Bicycles and Mechanical Advantage

[UMath1]

I think I understand both cases now. So before the bike begins accelerating, the net force and torque on the wheel is supplied by the chain which overcomes the force of static friction to male the wheel begin angularly accelerating. However, since the wheel is a part of the bike, the only net force on the bike comes from the force of static friction causing it to accelerate forward.

On the other hand when the bike is moving at a constant speed, there is no net torque or net force. Thats what I have a question about though. On a level surface, why do you have to keep pedaling? To keep moving at a constant speed, you can't have a net force. But if you pedal, wouldn't you receive a net force from static friction? Is there some other force I have not considered. This also makes me wonder, why can't wheels roll forever?

 

[russ_watters]

Is there some other force I have not considered. This also makes me wonder, why can't wheels roll forever?

Wind, rolling resistance and drivetrain friction.

 

[sophiecentaur]

when the bike is moving at a constant speed, there is no net torque or net force.

But you can only have a zero net force if the cyclist provides a force to overcome air/road friction, even on the flat.
Here's another thought. The force on the bike from the road does not act through the centre of mass, that means there is an unbalanced torque, which could send you over backwards (a wheelie), were it not for the restoring torque, provided by the fact that the CM is forward of the back wheel plus there is a front wheel to help keep you from falling forwards without consciously balancing fore and aft. If you ride a unicycle, you have to do this, of course.

 

[A.T.]

the net force and torque on the wheel is supplied by the chain which overcomes the force of static friction

Net force is the sum of all forces, not one force that overcomes some other force. Same for net torque.

 

[UMath1]

Wind, rolling resistance and drivetrain friction.

What is rolling resistance? Theres no friction so what creates the rolling resistance?And how exactly does drivetrain friction occur? All the parts of the bicycle move in a synchronized motion..there is no sliding of parts on one another..right?

 

[russ_watters]

What is rolling resistance? Theres no friction so what creates the rolling resistance?

Tires are made of rubber, which deforms when it contacts the ground. Since it isn't perfectly elastic, this continuous shape-changing dissipates energy:
https://en.wikipedia.org/wiki/Rolling_resistance

And how exactly does drivetrain friction occur? All the parts of the bicycle move in a synchronized motion..there is no sliding of parts on one another..right?

The chain slides on an off the sprockets (and not always very smoothly) and the links in the chain rotate on pins between them. That's why a chain is greased.

Those two are probably small (20% maybe?) of the losses when traveling at medium speed on a bike. Wind resistance dominates. In a car, though, drivetrain losses are a much bigger fraction due to the complexity of the drivetrain.

 

[sophiecentaur]

Theres no friction so what creates the rolling resistance?

A wheel is not diamond hard. It is constantly deforming because the footprint is flat against the ground. Effectively, the bike / car is constantly pulling itself 'uphill' to force the slack bit of rubber down against the road. Deforming rubber constitutes Work (and wheels get pretty hot don't they?)

how exactly does drivetrain friction occur? All

Gears / chains on sprockets and bearings all have sliding between surfaces. Even a ball bearing will have some sliding because the radii of inner and outer tracks are different radii. Find yourself a book on Mechanical Engineering ( or search on line) and look up gear design. It's fiendishly hard to get a really low loss transmission geometry.

@ Russ: snap. You got there just before me.

 

[UMath1]

I tink I still have some confusion regarding the free body diagram at a constant velocity.

Which way would the force of the chain go? Or does it cancel itself out while still providing a torque?

 

[sophiecentaur]

Which way would the force of the chain go? Or does it cancel itself out while still providing a torque?

I think your diagram has the right idea except for the "F" underneath the back sprocket. The chain is more or less slack there, supporting only its own weight and you can neglect this force and there is no "cancellation" or there would be no net torque. Cast your memory back to a normal bicycle. The chain is usually in tension along the top and it is in front of the back wheel. The chain is pulling against the top of the back sprocket, producing a force on the ground in a backwards direction (a continuous lever, using the spindle as a fulcrum). Friction with the ground provides a reaction force (against the tyre force) which is Forwards from the Bike's point of view. Rolling resistance appears as a force which will slow down the bike (i.e. a backwards force) but it is quite likely not to be horizontal, but to act at an angle to the ground (as I said earlier, the tyre undergoes a constant process of being squashed and having to be driven , effectively uphill all the time. The footprint always has a finite area so the tyre has a flat part in contact and the surface of the road will always provide a series of tiny hills for the wheel to climb. The torque, resulting from this, will be against the torque produced by the chain. There will also be drag from the air and friction inside the bearings. All of that constitutes rolling resistance.

 

[A.T.]

Which way would the force of the chain go? Or does it cancel itself out while still providing a torque?

As sophiecentaur said, the bottom force at the sprocket doesn't make sense. A chain cannot push, just pull. So if anything it would go the other way, but be much smaller than the top force, merely supporting the weight of the lower chain.

Also, rolling resistance cannot be localized at the contact point. The way you drew it would create a torque that drives the wheel forward. I would rather draw it at the center of the wheel.

 

[A.T.]

Rolling resistance appears as a force which will slow down the bike (i.e. a backwards force) but it is quite likely not to be horizontal,

The rolling resistance force that you compute using the rolling resistance coefficient is horizontal (parallel to surface):
https://en.wikipedia.org/wiki/Rolling_resistance#Rolling_resistance_coefficient

 

[sophiecentaur]

The rolling resistance force that you compute using the rolling resistance coefficient is horizontal (parallel to surface):
https://en.wikipedia.org/wiki/Rolling_resistance#Rolling_resistance_coefficient

I would disagree. The resistance is a torque, rather than a force. As I mentioned, a lot of the loss is because of the distortion of the tyre. Why should that just be a horizontal force? In the case of bearing resistance, it is definitely not a 'backward' force. Having said that, it may be convenient to characterise the resistance as a single force but that is really just a short cut.

 

[UMath1]

But if the force of the chain is in the same direction of static friction..wouldn't there be a net force on the wheel? Does this mean that the rolling resistance is equivalent to the combined force of static friction and the force of the chain?

If this is the case, what exactly would happen had there been no static friction? The force of the chain would provide both a net torque and a net force. Why then is static friction necessary?

 

[A.T.]

I would disagree.

With what? This is simply how the commonly used rolling resistance coefficient is defined. It's of course the simplest model.

 

[A.T.]

But if the force of the chain is in the same direction of static friction..wouldn't there be a net force on the wheel?

You forgot the force on the axle, from the frame.

 

[UMath1]

You forgot the force on the axle, from the frame.

Wait so if the force of the chain is canceled out by the force of the axle. How then does produce a torque?

 

[Orodruin]

The force on the axle has zero moment arm ...

 

[A.T.]

Wait so if the force of the chain is canceled out by the force of the axle. How then does produce a torque?

https://en.wikipedia.org/wiki/Couple_(mechanics)

 

[UMath1]

So although the chain produces no net force, it provides a net torque when the bike accelerates to start moving. But that brings another question. If the chain provides a torque greater than that of static friction. Then that must mean that the force imparted from the bicycle wheel on the ground must be greater that the force of static friction on the wheel. But wouldn't that go against Newton's 3rd Law?

Also, when the force of the wheel on the ground is partially downwards, and partially to the left right? Does this mean that there both a reaction force from the normal force and the force of static friction?

 

[Orodruin]

If the chain provides a torque greater than that of static friction.

What do you even mean by this, force and torque are not the same thing. And if you refer to static friction, static friction where? If the wheel is rolling without slipping, static friction between the wheel and ground is never overcome. If it was, the wheel would start spinning.

I am sorry to say so, but some of your questions suggest that you need to go back and obtain a deeper understanding of the basic concepts of forces and torques before working on the forces related to a bicycle wheel.

 

[A.T.]

So although the chain produces no net force...

"Net force" is the sum of all forces, which is zero for constant speed. The total chain force on the wheel is not necessarily zero.

If the chain provides a torque greater than that of static friction. Then that must mean that the force imparted from the bicycle wheel on the ground must be greater that the force of static friction on the wheel.

No it doesn't mean that.

Also, when the force of the wheel on the ground is partially downwards, and partially to the left right? Does this mean that there both a reaction force from the normal force and the force of static friction?

The ground force on the wheels has a vertical component that balances the weight of the bike.

 

[sophiecentaur]

It's of course the simplest model.

Yes, but the simple model actually seems to make it difficult to understand ( this thread and others testify to that), although it allows you to add just three forces together and to assess what the acceleration will be.
It seems to me that people view the rolling resistance as something that acts, in a rather magical way, at the 'point of contact with the ground. In fact, there is no point of contact and the component of rolling resistance (an all embracing term, in fact) that operates at the bottom of the wheel is only there because of the fact that the geometry is not 'ideal' in the contact region. The losses caused by actual wheel / ground contact must be due to some form of slippage / distortion and hysteresis in the tyre or road surface. You need a Force times a Distance to dissipate energy and, without a distance, no work is done and there will be no loss. This point always seems to be ignored and causes a lot of confusion.

It is true to say that the force accelerating the bike forwards is equal to the reaction force against static friction minus the net 'rolling resistance' but the rolling resistance has many components, which appear at many different locations in the mechanism. Many (most) of these forces are 'relayed' to the wheel contact point by intermediate torques on the way. When the (net) rolling resistance (backwards) equals the reaction force (forwards) against the static friction, the acceleration will be zero.

 

[UMath1]

What do you even mean by this, force and torque are not the same thing. And if you refer to static friction, static friction where? If the wheel is rolling without slipping, static friction between the wheel and ground is never overcome. If it was, the wheel would start spinning.

I understand that static friction is never overcome. The force of static friction must be present in order for the wheel to roll without slipping. What I mean is, the force of static friction has a moment arm on the wheel and thus provides a counterclockwise torque on the wheel. Therefore, in order for a stationary wheel to begin rolling forward a net clockwise torque must be provided. The only way this is possible is if the torque from the chain is greater than the torque from static friction, right?

Now my question is that if the chain's torque is greater than the torque from static friction, wouldn't that then mean that force imparted from the bicycle wheel on the ground would be greater than the force of static friction? But that can't be true, because the force imparted by the bicycle wheel on the ground must be equal to force from the ground on the wheel as per Newton's 3rd Law.

 

[Orodruin]

Now my question is that if the chain's torque is greater than the torque from static friction, wouldn't that then mean that force imparted from the bicycle wheel on the ground would be greater than the force of static friction?

No.

 

[sophiecentaur]

I understand that static friction is never overcome

You cannot assert that, I'm afraid. It is very common to get forward acceleration in the presence of wheel spin, where friction is less than the tangential wheel force. (Dragsters and speedway bikes are always experiencing this.)
The force that accelerates the vehicle / bike is a reaction to the Friction, static or dynamic. It would, of course, be a reasonable exercise to consider what happens with a 'rack and pinion' drive - like many mountain railway systems - in which the effect of unlimited static friction is achieved by teeth. That approach would eliminate one possible misconception.

 

[UMath1]

No.

Can you explain?

 

[Orodruin]

Can you explain?

Again, this is a very basic mechanical issue that suggests you need to go back to study the basics of mechanics of forces and torques before attacking the bicycle wheel. It is even unclear why you would think that the force on the ground from the wheel would be different from the force from the ground on the wheel. As you say, they are force pairs so they must be equal in magnitude and nobody has said anything to suggest that this is not the case in this thread.

 

[dean barry]

There is sliding friction in the chain bushes as the chain goes from straight to curved.

Rolling resistance (In Newtons) is the heat created in the deformation of the tyre, it is deemed to be a constant value regardless of the speed on the road, simply put, it is : m * g * Crr

Dont forget air drag as well, which becomes the dominating force eventually as the speed builds.

If you consider a fixed force (f, in Newtons) on your crank, then calculate the crank torque (t) from f * r
( r = crank radius in metres )
Say you have a crank with 40 teeth and a rear wheel with 20 teeth then the rear wheel torque (T) will be :
T = t * ( 20 / 40 )

The rear wheel driving force (F) you get from :
F = T / r
( r = rear wheel rolling radius )

You can apply the above to all the gear ratios you have.

 

[A.T.]

...wouldn't that then mean that...

No.

Can you explain?

It's actually up to you to justify the implications you suggest.

 

[russ_watters]

I understand that static friction is never overcome. The force of static friction must be present in order for the wheel to roll without slipping. What I mean is, the force of static friction has a moment arm on the wheel and thus provides a counterclockwise torque on the wheel. Therefore, in order for a stationary wheel to begin rolling forward a net clockwise torque must be provided. The only way this is possible is if the torque from the chain is greater than the torque from static friction, right?

Now my question is that if the chain's torque is greater than the torque from static friction, wouldn't that then mean that force imparted from the bicycle wheel on the ground would be greater than the force of static friction? But that can't be true, because the force imparted by the bicycle wheel on the ground must be equal to force from the ground on the wheel as per Newton's 3rd Law.

I'm not sure if you really understood the constant speed scenario, but you aren't applying what you learned. Why haven't you updated the diagram to show the acceleration forces?

 

[UMath1]

So here's my understanding of the constant speed scenario. There is no net torque or net force on the bicycle wheel.

Now in the case of the accelerating bicycle, the force needed for the bike to accelerate comes from static friction. However, the torque needed for the wheel to angularly accelerate comes from the chain.

The issue I am having is from my understanding the force the bicycle imparts on the road is dependent on the torque it receives from the chain. For example, if the gear is 5 m radius and the wheel has a 25 m radius. Then if you pedal with a force of 10 N, the bicycle wheel will impart a force of 2 N on the ground. Then what about when the torque from the chain is more than the torque from static friction. Say for example, the maximum static friction on the bike is 10 N and you pedal with a force of 100 N. Then the bicycle wheel would impart a force of 20 N. But that would go against Newtons 3rd Law.

 

[A.T.]

So here's my understanding of the constant speed scenario.

Rolling resistance force still has a wrong point of application.

Then what about when the torque from the chain is more than the torque from static friction.

For the accelerating case torques aren't balanced, because the angular velocity is changing.

Say for example, the maximum static friction on the bike is 10 N and you pedal with a force of 100 N. Then the bicycle wheel would impart a force of 20 N.

No, the wheel would eventually slip, if you pedal too hard.

 

[UMath1]

Exactly! The torques aren't bakanced. The torque of the chain is greater than the torque of static friction because angular velocity is changing. Now my question still remains: isn't force imparted by the bicycle wheel to the ground dependent on the torque it receives from the chain? If yes, then when the torque of the chain is greater than the torque of static friction, how does Newton's 3rd Law make sense? Wouldn't imply that the force of imparted by the bike is greater than the force imparted by the ground?

 

[russ_watters]

No. Force (or torque) pairs are always equal and opposite at the point of application. For the bike you are describing, there are two pairs:
Chain <> wheel
Wheel <> ground

Again, the parts of each pair are equal and opposite. But in a free body diagram, you only analyze what is happening to the object (the forces applied to it), not what it is doing to others. In the constant speed case, those forces (or torques) sum to zero, but in the accelerating case they do not, unless you add a term for inertia.

For example, if you push a block across the floor, the forces applied to it are:
Push + Friction = ma

Bringing inertia in for the bike is somewhat difficult because you have both linear and rotational inertia here.

 

[A.T.]

when the torque of the chain is greater than the torque of static friction, how does Newton's 3rd Law make sense?

What does Newton's 3rd Law have to do with this?

Wouldn't imply that the force of imparted by the bike is greater than the force imparted by the ground?

Can you explain the implication you are suggesting?