AP PHYSICS C: MECHANICS › UNIT 2, FORCE AND TRANSLATIONAL DYNAMICS › TOPIC 2.7
Kinetic and Static Friction
Two forces sharing one name. One has a formula and a fixed size; the other has an inequality and takes whatever value it needs. Confusing them is the most expensive mistake in this unit.
2.7.A Describe kinetic friction between two surfaces and calculate its size from the normal force.
2.7.B Describe static friction, state the inequality it obeys, and find the maximum value it can reach.
The whole topic on one sheet, the same one handed out in class. Click it to open the full-size version, which prints cleanly on a single page.
1. Two different forces with one name
Friction is a contact force exerted by a surface, parallel to that surface, and it comes in two kinds that behave completely differently.
Kinetic friction acts when the surfaces are sliding past each other. It always opposes the relative sliding, and its size is fixed by the surfaces and how hard they are pressed together:
Ff,k = μk Fn
Static friction acts when the surfaces are not sliding. It has no formula, because it does not have a fixed value. It takes whatever size and direction are needed to prevent sliding, up to a limit:
Ff,s ≤ μs Fnand the largest it can be is Ff,s,max = μsFn
Pull harder and harder on a block that is not moving. Watch the friction arrow copy you exactly, and watch the graph: a line at forty five degrees, then a cliff. The block does not slide until you pass the top of that line.
2. Why it drops when the block breaks loose
For a given pair of surfaces the static coefficient is usually larger than the kinetic one. So the force needed to start something moving is bigger than the force needed to keep it moving, and the instant it breaks loose the friction drops while your pull does not. The leftover becomes net force, and the block lurches.
Everyone has felt this pushing furniture. It is also why antilock brakes exist: a rolling tire grips through static friction, a skidding tire only has kinetic friction available, and the skidding one is worse.
3. What friction does not depend on
The area of contact. A brick on its side and the same brick on its end have the same friction, because the normal force is the same and only that appears in the formula. Wider tires help for reasons that are outside this model.
The mass, directly. Mass enters only through the normal force. Put the block on an incline, in an elevator, or under someone’s hand and the normal force changes without the mass changing at all, and the friction follows the normal force every time.
So the honest sequence is always: find the normal force first, from the perpendicular direction of the diagram, and only then compute friction.
4. Tilt it until it goes
Put a block on a surface and slowly raise one end. Two things change at once. The component of the weight along the slope, which is what friction has to resist, grows as mg sin θ. The normal force, which sets how much friction is available, shrinks as mg cos θ. They cross at exactly one angle, and there the block lets go:
mg sin θ = μs mg cos θso tan θ = μs
The mass cancels. A heavy block and a light block of the same material slide at the same tilt, which makes this the standard way of measuring a coefficient of static friction with nothing but a protractor.
Raise the ramp and watch the two friction readouts approach each other. The last one predicts where they meet. Change the roughness and the prediction moves, but the mass is not on the list of things that matter, and changing it would not move the angle at all.
5. Reading it wrong
Using μsFn as the static friction. That is the maximum, not the value. A block sitting on level ground with nobody pushing it has zero static friction on it, not μsmg.
Assuming the normal force equals the weight. Only on level ground with no vertical pull or push. On an incline it is mg cos θ, in an elevator it is m(g + a), and under a rope at an angle it is something else again.
Assuming friction always opposes motion. It opposes relative sliding of the surfaces. Static friction from the road is what pushes a car forward, in the direction it is going, and it is what stops your shoe sliding backward when you walk.
Check yourself
1. A 20 kg crate sits on level ground with μs = 0.6 and μk = 0.4. Nobody is pushing it. State the friction force on it, then state it again while somebody pushes with 80 N.
With nobody pushing, the friction is zero: there is nothing for it to oppose. With 80 N applied, the maximum static friction is 0.6 × 196 = 118 N, which is more than 80, so the crate does not move and friction is 80 N backward, matching the push exactly.
2. The same crate is pushed with 140 N. Find the friction force and the acceleration.
140 N exceeds the 118 N maximum, so it slides and friction switches to kinetic: 0.4 × 196 = 78.4 N. Net force 140 − 78.4 = 61.6 N, so a = 3.08 m/s². Using the static coefficient here is the usual error and gives a crate that never moves.
3. A block just begins to slide when a ramp is tilted to 27°. Find the coefficient of static friction, and say what a heavier block of the same material would do.
μs = tan 27° = 0.51. A heavier block slides at the same angle, because the mass appears on both sides of the comparison and cancels. That is what makes this a usable measurement: you do not have to know the mass, and you do not have to measure any force.
4. Explain why a car with locked, skidding wheels takes longer to stop than one whose wheels are still rolling, in terms of the two coefficients.
A rolling tire is not sliding across the road at its contact point, so the road can exert static friction, up to μsFn. A skidding tire is sliding, so only kinetic friction is available, and μk is smaller. Less backward force means a smaller deceleration and a longer stop, which is exactly what antilock braking is built to prevent.
5. A book is pressed against a vertical wall by a horizontal push and does not slide down. Name every force on the book, say which one balances the weight, and explain what happens as the push is reduced.
Four forces: the weight down, the normal force from the wall pointing horizontally out of the wall, the push horizontally into the wall, and static friction upward along the wall. Friction balances the weight, not the normal force. Reducing the push reduces the normal force, which reduces the maximum friction available, and when μsFn falls below the weight the book slides. Note that the friction here points upward, which is neither along nor against any horizontal motion.
Topic 2.8, Spring Forces. One more contact force, and the first one whose size depends on where the object is rather than on what is pushing it.