This physics grades 11-12 lesson on friction and inclined planes notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on friction and inclined planes and check yourself with the quiz.
Quick answer: Newton’s second law states that the resultant force equals mass times acceleration, ΣF = ma. Friction opposes sliding with f = μN, and on a slope of angle θ the normal reaction is mg cos θ while mg sin θ pulls the object down the slope. For connected bodies the acceleration is shared and the tension links the equations.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB A.2
Friction and inclined planes notes: the key ideas
What does Newton’s second law say?
The resultant force on an object equals its mass multiplied by its acceleration. In symbols, ΣF = ma. The resultant is found by adding all the forces as vectors, so friction, tension, weight and the normal reaction must all be included with their directions.
ΣF = ma. ΣF is the resultant force in newtons (N), m is mass in kilograms (kg) and a is acceleration in m/s². Weight W = mg with g = 9.81 N/kg.
How does friction work?
Friction is a contact force that opposes the relative motion, or the tendency to move, between two surfaces. Static friction can adjust itself up to a limit. Once the object slides, the friction has a nearly constant value.
The coefficient of friction μ is the ratio of the friction force to the normal reaction force N. For sliding, f = μk N. For the largest value before sliding starts, fmax = μs N. Static friction is at most μs N.
| Pair of surfaces | Typical μ | Comment |
|---|---|---|
| Rubber tyre on dry road | 0.7 to 0.9 | good grip |
| Wood on wood | 0.3 to 0.5 | depends on roughness |
| Steel on steel, dry | about 0.5 | varies with polish |
| Ice on ice | about 0.03 | very slippery |
A 40 kg crate is pushed with 180 N along a floor with μk = 0.35. N = mg = 40 × 9.81 = 392 N. Friction f = 0.35 × 392 = 137 N. Resultant = 180 − 137 = 43 N. a = 43 ÷ 40 = 1.07 m/s² in the direction of the push.
How do you analyse motion on a slope?
On a slope of angle θ the weight mg is split into a component mg sin θ pulling the object down the slope and a component mg cos θ pressing it into the surface. The normal reaction balances the second component, so N = mg cos θ, which is smaller than mg.
A 5.0 kg box slides down a 30° slope with μk = 0.20. Down the slope: mg sin θ = 5.0 × 9.81 × 0.500 = 24.5 N. N = 5.0 × 9.81 × 0.866 = 42.5 N and f = 0.20 × 42.5 = 8.5 N. Resultant = 24.5 − 8.5 = 16.0 N, so a = 16.0 ÷ 5.0 = 3.2 m/s². Mass cancels: a = g(sin θ − μ cos θ).
How do you treat connected bodies?
If two bodies are joined by a taut string, they have the same acceleration. Write ΣF = ma for each body separately, then solve the two equations together. The tension in the string is the same throughout a light string over a smooth pulley.
The 2.0 kg trolley and the 0.50 kg hanging mass are on a smooth table. For the whole system: 0.50 × 9.81 = (2.0 + 0.50) × a, so a = 4.905 ÷ 2.5 = 1.96 m/s². For the trolley: T = 2.0 × 1.962 = 3.92 N, which is less than the weight of the hanging mass.
Students write N = mg on a slope and put friction as μmg. The normal reaction is mg cos θ, so friction is μmg cos θ. Also, friction opposes the motion; it does not always act backwards, it acts against the sliding direction.
Draw the free-body diagram first and choose axes along and perpendicular to the slope. Write one equation for each axis, and the algebra almost solves itself.
- ΣF = ma; weight W = mg, g = 9.81 N/kg
- Friction f ≤ μN, sliding f = μk N
- On a slope: N = mg cos θ, driving force mg sin θ
- Slope acceleration: a = g(sin θ − μk cos θ)
- Connected bodies share the acceleration; tension links the equations
Forces, friction and motion on slopes: frequently asked questions
What is the difference between static and kinetic friction?
Static friction acts when there is no sliding and can take any value up to a maximum of μs N. Kinetic friction acts when the surfaces slide and is roughly constant at μk N. Normally μs is slightly larger than μk.
Why is the normal reaction smaller on a slope?
Only the component of the weight perpendicular to the surface, mg cos θ, presses the object into the slope. The surface balances this component. The component along the slope, mg sin θ, does not press on the surface.
Does mass affect how fast an object slides down a slope?
For a given slope and coefficient of friction, no. Both the driving force and the friction are proportional to mass, so mass cancels: a = g(sin θ − μ cos θ). Heavy and light objects accelerate equally.
How do you find the tension in a string joining two bodies?
First find the acceleration of the whole system. Then apply ΣF = ma to one body only, including the tension as the unknown force. The tension is the same at both ends of a light string over a smooth pulley.
How can you measure the coefficient of static friction with a slope?
Raise one end of a board until the block just starts to slide. At that angle μs = tan θ, because the friction limit equals the component of weight along the slope. The mass of the block is not needed.
