EASA / CAR Part-66 · Module 2 · 2.2 Mechanics · Cat A

Friction — Static & Kinetic

Objective: calculate friction as μN, find the force needed to break a load free, and see why friction rises to match a small push but then drops the instant the load starts to slide.

Block on a surface, all four forcesdrag to rotate

Friction against applied force

Applied force
0N
Break-free force μs N
0N
Friction now
0N
State
0
Ffriction = μ × N  ·  N = mg  ·  it breaks free once the push exceeds μsmg  ·  here μk = 0.8 μs

Held by friction

Static friction is matching the push exactly, so the block does not move.

Friction resists sliding. It depends on the pair of surfaces, through the coefficient μ, and on how hard they are pressed together, through the normal force N — here simply the weight, mg. Notice what it does not depend on: the contact area. A block on its side and the same block on its end need the same force to move.

While the push is small, static friction matches it exactly and nothing moves; friction is a reaction, not a fixed number. It can only grow up to a limit of μsN. Push past that limit and the block breaks free, and friction immediately drops to the lower kinetic value μkN — which is why a stuck component lurches once it finally moves. This simulation takes μk as 0.8 μs, a typical ratio.

In maintenance, friction is wanted in some places and unwanted in others. It is what makes a torqued bolt hold, what makes brakes and clutches work, and what tyres need on a wet runway. It is also what wears bearings and wastes engine power as heat, which is why lubrication lowers μ wherever motion is meant to be free.