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Re: [Phys-l] Another tire question



Here's my take on the situation. I think it's good to first order.

Before you put the tire on the wheel, the radius of each bead is smaller
that the radius of the part of the rim on which it is designed to
reside. When you put the tire on the wheel, assuming the bead is in its
ultimate resting place on the wheel, the bead will be stretched.
Assuming no bunching, there will be uniform tension in the bead. That
tension will remain uniform at one and the same value throughout the
rest of this discussion because the length of the bead will not change
and the tension in any segment of the bead is determined by its length
and its spring constant.

Considering forces only in the plane of the circle formed by the bead,
the wheel will be exerting a radially outward force-per-length loading
on the bead. That loading times the length of any infinitesimal segment
of the bead is the "in the plane of the circle formed by the bead"
component of the normal force on that segment of the bead. The Newton's
3rd law partner to this loading is a radially inward directed loading
exerted on the wheel by the bead. It pushes inward on the surface of
the wheel. The bead can only push on the wheel. The bead is not
trapped.

When you inflate the tire, the sidewalls of the tire exert a radially
outward directed loading on the bead. The net radially outward loading
on the bead never changes, rather, the radially outward loading exerted
on the bead by the wheel decreases as the radially outward loading
exerted on the bead by the sidewalls increases. This normal loading
exerted on the bead by wheel adjusts itself to whatever it has to be to
maintain the bead in the shape of a circle with the radius of the wheel.

When you lower the car down to the floor of the garage so that it is
resting on the floor, the radially outward loading exerted by the
sidewalls on the bead becomes lesser on the parts of the bead on the
lower half of the wheel for reasons already mentioned in this thread.
The net radially outward loading on the bead is the same on all parts of
the bead as it always was but on the lower half of the bead, more of
that loading is being provided by the wheel (than was being provided by
the wheel after inflation but prior to setting the car on the floor).
The wheel is pushing downward on the lower half of the bead harder than
the wheel is pushing upward on the upper half of the bead. By Newton's
third law, the bead is pushing upward on the lower half of the wheel
harder than the bead is pushing downward on the upper half of the wheel.
The net force exerted by the bead on the wheel is thus upward. This
counteracts the downward force exerted on the wheel at the center of the
wheel by the axel.

The wheel is supported because it is being pushed upward by the lower
half of the bead harder than it is being pushed downward by the upper
half of the bead.

The (wheel+bead) is being supported because it is being pulled upward by
the sidewalls in contact with the upper half of the (wheel+bead) harder
than it is being pulled downward by the sidewalls in contact with the
lower half of the (wheel+bead).