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To: MV=PY

“The coefficient of friction changes with normal force (weight).”

Not really all that much over actual vehicle load variation.


89 posted on 04/21/2022 4:50:19 PM PDT by Paladin2
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To: Paladin2

Sure, but aren’t we talking about making the vehicle light enough to reduce mpg?

I see how this would be effective in an aircraft (brilliant, I think), but the “buoyancy” would impact a car differently because it interacts with the road. Many attributes of a vehicle are designed around the weight.

What if it weighed 100 pounds? Great for aircraft, not for cars.

(Not arguing - please don’t make me do the math!)

;)


107 posted on 04/21/2022 6:54:09 PM PDT by MV=PY (The Magic Question: Who's paying for it?)
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To: Paladin2

Oh, I just realized that I was incorrect in my first post and your statement is correct.

Force of friction = normal force * coefficient of friction.

You are correct, coefficient of friction does not change signicantly in this case.

But as normal force is reduced, force of friction reduces proportionally.

A one ton car that “weighs” 100 pounds because it has a balloon in it has mass, but can’t accelerate or decelerate very well.

Now winding up with my last whiskey this evening.

:>)


109 posted on 04/21/2022 7:12:58 PM PDT by MV=PY (The Magic Question: Who's paying for it?)
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