Car's maximum acceleration on a road is proportional to what?

[rudransh verma]

-- the tires push the car forward and the car pushes back on the tires. (Newton's third law!)

How does the tire push the car? Are you talking about the force that engine delivers?

 

[Doc Al]

How does the tire push the car?

The tires are connected to the car. The road (via friction) pushes on the tires and the tires push on the rest of the car. (And the rest of the car pushes back on the tires, of course.)

Are you talking about the force that engine delivers?

No.

 

[bob012345]

Why? What about the torque delivered by engine to the wheels?

The torque turns the wheel but because of the static friction between the wheel and the road, the wheel pushes on the road and the road pushes back on the wheel according to Newton's Third Law. The wheel then transmits that force to the rest of the car and the whole car accelerates together.

Again, the torque is an internal torque between the wheel and the rest of the car. It cancels out when you treat the car as one single unit. All you have left with is the whole car of mass $m$ and the friction force which accelerates the car. You actually had the correct answer above in post #15 but you gave the wrong reasoning.

Note that the only time you might have kinetic friction between the tires and the road is if you are burning rubber or sliding on a wet or icy road.

 

[jack action]

Yes but even this is leaving out the force between the body of the car and the wheel which I am in favor of

No, it doesn't, it is applied at the axle on the shown figure. That is the applied force to the body that the OP is looking for; assuming the wheel mass is included with the body of the car for simplification.

but we should not say this is the diagram of the wheel alone but that it represents the whole car in this problem.

No, there is no torque applied to consider when analyzing the whole car.

The wheel torque explains why the friction force seems to be "backward" compared to a sliding block. This is what the OP doesn't seem to understand. The wheel mechanism reverses the friction force direction, thus propelling the car instead of braking it.

 

[rudransh verma]

The torque turns the wheel but because of the static friction between the wheel and the road, the wheel pushes on the road and the road pushes back on the wheel according to Newton's Third Law. The wheel then transmits that force to the rest of the car and the whole car accelerates together.

Edited:
Ok! I got it now! So as I was thinking $F_a(applied force by the wheel)=f_s$.static friction. Both $F_a$ and $f_s$ are necessary for acceleration of the car. Without $F_a$ there is no $f_s$ and without $f_s$ car would not accelerate.
So, $$f_s=ma$$
M is mass of car body and m is mass of tyre.
And the rest is same in #post15.
Newtons laws hold true!
Thanks everyone!

 

[rudransh verma]

If we assume that this is a four-wheel drive vehicle with all four wheels contributing equally then each wheel will contribute 1/4 of the force required to propel the body of the car forward. By Newton's second law, the total forward force from wheels on car body must be Ma. By assumption, each wheel contributes 1/4 of that. That is, the force between each wheel and the body of the car will have magnitude Ma4. Or (M/4)a if you prefer.

…$$ma=(M/4)a$$…

 

[jbriggs444]

$$ma=(M/4)a$$

Incorrect. Please provide your reasoning for this assertion.

 

[rudransh verma]

Incorrect. Please provide your reasoning for this assertion.

Four frictions ma provide four forces to the wheels and that inturn provide necessary force Ma to the body of the car.

 

[jbriggs444]

Four frictions ma provide four forces to the wheels and that in turn provide necessary force Ma to the body of the car.

No. The force with which the road pushes on the wheels is not identical to the force with which the wheels push on the car's body.

Let us go back to the assertion that you had made:$$ma = (M/4)a$$We divide by $a$ to get$$m=M/4$$But that is an assertion that each wheel has mass equal to 1/4 of the car's mass.

No. They do not.

 

[rudransh verma]

But that is an assertion that each wheel has mass equal to 1/4 of the car's mass.

Yeah! Right! Then?

 

[jbriggs444]

Yeah! Right! Then?

Well, the point is that you are working the problem incorrectly, jumping boldly to unjustified conclusions. The proper response to that would be to solve the problem correctly.

Pick variable names for the relevant quantities and document them. Enumerate the forces acting on each free body. Write down equations that you can justify, typically using Newton's second or third laws. Include the justification along with the equations. Solve the equations. Show your work. Check your answers to see if the solution you derive actually fits the givens of the problem.

 

[rudransh verma]

correctly.

Pick variable names that you have documented. Enumerate the forces acting on each free body. Write down equations that you can justify,

Hmm! Right but now I am tired

 

[bob012345]

No, it doesn't, it is applied at the axle on the shown figure. That is the applied force to the body that the OP is looking for; assuming the wheel mass is included with the body of the car for simplification.No, there is no torque applied to consider when analyzing the whole car.

The wheel torque explains why the friction force seems to be "backward" compared to a sliding block. This is what the OP doesn't seem to understand. The wheel mechanism reverses the friction force direction, thus propelling the car instead of braking it.

Frankly, I'm a bit confused looking at this free-body diagram. Is it for just the wheel or the whole car? It seems it would be hard to understand in either case since all the forces apparently cancel out even though it assumes the case of acceleration.

 

[rudransh verma]

Frankly, I'm a bit confused looking at this free-body diagram. Is it for just the wheel or the whole car? It seems it would be hard to understand in either case since all the forces apparently cancel out even though it assumes the case of acceleration.

This diagram is confusing me too since friction>force applied by the wheel.

 

[bob012345]

This diagram is confusing me too since friction>force applied by the wheel.

Ignore the green friction. That is something called rolling friction which is a different kind of resistance to rolling motion and is not considered in your problem. The only friction you need for this problem is static friction, what they gave you, because the wheel is not slipping with respect to the ground. The wheel applies a force to the ground and the ground applies a force to the wheel. The wheel accelerates forward and the ground (road) accelerates backward (again Newton's Third Law) but at such an imperceptibly small rate as to be considered zero.

Also note, you were not given the mass of the wheels, just the car as a whole. That is a big clue you don't need to treat this as a complicated problem with multiple parts. Just one car as a whole of total mass $m$.

 

[rudransh verma]

Ignore the green friction. That is something called rolling friction which is a different kind of resistance to rolling motion and is not considered in your problem. The only friction you need for this problem is static friction, what they gave you, because the wheel is not slipping with respect to the ground. The wheel applies a force to the ground and the ground applies a force to the wheel. The wheel accelerates forward and the ground (road) accelerates backward (again Newton's Third Law) but at such an imperceptibly small rate as to be considered zero.

I may have missed something before but I am again asking if it’s static friction between the road and tyre how is the tyre/car accelerating?

 

[bob012345]

I may have missed something before but I am again asking if it’s static friction between the road and tyre how is the tyre/car accelerating?

It is static friction that allows the car to roll. Rolling is not slipping. As a wheel rolls each point comes in contact with the road without sliding. Try it yourself with a small wheel. Mark a point and watch it as you roll it. The point does not slide as in kinetic friction.

 

[rudransh verma]

It is static friction that allows the car to roll. Rolling is not slipping. As a wheel rolls each point comes in contact with the road without sliding. Try it yourself with a small wheel. Mark a point and watch it as you roll it. The point does not slide as in kinetic friction.

But this static friction = applied force. Right?

 

[jbriggs444]

But this static friction = applied force. Right?

What do you mean when you say "applied force"? If you do not tell us what force you are talking about, we cannot tell whether it is static friction.

 

[rudransh verma]

What do you mean when you say "applied force"? If you do not tell us what force you are talking about, we cannot tell whether it is static friction.

The torque

 

[jbriggs444]

The torque

No. The torque is not equal to the force of static friction. The two do not even have the same units. Torque is also not what I would have expected "applied force" to have meant.

 

[rudransh verma]

It is static friction that allows the car to roll.

Are you saying the tyres do not accelerate. It’s the rolling not sliding that is going on. Static friction is causing the roll. And that is creating the acceleration in the car.

 

[haruspex]

Yes, It seems to me we are overcomplicating a very simple problem and it is not helping the OP understand the physics in my view.

It is the OP that has complicated it by worrying about the wheels separately. See post #15.

 

[bob012345]

Are you saying the tyres do not accelerate. It’s the rolling not sliding that is going on. Static friction is causing the roll. And that is creating the acceleration in the car.

Of course they accelerate because they move along with the car.

I'll try something else here. Take your finger and push against a surface. Do not let it slip[. If the surface is loose it will move, if it is rigid it will resist moving and you might move a bit opposite to your push. A tire is just like having an infinite set of fingers that can push against the road as each one comes in contact.

Or else, take a little wheel and roll it along a table. Apply torque with your hand and feel how the wheel wants to move.

 

[jbriggs444]

Are you saying the tyres do not accelerate. It’s the rolling not sliding that is going on. Static friction is causing the roll. And that is creating the acceleration in the car.

The tires do accelerate. Otherwise they would be left behind, begging for the car to "wait up".

Yes, it is rolling rather than sliding. This allows the bulk of the tire to move forward while the contact patch remains (nearly) stationary while it is in contact with the stationary road.

Static friction does not cause the roll. It resists the roll. The force of static friction forward on the bottom of the tire tends to reduce the tire's roll rate. The engine (and transmission and differential) forces the roll. The roll causes the strain that causes the static friction.

 

[haruspex]

It resists the roll.

I worry that the various contributors are now causing destructive interference with slightly different ways of expressing things.
It resists sliding, but neither would I say it "allows" the roll. It encourages rolling rather than sliding.

 

[rudransh verma]

The roll causes the strain that causes the static friction.

If there is static friction where is the force that generates it? Please answer this one question.

 

[bob012345]

It is the OP that has complicated it by worrying about the wheels separately. See post #15.

Please see post#14.

 

[bob012345]

I worry that the various contributors are now causing destructive interference with slightly different ways of expressing things.
It resists sliding, but neither would I say it "allows" the roll. It encourages rolling rather than sliding.

I agree and I'd be willing to work privately with the OP to resolve the issues if you like. Or would you rather?

 

[jack action]

I may have missed something before but I am again asking if it’s static friction between the road and tyre how is the tyre/car accelerating?

Looking at the following figure:

• Ignore the green arrows;
• The wheel torque $T_a$ is shown in blue (torque direction). That is the applied torque coming from the motor;
• The reaction force to this torque is the Friction force $F_f$. It is pushing the wheel forward;
• The blue force at the axle (center of the wheel) is equal and opposite to the friction force $F_f$. The distance between the two blue forces (wheel radius $r$) times the force's magnitude represents a couple that opposes the wheel torque;
• $F_f = \frac{T_a}{r}$. The maximum possible value is equal to the static friction force;
• If you would do the free body diagram for the car's body, that axle force would show as equal and opposite, thus pushing the car's body forward - so with the same direction as $F_f$.

The lesson in that is that the direction of the torque dictates the direction of the friction force at the tire contact patch and that friction force is the "force applied" that you talk about and pushes the entire car, coming directly from the torque applied at the wheel.

 

[bob012345]

Looking at the following figure:

View attachment 296562​

• Ignore the green arrows;
• The wheel torque $T_a$ is shown in blue (torque direction). That is the applied torque coming from the motor;
• The reaction force to this torque is the Friction force $F_f$. It is pushing the wheel forward;
• The blue force at the axle (center of the wheel) is equal and opposite to the friction force $F_f$. The distance between the two blue forces (wheel radius $r$) times the force's magnitude represents a couple that opposes the wheel torque;
• $F_f = \frac{T_a}{r}$. The maximum possible value is equal to the static friction force;
• If you would do the free body diagram for the car's body, that axle force would show as equal and opposite, thus pushing the car's body forward - so with the same direction as $F_f$.

The lesson in that is that the direction of the torque dictates the direction of the friction force at the tire contact patch and that friction force is the "force applied" that you talk about and pushes the entire car, coming directly from the torque applied at the wheel.

Is the blue force at the axle on the wheel or not?

 

[jack action]

Is the blue force at the axle on the wheel or not?

Yes. It is the force at the axle bearing.

 

[Steve4Physics]

@rudransh verma, maybe this helps...

In this problem the internal forces and torques between parts of the car don’t matter - they are in Newton 3rd law pairs and cancel overall.
_____________

Q: You sit in your 4-wheel drive car, on perfectly smooth ice, and try to drive off. What happens and why? [Click fuzzy area to check your answer.]

A: The wheels turn but the car doesn’t move. That’s because there is zero friction (no ‘grip’) between road and tyres. There is no horizontal applied force ($F_a=0$) on the car to cause acceleration.
_____________

Q: You do the same on a rough road. What happens and why? [Click fuzzy area to check your answer.]

A: The wheels turn, the tyres exert a backwards frictional force on the ground. Because of Newton’s 3d law, the ground exerts a forwards frictional force on the tyres. This is the force that makes the car accelerate.

In this situation, the applied force ($F_a$) which causes the car’s acceleration is just another name for the frictional force ($f$) of the ground on the tyres. They are not two different forces.

All things being equal, the frictional force on each of the tyre is ¼ of the total. (You might want to consider if this is still true for a 2-wheel drive car.

 

[bob012345]

Yes. It is the force at the axle bearing.

Then how can the wheel accelerate if the two blue forces are equal and opposite and both act on the wheel?

 

[rudransh verma]

The lesson in that is that the direction of the torque dictates the direction of the friction force at the tire contact patch and that friction force is the "force applied" that you talk about and pushes the entire car, coming directly from the torque applied at the wheel.

Edit: So $f_r-Torque=ma$ is wrong or the eqn $f_r-F_a=ma$. Torque is the one which generates static friction and this friction accelerates the car forward.