every conceptual error on this topic shows up as a wrong sketch. A graph that keeps rising past the peak treats static friction as an equality. One that starts flat at $\mu_k F_N$ applies kinetic friction to a stationary object. One with no drop ignores EK 2.7.B.3. Sketching it once beats memorizing three separate rules.

The inclined plane simulator shows the same transition on a ramp: raise the angle and watch the friction force track the downhill pull until the block breaks loose.

How Topic 2.7 shows up, and where friction goes next

The CED's sample instructional activity for this topic is a desktop experiment: measure the coefficient of static friction between your own shoe and a wood plank or metal track, at three equipment levels. Level 1 uses a spring scale. Level 2 uses a pulley, a spring, a toy bucket, and an electronic balance. Level 3 uses nothing but a protractor, which is the tilt-until-it-slips measurement. Being able to design that third version from scratch is exactly what skill 3.A rewards.

In the CED's sample free-response set, Question 1, the Mathematical Routines question, aligns to LO 2.7.A among several other objectives, with skills 1.A, 2.A, 3.B, and 3.C. Friction there is one force inside a larger dynamics problem rather than the whole question, which is how it usually arrives.

Friction does not stay inside Unit 2 either. It is the force that removes mechanical energy in Unit 3 problems, it is what lets a wheel roll without slipping in Unit 6, and it is what turns a car on a flat curve in Topic 2.9. In that last case the tires are not skidding, so the friction doing the turning is static, and the inequality is what sets the fastest safe speed.

One crate, two pushes: the ceiling decides

A 12 kg crate sits at rest on a level floor. The coefficient of static friction between crate and floor is $\mu_s = 0.45$ , and the coefficient of kinetic friction is $\mu_k = 0.32$ . A worker pushes horizontally with 40.0 N, and later with 60.0 N. For each push, find the friction force on the crate and its acceleration.

Set the convention. Take the direction of the push as positive $x$ and up as positive $y$ . The crate never accelerates vertically, and the push is horizontal, so nothing disturbs the vertical balance: $F_N = mg = (12 \, \mathrm{kg})(9.8 \, \mathrm{m/s^2}) = 117.6 \, \mathrm{N}$ , carried unrounded.
Find the ceiling once, since $F_N$ is the same for both pushes: $F_{f,s,\text{max}} = \mu_s F_N = (0.45)(117.6 \, \mathrm{N}) = 52.92 \, \mathrm{N}$ , which is 53 N to two significant figures.
First push, 40.0 N. The requirement, 40.0 N, is below the 52.9 N ceiling, so static friction holds. It supplies exactly what equilibrium needs: $F_{f,s} = 40.0 \, \mathrm{N}$ in the negative $x$ direction, and $a_x = 0$ . It is not 53 N, and there is no way to get 53 N out of this scenario.
Second push, 60.0 N. Now the requirement exceeds the ceiling, so the crate breaks loose and kinetic friction takes over: $f_k = \mu_k F_N = (0.32)(117.6 \, \mathrm{N}) = 37.632 \, \mathrm{N}$ , or 38 N to two significant figures.
Apply Newton's second law along $x$ with the unrounded value: $F_{net,x} = 60.0 \, \mathrm{N} - 37.632 \, \mathrm{N} = 22.368 \, \mathrm{N}$ , so $a_x = 22.368 \, \mathrm{N} / 12 \, \mathrm{kg} = 1.864 \, \mathrm{m/s^2}$ .
Sense check. Raising the push by 20.0 N changed the friction force from 40.0 N to 38 N, a drop rather than a rise. That is the graph from the section above: friction climbs with the push right up to the peak, then falls to a constant once sliding starts.

With the 40.0 N push, friction is 40.0 N backward and the crate stays at rest, so the acceleration is zero. With the 60.0 N push, friction is 38 N backward and the crate accelerates at $1.9 \, \mathrm{m/s^2}$ in the direction of the push. The two coefficients are given to two significant figures, so the answers stop there.

A push angled downward changes the normal force

A 6.0 kg carton is already sliding across a level floor. A worker keeps it moving by pushing with a 25 N force directed $37^\circ$ below the horizontal, in the direction of travel. The coefficient of kinetic friction is $\mu_k = 0.20$ . Find the normal force, the kinetic friction force, and the acceleration.

Set the convention. Positive $x$ is the direction of travel, positive $y$ is up. Resolve the push: its horizontal component is $F\cos 37^\circ$ forward, and its vertical component is $F\sin 37^\circ$ pressing down into the floor.
Use real trigonometry first: $\cos 37^\circ = 0.7986$ and $\sin 37^\circ = 0.6018$ , so $F_x = (25 \, \mathrm{N})(0.7986) = 19.97 \, \mathrm{N}$ and the downward component is $(25 \, \mathrm{N})(0.6018) = 15.05 \, \mathrm{N}$ .
Vertical balance gives the normal force. There is no vertical acceleration, so the floor must cancel the weight and the downward push together: $F_N = mg + F\sin 37^\circ = (6.0)(9.8) + 15.05 = 58.8 + 15.05 = 73.85 \, \mathrm{N}$ . That is well above the 58.8 N weight, which is the point of the example.
Kinetic friction follows: $f_k = \mu_k F_N = (0.20)(73.85 \, \mathrm{N}) = 14.77 \, \mathrm{N}$ , opposing the motion, so in the negative $x$ direction.
Newton's second law along $x$ : $F_{net,x} = 19.97 - 14.77 = 5.20 \, \mathrm{N}$ , so $a_x = 5.20 \, \mathrm{N} / 6.0 \, \mathrm{kg} = 0.866 \, \mathrm{m/s^2}$ . Rounding $F_N$ to 74 N and $f_k$ to 15 N before this step would have given $0.83 \, \mathrm{m/s^2}$ instead, which is why intermediate values stay unrounded.
Now redo it with the AP sheet's values for the 3-4-5 triangle, $\cos 37^\circ = 4/5$ and $\sin 37^\circ = 3/5$ : $F_x = 20.0 \, \mathrm{N}$ , downward component 15.0 N, $F_N = 73.8 \, \mathrm{N}$ , $f_k = 14.76 \, \mathrm{N}$ , net force 5.24 N, and $a_x = 0.873 \, \mathrm{m/s^2}$ . Both routes round to the same two-figure answer, and both earn credit.

The normal force is about 74 N, kinetic friction is about 15 N, and the acceleration is $0.87 \, \mathrm{m/s^2}$ forward. The normal force runs 26 percent above the carton's 58.8 N weight because the push has a downward component, so the friction it must overcome is larger too. Pushing at $37^\circ$ above the horizontal instead would have cut the normal force to 43.8 N.

Frequently asked questions

What does AP Physics 1 Topic 2.7 cover?

Two learning objectives. 2.7.A asks you to describe kinetic friction between two surfaces: when it occurs, that it points opposite the relative motion of the surfaces, that it does not depend on contact area, and that its magnitude is the coefficient of kinetic friction times the normal force. 2.7.B asks you to describe static friction, which takes the value and direction needed to prevent slipping, up to a maximum of the static coefficient times the normal force.

Why is static friction written as an inequality?

Because it has no single value. Static friction supplies exactly as much force as is needed to stop the surfaces from sliding, so it changes when the applied force changes. The inequality records that it can be anything from zero up to a ceiling. Kinetic friction has no such freedom: once the surfaces slide, the force sits at the coefficient times the normal force and stays there.

How do I decide between the static and the kinetic coefficient?

Ask whether the two surfaces are sliding across each other at the instant the problem describes. If they are, use kinetic friction at its fixed value. If they are not, use static friction, and remember it equals the maximum only at the exact verge of slipping. A wheel rolling without slipping counts as not sliding, because the contact patch is momentarily at rest against the road.

Does friction depend on how much surface area is touching?

Not in this model. 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. Turning a block onto a narrower face changes nothing. What does change the friction force is the normal force, which is why mass, ramp angle, and any vertical component of an applied force all matter.

Is the normal force always equal to mg in a friction problem?

Only on a level surface with no other vertical forces. The CED defines the normal force as the perpendicular component of the force the surface exerts on the object, directed away from the surface. On a ramp it is mg cos theta. A push angled downward increases it, a pull angled upward reduces it, and an accelerating elevator changes it again. Since both friction expressions take the normal force as their input, that step comes first.