B. Dynamics · Dynamics: Forces and Motion in a PlaneFriction: Static and Kinetic Coefficients
B2.3 B3.2 B2.1About 70 minutes
Learning goals- I can explain the difference between static and kinetic friction and why static friction can take any value up to a maximum.
- I can use Fs ≤ μs FN and Fk = μk FN to decide whether an object moves and to find its acceleration.
- I can calculate stopping distances and maximum accelerations limited by friction.
- I can explain advantages and disadvantages of static and kinetic friction in everyday technologies.
Words to know
static friction (Fs)
The friction force between two surfaces that are not sliding relative to each other. It matches the applied force up to a maximum value, μs FN.
kinetic friction (Fk)
The friction force between two surfaces sliding past each other: Fk = μk FN, roughly constant.
coefficient of static friction (μs)
The ratio of the maximum static friction to the normal force. It has no units.
coefficient of kinetic friction (μk)
The ratio of kinetic friction to the normal force. For most pairs of surfaces μk < μs.
lubricant
A substance (oil, grease, wax, a film of water) placed between surfaces to reduce friction.
Grip and slip
On an icy February morning, the difference between a safe stop and a slide into the intersection is the difference between static and kinetic friction. A tire that keeps rolling grips the road with static friction. A locked, skidding tire slides on it with kinetic friction, which is weaker. Anti-lock brakes exist because of that one fact.
In SPH3U you measured coefficients of friction and used Ff = μFN. In this lesson you will look closely at how static friction behaves, why it is "smart", and when friction helps or hurts.
Static friction is not a fixed number
Push gently on a heavy crate. It does not move, so the net force is zero: static friction exactly matches your push. Push harder, and static friction grows to match again. This continues until static friction reaches its maximum:
Fs ≤ μs FN (the equals sign holds only at the point of slipping)
Once the crate starts sliding, the friction drops to the kinetic value:
Fk = μk FN (roughly constant, whatever the speed)
Two surprising facts, to a good approximation: friction does not depend on the area of contact, and kinetic friction does not depend much on speed. It depends on the kinds of surfaces and on how hard they are pressed together (FN).
Where friction comes from
Even a polished surface is rough at a microscopic scale. Two surfaces touch only at the tips of their tiny bumps, and the real contact area is a very small fraction of what it looks like. At those points, the atoms of the two surfaces form weak bonds. Static friction is the force needed to break all those bonds at once. Once sliding begins, the bonds have less time to form, so kinetic friction is usually smaller. Pressing the surfaces together harder (a bigger FN) squashes the bumps and creates more contact points, which is why friction is proportional to FN.
| Surfaces | μs (approx.) | μk (approx.) |
|---|
| rubber on dry concrete | 1.0 | 0.8 |
| rubber on wet concrete | 0.7 | 0.5 |
| steel on steel (dry) | 0.74 | 0.57 |
| wood on wood | 0.25 to 0.5 | 0.2 |
| ice on ice | 0.1 | 0.03 |
| Teflon on Teflon | 0.04 | 0.04 |
Worked exampleExample 1: Will it move?
A 40 kg crate rests on a level floor (μs = 0.50, μk = 0.35). (a) A person pushes horizontally with 150 N. What is the friction force? (b) The push is increased to 250 N. Find the acceleration.
- FN = mg = 40 × 9.8 = 392 N. Maximum static friction = 0.50 × 392 = 196 N.
- (a) 150 N < 196 N, so the crate stays at rest. Static friction balances the push: Fs = 150 N (not 196 N!).
- (b) 250 N > 196 N, so the crate slides. Now friction is kinetic: Fk = 0.35 × 392 = 137.2 N.
- a = (250 − 137.2) ÷ 40 = 112.8 ÷ 40 = 2.8 m/s².
(a) 150 N of static friction; no motion. (b) 2.8 m/s².
Worked exampleExample 2: Rolling versus skidding
A car travels at 25 m/s on a wet road. For its tires, μs = 0.70 and μk = 0.50. Compare the stopping distance if (a) the wheels lock and skid, and (b) anti-lock brakes keep the wheels rolling at the edge of slipping.
- For any car on level ground, FN = mg, so the friction force is μmg and the deceleration is a = μg. The mass cancels.
- (a) Kinetic: a = 0.50 × 9.8 = 4.9 m/s². d = v² ÷ (2a) = 625 ÷ 9.8 = 64 m.
- (b) Static (the tire surface is momentarily at rest on the road): a = 0.70 × 9.8 = 6.86 m/s². d = 625 ÷ 13.72 = 46 m.
(a) About 64 m (b) about 46 m. Keeping the tires rolling saves about 18 m, and the driver can still steer, because a rolling tire can grip sideways too.
Worked exampleExample 3: The fastest possible start
Tires and a dry road have μs = 0.80. What is the greatest acceleration an all-wheel-drive car can have on level ground, and the least time to reach 100 km/h?
- The road pushes the car forward with static friction, at most μs FN = μs mg. So a(max) = μs g = 0.80 × 9.8 = 7.84 m/s².
- 100 km/h = 27.8 m/s. Δt = 27.8 ÷ 7.84 = 3.5 s.
7.8 m/s² and about 3.5 s. Without downforce or special tires, more engine power would only make the wheels spin.
Friction: helpful and harmful
Advantages of static friction: it lets you walk without slipping, lets tires push a car forward and steer it around curves, holds nails and screws in wood, keeps a knot tied and a belt drive turning without slipping.
Advantages of kinetic friction: brakes turn kinetic energy into thermal energy; sandpaper, files and grinding wheels shape materials; a match lights.
Disadvantages: friction wears out tires, brake pads, shoes and engine parts; it turns useful energy into waste heat (a significant share of a car engine's output goes to overcoming friction); a large static friction can make heavy objects hard to start moving.
Controlling it: lubricants (oil, grease), ball bearings and wheels reduce it; sand and grit on icy roads, winter tire treads and studs increase it. Skiers wax their skis to lower kinetic friction, and curlers sweep the ice in front of a stone to warm and polish it, so the stone slides farther and curls less.
Just for funI tried to write a joke about friction, but it kept dragging on.
Then I tried to start one about static friction, but I just could not get it moving.
Hands-on activityMeasure μs and μk with a shoe
You need: a sneaker, a luggage scale or spring scale, string, a few books or a water bottle to add mass, and two or three different surfaces (wooden floor, tile, carpet, a cutting board).
- Weigh the shoe with the scale (hang it) to find its weight, mg.
- Tie the string to the shoe. Pull horizontally and slowly increase the pull. Note the largest reading just before the shoe starts to slide: that is μs mg.
- Keep the shoe sliding at a steady speed and read the scale: that is μk mg.
- Calculate μs and μk. Repeat 3 times and average. Record in a table.
- Add mass (put a bottle in the shoe). Does μ change? (It should stay about the same while the forces grow.)
- Try different surfaces. Which shoe-surface pair grips best? Is μk less than μs every time?
- Bonus: pull a hardcover book lying flat, then on its edge (keep it upright with a light touch). Does the contact area change the friction?
Your turn
1A 20 kg box sits on a floor with μs = 0.60. What is the smallest horizontal force that will start it moving?
Fs(max) = 0.60 × 20 × 9.8 = 118 N.
2Once the box above is moving, μk = 0.45. What horizontal force keeps it moving at a constant velocity?
F = μk mg = 0.45 × 196 = 88 N.
3If you push the box (at rest) with 100 N, what is the friction force?
100 N is less than 118 N, so the box stays put and static friction is 100 N.
4If you push the moving box with 150 N, what is its acceleration?
a = (150 − 88.2) ÷ 20 = 3.1 m/s².
5A 1200 kg car skids with locked wheels from 20 m/s on a road with μk = 0.60. Find the stopping distance and time.
a = 0.60 × 9.8 = 5.88 m/s². d = 20² ÷ (2 × 5.88) = 34 m. t = 20 ÷ 5.88 = 3.4 s. (The mass did not matter.)
6Explain, in terms of surfaces, why μk is usually smaller than μs.
Surfaces touch at tiny high points where weak bonds form. When surfaces sit still, bonds have time to form at many points; when they slide, there is less time for bonds to form, so less force is needed to keep breaking them.
More practice: warm-up
1Does friction usually depend on the area of contact?
No (to a good approximation).
2Which is usually larger, μs or μk?
3What units does a coefficient of friction have?
None: it is a ratio of two forces.
4A box at rest is pushed with 30 N and does not move. What is the friction force?
30 N (static friction matches the push).
5Name one way to increase friction on an icy sidewalk.
Spread sand or grit, or wear boots with deep treads or ice cleats.
More practice: on your own
1It takes 30 N to start a 15 kg sled moving on snow and 20 N to keep it moving at a constant speed. Find μs and μk.
mg = 147 N. μs = 30 ÷ 147 = 0.20. μk = 20 ÷ 147 = 0.14.
2A puck leaves a stick at 8.0 m/s and slides 25 m before stopping. Find μk between the puck and the ice.
a = v² ÷ (2d) = 64 ÷ 50 = 1.28 m/s². a = μk g, so μk = 1.28 ÷ 9.8 = 0.13.
3For an all-wheel-drive car on a dry road with μs = 0.90, find the greatest possible acceleration and the least time to reach 100 km/h.
a = 0.90 × 9.8 = 8.8 m/s². t = 27.8 ÷ 8.82 = 3.2 s.
4You press a 1.5 kg book against a wall with a horizontal force. μs = 0.40 between the book and the wall. What is the smallest force that keeps the book from sliding down?
Your push is the normal force. Static friction (up) must hold the weight: μs F ≥ mg, so F ≥ 1.5 × 9.8 ÷ 0.40 = 37 N.
5Give two advantages and two disadvantages of friction on a bicycle.
Advantages: tire grip lets you accelerate, brake and turn; brake pads stop you by friction on the rim or disc. Disadvantages: friction in the chain and bearings wastes energy (oil reduces it); tires and brake pads wear out. (Other good answers are possible.)
Challenge
1A car travels at 30 m/s on a road with μs = 0.80 and μk = 0.60. How much shorter is the stopping distance with anti-lock brakes (static friction) than with locked wheels (kinetic friction)?
Static: d = 900 ÷ (2 × 7.84) = 57.4 m. Kinetic: d = 900 ÷ (2 × 5.88) = 76.5 m. Difference ≈ 19 m, about four car lengths.
2You push a 30 kg crate (μs = 0.50) with a force directed 30° below the horizontal. Find the smallest push that starts it moving. Then show that if the push is directed more steeply than about 63° below the horizontal, no push will move it.
At the point of slipping: F cos 30° = 0.50(294 + F sin 30°) → 0.866F − 0.250F = 147 → F = 239 N. In general F(cos θ − μs sin θ) = μs mg, which needs cos θ − μs sin θ > 0, i.e. tan θ < 1 ÷ μs = 2.0, so θ < 63.4°. Steeper than that, pushing harder only increases friction more than the push: the crate jams.
3A curling stone leaves the thrower's hand at 2.0 m/s and slides 25 m before stopping. Estimate μk between the stone and the ice. Why can sweeping change this distance?
a = v² ÷ (2d) = 4.0 ÷ 50 = 0.080 m/s². μk = a ÷ g = 0.080 ÷ 9.8 ≈ 0.008. Sweeping warms and polishes the ice in front of the stone, lowering μk, so the stone decelerates less and travels farther.
Quiz
1. A 10 kg box at rest has μs = 0.50. A horizontal push of 20 N acts on it. The friction force is:
2. Anti-lock brakes shorten stopping distances mainly because:
3. A crate slides across a floor. If its mass is doubled (same surfaces), the kinetic friction force:
4. A car skids to a stop on a level road. Its stopping distance from a given speed depends on:
5. Which is an advantage of kinetic friction?
6. A 5.0 kg block needs 20 N to start sliding on a level table. μs is about:
Big idea: Static friction adjusts to the situation up to a maximum μs FN, and kinetic friction is μk FN (B2.3). Students also weigh the advantages and disadvantages of each kind (B3.2).
How to help: Try the shoe activity together and ask your teen to predict before each pull. The moment the shoe breaks free (the scale reading drops) makes the difference between static and kinetic friction very concrete.
Common difficulties:
- Always writing Fs = μs FN, even when the object is not about to slip. That is only the maximum.
- Using μs for an object that is already sliding.
- Thinking that a bigger contact area means more friction.
- Forgetting that the normal force changes when a force acts at an angle (see the previous lesson).