Thread: friction

a cyclist and her bicycle have mass 75kg. she is riding on a horizontal road, and positions herself so that 60% of the normal contact force is on the back wheel and 40% on the front wheel. The coefficient of friction between the tires and the road is 0.8. What is the greatest acceleration she can hope to achieve?
Whilst riding at 6ms^-1, she applies both brakes to stop the wheels rotating. In what distance will she come to a stop?

My big problem here is I don't know which forces i am dealing with and what their directions are. I know how to calculate frictional force but i dont understand how I can solve these problems without knowing the force she applies to move the bicylce forward.

In the first situation, I'm assuming that the coefficient of friction you've quoted is that of static friction, correct? I think the idea here is, how much force can the cyclist apply before skidding? In the second situation, it's obvious the rider is skidding, which means you have to know the coefficient of kinetic friction in order to solve the problem. Or is this problem simplistically equating static with kinetic? In general, static coefficients are greater than their corresponding kinetic coefficients. Are you given both coefficients?

You are given the mass of the bicycle and rider. Therefore, in the first problem, if you were to know the force she applied, you would know the acceleration by Newton's 2nd law! Evidently, this is what you are supposed to find out. Some pieces of the puzzle that I deem important: for regular bicycles, only the rear wheel drives. So, in calculating the amount of force allowed in moving the bicycle forward without skidding, you can use 60% of the normal force in computing the force of friction.

The second problem is a fairly standard problem, complicated only by the 60/40 distribution of the rider's weight. Even then, I don't think the final result will be altered by that fact. You basically only have one force that is active in the direction of motion: friction. So, use Newton's Second Law and integrate.

Does this give you some ideas?

the force which moves the bicycle forward is the friction between the back wheel and the ground
edit: didnt notice the previous answer

what about the friction in the front wheel? static and kinectic friction are the same in this problem

The friction in the front wheel only plays a role in the second problem, when you're computing the amount of time to stop. Why? Because the brakes are applied to both front and rear wheels. In the first problem, however, you're only interested in the wheels that drive the bike. That would be the rear wheel only, in most bikes.