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[Phys-L] Re: centripetal force question



At 07:23 AM 11/15/2005, you wrote:
I've been thinking about this more after reading some of the insightful
responses. So then the rider pushes down (slanted) on the seat/cycle. This
adds a slanted downward force component to the cycle. So for the cycle,
this y-component is downward (making N larger), and the x-component goes
opposite to the frictional force, making the net horizontal force smaller.
Right?


As nobody seems to have mentioned it, the cross section of motor bike
tires is notably circular - compared with the flat surface
presented by auto tires. A bike rider will comment that he needs to
generate side traction when turning.



Brian Whatcott Altus OK Eureka!
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