# How can an engine provide more torque at the point of contact than the ground's response by static friction without slipping?

In response to the top answer here: Newtonian Mechanics - How does something like a car wheel roll?

I clearly fundamentally misunderstand something here. I can't reconcile there being an angular acceleration about the tires (and thus, a net torque) and the torque provided by the engine being equal to the torque in response to static friction. If the torque provided by the engine was greater than the torque in response to static friction, wouldn't the wheel slip?

update: My confusion was resolved in chat with @V.F., and for anyone else who might have had the same simple misunderstanding in this system, it came down to a misunderstanding of contact forces.

As stated, the response from static friction is equal to the force of the tyre on the road, but the force of the tyre on the road (and of the road of the tyre though static friction) is NOT equal to the force of the engine on the tyre. This is because the tyre delivers the force to the road through contact, meaning it must physically accelerate to bring the atoms closer to the atoms of the tyre on the road, resulting in both the tyre and the road experiencing a force in opposing directions via Coloumbs Law. Therefore, a little bit of the engine force necessarily goes into accelerating the tyre into the road, and the rest of it is is then delivered between the road and the tyre as a contact force, through static friction. This delta is what results in the constraint of rolling motion, as the tyre experiences both a clockwise rotational acceleration from the engine to bring into contact with the road, and then a forward translation via static friction.

This was easier to understand in the simpler translational case that is typically used to illustrate the difference between contact force and input force. Imagine two boxes, A and B, of equal mass, M, on a frictionless surface. If you exerted a force, F, on box A, it's clear that both boxes would equally accelerate as one system over the surface - but if you exert a force on box A, and box A is in contact with box B, wouldn't box B respond with an equal and opposite force? Then why would box A accelerate at all? i.e. why isn't the force, F, perfectly transmitted through box A to box B? Well, because box A and box B are, of course, not actually in contact. The atoms at their respective surfaces are at some mutual equilibrium caused by Columbus Law. So, to transmit force F to box B would require accelerating box A into box B - i.e. force F is actually distributed between box A and box B based on their respective masses, as "contact" in this sense requires the acceleration of box A into box B, and thus the contact force between box A and box B is necessarily less than the input force F.

• Where did the question come from? – Bob D Sep 1 '18 at 21:42
• @BobD apologies, what do you mean exactly? my question came in response to the answer linked – roozbubu Sep 1 '18 at 21:58

We can treat a driving wheel as just another gear that has to pass the torque down the line.

Ideally, the torque it passes should be equal to the torque it receives, but, since the wheel has some mass and, therefore, some moment of inertia, a little torque differential is needed to speed it up. At steady state, though, the torque on both sides should be the same.

The wheel may slip, if the drive torque exceeds the maximum torque that can be supported by static friction, which can happen, for instance, on a slippery surface (low static friction) or when the car is overloaded or somehow is blocked from moving or there is an attempt to accelerate it too fast.

• "Ideally, the torque it passes should be equal to the torque it receives, but, since the wheel has some mass and, therefore, some moment of inertia, a little torque differential is needed to speed it up. At steady state, though, the torque on both sides should be the same." – roozbubu Sep 1 '18 at 21:59
• ^ could you expand on this? I think this is where my understanding falls short. Even if the engine torque is slightly magnified by the wheel radius, wouldn't the response from static friction still come in response to this magnified torque (as this is at the point of contact)? I suppose my question is this: how can there exist a torque differential (against static friction) while the wheel doesn't slip? – roozbubu Sep 1 '18 at 22:02
• @roozbubu For the car to start moving, the wheel has to start spinning, i.e., it has to be accelerated. To accelerate the wheel, some net torque is needed in accordance with the Newton's second law for rotation: $(\tau=I\alpha)$. If the counter torque acting on the wheel due to static friction force was equal to the drive torque, the wheel would not accelerate. – V.F. Sep 1 '18 at 22:29
• ^ right, and this is exactly my point of confusion. I see why this has to be the case, but don't understand how there exists a net torque AND the wheel doesn't slip (relative to the ground). I feel like I must fundamentally misunderstand one of these concepts, any help is greatly appreciated. – roozbubu Sep 1 '18 at 22:34
• @roozbubu The wheel does not slip, because it can roll and, when it rolls, it acquires angular speed. – V.F. Sep 1 '18 at 22:37

This means that the torques do no balance out, which seemed to be the assumption that caused your whole confusion. The applied torque from the engine will not equal the torque from friction.

Friction in this case is supplying a torque, but that torque is less than the engine torque.

... if the wheel isn't slipping, which (seemingly) implies that the torque from the engine is equal to the torque from static friction

It does not. Static friction is just another force here. There's no reason to suppose that just because it's not slipping that the magnitude is equal to the opposing torque from the engine.

Assuming no slip/infinite grip, you can set up three equations. The first two don't depend on whether the wheel slips or not.

Net force (which in the x-direction is due entirely to friction) is equal to the mass times the acceleration of the vehicle. $F_f = ma$

Net torque is equal to the rotational acceleration of the wheel multiplied by the moment of inertia: $\tau_{engine} - R_{wheel}F_f = I\alpha$.

These will always be true. If the car is on ice and the wheel slips, then $F_f$ goes way down. This means the car doesn't accelerate and the torque of the engine is able to make the wheel spin much faster. Plug a zero in for it in the equations and see what happens.

The no slip condition ties these two equations together and says the vehicle acceleration is equal to the wheel's tangential acceleration: $a_{vehicle} = R_{wheel}\alpha$. Now you can't have a slow vehicle and a fast wheel.

I suppose the question is then why wouldn't the force due to static friction be equal and opposite to the torque applied by the engine at the wheel?

It seems unsatisfying, but "because it doesn't have to be".

If you take a wheel assembly and keep it from moving (mount the axle in a frame attached to the ground), then any torque you apply to the axle is counteracted by an equal amount of friction (up to the slip point). The friction rises as necessary to prevent the wheel's rotation.

But since the axle isn't fixed to the ground, it can rotate without slipping. The contact patch has no slip relative to the ground (which is what friction is maintaining), but the wheel is still increasing its angular rotation.

wouldn't the torque at the wheel be slightly greater than the torque at the axel, due to the radius/mass of the wheel creating a moment of inertia?

Just like the linear form $F_{net} = ma$, we can say $\tau_{net} = I\alpha$. There are only two torques on the wheel: the engine/transmission torque through the axle and the friction force via the tread. Since we are assuming the vehicle (and wheel) are accelerating forward, then there must be a net torque in that direction. Therefore the engine torque in that direction must be stronger than the friction torque in the opposite direction.

isn't there a difference between the torque the engine applies at the axle and the corresponding torque at the edge of the wheel? (due to the radius and mass of the wheel)

Ah... I see now what you're asking. Since the ground doesn't rotate, I think it's unusual to refer to the force that the wheel exerts on the ground as a torque.

But you're correct that it is unequal. We know the force the tire pushes on the ground is equal to the force the ground pushes on the tire. If the torque from the engine and the torque from the ground were equal, the tire's rotation would not accelerate.

So one way to think about it is that the engine torque is doing two jobs: it's accelerating the tire rotation and it's pushing on the ground. It can't push on the ground as much as it would if the tire were massless.

• I suppose the question is then why wouldn't the force due to static friction be equal and opposite to the torque applied by the engine at the wheel? And (perhaps related) wouldn't the torque at the wheel be slightly greater than the torque at the axel, due to the radius/mass of the wheel creating a moment of inertia? – roozbubu Sep 2 '18 at 0:59
• Here's diagram of how I currently understand the system: i.stack.imgur.com/3lW0z.jpg – roozbubu Sep 2 '18 at 1:14
• That's an odd diagram. I can guess "axle torque" means engine/transmission torque trying to accelerate the wheel. But what is "wheel torque"? What is creating that torque? – BowlOfRed Sep 2 '18 at 4:39
• Sorry if this was confusing - isn't there a difference between the torque the engine applies at the axle and the corresponding torque at the edge of the wheel? (due to the radius and mass of the wheel) – roozbubu Sep 2 '18 at 16:32
• It's unusual to call a force applied to an object that doesn't rotate a torque. So while the ground's force on the wheel can be termed a torque on the wheel, the force the wheel applies to the ground would not normally be called a torque. – BowlOfRed Sep 3 '18 at 5:27

Yes. If you apply too much torque to the wheels, they will spin.

Since the coefficient of dynamic friction is usually less than the coefficient of static friction, spinning the wheels can look and sound impressive but it actually produces worse acceleration of the car than not spinning them.

Most cars (not just high performance cars) will spin the wheels driving on loose surfaces like gravel, or on snow and ice, if you are too heavy-footed on the gas pedal.

• Apologies, it seems my question wasn't actually clear. To clarify: I'm not confused about the slip condition (where torque exceeds the limit of static friction, and the wheel slips, but kinetic friction still pushes the car forward). I'm confused about the no-slip condition - since the wheel itself does experience an angular acceleration, wouldn't this imply that there is a net torque acting upon it? But how is this possible if the wheel isn't slipping, which (seemingly) implies that the torque from the engine is equal to the torque from static friction (and thus, there is no net torque)? – roozbubu Sep 1 '18 at 21:56