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Re: [Phys-l] Another tire question



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/