Richard P. answered • 04/12/19
PhD in Physics with 10+ years tutoring experience in STEM subjects
Before you apply the brakes, weight is supported roughly equally by the front and rear wheel. In terms of external forces on the bike-rider system, there is an upward directed normal force W/2 exerted by the road on the front tire and an upward directed normal force W/2 exerted on the back tire. Here W is the combined weight of the bike and rider. These are the normal parts of the contact forces. Once you apply the brakes, the road imparts a frictional force (the other part of the contact force) directed horizontally to the rear on both tires. In addition to starting to slow the bike, these forces exert a torque on the rider’s body. The rider starts to pitch forward. This is countered by the force of the handle bars on the rider’s upper body. Concomitate with that, the rider’s center of mass is shifted forward. This has the effect of increasing the upward directed normal force on the front tire and decreasing the upward directed force on the rear tire. Since the force of friction is proportional the normal force, the force of friction exerted by the road on the rear tire lessens. Next, consider the torque acting on the rear wheel. The force of road friction exerts a positive torque while the brake caliper exerts a negative torque. As the force of friction lessens, the brake caliber force ‘wins’ and the rear wheel stops turning. This creates the skid of the rear tire. Once the skid starts, the force of friction is reduced further still (because the coefficient of dynamic friction is less than the coefficient of static friction). This is less likely to happen to the front tire because the increase in the frictional force means that the brake caliber is less likely to ‘win’. A similar mechanism explains why the front brakes on a car are more important in stopping power than the rear brakes.
