This AP Physics 1 lesson on friction and inclines notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on friction and inclines and check yourself with the quiz.
Quick answer: Friction is the contact force that resists sliding. Static friction rises to a maximum of μsN, and kinetic friction equals μkN. On an incline the normal force is mg cos θ, and connected objects share one acceleration, so a single net-force equation per object solves the tension.
Level: AP Physics 1 (algebra-based) · Curriculum: AP P1 Unit 2
Friction and inclines notes: the key ideas
What is friction and how is it calculated?
Friction is the contact force parallel to a surface that resists sliding between two surfaces. It acts opposite to the way the surfaces would slide or are sliding, and it depends on how hard the surfaces are pressed together, not on how large the contact area is.
Static friction acts between surfaces that are not sliding; its size adjusts up to a maximum fs,max = μsN. Kinetic friction acts between sliding surfaces and has the fixed size fk = μkN. The coefficient μ has no unit, and N is the normal force.
| Situation | Friction force | Typical μ |
|---|---|---|
| rubber on dry concrete | μs ≈ 0.8 | 0.6 to 1.0 |
| wood on wood | μs ≈ 0.4 | 0.25 to 0.5 |
| steel on ice | μk ≈ 0.03 | 0.02 to 0.05 |
Wanjiru pushes a 15 kg crate across a floor with μs = 0.50 and μk = 0.40. N = mg = 147 N.
Maximum static friction = 0.50 × 147 = 73.6 N. At a push of 50 N the crate does not move and the friction is 50 N.
At 80 N it slides: fk = 0.40 × 147 = 58.9 N, so a = (80 − 58.9) ÷ 15 = 1.4 m/s².
How do you solve an incline problem?
Tilt the axes: x along the slope, y perpendicular to it. Only the weight needs resolving: its component along the slope is mg sin θ and into the slope is mg cos θ. There is no motion perpendicular to the surface, so N = mg cos θ.
a = g (sin θ − μk cos θ) for sliding down
a = acceleration down the slope (m/s²); g = 9.81 m/s²; θ = slope angle; μk = kinetic coefficient. A block just starts to slide when tan θ = μs.
A 10 kg box slides down a 30° ramp with μk = 0.20.
a = 9.81 × (sin 30° − 0.20 × cos 30°) = 9.81 × (0.500 − 0.173) = 3.2 m/s².
Friction = 0.20 × (10 × 9.81 × cos 30°) = 17 N up the slope.
How do you analyse connected objects and springs?
Objects joined by a light string share the same acceleration size and, over a frictionless pulley, the same tension. Treat the whole system with one equation for the total mass, then isolate one object to find the tension. For a spring the force is F = kx, where k is the spring constant (N/m) and x the stretch or compression from the natural length (m).
In an Atwood machine m1 = 2.0 kg and m2 = 3.0 kg.
System: (m2 − m1)g = (m1 + m2)a, so a = 1.0 × 9.81 ÷ 5.0 = 1.96 m/s².
Light mass: T − m1g = m1a, so T = 2.0 × (9.81 + 1.96) = 23.5 N.
Students often use the weight mg instead of mg cos θ in f = μN on a slope, or assume the tension equals the weight of a hanging mass. The tension equals the weight only when the mass is not accelerating.
For connected objects, write one sentence per object: “net force on this object = its mass × a”. Add the equations so the tension disappears, then put the answer back to find T.
- Static friction: f ≤ μsN; kinetic friction: f = μkN.
- On a slope: N = mg cos θ and the weight component along it is mg sin θ.
- Sliding down: a = g (sin θ − μk cos θ).
- Atwood machine: a = (m2 − m1)g ÷ (m1 + m2).
- Spring force: F = kx, with k in N/m.
Friction, slopes, pulleys and springs: frequently asked questions
Does friction depend on the contact area?
In the simple model used in AP Physics 1, no. Friction depends on the normal force and on the surfaces through the coefficient μ, not on how large the area of contact is.
Why is the normal force less than mg on a slope?
The surface only has to balance the part of the weight pressing into it. That is the component mg cos θ, which gets smaller as the slope gets steeper.
How do you find the tension in a pulley problem?
Apply ΣF = ma to the whole system to find the acceleration, then apply it to one mass. For a hanging mass the tension is T = m(g − a) when it accelerates downward.
What does the spring constant tell you?
The spring constant k in N/m is the force needed per metre of stretch. A stiffer spring has a larger k, and F = kx holds while the spring stays within its elastic limit.
How can you measure μs with a tilting board?
Raise the board slowly until the block just slides and measure the angle θ. At that point μs = tan θ, because the weight component along the slope equals the maximum static friction.
