Calculate static and kinetic friction from coefficients and normal force, and use friction in constant-velocity, accelerating and stopping problems

Learning goal: Explain the origin of friction in the electric forces between the atoms of two surfaces; use F_k = μ_k F_N for kinetic friction and F_s ≤ μ_s F_N with F_s,max = μ_s F_N for static friction; decide whether an object slides; draw free-body diagrams and solve problems in which friction acts on an object at constant velocity, on an accelerating object, and as the only force slowing an object to a stop; and design an investigation to measure a coefficient of friction.

Where friction comes from. Two surfaces that look smooth are rough at the atomic scale, and where their high points touch, the atoms of one surface exert electric forces on the atoms of the other. Friction, the normal force and every other contact force are, at bottom, electric forces between charged particles in the two objects. Pressing the surfaces together harder brings more atoms into contact, so friction grows with the normal force, the force pressing the surfaces together perpendicular to them. In the simple model used here, friction depends on the materials and on the normal force, and on nothing else: not the contact area, and not (for ordinary speeds) the sliding speed. Those two claims are worth testing rather than believing, and the investigation below does so.

Kinetic friction. When two surfaces slide past each other, the friction force on each surface opposes the sliding and has magnitude

F_k = μ_k F_N,

where μ_k is the coefficient of kinetic friction, a number without units that depends on the pair of materials, and F_N is the normal force. On a level surface with no other vertical forces, F_N equals the weight, mg; but if something pushes down or pulls up on the object, or the surface is tilted, F_N is different and must be found from the free-body diagram first. Kinetic friction acts opposite to the object's velocity relative to the surface.

Static friction. Sometimes friction prevents sliding altogether: a crate pushed gently stays put. Static friction adjusts to match the applied force, up to a maximum:

F_s ≤ μ_s F_N, with F_s,max = μ_s F_N.

As long as the other forces along the surface add to less than or equal to F_s,max, the object does not slide; for an object at rest, such as the crate below, the static friction force is exactly what is needed to make the net force along the surface zero, in the direction opposite the applied push. (Static friction can also accelerate things without slipping, as when it turns a car on a curve.) If the applied force exceeds F_s,max, the object starts to slide and the friction becomes kinetic, with μ_k usually smaller than μ_s. That is why a crate needs a harder shove to start than to keep moving, and why a car's wheels grip best when they roll without skidding. The most common error is writing F_s = μ_s F_N for an object at rest: static friction equals its maximum only at the instant sliding is about to begin.

A fully worked model. A 20 kg crate rests on a floor with μ_s = 0.50 and μ_k = 0.35 (values chosen for the example). Weight 196 N; normal force 196 N. Maximum static friction: 0.50 × 196 = 98 N. Push with 80 N: since 80 N < 98 N, the crate does not move, and the static friction is 80 N backward, not 98 N; the net force is zero. Push with 110 N: 110 N > 98 N, the crate slides, and friction becomes kinetic, 0.35 × 196 = 68.6 N. Net force 110 − 68.6 = 41.4 N; acceleration 41.4 ÷ 20 = 2.1 m/s². To keep the crate sliding at constant velocity, the push need only equal the kinetic friction, 68.6 N; a push of 110 N keeps accelerating it.

Friction as the only horizontal force. A hockey puck of mass 0.16 kg slides at 8.0 m/s across ice with μ_k = 0.10. Only friction acts horizontally: F_k = 0.10 × 0.16 × 9.8 = 0.157 N, and the acceleration is −0.157 ÷ 0.16 = −0.98 m/s². Notice that a = −μ_k g regardless of the mass, because both the friction and the inertia scale with m. Stopping distance from v_f² = v_i² + 2aΔx: Δx = 64 ÷ (2 × 0.98) = 33 m; stopping time from v_f = v_i + at: 8.0 ÷ 0.98 = 8.2 s. A heavier puck at the same speed would stop in the same distance.

Braking and road conditions. For a car whose tires roll without skidding, the friction that slows it is static friction, up to μ_s F_N; if the wheels lock and skid, it drops to kinetic. Suppose (as example values, not measured data) that a road gives μ = 0.70 when dry, 0.40 when wet and 0.10 when icy. From 25 m/s, the maximum braking acceleration is μg: 6.9, 3.9 and 0.98 m/s², giving braking distances of 25² ÷ (2μg) = 46 m, 80 m and 320 m. Add the distance covered during a 1.0 s reaction time, 25 m, to each. Weather does not change the car's mass or speed; it changes the coefficient, and the stopping distance scales with 1/μ.

Investigation: measure μ_k and μ_s. Materials: a wooden block with a hook, a spring scale in newtons, slotted masses, a level table, a second surface (sandpaper, cloth), and a board that can be tilted with a protractor. Kinetic: pull the block along the table at a steady speed, as steadily as you can, and read the scale; the pull equals the kinetic friction because the net force is zero at constant velocity. Add 0.50 kg at a time to change the normal force and repeat. Record: normal force (block plus added mass, times 9.8) and friction. Analyze: plot friction against normal force; the slope is μ_k. For example, normal forces of 4.9, 9.8, 14.7 and 19.6 N with frictions of 1.5, 2.9, 4.4 and 5.9 N give a line of slope about 0.30. Static: with the block at rest, increase the pull slowly until it just slips and record the largest reading; F_s,max ÷ F_N gives μ_s, typically larger than μ_k. Then test the two claims: stand the block on its narrow face (smaller area, same normal force) and repeat; pull at a clearly faster steady speed and repeat. A result that supports the model: the same friction within measurement scatter for both areas and both speeds. Compare your coefficients with other groups using the same materials and discuss the spread: scale readings jitter while pulling, the speed is never quite constant, and surfaces pick up dust. Safety: keep the masses from falling off the block and off the table edge. What the model cannot show: a tilted-board method also measures μ_s (the block slips at the angle whose tangent equals μ_s), but explaining why needs the force components taught in the two-dimension lesson; and real tires and very rough surfaces do show some dependence on speed and area that this simple model omits.

Worked example 1 — constant velocity. A 12 kg box is dragged across a level floor at constant velocity by a horizontal rope. The coefficient of kinetic friction is 0.25. Free-body diagram: weight 117.6 N down, normal force 117.6 N up, rope tension forward, kinetic friction backward. Constant velocity means zero net force, so the tension equals the friction: 0.25 × 117.6 = 29 N. If the rope pulled with 40 N, the box would accelerate at (40 − 29.4) ÷ 12 = 0.88 m/s².

Worked example 2 — does it slide? A 5.0 kg book rests on a desk with μ_s = 0.40. A student pushes horizontally with 15 N. Maximum static friction is 0.40 × 49 = 19.6 N. Because 15 N is less than 19.6 N, the book stays put, and the static friction is 15 N opposite the push. A push of 20 N would exceed the maximum and start the book sliding; from then on the friction would be kinetic and smaller.

Worked example 3 — a sled coasting to a stop. A 30 kg sled and rider slide onto level snow at 6.0 m/s where μ_k = 0.15. Friction is the only horizontal force: F_k = 0.15 × 294 = 44.1 N, a = −44.1 ÷ 30 = −1.47 m/s². Stopping distance: Δx = 6.0² ÷ (2 × 1.47) = 12 m. Stopping time: 6.0 ÷ 1.47 = 4.1 s. The mass cancels: a = −μ_k g for any sled on that snow.

Worked example 4 — accelerating with a downward push. A 10 kg crate is pushed along a floor (μ_k = 0.30) by a force of 60 N directed horizontally, while a second student presses straight down on it with 40 N. Normal force: the floor must support the weight and the extra push, 98 + 40 = 138 N. Friction: 0.30 × 138 = 41.4 N. Net horizontal force: 60 − 41.4 = 18.6 N; acceleration 1.9 m/s². Without the downward press the friction would be 29.4 N and the acceleration 3.1 m/s²: pressing down on a sliding object makes it harder to move.

Representation task: three free-body diagrams with numbers. For the 20 kg crate in the lesson (μ_s = 0.50, μ_k = 0.35), draw and fully label free-body diagrams for three cases: pushed with 80 N and at rest, pushed with 110 N and accelerating, and pushed with 68.6 N and sliding at constant velocity. Give the size of every force (weight 196 N, normal 196 N, push, friction) and the net force, and under each diagram write one sentence stating whether the friction is static or kinetic and why.

Independent performance: a stopping-distance report. Carry out the friction investigation to find μ_k for one pair of surfaces, then use it to predict the stopping distance of the block launched across the same surface at a measured speed (time it over 0.50 m with photogates or video). Test the prediction with three launches. Success criteria: the friction vs. normal force graph has at least four points with a fitted line and its slope reported to two significant figures; the free-body diagram for the coasting block shows friction as the only horizontal force; the predicted stopping distance is derived from v² = 2μ_k g Δx with the algebra shown; measured distances are compared as a percentage; and the report names two sources of uncertainty and the direction in which each one shifts the result.