AP Physics C: Mechanics · Topic 2.7

Topic 2.7: Kinetic and Static Friction

Unit 2: Force and Translational Dynamics20-25% of the multiple-choice section

Kinetic friction acts while surfaces slide and has a fixed magnitude: the coefficient of kinetic friction times the normal force. Static friction takes whatever value prevents sliding, up to a maximum. AP Physics C states both as AP Physics 1 does; the C exam wants symbols, not numbers.

AP Physics: Unit 2 (topics 2.7 Kinetic and Static Friction). AP Physics C: Mechanics Unit 2, Topic 2.7. Two learning objectives. 2.7.A, describe kinetic friction between two surfaces, supported by 2.7.A.1 (kinetic friction occurs when two surfaces in contact move relative to each other), 2.7.A.1.i (exerted opposite the motion of each surface relative to the other surface), 2.7.A.1.ii (the force of friction does not depend on the size of the surface area of contact), 2.7.A.2 (the magnitude of kinetic friction is the product of the normal force and the coefficient of kinetic friction, printed as an EQUALITY |F_f,k| = |mu_k F_N|), 2.7.A.2.i (the coefficient depends on the material properties of the surfaces in contact) and 2.7.A.2.ii (normal force is the perpendicular component of the force exerted on an object by the surface with which it is in contact, directed away from the surface). 2.7.B, describe static friction between two surfaces, supported by 2.7.B.1, 2.7.B.2 (static friction adopts the value and direction required to prevent an object from slipping or sliding on a surface, printed as an INEQUALITY |F_f,s| <= |mu_s F_n|), 2.7.B.2.i (slipping and sliding refer to situations in which two surfaces are moving relative to each other), 2.7.B.2.ii (there exists a maximum value; DERIVED equation F_f,s,max = mu_s F_N) and 2.7.B.3 (the coefficient of static friction is TYPICALLY greater than the coefficient of kinetic friction for a given pair of surfaces; the hedge is in the original). No boundary statement, in either course. This topic's content is IDENTICAL to AP Physics 1 Topic 2.7, same eleven statements and same numbering, and no calculus is involved. Verified against the Table of Information appendix: the C: Mechanics sheet prints ONE friction line for both kinds, |F_f| <= |mu F_N| with no subscript on mu, identical to the AP Physics 1 sheet; neither the kinetic equality nor the derived maximum-static equation is printed. Suggested skills are 1.B, 2.A, 2.B, 3.A and 3.B, five in total, one of the largest counts in the unit. The CED's sample multiple-choice question 7 aligns to 2.7.B and 2.7.B.2 with skill 3.C, answer B: two blocks on a board that is slowly raised, with mass and contact area given as distractors and the smaller coefficient of static friction as the answer. Sample free-response question 2 also aligns to 2.7.B. One of the unit's six optional sample instructional activities is on Topic 2.7.

Same content, and one equation the sheet compresses

The eleven essential-knowledge statements of Topic 2.7 are the same in AP Physics 1 and AP Physics C: Mechanics, with the same numbering, the same two relevant equations, the same derived equation and no boundary statement in either course. There is no derivative and no integral here.

So the useful thing to spend a Physics C page on is what the exam does with it, and one detail about the equation sheet that catches people out.

The framework writes friction as two separate relations. Statement 2.7.A.2 gives kinetic friction as an equality:

\lvert \vec{F}_{f,k}\rvert = \lvert \mu_k \vec{F}_N\rvert

and statement 2.7.B.2 gives static friction as an inequality:

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

The AP Physics C: Mechanics equation sheet prints neither of those. It prints one line for both kinds of friction, with no subscript on the coefficient at all:

\lvert \vec{F}_f\rvert \leq \lvert \mu \vec{F}_N\rvert

That is checked against the Table of Information appendix. The AP Physics 1 sheet prints the identical line. So the sheet gives you the inequality and leaves you to know that it is an equality for kinetic friction and only becomes one for static friction at the point of slipping, which is what the derived equation of 2.7.B.2.ii, $F_{f,s,\max} = \mu_s F_N$ , records. That derived equation is not on the sheet either.

If you take one thing from this page, take that: the sheet's friction line is an inequality, and knowing when it is tight is the whole topic.

What the CED requires of Topic 2.7

Two learning objectives, eleven statements between them, and five suggested skills, one of the largest counts in the unit: 1.B, 2.A, 2.B, 3.A and 3.B.

Objective 2.7.A: describe kinetic friction between two surfaces.

2.7.A.1: kinetic friction occurs when two surfaces in contact move relative to each other.
2.7.A.1.i: the kinetic friction force is exerted in a direction opposite the motion of each surface relative to the other surface.
2.7.A.1.ii: the force of friction between two surfaces does not depend on the size of the surface area of contact.
2.7.A.2: the magnitude of the kinetic friction force exerted on an object is the product of the normal force the surface exerts on the object and the coefficient of kinetic friction, with the equation above.
2.7.A.2.i: the coefficient of kinetic friction depends on the material properties of the surfaces that are in contact.
2.7.A.2.ii: normal force is the perpendicular component of the force exerted on an object by the surface with which it is in contact; it is directed away from the surface.

Objective 2.7.B: describe static friction between two surfaces.

2.7.B.1: static friction may occur between the contacting surfaces of two objects that are not moving relative to each other.
2.7.B.2: static friction adopts the value and direction required to prevent an object from slipping or sliding on a surface, with the inequality above.
2.7.B.2.i: slipping and sliding refer to situations in which two surfaces are moving relative to each other.
2.7.B.2.ii: there exists a maximum value for which static friction will prevent an object from slipping on a given surface. Derived equation $F_{f,s,\max} = \mu_s F_N$ .
2.7.B.3: the coefficient of static friction is typically greater than the coefficient of kinetic friction for a given pair of surfaces.

Note the hedge in 2.7.B.3: typically greater, not always. That word is in the framework and it is worth keeping, because a question is entitled to hand you a pair of surfaces where it is not true.

Note also what 2.7.A.1.ii says and does not say. Friction does not depend on the size of the contact area. It says nothing about the area being irrelevant to anything else, and it certainly does not say the pressure is irrelevant to the real world; it is a statement about this model.

Static friction has no formula, it has a ceiling

This is an expensive misunderstanding, and statement 2.7.B.2 says the correct thing in one sentence: static friction adopts the value and direction required to prevent an object from slipping or sliding on a surface.

So static friction is an unknown you solve for, not a number you compute. You find it by writing the equilibrium condition and asking what value friction must take to satisfy it. Only then do you check that value against the ceiling $\mu_s F_N$ from 2.7.B.2.ii.

The procedure, in order:

Assume the object does not slip, so its acceleration along the surface is whatever the rest of the system requires (usually zero).
Write the equation along the surface and solve for the friction force $f$ it demands.
Compute the ceiling $f_{\max} = \mu_s F_N$ from the perpendicular equation.
If $f \leq f_{\max}$ , the assumption holds and $f$ is the answer. If $f > f_{\max}$ , the object slips, and you start over with kinetic friction at the fixed value $\mu_k F_N$ directed opposite the relative motion.

Step 4 is what the inequality on the equation sheet is telling you to do.

The standard failure is to skip straight to $f = \mu_s F_N$ for a block that is nowhere near slipping. A 10 kg crate on a floor with $\mu_s = 0.6$ , pushed with 20 N and not moving, has 20 N of static friction on it, not 59 N. Setting it to the maximum would make the crate accelerate backwards.

Kinetic friction is the easy one by comparison. Statement 2.7.A.2 is an equality: once the surfaces are sliding, the magnitude is $\mu_k F_N$ , full stop, and 2.7.A.1.i fixes the direction as opposite the relative motion of the surfaces. It does not depend on how fast they slide, and it does not adjust itself to anything.

The static versus kinetic friction guide works through the distinction with more cases, and the coefficient of friction guide covers measuring it.

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

Both friction relations are built on $F_N$ , so every friction error is often really a normal-force error.

Statement 2.7.A.2.ii defines the normal force as the perpendicular component of the force exerted on an object by the surface with which it is in contact, directed away from the surface. That is a definition, not a formula. There is no equation for $F_N$ anywhere in the framework or on the equation sheet, because there cannot be one: you get it from the perpendicular component of Newton's second law, every time.

How it changes:

Situation	Normal force
block on a level floor, nothing else vertical	$mg$
block on an incline at angle $\theta$	$mg\cos\theta$
push at angle $\phi$ below horizontal	$mg + F\sin\phi$
pull at angle $\phi$ above horizontal	$mg - F\sin\phi$
in a lift accelerating up at $a$	$m(g + a)$
car on a flat road with aerodynamic downforce $F_d$	$mg + F_d$

The last row is the CED's own sample free-response question 2, where race cars on a flat horizontal road experience a downward aerodynamic force $F_{\text{down}} = bv^2$ . That force does not slow the cars; it presses them into the road, which raises the normal force, which raises the maximum static friction available, which raises the speed at which they can corner. Part B of that question asks students to derive the critical radius at which a car could travel at any speed without slipping, in terms of $m$ , $\mu$ , $b$ and physical constants, and it explicitly instructs them to begin the derivation by writing a fundamental physics principle or an equation from the reference information.

The normal force guide and the normal force glossary entry cover the routine, and weight versus normal force covers the confusion directly.

One more consequence of 2.7.A.1.ii, the area statement. Since friction depends on $\mu$ and $F_N$ only, two blocks of the same mass and material on the same surface experience the same friction whichever face they rest on. Doubling the contact area halves the pressure and leaves the product unchanged, which is the physical reason behind the model.

Direction, and the cases where friction points the way you do not expect

Statement 2.7.A.1.i says the kinetic friction force is exerted in a direction opposite the motion of each surface relative to the other surface. Relative, not absolute. That word settles every case that feels backwards.

Walking. Your shoe pushes backward on the ground, so the ground's static friction on your shoe points forward. Without it you go nowhere, which is why ice is hard to walk on.

Driving. The driven wheel's contact patch pushes backward against the road, so friction on the car points forward. That is what accelerates a car, and it is static friction as long as the tyre is not spinning.

A box on an accelerating truck bed. The box is not sliding relative to the bed, so the friction is static, and it points forward, in the direction of the acceleration, because it is the only horizontal force available to accelerate the box.

A block on a block. In the second worked example below, the friction on the top block from the bottom one points forward. Its third-law partner points backward on the bottom block.

Rolling without slipping. The contact point of a rolling wheel is instantaneously at rest relative to the ground, so the friction there is static even though the wheel is moving. That is Unit 6 material, but the reason lives in 2.7.B.1: static friction may occur between the contacting surfaces of two objects that are not moving relative to each other.

Traps worth naming while you are here.

Friction always opposes motion. No: it opposes relative motion of the surfaces. In the truck-bed and walking cases it is what causes the motion.

Static friction is always at its maximum. No: it is at its maximum only at the verge of slipping.

Bigger objects have more friction because there is more contact. No, by 2.7.A.1.ii. They have more friction because $F_N$ is larger.

$\mu_s$ is always bigger than $\mu_k$ . Statement 2.7.B.3 says typically, not always.

Friction depends on speed. Not in this model. Kinetic friction is $\mu_k F_N$ at any sliding speed. A speed-dependent resisting force is a resistive force, which is Topic 2.9, a different topic with a different equation and a differential equation to solve.

If you want the algebra-based version of this topic

The content is the same, so choose by course. If you are taking AP Physics 1, AP Physics 1 Topic 2.7: Kinetic and Static Friction covers the same eleven statements with algebra-based worked examples. If you are taking AP Physics C: Mechanics, stay here for the symbolic and comparative framing the C exam uses.

	AP Physics 1 Topic 2.7	AP Physics C Topic 2.7
Objectives	2.7.A, 2.7.B	2.7.A, 2.7.B
Statements	eleven, same numbering	eleven, same numbering
Kinetic friction	equality at 2.7.A.2	identical
Static friction	inequality at 2.7.B.2	identical
Derived equation	$F_{f,s,\max} = \mu_s F_N$	identical
Boundary statement	none	none
Sheet line	$\lvert\vec{F}_f\rvert \leq \lvert\mu\vec{F}_N\rvert$	identical
Calculus involved	none	none
Typical expected answer	a number	a symbolic expression or a ranking

The suggested skills do differ. AP Physics C lists 1.B, 2.A, 2.B, 3.A and 3.B, five of them. Skill 2.A, derive a symbolic expression, is the notable addition, and it is why the first worked example below produces $\tan\theta = \mu_s$ before it produces a number.

Also on this site: the coefficient of friction table for typical values, the friction calculator, the inclined plane walkthrough, and the friction practice set.

How Topic 2.7 is tested

Unit 2 carries 20 to 25 percent of the multiple-choice section across about 15 to 25 class periods, the highest minimum weighting of any unit in AP Physics C: Mechanics (only Unit 3, at 15 to 25 percent, reaches the same 25 percent ceiling). Topic 2.7 holds two of the unit's nineteen learning objectives and, with five suggested skills, one of the largest skill sets in the unit.

Those skills are 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.B, calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway; 3.A, create experimental procedures that are appropriate for a given scientific question; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

The CED's sample multiple-choice question 7 is a Topic 2.7 question, aligned to learning objective 2.7.B and essential knowledge 2.7.B.2, with skill 3.C. It is worth studying because of what it is not: it contains no calculation at all. Two blocks of different materials sit on the same wooden board, and a table gives each block's mass, its contact area and its coefficient of static friction. One end of the board is slowly raised, and the question asks which block starts sliding first and why. The correct answer is the block with the smaller coefficient of static friction, and both the mass and the area in the table are there as distractors. The mass cancels out of the slip condition, and 2.7.A.1.ii says the area does not enter at all.

That question is a compact test of the first worked example below, which derives why the mass cancels.

Skills 1.B and 3.A point at the laboratory version. The CED lists an optional sample instructional activity on Topic 2.7: students are asked to find the coefficient of friction, static or kinetic, of a shoe or other object, and the activity can be run as a competition where the team with the simplest procedure, or the one that uses the least equipment, wins. The tilt method of the first worked example is about as simple as it gets: raise the surface until the object slips and take the tangent of the angle.

Sample free-response question 2 also aligns to 2.7.B among its seven learning objectives, so friction appears in both halves of the CED's sample exam.

The slip angle, and why the mass cancels

A block rests on a board that is slowly tilted from horizontal. The coefficient of static friction between block and board is $\mu_s = 0.42$ and the coefficient of kinetic friction is $\mu_k = 0.31$ . Derive the angle at which the block begins to slide, evaluate it, and then find the block's acceleration immediately after it starts sliding, at that same angle. Use $g = 9.8\ \mathrm{m/s^2}$ .

Draw the free-body diagram: three arrows on a dot, gravity straight down, the normal force perpendicular to the board, static friction along the board pointing up-slope. Per the Topic 2.2 boundary statement, no components are drawn.
Choose axes along and perpendicular to the board, per 2.2.B.4, and declare the positive direction along the board as down the slope.
While the block is not sliding, both axes are in equilibrium. Perpendicular: $F_N = mg\cos\theta$ . Along the board: $f_s = mg\sin\theta$ . Note that the static friction is not $\mu_s F_N$ here; it is whatever value the equilibrium demands, which is 2.7.B.2.
The block slips at the angle where the demanded friction first exceeds the ceiling of 2.7.B.2.ii. Set them equal: $mg\sin\theta = \mu_s\, mg\cos\theta$ .
Divide through by $mg\cos\theta$ . Both $m$ and $g$ cancel, and the result is $\tan\theta = \mu_s$ . That is the symbolic answer, and it is the point of the exercise: the slip angle depends on the coefficient of static friction and on nothing else at all. Not the mass, and by 2.7.A.1.ii not the contact area either.
Evaluate: $\theta = \arctan(0.42) = 22.8^\circ$ .
This is exactly the reasoning behind the CED's sample multiple-choice question 7. Given two blocks with different masses, different contact areas and different coefficients, the one that slips first is the one with the smaller $\mu_s$ , because that is the only variable in $\tan\theta = \mu_s$ .
Now the moment after slipping. The friction switches to kinetic and drops to the fixed value $\mu_k F_N$ , directed up the slope because the block is now sliding down. The angle has not had time to change.
Along the board, with down-slope positive: $ma = mg\sin\theta - \mu_k\, mg\cos\theta$ , so $a = g(\sin\theta - \mu_k\cos\theta)$ . The mass cancels again.
At $\theta = 22.8^\circ$ : $\sin\theta = 0.3872$ and $\cos\theta = 0.9220$ , so $a = 9.8\left(0.3872 - 0.31 \times 0.9220\right) = 9.8(0.3872 - 0.2858) = 9.8(0.1014) = 0.994\ \mathrm{m/s^2}$ down the slope.
Cross-check with a tidier route. Since $\tan\theta = \mu_s$ at this angle, $\sin\theta = \mu_s\cos\theta$ , so $a = g\cos\theta(\mu_s - \mu_k) = 9.8(0.9220)(0.42 - 0.31) = 9.8(0.9220)(0.11) = 0.994\ \mathrm{m/s^2}$ . The two agree.
Read the second form. The acceleration at the moment of slipping is proportional to the gap between the two coefficients. That is why a block sitting on the verge of sliding lurches into motion rather than easing into it: the resisting force drops discontinuously from $\mu_s F_N$ to $\mu_k F_N$ the instant it moves.

$\tan\theta = \mu_s$ , so the block slips at $\theta = 22.8^\circ$ , independent of its mass and contact area. Immediately after, $a = g\cos\theta(\mu_s - \mu_k) = 0.99\ \mathrm{m/s^2}$ down the slope.

Stacked blocks: how hard can you push before the top one slides

A $2.0$ kg block sits on top of a $5.0$ kg block, which rests on a frictionless floor. The coefficient of static friction between the two blocks is $\mu_s = 0.35$ . A horizontal force is applied to the lower block. Find the largest force that can be applied without the upper block sliding, and the friction force on the upper block at that point.

Set up the two free-body diagrams separately, because that is the only way to see which forces act where. Declare rightward positive, in the direction of the applied force.
Upper block: gravity down, the normal force from the lower block up, and static friction from the lower block horizontally. Nothing else touches it, so friction is the only horizontal force on it.
Lower block: gravity down, the normal force from the upper block pressing down, the normal force from the floor up, the applied force horizontally, and the third-law partner of the friction on the upper block, pointing backward.
As long as they do not slip, both blocks share one acceleration. That is the condition to use, and it is what makes the problem solvable.
Apply the second law to the upper block alone: $f_s = m_t a$ . The friction is the only thing accelerating it, so the acceleration the pair can reach is limited entirely by how much friction the interface can supply.
The ceiling on that friction is 2.7.B.2.ii, $f_{s,\max} = \mu_s F_N$ , where $F_N$ is the normal force between the blocks. From the upper block's vertical equation, $F_N = m_t g$ .
Combine: $m_t a_{\max} = \mu_s m_t g$ , so $a_{\max} = \mu_s g$ . The upper block's mass cancels, which is worth noticing: the maximum shared acceleration does not depend on how heavy the top block is.
Numerically, $a_{\max} = (0.35)(9.8) = 3.43\ \mathrm{m/s^2}$ .
Now the applied force. Treat the two blocks as one system, in which the friction between them is internal and drops out by 2.3.A.2, and the floor is frictionless: $F_{\max} = (m_t + m_b)a_{\max} = (7.0)(3.43) = 24.01$ N, so $24$ N to two significant figures.
Friction on the upper block at that point: $f_s = m_t a_{\max} = (2.0)(3.43) = 6.86$ N, forward. Check it against the ceiling: $\mu_s m_t g = (0.35)(2.0)(9.8) = 6.86$ N. They are equal, as they must be at the verge of slipping.
Notice the direction. The friction on the upper block points forward, in the direction of motion, because it is the only force available to accelerate it. Friction opposes relative sliding of the surfaces, not motion, and here the two surfaces are not sliding at all.
One extension worth doing in your head: if the force were applied to the upper block instead, the analysis changes completely, because friction would then have to accelerate the heavier lower block. The maximum acceleration is the same $\mu_s g$ , but the force that produces it acts on a different body, so the answer differs.

$a_{\max} = \mu_s g = 3.43\ \mathrm{m/s^2}$ , giving $F_{\max} = (m_t + m_b)\mu_s g = 24$ N. At that point the static friction on the upper block is $6.86$ N, forward, exactly at its maximum value.

Frequently asked questions

Is friction different in AP Physics C than in AP Physics 1?

No. Topic 2.7 has the same two learning objectives and the same eleven essential knowledge statements in both course and exam descriptions, with the same equality for kinetic friction, the same inequality for static friction, the same derived equation for maximum static friction, and no boundary statement in either course. No calculus is involved. What differs is the exam. AP Physics C lists five suggested skills for this topic, including 2.A, derive a symbolic expression, which carries 25 to 30 percent of the multiple-choice section, so a C question is more likely to want an expression such as the tangent of the slip angle equalling the coefficient of static friction than a number.

Why does the AP Physics equation sheet show friction as an inequality?

Because it prints one line for both kinds of friction. The AP Physics C: Mechanics Table of Information gives the magnitude of the friction force as less than or equal to the magnitude of mu times the normal force, with no subscript on mu, and the AP Physics 1 sheet prints the identical line. The framework itself is more specific: essential knowledge 2.7.A.2 gives kinetic friction as an equality, and 2.7.B.2 gives static friction as an inequality, becoming an equality only at the maximum value recorded in the derived equation of 2.7.B.2.ii. Neither of those more specific forms is printed, so knowing when the inequality is tight is left to you.

How do you find static friction if it has no formula?

Solve for it from the equilibrium condition, then check it against the ceiling. Essential knowledge 2.7.B.2 says static friction adopts the value and direction required to prevent an object from slipping or sliding on a surface, so it is an unknown rather than something you compute directly. Write Newton's second law along the surface with the acceleration the situation requires, solve for the friction force that makes the equation true, and then compare that value with the maximum from the derived equation of 2.7.B.2.ii, the coefficient of static friction times the normal force. If your value is at or below the maximum the object does not slip. If it exceeds it, the object slips and you redo the problem with kinetic friction.

At what angle does a block start to slide down a ramp?

When the tangent of the angle equals the coefficient of static friction. Setting the demanded static friction equal to its maximum gives mg sin theta equal to mu times mg cos theta, and both the mass and g cancel, leaving the tangent of theta equal to the coefficient of static friction. So the slip angle depends only on the two materials in contact. It does not depend on the mass of the block, and by essential knowledge 2.7.A.1.ii it does not depend on the contact area either. This is the reasoning the CED's own sample multiple-choice question 7 tests, where mass and area are supplied as distractors.

Does friction depend on the area of contact?

Not in this model. Essential knowledge 2.7.A.1.ii of the AP Physics C: Mechanics framework states that the force of friction between two surfaces does not depend on the size of the surface area of contact. The friction force is set by the coefficient and the normal force only, so a brick on its side and the same brick on its end experience the same friction on the same surface. Physically, doubling the area halves the pressure at the interface and the product stays the same. Contact area is a common distractor in exam questions and can be ignored whenever this model applies.

Can friction point in the direction of motion?

Yes, often. Essential knowledge 2.7.A.1.i says kinetic friction is exerted in a direction opposite the motion of each surface relative to the other surface, and the key word is relative. When you walk, your shoe pushes backward on the ground and the ground's static friction on your shoe points forward, which is what moves you. When a box rides on an accelerating truck bed without sliding, the static friction on the box points forward, because it is the only horizontal force available to accelerate it. Friction opposes relative sliding of the surfaces, not motion through space.

Is the coefficient of static friction always greater than kinetic?

Usually, but the framework hedges. Essential knowledge 2.7.B.3 of the AP Physics C: Mechanics course and exam description says the coefficient of static friction is typically greater than the coefficient of kinetic friction for a given pair of surfaces. The word typically is in the original, so a question is entitled to give you a pair where they are equal or reversed, and you should use the numbers you are given rather than the general rule. When the static coefficient does exceed the kinetic one, the consequence is that an object on the verge of sliding lurches into motion, because the resisting force drops the instant it starts to move.