Question

A cyclist of mass 60 kg exerts a force of 300 N to move his bicycle with an acceleration of 6
m/s . How much is the force of friction between the road and the tyres?

Answer 1

Answer:

Since any acceleration of the bicycle is ruled by

F = ma

where bicycle+rider mass m = 60 kg

and its acceleration a = 6 m/s^2 (not 6 m/s),

the net motive force on the bicycle

F = 360 N.

And since the motive force is applied by static friction between rear tire and road, the answer must be 360 N.

So how can that be, when we are told that the rider exerted 300 N? Well, consider the mechanical advantage of bicycle pedal crank and gears. The work performed by the rider is the force 300 N, multiplied by the circumferential distance traversed by the rider’s feet:

C = 2π(pedal-crank radius)(crank turns)

Given some work wasted by friction inside the gears and chain, plus air resistance, the net work increases the bicycle’s kinetic energy. Assuming constant acceleration a, after the rider’s feet have moved distance C, the bicycle has travelled road distance S, so work exerted by the rider

W = 300*C

and bicycle+rider kinetic energy

E = F*S = 360S -(wasted energy)

Therefore E < W, and

C/S > 360/300

meaning the rider’s feet (while pushing) move at least 20% farther than the bicycle moves. Of course, in a real bicycle, the rider changes gears for optimal rider force, and acceleration may be high but never of long duration