The big idea: Friction is the force that resists sliding between two surfaces in contact. It always acts to oppose the motion (or the attempted motion).
There are two kinds: static friction holds a still object in place, and dynamic (sliding) friction acts once it is moving.
Static friction (not moving)
- Acts while the object is still stuck
- Grows to match whatever pushes it
- Has a maximum of μs R — beyond that it slips
- Rule: F_f ≤ μ_s R
Dynamic friction (sliding)
- Acts while the object is sliding
- Has a fixed size
- Usually smaller than the max static friction
- Rule: F_f = μ_d R
[Diagram: phys-free-body] - Available in full study mode
Spot the difference: Static = stuck (friction can be anything up to μs R). Dynamic = sliding (friction is exactly μd R).
The normal force R (also written F_N) is the surface pushing back, at a right angle to the surface.
Both friction rules are given in the data booklet. The normal force R is how hard the surface pushes back; on flat ground R is just the object's weight, R = mg.
- friction force, parallel to the surface (N)
- coefficient of STATIC friction (no unit — just a number)
- normal force — the support push, perpendicular to the surface (N). Also written F_N.
- friction force while sliding, parallel to the surface (N)
- coefficient of DYNAMIC (kinetic) friction (no unit — just a number)
- normal force — the support push, perpendicular to the surface (N). Also written F_N.
Why is μ just a number?: Rearrange Ff = μR to μ = F_f ÷ R — that's a force ÷ a force. The newtons on top and bottom cancel, so μ has no unit: it is dimensionless (a pure number, usually between 0 and 1).
[Diagram: phys-formula-triangle] - Available in full study mode
Worked example — friction while sliding
A crate of mass 20 kg is dragged across a flat floor at steady speed. The coefficient of dynamic friction is μd = 0.30. Take g = 9.8 m s⁻². Find the friction force. (On flat ground the normal force R = mg.)
Solution
- First find the normal force R — on flat ground it equals the weight:
- Start with the given formula for sliding friction:
- Put in the numbers (μd = 0.30, R = 196):
- Work it out — keep the unit:
Final answer
friction force = 59 N (to 2 s.f.).
Practice with real exam questions
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How this is tested: Friction shows up in force and motion questions.
- Paper 1A: quick concept — e.g. outline why μ is dimensionless, or static vs dynamic. - Paper 1B / Paper 2: the classic 'show that' — find the minimum force to start an object moving.
Classic trap: to start motion you must beat the largest static friction (μ_s R), not the smaller sliding friction (μd R). Use μ_s, not μd.
Minimum force to start moving: An object stays still until your push beats the biggest static friction it can muster. So the minimum force to start it sliding equals μ_s R — the maximum value of Ff ≤ μs R.
IB-style question — minimum force to start a box
A box of mass 1.2 kg sits on a flat bench. The coefficient of static friction between the box and bench is μs = 0.35. Take g = 9.8 m s⁻². Show that the minimum horizontal force needed to start the box moving is about 4 N.
Solution
- On the flat bench the normal force R equals the weight:
- To start moving, the push must reach the largest static friction (the maximum of the given rule Ff ≤ μs R):
- Put in the numbers (μs = 0.35, R = 11.76):
- Work it out — keep the unit:
Final answer
minimum force ≈ 4 N — and it must use μs (static), because that is the friction you have to overcome to get it moving.