Friction ON wheels (say a car driving or accelerating)

[alkaspeltzar]

I know it takes friction for a car/wheels to move forward. However, I am confused by the free body diagram.

As a torque is applied to the wheel, the wheel applies a force to the road, and as a reaction, the road puts a force on the wheel(this is friction). Then wouldn't the torque of friction on the wheel(Fs X Radius of wheel) cancel out the torque of the axle on wheel, thus no movement?

I also don't see how the force of friction on the wheel from the road causes it to act at the center of wheel driving it forward.

Thanks for the help. Below were some sites I was using for help
http://www.geocities.ws/fcsonic/physics/FrictiononWheels.htmhttp://www.lsop.colostate.edu/2018/07/16/everything-you-need-to-know-about-force-motion/

 

[Dale]

Then wouldn't the torque of friction on the wheel(Fs X Radius of wheel) cancel out the torque of the axle on wheel, thus no movement?

The torque from the road definitely reduces the net torque (the wheel's angular acceleration is much higher on ice than on asphalt for the same motor torque), but it does not have the same magnitude as the motor torque so it does not cancel out. Is there any reason why you think that the motor and the road torques would be equal magnitude?

I also don't see how the force of friction on the wheel from the road causes it to act at the center of wheel driving it forward.

It doesn't act at the center of the wheel. It acts at the bottom surface of the wheel. If it acted at the center then it would not produce a torque on the wheel, and as we already discussed it clearly does produce a torque.

 

[alkaspeltzar]

Dale, I am probably confusing the torque. For right now I am going to ignore that question.

I guess what has me stumped is in my physics book, there is a free body diagram of the car similar to the what I attached above. It shows the only forces acting on the car are its weight and the normal force. These cancel out. And then the force of static friction, which is the reaction to the wheel pushing horizontally back against the ground.

They say it is the friction force from the ground on the wheel that propels the car forward. How? I see the friction force pushing that the bottom of the tire, but it is static friction. The cars tire doesn't move relative to the ground. Yet the entire car moves and accelerates forward.

So my question is how does the friction force pointing in the direction of the cars motion, but it only acts on the bottom of the tire, actually do so?

 

[russ_watters]

I know it takes friction for a car/wheels to move forward. However, I am confused by the free body diagram.
View attachment 249807

That's not a free body diagram. A free body diagram shows all of the forces acting on an object only that object. That diagram shows a Newton's 3rd law force pair between the tire and road. A free body diagram would show a torque applied by the axle and a backwards force applied by the shaft bearing. If the forces and torques do not sum to zero, the car (wheel) accelerates.

So my question is how does the friction force pointing in the direction of the cars motion, but it only acts on the bottom of the tire, actually do so?

Think about what happens if instead of a tire, the forces are applied to a stick (or paddle wheel on a boat) in the same configuration. What does the stick do and why?

 

[Dale]

It shows the only forces acting on the car are its weight and the normal force. These cancel out. And then the force of static friction, which is the reaction to the wheel pushing horizontally back against the ground.

They say it is the friction force from the ground on the wheel that propels the car forward. How?

What does Newton's 2nd law say happens when you have a force pointing forward and no force pointing backwards?

I see the friction force pushing that the bottom of the tire, but it is static friction.

Does Newton's 2nd law have an exception for static friction forces? Is it $F_{\text{net}} = ma$ or is it $F_{\text{net}}-F_{\text{static friction}}=ma$?

 

[alkaspeltzar]

Think about what happens if instead of a tire, the forces are applied to a stick (or paddle wheel on a boat) in the same configuration. What does the stick do and why?

Okay let's use the paddle boat example. It pushes the water back, and the water pushes the wheel. But the point of contact doesn't move but yet the boat does. I guess why is that the case? We can see it happening but don't understand why if the water is pushing at the bottom of the paddle, the enter wheel/boat moves forward

 

[alkaspeltzar]

What does Newton's 2nd law say happens when you have a force pointing forward and no force pointing backwards?

IT will accelerate.

Does Newton's 2nd law have an exception for static friction forces? Is it $F_{\text{net}} = ma$ or is it $F_{\text{net}}-F_{\text{static friction}}=ma$?

No it does not. But hear what I am saying. The force is at the bottom of the tire, but the entire car moves forward. how does the force of static friction cause/or translate itself to make the entire car move? When it is applied to the bottom of the tire, the tire doesn't slide forward. No it rolls and the car accelerates. That is my question

 

[anorlunda]

Okay let's use the paddle boat example.

OK, but lets' simplify it to a single paddle. That's analogous to a canoe paddle. When I paddle my canoe, I move it in a semi-circular motion. Do you have difficulty understanding a canoe?

 

[Dale]

The force is at the bottom of the tire, but the entire car moves forward. how does the force of static friction cause/or translate itself to make the entire car move?

There is no need for the force to translate itself. Regardless of where the force is applied the acceleration of the center of mass of the system is the same.

It appears to me that you have a misunderstanding about Newton’s 2nd law. You seem to be under the impression that a force must be applied at the center of mass in order to accelerate the center of mass. That is not the case. Regardless of where a force is applied it will cause the center of mass to accelerate.

If the object is not rigid then you need to know the internal forces in order to know how the various parts move relative to each other.

When it is applied to the bottom of the tire, the tire doesn't slide forward. No it rolls and the car accelerates.

That essentially happens due to internal forces within the car. You can solve Newton’s equations at the wheel, at the axle, at the crankshaft, at the piston, etc, to get the relative motions of the various parts of the car.

 

[alkaspeltzar]

If the object is not rigid then you need to know the internal forces in order to know how the various parts move relative to each other.

That essentially happens due to internal forces within the car. You can solve Newton’s equations at the wheel, at the axle, at the crankshaft, at the piston, etc, to get the relative motions of the various parts of the car.
[/QUOTE]

So basically, Dale what you and [U]Anorlunda[/U] are saying is that due to internal forces, that is why the force moves the car as it does. Which makes sense, as we looks at the car/tire as the whole system and do a Free Body Diagram, we don't worry about breaking it down into pieces to see how all the individual parts move.

Using the canoe example above, there is a reaction force on the canoe paddle, but as it flows thru the body and internal muscle forces, it makes the person/boat etc as a unit move forward. When I think about canoing it makes sense because as I push with the paddle, I am rigid. So any reaction goes to me as system

Correct?

 

[alkaspeltzar]

There is no need for the force to translate itself. Regardless of where the force is applied the acceleration of the center of mass of the system is the same.

It appears to me that you have a misunderstanding about Newton’s 2nd law. You seem to be under the impression that a force must be applied at the center of mass in order to accelerate the center of mass. That is not the case. Regardless of where a force is applied it will cause the center of mass to accelerate.

This makes sense, I guess I wasn't realizing it. In space, if you push a potatoe from the side or center, it would accelerate the same. So many times, the item we are pulling or pushing is pinned or under other forces, so it is hard to see this. THanks for clarification

 

[Mister T]

Okay let's use the paddle boat example. It pushes the water back, and the water pushes the wheel. But the point of contact doesn't move but yet the boat does. I guess why is that the case? We can see it happening but don't understand why if the water is pushing at the bottom of the paddle, the enter wheel/boat moves forward

The entire boat moves relative to the water. Whether you insist that it's the boat that's moving and not the water, you are simply making a choice of a frame of reference to view things from. If you were co-moving through the water along with the boat, the boat would be at rest and the water would be moving.

 

[russ_watters]

The entire boat moves relative to the water. Whether you insist that it's the boat that's moving and not the water, you are simply making a choice of a frame of reference to view things from. If you were co-moving through the water along with the boat, the boat would be at rest and the water would be moving.

Yes, this is why I like the paddle-wheel example. What is moving and what is stationary depends on the reference frame, and it just feels so much easier to swallow that the paddle is pushing the water backwards instead of the water pushing the paddle forwards. But it's really the same thing.

 

[A.T.]

If the object is not rigid then you need to know the internal forces in order to know how the various parts move relative to each other.

Or the other way around: You know that the axle cannot accelerate faster than the car frame, which allows you to solve for the force between the axle and the frame.

But to know the acceleration of the car's COM you only need to now the external forces on the car as a whole.

 

[alkaspeltzar]

So basically when looking at the acceleration of the car, we look at only the external forces acting on the car as a whole.

Similarly to the canoe example. And the reason the canoe moves not the paddle is due to the balance of internal forces of the paddler in the canoe

 

[anorlunda]

And the reason the canoe moves not the paddle

You're confusing things with flawed examples. In the frame of the lake, both the canoe and the paddle move.

A better example is a kids kick scooter. In the frame of the Earth, the scooter moves and the kid's foot is stationary when it is contact with the ground.

 

[alkaspeltzar]

I just want someone to confirm my confirmation in post #10. I feel like we are getting off subject between what I asked and now looking at movements relavant to reference frames

 

[A.T.]

I just want someone to confirm my confirmation in post #10.

Post #10 contains some correct statements, and some unclear statements. It cannot be confirmed as a whole.

 

[alkaspeltzar]

Post #10 contains some correct statements, and some unclear statements. It cannot be confirmed as a whole.

That doesn't help.

 

[jbriggs444]

When I think about canoing it makes sense because as I push with the paddle, I am rigid

Among the other unclear statements, post #10 contained the above.

When I push with a paddle, I am decidedly non-rigid. Among other things, my arms are moving.

My typical stroke is from the rear seat. The paddle is inserted near vertically and, if paddling right-handed, canted at an angle 20 or 30 degrees left of perpendicular with the craft's course. The left hand is on the top of the paddle, the fingers of the right just on the top of the blade. The left arm acts primarily as a brace and the right hand is drawn back. The right hand controls the can't of the blade in the water, providing the steering input. The drawing back of the right arm is the primary source of motive power. Additional power is generated from body rotation and progressing from a forward lean at the beginning of the stroke into a slight reclining posture at the end. The blade leaves the water nearly horizontal, making the end of the stroke markedly less efficient than the beginning. A bent blade paddle is sometimes used to alleviate this problem. [Hard to find if renting gear]

 

[alkaspeltzar]

My original question was this:

Let's use canoe example. We apply with the canoe paddle a force against the water. The water, applies a force to the paddle/us thanks to Newton's third law. My question is why doesn't just the canoe paddle feel the reaction? In reality, the reaction force on the paddle makes me move forward.

 

[jbriggs444]

My question is why doesn't just the canoe paddle feel the reaction? In reality, the reaction force on the paddle makes me move forward.

Only the canoe paddle feels the force of water on paddle. The canoer's hands feel the force of paddle on hands.

The canoer's arms feel the force of hand on arm. The canoer's shoulders feel the force of arm on shoulder. The canoer's body feels the force of shoulder on body. The canoe feels the force of body on seat.

Often this idea will be explored with an exercise where a string pulls on a block and another block is stacked on top or beneath.

 

[alkaspeltzar]

So isn't that the same as saying the reaction force of the water on the paddle is then thru the action/reaction pairs of the body transferred to make the person/canoe move forward with respect to the water?

 

[jbriggs444]

So isn't that the same as saying the reaction force of the water on the paddle is then thru the action/reaction pairs of the body transferred to make the person/canoe move forward with respect to the water?

Force is not a conserved quantity. What is applied on one end of a body is not always equal to what comes out the other. I shy away from using the term "transferred" for this reason. If you must reason about what comes out the other end, Newton's second law is your friend.

 

[alkaspeltzar]

Force is not a conserved quantity. What is applied on one end of a body is not always equal to what comes out the other. I shy away from using the term "transferred" for this reason. If you must reason about what comes out the other end, Newton's second law is your friend.

I was talking about the ideal case but okay.

 

[anorlunda]

Come on @alkaspeltzar , can't you see that the paddler's arms are relatively rigid during the stroke analogous to the spokes of a wheel, and the paddler's shoulders rotate giving a torque analogous to the axle of a wheel?

 

[alkaspeltzar]

Yes I do. But that is not my question. My question is this:

If a swimmer pushes away from a wall, the wall pushes back. It's her, the whole body that moves away, why? Isn't the feet the area receiving the reaction force from wall.

I keep trying to ask it simply and keep getting more variations. Sorry

 

[A.T.]

If a swimmer pushes away from a wall, the wall pushes back. It's her, the whole body that moves away,

Try to be more precise: the center of mass of the whole body accelerates.

 

[jbriggs444]

I keep trying to ask it simply and keep getting more variations.

Because any correct answer is going to contradict the simplifying assumptions implicit in the question. We have to make it complicated to make it correct. That means adding details that were left out of the question. Which details get added tend to vary with each person providing an answer.

 

[PeroK]

Yes I do. But that is not my question. My question is this:

If a swimmer pushes away from a wall, the wall pushes back. It's her, the whole body that moves away, why? Isn't the feet the area receiving the reaction force from wall.

I keep trying to ask it simply and keep getting more variations. Sorry

If her feet were loose and could come away from her legs, then the feet might leave the rest of the body behind.

Also, you might as well ask why not only the skin on the soles of her feet move? Her bones are not actually in contact with the wall either. Why do the bones and everything else inside her feet move?

To answer my own questions, it is because everything is connected physically.

In general, things that are physically connected cannot exhibit independent motion. Whether they are fully rigid or not.