Lesson 6: Friction, Inclines, and Free-Body DiagramsAP

Unit 2's hardest-working ideas are friction and the free-body diagram (FBD), because they force precision: which forces exist, on which object, in which direction, and how large. This lesson turns the FBD from a sketch into a solving machine, then puts it to work on the two classic AP set-ups — inclines and Atwood machines.

Friction: one interaction, two behaviours

Friction is the contact force parallel to a surface, arising from microscopic interactions. Its direction always opposes relative sliding — or the sliding that would occur without it.

• Static friction acts when there is no sliding. It is adjustable: it matches the applied tendency up to a maximum f_s,max = μ_s N. Writing f_s = μ_s N as an equality when nothing is on the verge of slipping is a classic error.
• Kinetic friction acts during sliding with the steadier magnitude f_k = μ_k N.

Usually μ_k < μ_s, which is why a sliding box is easier to keep moving than to start, and why anti-lock brakes keep wheels rolling — static friction at the contact patch can be larger than kinetic.

The coefficients depend on the pair of surfaces and (to a good approximation in this course) not on contact area or speed. Notice that friction depends on the normal force, and the normal force is whatever it must be to satisfy ΣF⊥ = 0 for the surface direction — it is not automatically mg.

Free-body diagrams: the method

• Isolate one object. Draw it as a dot or box.
• Add every force exerted on it by something else: gravity (always), normal force (any touching surface), friction (parallel to a touching surface), tension (any rope, pulling away), applied pushes/pulls.
• Choose axes along and perpendicular to the expected acceleration; resolve any angled force into components.
• Write ΣF_x = ma_x and ΣF_y = ma_y. Solve.

Inclines: tilt the axes

On a slope of angle θ, gravity splits into mg sinθ down the slope and mg cosθ into the surface. The normal force is therefore N = mg cosθ (when nothing else presses on the object), and friction magnitude caps at μN. A block slides down unaided only if the driving component beats maximum static friction: tanθ > μ_s. The angle at which slipping just begins (the angle of repose) satisfies μ_s = tanθ — a measurement method, not just a formula.

Atwood machines: coupled motion

Two masses over a light, frictionless pulley share one rope, so they share one acceleration magnitude; the tension is the same throughout the rope (ideal pulley). Choosing each mass's downward/upward positives carefully and adding the two ΣF = ma equations eliminates T: a = (m_2 − m_1)g/(m_1 + m_2) for m_2 > m_1. Choosing the two masses plus rope as one system gives the same result in one line, because tension becomes internal.

Worked example 1 — block on an incline

A 2.0 kg block slides down a 30° incline with μ_k = 0.20 (g = 10 m/s²). Find its acceleration. Given: θ = 30°, forces along slope: mg sinθ down, f_k = μ_k mg cosθ up.

N = mg cosθ = 2.0 × 10 × 0.866 = 17.3 N
f_k = 0.20 × 17.3 = 3.46 N
ΣF_along = mg sinθ − f_k = 20 × 0.5 − 3.46 = 10 − 3.46 = 6.54 N
a = 6.54 / 2.0 ≈ 3.3 m/s² down the slope

Check: with friction removed the result would be g sinθ = 5.0 m/s²; friction reduces it to 3.3 m/s² — direction and size both sensible. Equivalently a = g(sinθ − μ_k cosθ).

Worked example 2 — Atwood machine

Masses of 3.0 kg and 5.0 kg hang from a light string over a frictionless pulley (g = 10 m/s²). Find (a) the acceleration, (b) the tension. Given: m_1 = 3.0 kg (rises), m_2 = 5.0 kg (falls).

(a) a = (m_2 − m_1)g / (m_1 + m_2) = (2.0 × 10)/8.0 = 2.5 m/s²
(b) For m_1: T − m_1 g = m_1 a → T = 3.0(10 + 2.5) = 37.5 N

Check: the tension (37.5 N) lies between the two weights (30 N and 50 N), as it must — the lighter mass needs a net upward force, the heavier a net downward one. Both masses get equal-magnitude accelerations from the same T, confirming consistency.

Worked example 3 — will it slip, and how hard does friction push?

A 5.0 kg box sits on a horizontal floor with μ_s = 0.40, μ_k = 0.25 (g = 10 m/s²). A horizontal push of 15 N is applied. Find the friction force and the acceleration. Method: compare the push with maximum static friction first.

f_s,max = μ_s N = 0.40 × 50 = 20 N
Push (15 N) < f_s,max (20 N) → the box does not move.
Friction = 15 N (it matches the push), a = 0.

Check / trap avoided: blindly writing f = μ_s N = 20 N would predict the box accelerating backwards under friction alone — an absurdity that reveals the error. Had the push been 25 N, the box would break free and kinetic friction (12.5 N) would apply, giving a = (25 − 12.5)/5.0 = 2.5 m/s².

Real-world application: tyres, brakes and road design

A car's maximum braking force is static friction between tyre and road (the contact patch is instantaneously at rest), so braking distances are governed by μ_s; locked, skidding wheels drop to μ_k, lengthening the stop and removing steering — precisely what ABS prevents. Road crews bank and roughen curves and post rain advisories because water can halve μ. Cleats, climbing-shoe rubber and chalk all exist to raise μ_s for the surface pair involved.

Experimental design note

Measure μ_s with nothing but an incline and a protractor: raise the slope until the block just begins to slide; μ_s = tanθ at that angle. Repeat with different masses to test the model's prediction that μ_s is mass-independent, and average several trials because "just begins" is a judgement call — naming that uncertainty and how repeats reduce it is exactly what experimental-design scoring rewards.

Common mistakes (from College Board scoring commentary)

• Writing friction = μN as an equality for static situations below the maximum.
• Using N = mg on inclines (it is mg cosθ) or when other vertical forces act.
• Drawing friction in the direction of motion rather than opposing the (relative) sliding.
• In Atwood problems, giving the two masses different tensions or different acceleration magnitudes.

AP Exam Experience

MCQ 1. A block rests on an incline. As the incline angle is slowly increased, the static friction force on the block:

• (A) stays zero until slipping
• (B) stays equal to μ_s N
• (C) increases, matching mg sinθ, until slipping begins
• (D) decreases as the normal force decreases

Answer: (C). Static friction is adjustable — equilibrium along the slope requires f_s = mg sinθ, which grows with θ until it reaches its cap μ_s N and slipping starts.

MCQ 2. A 4.0 kg block slides down a 37° incline at constant velocity (sin 37° ≈ 0.6, cos 37° ≈ 0.8). The coefficient of kinetic friction is:

• (A) 0.60
• (B) 0.75
• (C) 0.80
• (D) 1.33

Answer: (B). Constant velocity means zero net force: mg sinθ = μ_k mg cosθ, so μ_k = tanθ = 0.6/0.8 = 0.75.

MCQ 3. In an Atwood machine with masses 2.0 kg and 6.0 kg (g = 10 m/s²), the tension in the string is:

• (A) 20 N
• (B) 30 N
• (C) 40 N
• (D) 60 N

Answer: (B). a = (6 − 2)(10)/(8) = 5.0 m/s². For the 2.0 kg mass: T − 20 = 2.0 × 5.0, so T = 30 N — between the two weights (20 N, 60 N), as it must be.

FRQ. A 3.0 kg block sits on a 37° incline connected by a light string over a pulley at the top to a hanging 4.0 kg mass. The coefficients of friction between block and incline are μ_k = 0.25 and μ_s = 0.40 (sin 37° ≈ 0.6, cos 37° ≈ 0.8, g = 10 m/s²). (a) Determine whether the system moves when released, and if so, in which direction. Justify your answer numerically. (b) Calculate the acceleration of the system. (c) Calculate the tension in the string. (d) Explain how the acceleration would change if the incline were made frictionless. Scoring guideline. (a) 2 points: downslope component of the block's weight = 30 × 0.6 = 18 N; hanging weight = 40 N; maximum static friction = μ_s N = 0.40 × (30 × 0.8) = 9.6 N. The imbalance (40 − 18 = 22 N) exceeds 9.6 N, so the system moves: block up the slope, hanging mass downward (1 point for the comparison, 1 for the direction). (b) 2 points: kinetic friction f_k = 0.25 × 24 = 6.0 N opposes motion (acts down the slope); treating the system as a whole, net driving force = 40 − 18 − 6 = 16 N over total mass 7.0 kg, so a ≈ 2.3 m/s² (1 point method, 1 point value). (c) 1 point: for the hanging mass, 40 − T = 4.0 × 2.3, giving T ≈ 31 N (accept 30–32 N). (d) 1 point: without friction the net driving force rises to 22 N, so a = 22/7 ≈ 3.1 m/s² — larger, in the same direction.

Worked example 4 — stopping distance from friction alone

A box sliding at 8.0 m/s on a floor with μ_k = 0.40 stops under friction alone (g = 10 m/s²). Find (a) the stopping distance, (b) the stopping time.

Friction is the only horizontal force: a = −f_k/m = −μ_k g = −4.0 m/s² (mass cancels!)
(a) v² = v_0² + 2aΔx → 0 = 64 − 8Δx → Δx = 8.0 m
(b) t = (0 − 8.0)/(−4.0) = 2.0 s

Check: the mass cancellation is the physics headline — doubling the mass doubles both the friction force and the inertia, so any mass stops in the same 8.0 m at this μ_k. This is why braking-distance tables are published per speed, not per car model (tyre compound and road state change μ, not m).

Worked example 5 — modified Atwood: mass on a table

A 2.0 kg block on a frictionless table is tied over a pulley to a hanging 1.0 kg mass (g = 10 m/s²). Find the acceleration and tension.

System view: driving force = hanging weight = 10 N; total mass = 3.0 kg
a = 10/3.0 ≈ 3.3 m/s²
For the hanging mass: 10 − T = 1.0 × 3.3 → T ≈ 6.7 N

Check: T (6.7 N) is less than the hanging weight (10 N) because the mass accelerates downward; applied to the table block, T = 6.7 N gives a = 6.7/2.0 ≈ 3.3 m/s² — consistent from both sides of the pulley ✓.