Traction Torque Effect On A Rotating Wheel In the figure is shown an electric-car's wheel . If we made a dynamic force analysis to this wheel (which is considered to rotate in the anti-clockwise direction) ; We will find that a traction force is opposing the force that the wheel pushes with at the ground (here represented with a red arrow); the torque of this traction torque should act in the clock-wise direction which is actually opposing the wheel's driving torque from the motor. Now , Won't that affect the wheel's acceleration; since the driving torque will do a positive torque on the wheel trying to accelerate it while the traction torque is opposing this acceleration. Also , engineers usually say that the wheels driving torque is always equal to the traction force times the radius of the wheel. If that's what's happening ; Then :
1) This wheel would never accelerate as two opposing torques of equal magnitude are acting on the wheels such they should cancel each other creating ZERO acceleration.
2) I'll suppose that this traction torque was less than the driving torque ; Won't that also affect the main-ring-gear acceleration (The big one connected to the wheel directly)? Then how would the gold shaft which is directly connected to the motor respond to this opposing torque? Doesn't that mean that the traction torque is always trying to make the ring gear drive the gold shaft? This will absolutely affect the gear's velocity ratio . (like think if the wheels were not connected to the gear's shaft) will the velocity output of the ring gear remain the same as it was after connecting the wheel to it? What is the traction torque effect on the ring gear's speed? I think it would reduce its original speed before connecting it with the wheels . (And assuming that the motor is providing a constant speed to the gold gear; Hence the velocity ratio should change and it is impossible for two meshed gears to have variable gear ratio this is because the number of teeths are fixed).

 A: 
engineers usually say that the wheels driving torque is always equal to the traction force times the radius of the wheel.

That's a simplification where the mass of the wheel is insignificant compared to the mass of the vehicle.  In fact the driving torque must be larger than the friction force to allow the (massive) wheel to accelerate.

Won't that also affect the main-ring-gear acceleration

It affects any component of the transmission with mass.  When comparing the acceleration of a 1000kg car and a 5kg gear, we might ignore the acceleration of the gear.  But for it to work, there must be a slightly greater torque on its input (as it's accelerating) than there is on its output.  That just means the torque produced by the engine is not 100% equal to the torque on the wheel.  There is some loss.
A: 

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*This wheel would never accelerate as two opposing torques of equal magnitude are acting on the wheels such they should cancel each other creating ZERO acceleration.


No. You need to learn what free body diagrams are and start drawing them. You mixing and confusing  forces applied by the body and those being applied to the body. That is why free body diagrams only have the forces acting on the body in them.
Only the forces acting on the body can contribute to acceleration. i.e. Your legs do not push yourself forward. Your legs push the ground and it is the ground that pushes you forward.



*I'll suppose that this traction torque was less than the driving torque ; Won't that also affect the main-ring-gear acceleration (The big one connected to the wheel directly)? Then how would the gold shaft which is directly connected to the motor respond to this opposing torque? Doesn't that mean that the traction torque is always trying to make the ring gear drive the gold shaft? This will absolutely affect the gear's velocity ratio . (like think if the wheels were not connected to the gear's shaft) will the velocity output of the ring gear remain the same as it was after connecting the wheel to it? What is the traction torque effect on the ring gear's speed? I think it would reduce its original speed before connecting it with the wheels . (And assuming that the motor is providing a constant speed to the gold gear; Hence the velocity ratio should change and it is impossible for two meshed gears to have variable gear ratio this is because the number of teeths are fixed).


That gear setup is a differential. They allow the wheels to turn at different speeds so the wheels do not skid in a turn. There is more going on there that can be described effectively in text. But if the force applied to both wheel edges is the same, the torque applied by the engine to both wheels is equal. If zero force is applied to one wheel then all engine torque is applied to that wheel and the other wheel gets no torque. Very bad if you lifted a wheel on a rock.
Look up animations on youtube:
https://www.youtube.com/watch?v=SOgoejxzF8c
A: *

*If the torque due to traction equals that of the motor the angular acceleration is indeed zero. Obviously when a vehicle accelerates the driving force of the motor overcomes that of the friction. It can reach a point where the driving force is equal to all other opposing forces, such as torque due to traction and air resistance then acceleration ceases and the vehicle cruises at constant velocity. I don't know what engineers usually say but that's how I understand it and I hope it answers the question


*Assuming these are simple ordinary gears the velocity ratio should stay constant regardless of the forces acting on it
$$ \frac{R_1}{R_2} = r \to R_1 = r R_2 \to \omega_1 = r \omega_2$$
where $r$ is the ratio of the gears radii.
When traction acts on the wheel it provides a torque opposite to the direction of the motor torque, which affects the gear attached to the wheel since they are effectively one rigid body. The motor's gear in turn "feels" the same torque and decelerates accordingly
A: For any mechanical system, every piece of it will obey Newton's laws.  For a system like the one you've shown, each colored part will obey Newton's laws during operation (this analysis is usually done with a "free body diagram"), but also, any arbitrary piece of each of these parts, such as a small cube conceptually isolated within the tire, will also obey Newton's laws (this analysis is usually called "finite element analysis").
So, to answer your questions, just consider all the forces and torques on the component (or combination of components) you're interested in and work it out.  The only conceptually difficult part is to be rigorous about finding all applied forces and torques.
For example, consider the wheel.  The axle applies a downward force (eg, part of the weight of the car) and a horizontal force (eg, resistance to breaking), and a torque.  The ground applies an upward force (through solid body contact) and horizontal force (through friction), and the frictional force also causes a torque about the axle (while the upward force doesn't since it is aligned with the center of the wheel).
So, just considering the wheel on its own, the resultant net linear horizontal force on the wheel will accelerate the wheel horizontally, and the resultant net torque will produce an angular acceleration of the wheel.  Since the wheel is solid, and we'll assume won't slip, these must also scale together by the radius of the wheel.  It may be that the weight of the wheel and moments of inertia of the wheel are small, and then the net torque on the wheel and net horizontal force on the wheel are small, and can be ignored when considering the motion of the vehicle, in which case you could just consider the horizontal frictional force that pushes the rest of the car to be $F_t$.  In general, though, these need to be worked out (or guessed) to know for any particular situation.  Usually, the weight of the wheel is less important than it's moment of inertia, so considering the toques and angular acceleration of the wheel is the most important first correction to make.
The same calculation is done for the gear.  You could also, for example, consider cutting the bar, and, then examine the forces that would be present to hold it together, since all parts obey Newton's laws at all times.
A: Here is a much simplified schematic diagram of the rear end of the transmission chain as a car is accelerating towards you.

For the transmission shaft the net torque on it, $\tau_{\rm TE}-\tau_{\rm TF}-\tau_{\rm TR}'$, produces the angular acceleration of the transmission shaft and a reduction in the torque transmitted to the rear assembly.
The crown and pinion acts as a torque convertor increasing it in the ratio of the teeth on the crown to pinion.
For the rear assembly the net torque is $(\tau_{\rm RT}-\tau_{\rm RF}-\tau_{\rm RG}$ and this produces the angular acceleration of the rear assembly and there is a reduction in the torque transmitted to the wheels.
You can imagine $\tau_{\rm RG}$ as being produced by forces on the wheels towrds you, due to the ground at the points of contact between the wheels and the ground, causing the centre of mass of the car to accelerate.
If there is no slipping then there are also equal in magnitude but opposite in direction forces on the ground due to wheels, ie away from you.
If the car is on gravel then these forces can be seen to move gravel backwards relative to the forward acceleration of the car.
