# I How does a bike turn?

1. Mar 7, 2017

### yosimba2000

A bike turns when you turn the front wheel, and the front wheel only turns in the direction it's aligned in.

The pedaling motion provides a force in the up, or +Y direction.

If the wheels are aligned up, the bike moves along +Y. If the front wheel is turned 90 degrees, the bike won't move. If you turn the wheel 45 degrees to the right, the bike turns to the right.

But if the wheel is aligned 45 degrees to the right, that means the wheel can only turn along that direction, meaning there must be forces in both the +X and +Y direction to turn that wheel.

But where is the +X force coming from? Pedaling provides force in +Y, so even if the wheel is turned, friction works against the wheel in the -Y direction, so what is pushing the front wheel with a force in the +X direction?

Some thoughts: leaning when turning the bike would provide a force in the required direction. But that leads me to assume that if there is 0 lean on the bike, then the bike won't move if the wheel isn't aligned straight, because if the wheel is turned, it must need some X force to turn the wheel in that direction, and if pedaling provides force in only +Y, then there is no way to obtain a force in X.

2. Mar 7, 2017

### willem2

The force to turn the bike is provided by friction with the ground. There's no need to lean the bike to get a friction force, but the bike will have to lean, so the torque provided by gravity, and by the sideways friction force will cancel and the bike won't fall over.

3. Mar 7, 2017

### A.T.

I see no reason to assume that.

4. Mar 7, 2017

### rcgldr

Pedaling ultimately results in the tire exerting a -Y force onto the pavement, which is part of a Newton third law pair, the other part being the pavement exerting a +Y force onto the tire, which in turn exerts a +Y force onto the rest of the bike. The steering to the right also involves a Newton third law pair, an outwards force exerted by the tires onto the pavement, an inwards centripetal force exerted by the pavement onto the tires. Part of the pavements inwards centripetal force would be in the +X direction.

Last edited: Mar 7, 2017
5. Mar 7, 2017

### yosimba2000

Could you elaborate more on this? If pedaling produces a Y force, how does turning a wheel convert that Y force/friction pair into X directions?

Where does outward force come from, because it seems this outward force already has some force in the Y and X directions. Pedaling should produce Y only forces, right?

And in a similar vein: For a turning wheel with a free axle (not secured to wall), the wheel will move forward as well as slow down in rotational speed, right? So there is rotational force and translational force on the wheel, both provided by friction. Is there a way to separate how much of friction goes into slowing the rotational speed and how much of friction goes into providing translational movement?

Last edited: Mar 7, 2017
6. Mar 7, 2017

### A.T.

When you drive your bike into a wall at 45°, the wall will also exert forces on you that have components perpendicular to your driving direction. Whatever force is needed to prevent you from passing though the wall, will be exerted on you. Same with static friction at the turned wheel. Whatever force is needed to prevent it from simply sliding along Y only will be exerted on it.

On the atomic level these contact forces adjust their direction and magnitude due to local deformation at the contact patch.

Last edited: Mar 7, 2017
7. Mar 7, 2017

8. Mar 7, 2017

### A.T.

It is complicated regarding the balance of the bike. But the OPs confusion about steering and the direction of static friction is more basic, and equally applies to cars, where balance is not a key issue.

9. Mar 7, 2017

### rcgldr

This force pair is related to steering, not pedaling. The outwards force exerted by the tires onto the pavement is a reaction to the inwards centripetal acceleration related to the inwards centripetal force exerted by the pavement onto the tires.

10. Mar 7, 2017

### sophiecentaur

The two wheels are not aligned in the same direction when you are turning a corner. If you are taking your XY frame to be the front wheel's plane and spindle axis then the rear wheel force has a Y component. But you don't need to pedal in order to turn so it isn't necessary to include that - it just adds to the problems. The direction of motion is what counts here and the wheel is steered away from that direction, there is a centripetal force (reaction against the Y axis (lateral) force against the ground) to cause a circular motion. It is necessary to lean on a bike but that's for purely practical reasons of staying upright. A tricycle steers fine and you cannot lean it.
My comments contain what rcgldr has said too but it is important to eliminate the superfluous factors in order to get down to the basic steering process.

11. Mar 7, 2017

### Staff: Mentor

12. Mar 8, 2017

### lychette

13. Mar 8, 2017

### sophiecentaur

The principle of resolving forces is A level but I haven't come across explicit steering as such at that level. A bike is the worst possible thing to start with when explaining steering because of the balancing issue.

14. Mar 9, 2017

### A.T.

Yes, see post #8.