Originally Posted by number_2
my question is, if f = m x a (which obviously it does), what force f is being measured? torque? torque divided by the wheel's radius, as mentioned above? and is that an oversimplification? obviously it's negating outside factors like wind resistance, etc. but as far as the actual mechanics involved, what is the force which, when divided by the mass of the vehicle (or would it be the mass of the engine?), gives you your acceleration?
and did any of that even make sense?
you are correct.
the driving "force" accelerating the vehicle (which i shall call thrust) is the force the road exerts on the tires (which is equal in magnitude and opposite in direction to the force the tires exert on the road--newton's third law thing).
basically as the wheel turns (via torque from the axle), it exerts a force on the road tangental to the surface of the tire (that is, perpendicular to the RADIUS of the wheel). the torque the axle delivers is determined by the torque the engine produces, multiplied by the mechanical advantage of the gearing, and subtracting any frictional losses through the drivetrain.
the car is accelerated by the force the road exerts on the tire, and this is equivalent to the torque the tire exerts on the road (again, by newton's 3rd law), divided by the radius of the tire (which is equivalent to the force vector portion of the torque).