Lessons › 2 Mechanics › 2.3 ForcesPrevious lessonNext lesson

2.3.4 Friction

By the end of this lesson you can:

  • Distinguish static and kinetic friction
  • Use the coefficient of friction to calculate the friction force
  • Solve friction problems on flat surfaces and slopes
Find your course: NSW Year 11: Forces and motionNSW Year 11: Motion on inclined planesNSW Year 12QLD Unit 2: Linear motion and forceVIC Unit 2: MotionVIC Unit 3: Motion in two dimensionsAP Physics 1: Unit 2IB: A.2 Forces and momentum
🎥 Watch📝 Practise🕹️ Explore🎯 Exam ready🍎 Teachers
🎥Watch
Learn the idea: a short explanation, then a check.
So far, we have only examined explicitly applied contact forces acting on objects.

In reality, whenever two surfaces come into contact, microscopic irregularities called asperities interlock and resist sliding. This gives rise to a contact force parallel to the surface: friction.

Simply put, friction is the force that opposes the relative motion (or attempted motion) between contacting surfaces.

We categorize friction into two primary regimes:

  • Static Friction (fs): Prevents an object from starting to move (fs ≤ μsFN).
  • Kinetic Friction (fk): Opposes an object already in motion (fk = μkFN).

The magnitude of friction depends directly on the normal force (FN) pressing the surfaces together and the coefficient of friction (μ), which quantifies the physical roughness and material properties of the contacting surfaces.

Theory
​In this video we examine the principles of what friction is, the types of friction, and the mathematical models were use to calculate fictional forces.
📝Practise
Worked examples first, then problems to try, then worksheets.
1. A boulder of mass 45 kg is pushed on a surface with a coefficient of sliding friction of 0.85. What force has to be applied to produce an acceleration of 0.20 m/s2?

Given: m = 45 kg, μk = 0.85, a = 0.20 m/s2, g = 9.8 m/s2

Step 1 (Calculate Normal Force): FN = mg = 45 × 9.8 = 441 N

Step 2 (Calculate Friction Force): fk = μkFN = 0.85 × 441 = 374.85 N

Step 3 (Apply Newton's Second Law): Fnet = Fapp - fk = ma

Fapp = ma + fk = (45 × 0.20) + 374.85 = 9 + 374.85 = 383.85 N

Answer: 380 N (or ~384 N)
2. A horizontal force of 30 N is required to slide a 12 kg wooden crate across the floor at a constant velocity. What is the coefficient of kinetic friction between the crate and the floor?

Given: Fapp = 30 N, m = 12 kg, a = 0 m/s2 (constant velocity), g = 9.8 m/s2

Step 1 (Determine Friction Force): Since a = 0, Fapp = fk = 30 N

Step 2 (Calculate Normal Force): FN = mg = 12 × 9.8 = 117.6 N

Step 3 (Calculate Coefficient of Friction): μk = fk / FN

μk = 30 / 117.6 ≈ 0.255

Answer: 0.255
3. A sled is travelling at 4.00 m/s along a horizontal stretch of snow. The coefficient of kinetic friction is μk = 0.0500. How far does the sled go before stopping?

Given: vi = 4.00 m/s, vf = 0 m/s, μk = 0.0500, g = 9.8 m/s2

Step 1 (Find Deceleration): a = -μkg = -(0.0500 × 9.8) = -0.49 m/s2

Step 2 (Calculate Stopping Distance): vf2 = vi2 + 2ad

0 = (4.00)2 + 2(-0.49)d → 0 = 16 - 0.98d → d = 16 / 0.98 ≈ 16.33 m

Answer: 16.3 m
Want further practice? Here is another worksheet involving force and friction.
🕹️Explore
Test your predictions in a simulation.

Traditionally, forces have been categorized as either contact forces (where two objects physically touch) or non-contact forces (action-at-a-distance forces like gravity, magnetism, and electrostatic attraction).


At the atomic level, however, "true contact" never actually occurs. Electromagnetic repulsion between the outer electron clouds of opposing atoms prevents surfaces from touching. The resistance we perceive as friction is driven by these electrostatic repulsive forces combined with microscopic surface irregularities. See the animation below

Note: Because atomic-scale electromagnetic interactions are highly complex, macroscopic physics models continue to treat forces as contact or non-contact for practical analysis.