I'm studying rolling without slipping, I want to find the net force making a wheel go forward when applying a torque about the axle of the wheel.

In this page https://www.school-for-champions.com/science/friction_rolling_starting.htm#.XTz4vpNKgWo

It says that the force causing forward motion is

$F = F_e − F_f$

Where $F_e$ is the force at the edge/ground caused by the applied torque and $F_f$ if the static friction force.

My question is: what if friction $F_f$ is very small (like on ice). Woudn't this mean $F ≈ F_e$ and $F$ would be very big? A wheel on ice should have a smaller $F$, not bigger. Something doesn't add up... what makes $F$ be small when $Ff$ is also small?


3 Answers 3


I think the page is not written very clearly. I would have described it differently.

Initially it is considering a portion of a wheel on the edge (such as one of the tread pieces). If we imagine a wheel in the air, without any forces on it, the block will stay stationary. Then by applying a torque on the axle, a force will be transmitted to the tread.

$$F_{net} = F_e$$

Since the force is non-zero, the block will accelerate (the wheel will spin).

When we put the wheel on the ground, a friction force can develop. Under most circumstances, the static force of friction will be sufficient to keep the bottom of the wheel motionless against the ground. If it stays motionless, we can assume it is not accelerating and therefore has zero net force.

$$F_{net} = 0 = F_e + F_f$$ or $$F_f = -F_e$$

The force from torque is an internal force. Any momentum it puts on the tread, it puts opposite momentum on something else (like the axle or another bit of tread). The force from friction is an external force and affects the total momentum of the car.

what if friction Ff is very small (like on ice). Woudn't this mean F≈Fe and F would be very big?

Yes. This means that without friction, if you hit the accelerator, the tread on the wheel will accelerate because the wheel will quickly rotate. The car won't go forward though.

  • $\begingroup$ How can the wheel accelerate without a net force acting on it? $\endgroup$ Jul 28, 2019 at 3:44
  • $\begingroup$ @AaronStevens, can you explain your question in more detail? The wheel does not accelerate without a net force. I don't think I suggest that it does. It can rotate only from the non-zero axle torque, or it can move forward only from the non-zero force of friction. $\endgroup$
    – BowlOfRed
    Jul 28, 2019 at 3:58
  • $\begingroup$ Sorry. I really just meant to ask why you assume the wheel has no net force acting on it yet expect it to be able to move. $\endgroup$ Jul 28, 2019 at 4:04
  • $\begingroup$ @AaronStevens I nowhere assume that. Perhaps you are mistaking when I refer to the net force on a section of tread (one part of the wheel) and when I refer to the force on the wheel itself. It is the tiny piece of tread that is touching the ground that has no net force (and which isn't moving at that point). The wheel as a whole has a net force (equal to friction) acting on it. $\endgroup$
    – BowlOfRed
    Jul 28, 2019 at 4:36
  • $\begingroup$ Oh ok so you mean to say there is no net horizontal force acting on the part of the wheel that touches the ground. Thanks. $\endgroup$ Jul 28, 2019 at 4:43

Consider the dynamics of the wheel. It's angular acceleration is given by the net torque on it and it's angular inertia. That net torque in turn is the torque on the axle minus the torque due to the force at the rim.

That force at the rim is the net force accelerating the vehicle. How big is it? It's given by F = ma for the vehicle.

OK, we know the mass of the vehicle, but what's its acceleration? That's where rolling without slipping comes in: It has to be the same as the radius times the angular acceleration of the wheel so that it keeps rolling without slipping.

So now you can construct the set of equations to solve for whatever items you need.


Your equation $F = F_e − F_f$ gives the net force acting on the wheel. If $F_f$ is really small, then the net force will be nearly equal to $F_e$ since that is essentially the only force acting on the wheel.

Notice that this doesn't say anything about the surfaces in question. You haven't included any assumptions about that in your analysis so far. You are just saying "If I apply some force $F_e$ and I know the friction force is $F_f$, then the net force is $F=F_e-F_f$."

Also note that for a car wheel on ice, there isn't a singular applied force pushing on the edge of the wheel. The mechanisms of the car apply a net torque without a net force on the tires. Therefore this analysis isn't correct for a car wheel. This would instead be the case if you had a wheel on the ground and then applied a force on the top edge with your hand, or if you pushed from the axel itself. Really this is for any case where you are applying a single, horizontal force to some part of the wheel. This might be where some of your confusion lies.

  • $\begingroup$ I see, that makes sense. How do I analyze the case where a net torque is applied on the wheel without a force applied on the edge? Can you point me to a post or article? Thanks! $\endgroup$ Jul 28, 2019 at 2:48
  • $\begingroup$ @carlosdubusm Just use $F=ma$, $\tau=I\alpha$, and your non-slip condition $a=\alpha R$. $\endgroup$ Jul 28, 2019 at 2:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.