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If a battery-operated toy car is moving with constant velocity across a floor without slipping, how should one draw and label its force diagram?

Obviously, when moving with constant velocity, the net force in the direction of motion equals zero, but if static friction between the tire and floor is responsible for the non-slipping forward motion, what is the force that balances that? It obviously can't be air resistance, as this would, presumably, also work on the moon or in a vacuum chamber.

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If you're considering an idealized situation with no rolling resistance, then once the wheels are rolling with uniform velocity/angular velocity such that $v = rw$, the force due to static friction on the wheels would become zero since the speed of the point of contact would be zero (principle of superposition)

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  • $\begingroup$ This is true. But it begs the question: if I shut off the engine and allow the car to coast, what is the force that slows the car? This force would be static friction between the road and the tire. So static friction applies both a forward force and a backwards force, and at constant speed the net of these two forces is zero. This is conceptually odd. Do I draw one arrow having zero length to represent friction, or do I draw two arrows, both representing the same phenomenon, but one pointing forward and one pointing backwards? It's arguable. $\endgroup$ – garyp Mar 15 '17 at 13:43
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    $\begingroup$ @garyp I think it's clearer to call that force rolling resistance. Calling it static friction is misleading since the cause of the force is the non-rigidness of any body. However, I don't feel like the OP wants to take rolling resistance into account. I feel like he's trying to sketch the free body diagram in an idealized scenario and is under the false assumption that static friction acts even when a body moves with constant $v$ and $\omega$ $\endgroup$ – xasthor Mar 15 '17 at 14:41
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    $\begingroup$ I think you are right. The internally generated power from the engine is used to overcome internal frictions. The force of static friction between the tires and the road is zero. $\endgroup$ – garyp Mar 15 '17 at 18:20
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The key point is that the thrust from the electric motor is not constant. As the speed (revs) of an electric motor increases, the torque (and therefore also thrust) decreases.

Initially the torque/thrust is much higher than resistance, so the toy car accelerates. But as the speed of the car (and the revs of the electric motor) increases, the thrust decreases, and at some point it becomes equal the external forces of resistance, which are approximately constant.

So what external forces of resistance are there? Air resistance remains quite low and insignificant for a toy car. (Unless there is a head- or tail-wind.) As Floris points out, there is also static friction on the non-driving wheels and 'rolling resistance' on all wheels.

While the car is accelerating, the thrust force has to push against the translational inertia of the car but also the rotational inertia of the non-driving wheels which rest on the ground. Static friction from the ground (acting backwards) opposes any acceleration of these wheels. So as Floris states, static friction acts in the opposite direction on driving wheels and free wheels. Friction in the bearings also resists all rotation of these wheels; so when the car is moving at constant speed the static friction force is only required to overcome the friction in these bearings.

Friction in the bearings also resists motion of drive wheels, but technically this decreases the thrust they supply to the ground. (You could draw 3 separate FBDs for the motor/chassis, the wheels/bearings and the road. Friction in the bearings of all wheels would then be external to the motor/chassis.)

Rolling resistance arises from the asymmetry of forces which deform the wheels or ground (or both) when the wheels move. It could also include any tendency of the wheels to stick to the ground.

As garyp says, all of the resistance forces (including air resistance) are usually summarised in a single force-arrow on the FBD, in a direction which opposes relative motion between the car and the ground.

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This is an interesting question.

Let's ignore rolling friction. The forces that would slow the car are friction in the bearings. If we take the system to be the car and all of its parts, these forces occur inside the system. It's hard to see how they can be considered to be an external force on the car.

Ultimately, however, the drag on the wheels is transferred to friction between the tires and the road in exactly the same way as the forward force generated by the motor is transferred to friction. If you imagine the car to be moving at a constant speed and think about the interface between the tires and the road, it is sensible that the net force of static friction at the point of contact is zero.

One of the reasons that this is hard to think about is that the system has a source of energy and can do work, which implies that the system can generate a force. The force that accelerates the car is the reaction force to the force that the car itself generates.

So for your free body diagram you will have two arrows. One pointing forward, the other pointing backwards. Both are friction forces. They might need some better labeling to better identify them.

Update

I think the point of view taken in the answer by @xasthor is a better analysis. I'll leave answer up because the discussion of the internal parts of the car might be useful to someone.

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    $\begingroup$ Including rolling friction makes the diagram much more interesting - the force on the front and rear will not be the same, for example (especially if it is two wheel drive only) $\endgroup$ – Floris Aug 28 '16 at 16:15

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