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Paul,

Although off topic, I cannot resist attempting the bicycle puzzle that you mentioned in the first post of this thread. So back to my schoolboy physics. If Frank, as moderator, you permit this post, maybe you, as a former physics lecturer, could put me right if I am wrong!

I feel that one needs to know one further piece of information. With the cycle held off the ground and a pedal moved horizontally backwards (say) 1cm, how far does a point at the base of the driven tyre move? This is ‘gearing’ and, for now, I assume that such point moves 2cm backwards.

For the cycle to move at all, there must be a net external force on the structure. The only horizontal forces involved here are, say, 10 Newtons, applied externally backwards to the pedal and the force experienced by the tyre from the road via the gearing as a result of the pedal effort. With 2:1 gearing as above, such force will be 5 Newtons forwards, or slightly less due losses. So, the net force on the cycle is 5 Newtons backwards and that is the way it will move.

If the cycle had a very low gearing so that the pedal movement was greater than the tyre rim movement, the force from the road would be greater than 10 Newtons and the cycle would move forwards.

Of course, it would be possible to devise gearing so that the force from the road was exactly equal to the applied force. The cycle would then remain stationary unless the applied force was sufficient to cause skidding.  

Bill Ritchie