2.7 Kinetic and Static Friction

Picture stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last.

1. The box that will not budge

Read the story as text

You are moving a heavy 40 kg box of books across a wooden floor. You push gently. Nothing happens. You push harder. Still nothing. Then, all at once, the box breaks free and lurches forward. Strangely, once it is moving, keeping it moving feels easier than getting it started.

What force holds the box still, why does it suddenly give up, and why is sliding easier than starting?

2. A heavier load rubs harder

3. Sliding to a stop

Quick check. Leo pushes a heavy crate with 50 N and it does not move. How big is the friction on the crate?

4. Friction helps you walk

5. Will the block slip down the ramp?

6. Fast or slow, the same friction

Quick check. Kofi starts walking forward on a dry sidewalk. Which way does the ground's friction push on his shoe?

7. Check yourself

Think of your answer first, then tap to see it.

a) A box sits on a level floor and nobody pushes it. How much friction acts on it?

Show answer

None. Static friction only appears to oppose a push. With no sideways push, friction is zero.

b) Which is usually bigger: the most static friction can give, or sliding (kinetic) friction?

Show answer

Maximum static friction is usually bigger. That is why starting a box is harder than keeping it sliding.

c) A 10 kg crate slides on a level floor with μk = 0.25. How big is the friction?

Show answer

N = 10 kg × 9.8 m/s² = 98 N, so f = 0.25 × 98 N = 24.5 N, against the motion.

Already know this?

Three quick questions. Get all three right on the first try and you can skip ahead to the simulation.

1. A box slides across a floor at a steady speed. You make it slide twice as fast, still steady. The sliding friction on it is now:

2. A 20 kg box rests on a level floor with μs = 0.40. You push sideways with 60 N. What is the friction on the box?

3. You stack books inside a box so that the floor's normal force on it doubles. The largest static friction the floor can give:

The idea in plain words

Friction is the force a surface exerts along the surface, opposing sliding (or the tendency to slide) between the two surfaces. It always acts parallel to the surface.

Static friction acts when the surfaces are not sliding over each other. It is a "matching" force: it becomes exactly as big as it needs to be to stop sliding, up to a maximum.

fs ≤ μs FN

Kinetic friction acts when the surfaces are sliding. Its size is fixed by the normal force and does not depend on speed or on how hard you push:

fk = μk FN

applied push F friction f μs FN (largest static) μk FN (kinetic, constant) static: f = F sliding

Worked with numbers: the 40 kg box (μs = 0.50, μk = 0.30)

Read the steps as text
  1. Normal force on a level floor with a horizontal push: FN = m g = 40 kg × 9.8 m/s² = 392 N. Vertically nothing accelerates, so the floor pushes up as hard as gravity pulls down.
  2. Largest static friction: fs,max = μsFN = 0.50 × 392 = 196 N. This is the push you must beat to get the box moving.
  3. Push with 150 N: 150 < 196, so the box stays put and static friction is 150 N backward (not 196 N). At rest, a = 0, so friction must cancel the push exactly.
  4. Push with 200 N: 200 > 196, so it slides. Now kinetic friction acts: fk = 0.30 × 392 = 118 N. Once sliding, use μk, not μs.
  5. Acceleration: a = (F − fk)/m = (200 − 117.6)/40 = 82.4/40 = 2.1 m/s² forward. Newton's second law with the net horizontal force (forward positive).
  6. To keep it sliding at a constant speed you only need a push of 118 N. Constant velocity means a = 0, so the push only has to match kinetic friction. 118 N < 196 N: easier than starting.

Friction on an incline

On a slope at angle θ, tilt your axes: one along the slope, one perpendicular. Gravity splits into m g sin θ (down the slope) and m g cos θ (into the slope). The normal force balances the second part: FN = m g cos θ.

Worked with numbers: block on a 30° slope, μk = 0.30

Read the steps as text
  1. sin 30° = 0.500, cos 30° = 0.866. The two components of gravity along and into the slope.
  2. Once sliding down: a = 9.8 (0.500 − 0.30 × 0.866) = 9.8 × 0.240 = 2.4 m/s² down the slope. Down-slope gravity part minus kinetic friction, divided by m (the m cancels).
  3. Would it start sliding on its own if μs = 0.40? tan 30° = 0.577 > 0.40, so yes. It starts slipping at θ = tan−1(0.40) = 21.8°. Static friction can hold it only up to tan θ = μs.

Try it: animations

Forward (to the right) is positive in the first animation. In the second, "down the slope" is positive.

1. Push a box harder and harder

Your push grows steadily from zero. Watch the friction graph: static friction matches your push until it hits μsFN, then the box breaks free and friction drops to μkFN. (If you set μk above μs, the animation uses μk = μs, because real surfaces do not behave that way.)

Free body diagram of the box

Before slipping, the friction arrow is exactly as long as the push arrow. After slipping, the push wins and the box accelerates.

2. Block on an incline with friction

Set the angle and the coefficients. Give the block a nudge down the slope (positive speed) or a shove up the slope (negative speed). Does it stay, slide, or stop?

Free body diagram of the block

Three forces: weight (straight down), normal force (perpendicular to the slope) and friction (along the slope, opposing sliding or the tendency to slide).

Examples

Example 1: a sliding crate basic

A 10 kg crate slides across a level floor. μk = 0.25. What is the friction force on it?

Show solution
Read the steps as text
  1. FN = m g = 10 × 9.8 = 98 N. Level floor, no vertical push or pull.
  2. fk = μkFN = 0.25 × 98 = 24.5 N, opposite to the sliding. It is sliding, so kinetic friction, whatever the speed.

Example 2: does it move? medium

The same 10 kg crate is at rest. μs = 0.40, μk = 0.25. (a) You push horizontally with 30 N. What is the friction force? (b) You push with 50 N. Find the acceleration.

Show solution
Read the steps as text
  1. fs,max = 0.40 × 98 = 39.2 N. First find the limit.
  2. (a) 30 N < 39.2 N: no sliding. Friction = 30 N backward. Static friction matches the push; it is NOT 39.2 N.
  3. (b) 50 N > 39.2 N: it slides. fk = 24.5 N. a = (50 − 24.5)/10 = 2.55 m/s². Once sliding, use μk.
Trap: writing fs = μsFN = 39.2 N for the crate that does not slide. Instead: μsFN is only the maximum; at rest, static friction just cancels the push: 30 N.

Example 3: on the slope AP

A 5.0 kg block sits on a 30° incline. μs = 0.60, μk = 0.40. (a) Does it slide on its own? (b) What is the friction force? (c) A small tap starts it sliding down. What is its acceleration?

Show solution
Read the steps as text
  1. (a) tan 30° = 0.577 < 0.60, so it stays at rest. Static friction can hold it as long as tan θ ≤ μs.
  2. (b) At rest, friction balances the down-slope part of gravity: fs = m g sin 30° = 5.0 × 9.8 × 0.5 = 24.5 N up the slope. (The maximum would be 0.60 × 5.0 × 9.8 × 0.866 = 25.5 N, so it is close to slipping.) Static friction is only as big as needed.
  3. (c) Sliding: a = 9.8 (0.500 − 0.40 × 0.866) = 9.8 × 0.154 = 1.5 m/s² down the slope. It speeds up. Kinetic friction (0.40 × 5.0 × 9.8 × 0.866 = 17.0 N) is now less than the 24.5 N gravity part.

Example 4: pulling a sled with a slanted rope AP

A child pulls a 20 kg sled with a rope at 30° above the horizontal. The tension is 100 N and μk = 0.20. Find the acceleration.

Show solution
Read the steps as text
  1. Vertical: FN + T sin 30° − m g = 0, so FN = 196 − 100 × 0.5 = 146 N. The rope lifts a little, so the ground pushes up less than m g.
  2. fk = 0.20 × 146 = 29.2 N. Friction uses the real normal force, not m g.
  3. Horizontal: a = (T cos 30° − fk)/m = (86.6 − 29.2)/20 = 57.4/20 = 2.9 m/s². Only the horizontal part of the tension pulls forward.
Trap: using FN = mg = 196 N when the rope pulls at an angle. Instead: the rope's upward part (T sin 30° = 50 N) lifts the sled, so FN = 146 N and friction is 29.2 N.

Example 5: a puck slides to a stop AP

A hockey puck leaves a stick at 8.0 m/s and slides on ice with μk = 0.10. How far does it go? Would a puck with twice the mass go farther?

Show solution
Read the steps as text
  1. a = −μk m g / m = −μk g = −0.98 m/s². Friction is the only horizontal force; the mass cancels.
  2. 0 = v₀² + 2aΔx → Δx = 8.0² / (2 × 0.98) = 64 / 1.96 = 33 m. Kinematics with constant acceleration.
  3. Twice the mass: same distance. Friction doubles, but so does the mass, so a is unchanged.

Practice (AP style)

1. A 25 kg box rests on a floor with μs = 0.50. A student pushes horizontally with 60 N and the box does not move. The friction force on the box is

Show answer

60 N. The box is at rest, so the net force is zero and static friction exactly matches the push. 123 N (= 0.50 × 245 N) is only the maximum static friction. 245 N is the weight.

2. A horizontal push on a box at rest increases slowly from zero until the box slides. Which describes the friction force as the push increases?

Show answer

Static friction matches the push up to μsFN. After slipping, kinetic friction μkFN is constant and (usually) smaller. Try it in animation 1.

3. A block placed on a board stays at rest until the board is tilted to 25°, when it starts to slide. The coefficient of static friction is closest to

Show answer

At the slipping angle, μs = tan θ = tan 25° = 0.47. 0.42 is sin 25°, 0.91 is cos 25°, 2.1 is 1/tan 25°.

4. Two boxes of the same material slide across the same floor with the same starting speed. Box B has twice the mass of box A. Which is true about their stopping distances?

Show answer

a = μkm g/m = μkg for both. Same acceleration and same starting speed give the same stopping distance.

5. You can pull a heavy suitcase across a floor with a strap that is horizontal or one that is angled 30° above horizontal, with the same tension. Why does the angled strap often accelerate it more?

Show answer

FN = m g − T sin θ, and friction = μkFN. The horizontal part T cos θ is actually a little smaller, but if friction drops enough, the net force is larger. μk depends only on the surfaces.

6. A block slides down a ramp at 20° at constant speed. The coefficient of kinetic friction is

Show answer

Constant speed: m g sin θ = μk m g cos θ, so μk = tan 20° = 0.36. The mass cancels. 0.34 is sin 20°.

7. A person walks forward on a sidewalk without slipping. The friction force from the ground on the person's shoe is

Show answer

The shoe pushes backward on the ground; the ground pushes the shoe forward (third law). The shoe does not slip, so it is static friction. It is the force that speeds the person up.

8. (Short free response: mathematical routines) A 2.0 kg book is pressed against a vertical wall by a horizontal force F. μs between book and wall is 0.50. (a) Draw (describe) the free body diagram. (b) Find the smallest F that keeps the book from sliding down. (c) If F is doubled, what happens to the friction force? Explain.

Show answer
  1. (a) Four forces: F (horizontal, into the wall), normal force from the wall (horizontal, away from the wall, equal to F), weight 19.6 N down, static friction up.
  2. (b) Friction must hold the weight: fs = m g = 19.6 N, and fs ≤ μsFN = 0.50 F. So F ≥ 19.6 / 0.50 = 39 N.
  3. (c) Friction stays 19.6 N. The book is still at rest, so friction only has to balance the weight. Doubling F doubles the maximum static friction, not the actual friction.

Scoring idea: 1 point for all four forces with correct directions, 1 for FN = F, 1 for fs = m g and the inequality, 1 for 39 N, 1 for "unchanged" with the reason.

9. (Experimental design) You have a wooden block, a board that can be tilted, a protractor, a meter stick and a stopwatch. Describe how to measure μs and μk between the block and the board.

Show answer
  1. μs: place the block on the board and tilt slowly until it just starts to slide. Read θslip with the protractor. μs = tan θslip. Repeat 5 times and average.
  2. μk: at a steeper angle θ, release the block from rest, measure the distance d along the board and the time t. a = 2d/t². Then μk = (g sin θ − a)/(g cos θ). Repeat and average; or use several angles and graph.
  3. Alternative for μk: find the angle where a tapped block slides at constant speed (equal times over two equal distances). Then μk = tan θ.

Scoring idea: 1 point for the slipping-angle method and tan θ, 1 for measuring a with d and t, 1 for the μk equation, 1 for repeated trials to reduce error.

10. (Short free response: translation between representations) A 10 kg box rests on a level floor (μs = 0.50, μk = 0.30). A horizontal push F grows slowly from 0 to 80 N. (a) Describe the graph of the friction force against F: give the key values. (b) Which description fits? (A) friction is 0 until the box moves; (B) friction equals F up to 49 N, then drops to 29.4 N and stays there; (C) friction is always 49 N; (D) friction grows along with F forever. (c) Calculate the acceleration when F = 80 N.

Show answer
  1. FN = mg = 98 N. Largest static friction: 0.50 × 98 = 49 N. Kinetic friction: 0.30 × 98 = 29.4 N.
  2. (a) From F = 0 to 49 N the box stays still and friction equals the push: a straight line at 45° (slope 1) from (0, 0) to (49 N, 49 N). Just past 49 N the box slips and friction drops to 29.4 N, then stays flat at 29.4 N up to 80 N.
  3. (b) (B).
  4. (c) a = (80 − 29.4) / 10 = 5.1 m/s².

Point guide (4 points):

  • 1 point: rising part where static friction equals the push.
  • 1 point: peak 49 N and drop to a flat 29.4 N.
  • 1 point: picks (B).
  • 1 point: a ≈ 5.1 m/s² using kinetic friction.

11. (Short free response: qualitative/quantitative translation) A block slides down a ramp at constant speed when the ramp is at angle θ. (a) Derive an expression for μk in terms of θ. (b) Without numbers: if the block's mass is doubled, does it still slide at constant speed? Explain using your expression. (c) Calculate μk for θ = 20°. (d) The ramp is raised to 30°. Calculate the block's acceleration.

Show answer
  1. (a) Constant speed means ΣF = 0. Along the slope: mg sin θ = μk mg cos θ, so μk = tan θ.
  2. (b) Yes. The mass cancels, so the condition depends only on θ and the surfaces. Doubling m doubles both the down-slope pull and the friction.
  3. (c) μk = tan 20° = 0.36.
  4. (d) a = g(sin 30° − μk cos 30°) = 9.8(0.500 − 0.364 × 0.866) = 1.8 m/s² down the slope.

Point guide (4 points):

  • 1 point: constant speed means zero net force along the slope.
  • 1 point: μk = tan θ.
  • 1 point: mass cancels, so yes, with the reason.
  • 1 point: 0.36 and 1.8 m/s².

Common mistakes

Using μsFN as the static friction

The mistake: writing fs = μsFN for an object at rest.

Why it is wrong: μsFN is the largest static friction can be. Usually it is smaller: just enough to keep things at rest.

How to spot it: if your "friction" is bigger than the push it opposes, the object would accelerate backward. Impossible: fix it.

Always using FN = m g

The mistake: using m g for the normal force on a slope or with an angled pull.

Why it is wrong: FN comes from the perpendicular force balance: m g cos θ on a slope, m g − T sin θ with a rope pulling up.

How to spot it: any time there is an angle, write the perpendicular equation before using μ.

Friction always opposes motion

The mistake: drawing friction backward on a walking person or on a box in an accelerating truck.

Why it is wrong: friction opposes the slipping of the surfaces. Sometimes that is forward.

How to spot it: ask "which way would the surfaces slip if there were no friction?" Friction points the other way.

Thinking friction depends on speed or contact area

The mistake: making kinetic friction bigger for faster sliding or a wider box.

Why it is wrong: in the AP model, fk = μkFN: only the surfaces and the normal force matter.

How to spot it: if speed or area appears in your friction formula, remove it.

Mixing up μs and μk

The mistake: using μs while the object slides, or μk while it is at rest.

How to spot it: first decide: are the surfaces sliding? Sliding → μk, equals. Not sliding → μs, at most.

Why it matters: using the wrong one changes whether the object moves at all.

Friction pointing the wrong way on a shoved block

The mistake: drawing friction up the slope for a block shoved up the slope.

Why it is wrong: while sliding up, friction points down the slope (it opposes the sliding), so it adds to gravity: a = g (sin θ + μk cos θ).

How to spot it: check the direction of the velocity first; kinetic friction is always opposite to it (relative to the surface).