If a car is moving with constant speed, the wheels of the car exerts a constant force on the floor. Since $F = ma$, the car should be accelerating rather than maintaining the same speed.
What is going on here?
Why cars don’t accelerate according to $F = ma$
accelerationforcesnewtonian-mechanics
Related Solutions
I had fun trying to make this as intuitive as possible. I hope I've succeeded without doing the physics of the situation much injustice.
When a car is driving straight ahead, the plane in which the wheels are rotating is aligned with the direction of movement. Another way of saying this is that the rotation axis is perpendicular to the momentum vector $\vec{p}=m\vec{v}$ of the car. So the friction merely makes it harder for the car to move, which is part of the reason why you need to put your foot on the gas pedal to maintain a constant speed. At the same time, the friction is what allows you to maintain that constant speed because the rotating tires sort of grab onto the ground, which is the intuitive picture of friction. The tires grab the ground and pull/push it backwards beneath themselves, as you would do when dragging yourself over the floor (if it had handles to grab onto). Those grabbing and pulling/pushing forces are what keeps you going.
Things change when the wheels are turned. The plane in which they are rotating now is at an angle with the direction of motion. Alternatively but equivalently, we could say the rotation axis now makes an angle with the momentum vector of the car. To see how friction then makes the car turn, think again in terms of the wheels grabbing onto the ground. The fact that they now make an angle with the direction of motion, means the force the tires are exerting is also at an angle with the direction of motion - or equivalently, the momentum vector.
Now, a force is a change in momentum$^1$ and so (because the wheels are part of the rigid body that is a car) this force will change the direction of the car's momentum vector until it is aligned with the exerted force. Imagine dragging yourself forward on a straight line of handles on the floor and then suddenly grabbing hold of a handle slightly to one side instead of the one straight ahead. You'll steer yourself away from the original direction in which you were headed.
$^1$ Mathematically: $$\vec{F}=\frac{d\vec{p}}{dt}$$
[...] and there's a large normal force on the system I've described, and it points forward, along the direction that would push the car forward
The normal force points vertically upwards (it comes into existence to balance out the weight). The force pulling sideways is not a normal force but static friction. You say it right later on, but messes up the words here.
But this scenario is only applying to the section of the wheel that is on the asphalt. So, as the wheel turns, different sections of the wheel have this effect. How does this produce a thrust forward for the car if the car keeps losing that segment of the tire that is producing a friction force with the asphalt?
How do you make a thrust when walking? You do that by pushing off from the ground with one foot, and then the other foot takes over. Each contributes with a portion of the forward thrust.
Now imagine a star. When a spike is touching the ground it pushes off from the ground in the same way. It creates thrust, acceleration, while it pushes. Then the next spike takes over.
Now imagine a star with more spikes. With many more spikes. So many that you can't distinguish them. So many placed so close that it looks like... a round object. A wheel. When this wheel rolls, each spike - which is now only a single point - pushes off from the ground during the infinitely short time it is touching.
It might be a very short time of thrust done by one point on the wheel, but it happens constantly, since another point takes over immediately. In total, there is constantly a thrust going on.
Is there a tangential force fighting against $Ff$ that I haven't included?
No. (In realistic circumstances there will be some energy loss in axles and gearing, in the flexing rubber and in a deforming surface - all such losses will slow down the wheel and are in one term called rolling resistance. But in ideal models these are neglected (just like air drag is), and only static friction is working horizontally.)
Because the motor is accelerating the tires radially, [...]
A clarification on the wording: The motor is accelerating the tires by applying a torque around the axle. This makes the tire turn. Because all the particles of the tire are stuck together, they hold on to each other inwards - radially. This is why the radial acceleration appears on them. Yes, a particle in circular motion must have a radial acceleration, but the motor is only indirectly causing that - the direct reason is rather the rigid wheel and the hold-on forces from inner particles.
[...] so each segment of the tire should have a tangential force on it that is going to be, at the moment I've described, opposite to friction?
And to the question: Yes, at the contact point where static friction happens there is a force oppositely. The wheel is applying a static friction on the ground. The ground replies with a static friction on the wheel in the other direction. This is Newton's 3rd law.
But there are no other forces on the particles of the wheel than this friction.
There are of course still the holding-on forces between particles as mentioned above, which pull every particle towards the centre - but this makes no difference in the motion, since at the contact point these are perpendicular to the friction force and direction of motion and have no influence.
In the same way there are of course also still weight and a normal force, but these also work at (or through) the contact point perpendicularly to the motion and have no influence.
What if there was no motor force on the wheels, and it just had some $\omega$? This scenario I've described in the latter sentence sounds like asking if there is a friction force acting on an object moving at a constant speed.
By "no motor force" you mean no motor torque. No motor torque means no forced push. The wheel touches the ground but does not try to push itself off from the ground, because it doesn't try to speed up the rotation. When there is no pushing off from the ground, there is no need for friction to appear to hold back against anything.
Conclusion: There is no static friction happening when the wheel rolls without accelerating. In an ideal case, it will roll with constant $\omega$ forever.
Best Answer
The car experiences not only the force made by the engine. There are also other forces: friction and air resistance. The air resistance force is proportional to $v^2$, hence it becomes more and more important at higher speed $v$. So the total force becomes smaller while the car accelerates, until (at a certain high speed) the forces cancel to zero and there is no more acceleration.