On the kart: brake pads on the rotor, clutch shoes on the drum, tires on the track, the chain on the sprocket teeth, every nut that stays tight. Friction is where force multipliers end: all the leverage and hydraulics of M01 and M02 exist to squeeze two surfaces together, and friction is what that squeeze buys.
The principle in one sentence
The sideways force two surfaces can resist before sliding is a fixed fraction of how hard they are pressed together.
The one equation
F_friction = mu x NN is the normal force, the push straight into the surface (lb). mu is the friction coefficient, a number for the pair of materials. F is the most force the surfaces can pass along the surface before they slip (lb). Rubber on asphalt has a high mu (around 1, meaning the sideways force can equal the weight); brake pad on steel around 0.4; steel on oiled steel 0.1.
Math: coefficients. mu has no units: it is a force divided by a force, so the pounds cancel and what's left is "how many times." mu = 0.4 means the surfaces resist 0.4 lb sideways for every 1 lb of squeeze. Coefficients are how physics writes "it depends on the material" as one number you can look up or measure. On a calculator: squeeze in pounds, times mu, gives the friction in pounds (1024 x 0.4 = 410).
Two versions of mu exist for every pair: static (not yet sliding) and kinetic (sliding). Static is a little higher. That's why a locked, sliding tire stops you worse than one at the edge of grip, and why a brake that has just locked keeps sliding easily.
Why it works
Surfaces are rough at a scale you can't see. Press them together and the high points touch and interlock; slide them and those points have to be sheared or climbed over. Press harder and more points touch, so the resistance goes up in proportion. That's mu x N. Rubber adds a second effect (it flows into the roughness and grips by deforming), which is why tires break the rule a little: lesson 4 and its addendum cover that.
On our kart, in numbers
Pads on the rotor. M02 delivered 1024 lb of squeeze across the pads. Pad mu 0.4 (typical): friction = 0.4 x 1024 = 410 lb, acting at the rotor's effective radius, about 3.0 in (typical; measure our rotor). That's a braking torque on the axle of 410 x 3.0 = 1230 lb-in.
From axle to tire: the lever in reverse. The tire is a lever too (M01): torque at the axle over the tire radius gives the force at the road. 1230 lb-in / 5.45 in = 226 lb of braking force at the rear contact patches. On 360 lb that is 226 / 360 = 0.63 g, if the tires can hold it. Notice the chain: 50 lb at the foot, x4 (pedal), x5.12 (hydraulics), x0.4 (friction), x3.0/5.45 (rotor over tire radius) = 226 lb at the road. Four machines in series, and every one of them is a ratio.
Can the tires hold it? The tires are friction too. Rear load mid-braking about 117-160 lb (lesson 3 Q7, lesson 7 Q4), tire mu 1.2 (typical): 1.2 x 159 = 191 lb before they lock. The brakes can deliver 226 lb; the tires can pass 191. So on this (typical) geometry the brakes are strong enough to lock the rear, and the limit is the tire, not the pedal. That is what lesson 7 found from the other direction: rear-only braking on this kart tops out near 0.6 g.
Glazed pads. Overheated pad material hardens into a glassy layer; mu drops from 0.4 to about 0.25 (typical). Same 1024 lb of squeeze, now 256 lb of friction instead of 410: braking force down 37%. The pedal feels the same (the hydraulics didn't change), the kart just doesn't stop. The fix is new pads, or sanding the glaze off, not more pedal.
Clutch shoes on the drum. Same law: the shoes are thrown outward (M04), N is that outward push, mu of the shoe lining (typical 0.3-0.4) sets the torque the clutch can pass before it slips.
Two things to notice
- Friction doesn't care about area. F = mu x N has no area in it. A wider pad doesn't brake harder for the same squeeze; it just spreads the heat and wears slower. (Tires are the exception, again because rubber isn't a rigid surface: lesson 4.)
- Friction is the only machine on this list that makes heat on purpose. Every other force multiplier passes energy through; friction turns the kart's motion into heat in the rotor, the drum, and the tire. Lesson 6's energy has to go somewhere, and this is where.
Problems
1. A softer pad. Swap to a pad with mu 0.5 (typical for a racing compound). Same 1024 lb of squeeze, same 3.0 in rotor radius, same 5.45 in tire: how many pounds of braking force at the road now? Would the tires (191 lb limit) still be the limit, or the brakes?
2. Half the foot. The driver brakes with 25 lb instead of 50 lb. Walk the chain (x4, x5.12, x0.4, x3.0/5.45): how many pounds at the road? Is that above or below the 191 lb the rear tires can hold? So somewhere between 25 lb and 50 lb of foot the rear locks: what does that say about how much of the driver's useful pedal effort is above the lock point?
3. Static vs kinetic. A locked tire slides with kinetic mu; a rolling tire at the edge of grip uses static mu. If static is 1.2 and kinetic is 1.0 (typical), by what percentage does braking force drop the instant the rear locks? Say in one sentence why that makes a lock self-reinforcing.
Go look: pull a brake pad out (two pins, per the caliper photo in the records) and look at its face. Dull matte grey is healthy; shiny or glassy is glazed. Then look at the rotor face for the same shine.