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Re: [Phys-l] Another tire question

Maybe it would help to think about this problem in the
limit - ie, when the tire is completely deflated. All
six vertical forces (rim on bead, bead on rim, bead on
tread, tread on bead, tread on road, road on tread)
are of equal magnitude. Right?

Then, how do these forces change, and what new forces
are present, while air is slowly added such that the
wheel rises at constant speed? In particular, what is
the situation when the bead & rim just lose contact
with the tread? If we mentally stop the action at this
point, it seems that the sidewall profile must play an
important role in the answer since the tire is much
flatter with low pressure inside than later at full
inflation. But I need to think a bit more on this.

John Barrere
Fresno Unified School District
--- Carl Mungan <mungan@usna.edu> wrote:

The May article makes things clearer. The 'trick'
here is to
separate the tire into a top half and a bottom
half. The tension of
the bead pulling on the wheel is uniform around the
wheel with the
wheel off the ground, but when on the ground two
things happen to
the bottom half of the tire. It has an upwards
force of the road (a
portion of the weight of the car) and the
deformation (bulging) of
the bottom of the tire changes the direction of the
pull on the
bottom half of the wheel and the vertical component
of the road
force. These two effects reduce the force of the
bottom half of the
tire compared to the top half by just the amount of
weight that the
tire must support.

That the road force can be considered to act only
on the bottom half
of the tire is, I assume, due to the flexibility of
the tire. That
is, if the tire were really rigid, we would have to
take the force
to be acting on the entire tire, but then the rigid
side-walls could
support the weight through compression.

Rick

Sorry, I must be slow, but I still don't fully get
it.

First, I don't understand what the red arrows are in
Michael E's
diagram. Could you please spell out exactly *where*
those forces are
acting, *on* what, and *by* what? In particular,
labeling the red
arrows from top to bottom as 1, 2, 3, and 4, am I
supposed to believe
that F2 - F3 is equal and opposite to the green
arrow? This does NOT
seem to agree with the numbers in Shurcliff's
article, where I
interpret F2 = 2000 lbs and F3 = 1900 lbs. Please
reconcile this
apparent conflict.

Second, Shurcliff's article also has baffling
features in it.
Shurcliff claims the different slope of the lower
portion of the tire
sidewall has a small *but nonzero* effect on the
problem. But he then
uses numbers that contradict this fact: including
the slopes, he
calculates the "downward force exerted by the lower
half of tire"
(I'm not keen on this choice of wording) to be 1900
- 500 = 1400 lbs.
But if I were to exclude the slopes, presumably I
would instead
calculate 2000 - 600 = 1400 lbs, exactly the same
answer! That is,
using his numbers the slopes have a *zero* effect.
What's the deal -
is it a zero effect or not?

So I conclude I don't understand this business of
slopes, neither in
Edmiston's nor in Shurcliff's exposition. I instead
propose
simplifying the analysis to the following:

Mentally cut the tire horizontally in two, as
Shurcliff suggests.
Consider the lower half. (To be explicit: consider
*only* the rubber
of the lower half tire to be our system, and *not*
the air in it, the
lower half of the bead, or the lower half of the
steel wheel.) Force
balance on this system:

(1) air exerts a downward force of 2000 lbs - how to
calculate this:
multiply gauge pressure of tire times width of tire
times diameter of
tire; thus 30 psi * 5" * 13-1/3"

(2) ground exerts an upward force of 600 lbs -
calculated as weight
of car divided by 4 (crudely assuming equal support
by each of the 4
tires); I don't understand Rick's concern here, as
the ground is
*only* in physical contact with lower half of tire

(3) lower half of bead must therefore exert a force
of 1400 lbs
upward since our system is in equilibrium

Using Newton's third law, we conclude that the
system must pull down
on the lower half of the bead with 1400 lbs of
force.

Repeat the above analysis for the upper half of the
tire. This time
we don't have force (2). Thus we conclude the upper
half tire must
pull up on the upper half of the bead with 2000 lbs
of force.

Putting these two statements together, we thus
conclude: the bead is
overall pulled upward by the tire with 600 lbs of
force. This
suppports the car because the bead runs around the
bottom of the
steel wheel and pulls up on the wheel with that same
force. -Carl
--
Carl E Mungan, Assoc Prof of Physics 410-293-6680
(O) -3729 (F)
Naval Academy Stop 9c, 572C Holloway Rd, Annapolis
MD 21402-5002
mailto:mungan@usna.edu
http://usna.edu/Users/physics/mungan/
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