Bicycles and Mechanical Advantage

[UMath1]

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 that's true, then what about when the force of the bike exceeds the maximum possible static friction?

 

[A.T.]

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.

 

[russ_watters]

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 that's 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 don't 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 can't see why.

 

[UMath1]

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?

 

[jbriggs444]

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.

 

[russ_watters]

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?

How many times do we need to tell you that they are equal before you will internalize it and stop asking? [/exasperated]

 

[rootone]

... what about when the force of the bike exceeds the maximum possible static friction?

The wheel slips, it loses traction.

 

[Orodruin]

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?

You keep saying this and you keep getting the answer that they are equal. You will not get any further by insisting on something that you have been repeatedly told is not true. Again, this all seems to stem from fundamental misuderstandings of basic mechanical concepts, which is why I have recommended going back to study the basics of free body diagrams, forces, and torques.

Since we will not get anywhere until that has happened and the OP has been answered several times over, this thread is closed.