How does static friction propel a car forward given a torque applied to the wheels?

  • #1
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Summary:
Curious about how static friction propels car forward given torque applied to wheel
Ok so please bear with me here and what is almost certainly a really stupid question, partly because I don't quite know how to ask it.

When you have a wheel such as one attached to a car, a torque is applied to it from the engine. It's my understanding that the wheel would slide over the road and the car would not move if it were not for static friction which is of course the concept of rolling without slipping.

The problems I am having is when I consider what the magnitude of the force exerted on the road from the wheel would be, or specifically the equal and opposite force exerted back onto the wheel from the road via static friction. The reason for this is because I want to say that it is just the torque exerted on the wheel by the engine divided by the radius of the wheel but that seems to lead to a ridiculous conclusion. If I follow that logic, then the equal and opposite force exerted on the wheel by the road, multiplied by the radius of the wheel to get the counter-torque, is equal to the torque exerted on the wheel by the engine but in the opposite direction. That of course would mean that there is no net torque and the wheels would never move (unless the force were greater than that available from static friction but then it would just slip) which means that the car would never be able to move!

Does the torque on the wheel become the force on the road, such that the only force the wheel feels is the reaction from the road on the tire propelling it forward?

So simply put, what am I missing that is almost certainly staring me in the face (again)? Clearly things roll and cars move so I know i'm going horribly and embarrassingly wrong somewhere.
 

Answers and Replies

  • #3
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Think about the difference between 'move' and 'accelerate.'
 
  • #4
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Draw a FBD; that should help.
 
  • #5
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IF I draw a freebody diagram on the wheel, all I see is a torque on the wheel and a torque from friction, which cancel out

1569338029321.png


I am missing something and icannot figure it out. which is why I am asking it again, but hopefully with more clarification
 
  • #7
A.T.
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...the equal and opposite force exerted on the wheel by the road, multiplied by the radius of the wheel to get the counter-torque, is equal to the torque exerted on the wheel by the engine but in the opposite direction.
Only at constant speed and ignoring rolling resistance. Otherwise the torque from the engine is different from the torque by static friction.
 
  • #8
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Only at constant speed and ignoring rolling resistance. Otherwise the torque from the engine is different from the torque by static friction.

okay, but then how does the force of static friction move the car forward?
 
  • #9
berkeman
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okay, but then how does the force of static friction move the car forward?
When you are climbing stairs, how does the force of the stair pushing back on your foot help you to move upward? :smile:
 
  • #10
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Okay, I understand that the force from the wheel on the road causes an equal and opposite force on the wheel by the road. This is Newtons thirds law. By that, the car accelerates much like the climbing stairs example.

What I am confused by is the torque. How does the torque applied by the axle on the wheel get to be a force on the road. Does the torque applied at the axle translate to the force on the road? So really the torque at the axle is just how the we get the force on the road?

I am trying to understand how the torque from the car/axle becomes the backward push on the road.
 
  • #11
A.T.
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What I am confused by is the torque. How does the torque applied by the axle on the wheel get to be a force on the road. Does the torque applied at the axle translate to the force on the road? So really the torque at the axle is just how the we get the force on the road?

I am trying to understand how the torque from the car/axle becomes the backward push on the road.
It's not clear what you mean by all those "get to be", "translate to" and "becomes".
 
  • #12
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It's not clear what you mean by all those "get to be", "translate to" and "becomes".

There is a torque applied the tire/wheel correct? It is large as it comes from the engine. But the tire applies a force back upon the road. I am trying to understand why/how we are applying a torque from the axle but yet the force is felt on the ground

Take a bike, turn it upside down. Turn the pedal and put your other hand on the wheel. You can feel the force from the wheel on your had. But we applied a torque thru the pedal. How/why does this work?
 
  • #14
rcgldr
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A small part of the torque from the engine is accelerating the drive train and tire, and the equal and opposing torque to this small part of the torque is related to the angular momentum inertia of the drive train times the rate of acceleration of the drive train, and doesn't contribute to the Newton third law pair of forces of the tire onto the pavement and of the pavement onto the tire. The net result is that the total torque from the engine is slightly greater than the opposing torque related to the forward force from the ground, and the drivetrain accelerates in correspondence to the rate of acceleration of the car.
 
Last edited:
  • #15
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A small part of the torque from the engine is accelerating the drive train and tire, and the equal and opposing torque to this small part of the torque is related to the angular momentum of the drive train times the rate of acceleration of the drive train, and doesn't contribute to the Newton third law pair of forces of the tire onto the pavement and of the pavement onto the tire. The net result is that the total torque from the engine is slightly greater than the opposing torque related to the forward force from the ground, and the drivetrain accelerates in correspondence to the rate of acceleration of the car.

So some torque is being used to accelerate drive train and wheels. I agree, that doesn't contribute to the N3L reaction/action pair of the road and tires

So that means the rest/majority of the torque from the engine is being applied as the force of the tires onto the road right? And then because of this, the car is pushed forward by the static friction.

Does this makes sense now, is that correct? Please and thank you all for the help
 
  • #16
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A small part of the torque from the engine is accelerating the drive train and tire, and the equal and opposing torque to this small part of the torque is related to the angular momentum of the drive train times the rate of acceleration of the drive train, and doesn't contribute to the Newton third law pair of forces of the tire onto the pavement and of the pavement onto the tire. The net result is that the total torque from the engine is slightly greater than the opposing torque related to the forward force from the ground, and the drivetrain accelerates in correspondence to the rate of acceleration of the car.


RCGLDR, is this website below correct as well. I think that explains where the majority of the torque is going, applying the force to the road.
https://www.school-for-champions.com/science/friction_rolling_starting.htm#.XYo7bBhOk0M
 
  • #17
Dale
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Summary: Curious about how static friction propels car forward given torque applied to wheel

The problems I am having is when I consider what the magnitude of the force exerted on the road from the wheel would be, or specifically the equal and opposite force exerted back onto the wheel from the road via static friction.
Have you tried drawing a free body diagram and applying Newton’s laws? You asked an almost identical question recently and received good answers. If those answers didn’t do it for you then most likely more answers will not help. You may need to sit down and work this out rigorously. Just draw the free body diagram and apply Newton’s laws.
 
  • #18
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Dale, I think that is part of the problem. If I draw a fbd on the ground, there is a force from the tire pushing backwards. If I draw a fbd on the tire, i see the torque of the axle and counter torque of friction. These cancel Then the car wouldn't move, but that makes no sense.

I was hoping someone could draw out what I am missing.
 
  • #19
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I get the free body diagram of the car. Only external force is the force of friction.

What I have been asking is how does the torque applied at the wheel via the axle get to be the force pushing the road back? No one can seem to answer that for me
 
  • #20
jbriggs444
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If I draw a fbd on the tire, i see the torque of the axle and counter torque of friction. These cancel
You say the torque of the axle on the tire and the torque of friction on the tire cancel. Can you share the reasoning leading to this [false] conclusion?

It is worth pointing out that these are torques. If you expect the car's linear momentum to change, you need a net force.
 
  • #21
Dale
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If I draw a fbd on the ground, there is a force from the tire pushing backwards. If I draw a fbd on the tire, i see the torque of the axle and counter torque of friction. These cancel Then the car wouldn't move, but that makes no sense.
After you draw the FBD then the next step is to write down the equations according to Newton’s laws and any other relevant laws. Can you show how those equations lead to the cancellation? Which equations are you using to determine that they cancel?
 
  • #22
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You say the torque of the axle on the tire and the torque of friction on the tire cancel. Can you share the reasoning leading to this [false] conclusion?

It is worth pointing out that these are torques. If you expect the car's linear momentum to change, you need a net force.
Doesn't the static friction of the ground pushing back on the tire create a torque? If so isn't this the same as the torque being applied to the wheel itself. So aren't those torques acting on the same obejct?

Yes I need a net force. I see it's from static friction from the ground on the car as a whole. But I don't understand how the torque we apply gets to be a backwards push on the ground. Do you see what I am asking?
 
  • #23
Dale
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Doesn't the static friction of the ground pushing back on the tire create a torque?
Yes

If so isn't this the same as the torque being applied to the wheel itself.
No. Can you show your cauulations or the equations that would lead you to believe they are equal? I think that attempting to do so will help you.

So aren't those torques acting on the same obejct?
Yes
 
  • #24
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I don't have a problem with numbers Dale, so I don't have the equations. Im just trying to see conceptually how the torque on the tire from the axle causes there to be a net linear force forward on the car.

When I think or draw out the tire, I see it has a torque from the axle. I see there is a torque from friction. The reminder net torque causes wheel to roll. But then what I read say it is the net external friction force that propels the car forward. Where does that come into play?
 
  • #25
jbriggs444
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Doesn't the static friction of the ground pushing back on the tire create a torque? If so isn't this the same as the torque being applied to the wheel itself. So aren't those torques acting on the same obejct?
If you are distinguishing between tire and wheel then no, the torque applied to the tire by the ground can be different from the torque applied to the wheel by the tire.

There is a relevant equation. It is the rotational analog of Newton's second law.
 

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