Bicycles and Mechanical Advantage

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  • #51
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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?
 
  • #52
Orodruin
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The force on the axle has zero moment arm ...
 
  • #54
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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?
 
  • #55
Orodruin
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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.
 
  • #56
A.T.
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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.
 
  • #57
sophiecentaur
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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.
 
  • #58
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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.
 
  • #59
Orodruin
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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.
 
  • #60
sophiecentaur
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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.
 
  • #62
Orodruin
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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.
 
  • #63
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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.
 
  • #64
A.T.
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  • #65
russ_watters
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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?
 
  • #66
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So here's my understanding of the constant speed scenario. There is no net torque or net force on the bicycle wheel.
Screenshot_2015-07-15-13-10-01.png


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.
Screenshot_2015-07-15-13-16-21.png


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.
 
  • #67
A.T.
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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.
 
  • #68
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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?
 
  • #69
russ_watters
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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.
 
  • #70
A.T.
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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?
 
  • #71
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What I mean is, wouldn't you be able to calculate the force the bicycle imparts on the ground by simply knowing the force applied on the pedal, the gear moment arm, and the bicycle moment arm? Now to get the bike to start accelerating, the torque generated by the chain needs to be more than the torque generated by static friction. Now if I tried to calculate the force imparted by the bicycle wheel it would come to be greater than the force of static friction. That's the issue I am facing.

See the way I understand it is that static friction is the force that reacts to the force of the bike wheel, not the other way around. If thats true, then what about when the force of the bike exceeds the maximum possible static friction?
 
  • #72
A.T.
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what about when the force of the bike exceeds the maximum possible static friction?
It can't. It will slip, as already explained to you.
 
  • #73
russ_watters
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What I mean is, wouldn't you be able to calculate the force the bicycle imparts on the ground by simply knowing the force applied on the pedal, the gear moment arm, and the bicycle moment arm? Now to get the bike to start accelerating, the torque generated by the chain needs to be more than the torque generated by static friction. Now if I tried to calculate the force imparted by the bicycle wheel it would come to be greater than the force of static friction. That's the issue I am facing.
What is your issue with that? I see nothing wrong there.
See the way I understand it is that static friction is the force that reacts to the force of the bike wheel, not the other way around. If thats true, then what about when the force of the bike exceeds the maximum possible static friction?
Youve been told the answer to that question several times now. I dont understand why you keep getting stuck on questions you already know the answers to. You seem to be stuck in some sort of a loop and i cant see why.
 
  • #74
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What is your issue with that? I see nothing wrong there.
The issue with that is, because the force of static friction and the force imparted by the bike on the ground are an action-reaction pair. Then if they are not equal, how does Newton's 3rd Law apply?
 
  • #75
jbriggs444
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The issue with that is, because the force of static friction and the force imparted by the bike on the ground are an action-reaction pair. Then if they are not equal, how does Newton's 3rd Law apply?
But they are equal and nothing anyone else has said conflicts with that.
 

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