Bicycles and Mechanical Advantage

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  • #26
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Why dont you just compute the torque needed as suggested? You should find that it does not depend on the moment arm. However, since the torque is constant, you need to supply less force on the chain with a longer moment arm.
Screenshot_2015-07-05-18-19-21.png

Is this correct? From what I understand the torque the chain applies on the wheel must equal the torque from static friction if the bicycle is traveling at constant velocity. Based on that I am able to determine the output force from the bicycle wheel.

But what about when the bicycle is trying to accelerate, is it friction or the chain that supplies the net force and net torque on the wheel? If friction is what supplies the net force, then wouldn't the bicycle roll backwards(friction would give a ccw torque)?

And what exactly happens when the bicycle slips down the hill? If you are in a small gear, then you have small output force which means you get less static friction. So then is it that because you get less static friction, gravity's x-component supplies a net force causing a downward acceleration?
 
  • #27
The tire exerts a force onto the ground equal to 50N. It can also be said that the ground exerts a 50N force on the tire. If this is the only force acting on the bike in the x-direction, then the bike will accelerate in the x-direction. If not, then you must sum all of the forces acting in the X-direction. The bike will accelerate if there is any net force. Opposition forces include drag, and rolling resistance.
 
  • #28
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The tire exerts a force onto the ground equal to 50N. It can also be said that the ground exerts a 50N force on the tire. If this is the only force acting on the bike in the x-direction, then the bike will accelerate in the x-direction. If not, then you must sum all of the forces acting in the X-direction. The bike will accelerate if there is any net force. Opposition forces include drag, and rolling resistance.
But isn't the force of the chain also acting on the wheel? And if the net force is provided by the ground, then how come the acceleration of the wheel is cw and not ccw?
 
  • #29
You just gave the force of the chain acting on the wheel. How are you still confused about it? The chain simply transmits the linear force from one gear to the other. The radius of the gear about its fixed axis makes it a torque applied to the wheel, and the radius of the wheel changes it back into a linear force.

Also, I don't know what cw and ccw are. So, I can't answer that question, yet.
 
  • #30
One of the things I think you're having problems with is your free body diagram. Free body diagrams are only meant to view a single body. I think you may be confusing the equal and opposite forces at work with some idea that they may be canceling each other out. They are not. In the end, the tire exerts a force on the ground at that force causes the acceleration of the bike as a system. There is no need to consider the ground, the chain, the foot on the pedal. Once you know that the force is being exerted out of the system, then you know that the system is accelerating.

There's a lot going on internally that gets that force to the tire, but looking at the system as a whole.... as a single body, it is exerting a force, and as long as it is greater than the net forces acting against it, the system is undergoing acceleration.

So starting from the beginning with a bike and its rider. The rider exerts a 10 Newton force on a pedal that is 0.5m from its axis, then there is 5Nm of torque on the drive gear. If the drive gear's radius is 1/10 of a meter, the gear puts a 50N force on the chain. The 50N force is transferred via the chain to the rear gear. The rear gear is .5m. So, the torque transmitted to the wheel is 25Nm. The wheel is 1m. So the force between the tire at the road is 25N. This assumes adequate traction between the tire and road surface to transmit that force.

To determine the acceleration on the system, you will have to do some more work and gather some more facts. For instance, you'll need to know the mass of the rider and his bike. You will have to know the center of mass of the system. You'll have to break down the 25Nm force into it's X and Y components. Once you have them, you can calculate the acceleration in each direction, the shift of weight from front to rear, etc...
 
  • #31
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I understand that, but in this I am considering the system to be the wheel, not the bike. So if the wheel is being considered, the force of the chain must be considered.

By cw and ccw, I mean clockwise and counterclockwise. So what I mean is, if the force of static friction from the ground is what provides the net force on the wheel, it must also therefore provide a net torque, right? And if that is true, wouldn't the acceleration of the wheel be counterclockwise instead if clockwise as it should be?
 
  • #32
A.T.
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if the force of static friction from the ground is what provides the net force on the wheel,
The net force on the wheel is the sum of all forces on the wheel.

it must also therefore provide a net torque, right?
No, same as above. The net torque on the wheel is the sum of all torques on the wheel.
 
  • #33
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Yeah, but in order for the bicycle wheel to being moving, it must accelerate, meaning there must be both a net torque and net force.
 
  • #34
A.T.
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Yeah, but in order for the bicycle wheel to being moving, it must accelerate, meaning there must be both a net torque and net force.
To accelerate, not to move at constant speed.
 
  • #35
russ_watters
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Yeah, but in order for the bicycle wheel to being moving, it must accelerate, meaning there must be both a net torque and net force.
Sure, but until you have the simple case figured out, it is best not to move on to a more complicated one: stick to constant speed, where the torques sum to zero for now.
 
  • #36
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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?
 
  • #37
russ_watters
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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.
 
  • #38
sophiecentaur
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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.
 
  • #39
A.T.
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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.
 
  • #40
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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?
 
  • #41
russ_watters
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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.
 
  • #42
sophiecentaur
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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.
 
  • #43
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I tink I still have some confusion regarding the free body diagram at a constant velocity.
Screenshot_2015-07-13-17-58-33.png


Which way would the force of the chain go? Or does it cancel itself out while still providing a torque?
 
  • #44
sophiecentaur
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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.
 
  • #45
A.T.
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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.
 
Last edited:
  • #47
sophiecentaur
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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.
 
  • #48
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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?
 
  • #49
A.T.
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I would disagree.
With what? This is simply how the commonly used rolling resistance coefficient is defined. It's of course the simplest model.
 
  • #50
A.T.
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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.
 

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