Still no dice; you’re right about everything you said about the normal force, but the normal force isn’t “hitting” the shaft directly. It still passes through the air in the tire, which will still transmit that pressure equally throughout.
I’ll post the solution this weekend, I just don’t have time at work to whip up some Paint illustrations. But the gist of it is:
Spoiler
When the shaft starts to fall down, the tire “sags”, for lack of a better term. This sagging increases the area below the shaft in contact with the air, while simultaneously decreasing the area above. The pressure is always the same above and below, but the shaft is now exposed with a greater area below it: the force from below starts to overpower the force from above.
When I show the illustration, I can explain in more detail what I mean by “sagging”.
[details=Spoiler]either you assume that mass changes over time, such that d(m)/d® because mass is now a function of the radius of the paper or that the mass is constant.
If we assume that mass does not change and it’s perpetual in weight then the problem can be simplified easier, but because they said changing force over time, force is now a function of time, so in reality d(F)/d(t), but that makes the problem hard, so lets assume that we are at the instant the frictional force is overcome.
assuming a free body diagram, where the black dot is the center, the normal force isn’t straight up, its at an angle theta. and you have two forces acting on the black dot, Fx and Fy, my other assumption as that gravity acts on the paper in one direction and doesn’t affect the x axis (which I think it does), and that you have to eventually solve for some force as inertia because of the unequal load distribution.
Fx = (F_g_x) + (F_friction_x) if all forces are acting on the body at a diagonal before the F_tension overcomes F_friction_y
Fy = (F_g_y) + (F_tension) - (F_friction)
when F_tension + F_g > sqrt([ (F_tension_x)^2 + (F_tension_y)^2), you will have frictional slipping. that’s as much as ill do[/details]
[details=Spoiler] the air is acting in all directions inside the tire, you have a pressure vessel to distribute the force equally, both acting on the rubber itself and the rim (most important thing)
as you increase the pressure of the tire, the force per unit area on the contact patch increases as you load the tire, but its also exerting that same force on the rim
when you connect a shaft to the tire, there are only two points on the shaft on which the weight of the vehicle is acting on the shaft. those are connecting points of sprung weight
if you assume that the shaft is part of the tire itself, and it acts as a load distributor from the car to the rims, the air pressure then exerts a force on the rim equal to the force it exerts on the contact patch, which is what is holding the car in place.
Ok here is one that I found pretty interesting (unfortunately SRK is not equipped with LaTex afaik, so forgive the not so great notation):
Tesla decides to go back to basics with their new line, the Entropica series. Leaving aside their rather successful venture into electric engines, Tesla pivots to embrace the heat engine, which uses two heat blocks of initial temperature, T+ and T- to drive an engine through a thermodynamic cycle.
Concretely, each cycle drives some heat dQ (this would typically be “delta Q”, but again, SRK isn’t equipped for that sort of thing) from the hot reservoir to the cold reservoir, and the engine performs some work dW to propel the car, such that the engine returns to its original state after each cycle. Clearly, each cycle lowers the temperature T+ , and raises the temperature T-, until T+=T-=T*, at which point the engine can do no further work.
Find T* (in deg Kelvin) at the point where the engine stops running.
Assumptions and Details
The entire process is thermodynamically reversible.
T+ = 423 K
T- = 300 K
Both heat reservoirs are metal blocks of heat capacity v
I struck out on my first attempt at this problem, but maybe with multiple people giving it a go we can figure it out.
if you assume that the engine is a heat exchanger, which it is since heat has to be exchanged in order to do work, LOGMean Temperature should give you an answer. Ideally you could say that the final temperature should be the average, except heat flux through a system is not linear.