Friction/heat kinetic energy - reactive force

[TonyCross]

Hi,
Could anyone please give me a little advice.
If we look at a disc brake on a vehicle, the disc brake pads apply a friction force on the disk rotor which causes the kinetic energy of the moving vehicle to be turned into heat.
Does this heat reduce the reactive force experienced on the disks? If there were no heat would there be an increase of lateral force on the brake pads?
My way of thinking is that the total kinetic energy of the vehicle must be divided into two components the creation of the heat, or the thermal component and the reactive lateral force experienced by the pads.
Thanks
Tony

 

[A.T.]

My way of thinking is that the total kinetic energy of the vehicle must be divided into two components the creation of the heat, or the thermal component and the reactive lateral force experienced by the pads.

Sounds like you are confusing energy and force.

 

[TonyCross]

Maybe I am but let's consider a squash ball, the kinetic energy of the ball which then hits the wall with a force, it then warms is this warming reducing the force it rebounds from the wall?

 

[TonyCross]

The formula for potential energy depends on the force acting on two objects, so they are linked, my question simply asks if the thermal effect changes the energy, in my mind it must otherwise it would violate the conservation of energy law.

 

[kuruman]

You are correct. A fraction of the mechanical energy in the form of kinetic energy is converted to heat, sound, etc. which means that when the ball rebounds, its mechanical energy will be reduced.

 

[TonyCross]

You are correct. A fraction of the mechanical energy in the form of kinetic energy is converted to heat, sound, etc. which means that when the ball rebounds, its mechanical energy will be reduced.

Thanks that is what I thought, do you know the formula to work out the loss in Kinetic energy, I know that the
units are joules, would you simply see how many joules were needed in the delta temp increase in the object then subtract this from the objects initial energy?

 

[russ_watters]

If we look at a disc brake on a vehicle, the disc brake pads apply a friction force on the disk rotor which causes the kinetic energy of the moving vehicle to be turned into heat.
Does this heat reduce the reactive force experienced on the disks? If there were no heat would there be an increase of lateral force on the brake pads?
My way of thinking is that the total kinetic energy of the vehicle must be divided into two components the creation of the heat, or the thermal component and the reactive lateral force experienced by the pads.

It all becomes heat. Think if it this way; if it didn't become heat, where would it go/what would it be??

 

[vela]

Does this heat reduce the reactive force experienced on the disks?

What's a reactive force? How is it different than than a plain old force exerted on the disks? I just ask because you keep referring to it as a reactive force as if "reactive" has some special meaning.

If there were no heat would there be an increase of lateral force on the brake pads?

Increase compared to what?

The formula for potential energy depends on the force acting on two objects, so they are linked, my question simply asks if the thermal effect changes the energy, in my mind it must otherwise it would violate the conservation of energy law.

That's like saying the amount of cash you have on hand is related to the money you earned at your job, so finding a $20 bill on the ground affects your job.

There is a well-known problem called brake fade, so the answer to your original question is yes.

 

[A.T.]

my question simply asks if the thermal effect changes the energy,

The problem was not with the heat energy, but with decomposing kinetic energy into heat energy and force. That makes no sense. You cannot add energy and force.

 

[TonyCross]

What's a reactive force? How is it different than than a plain old force exerted on the disks? I just ask because you keep referring to it as a reactive force as if "reactive" has some special meaning.Increase compared to what?That's like saying the amount of cash you have on hand is related to the money you earned at your job, so finding a $20 bill on the ground affects your job.

There is a well-known problem called brake fade, so the answer to your original question is yes.

Brake fade, yes of course that is a real world example of the effect. Reactive force is important as the object in motion applies a force to the object it collides with this is known as reactive force, my question relates directly to the reactive force.

 

[kuruman]

Thanks that is what I thought, do you know the formula to work out the loss in Kinetic energy, I know that the
units are joules, would you simply see how many joules were needed in the delta temp increase in the object then subtract this from the objects initial energy?

There is no formula per se because the energy loss depends on the material that makes up the ball (watch this very short video.) However, you can deduce the loss experimentally. Drop a ball from height $h_i$, let it bounce off the floor and measure the $h_f$ to which the ball rebounds. The mechanical energy loss is $mg(h_f-h_i)$.

 

[A.T.]

Reactive force is important as the object in motion applies a force to the object it collides with this is known as reactive force, my question relates directly to the reactive force.

A force is a force. The word "reactive" has no physical significance here.

 

[TonyCross]

The problem was not with the heat energy, but with decomposing kinetic energy into heat energy and force. That makes no sense. You cannot add energy and force.

Sorry I don't understand of course energy Ke and Force (Newtons) are different however they are most definitely linked reduce the Ke and the Force will be reduced increase the Force applied and the Ke will increase/

 

[TonyCross]

A force is a force. The word "reactive" has no physical significance here.

So if my ball hits a tin can and the tin can falls over what is this force called? Newtons 3rd every action has an equal and opposite REACTION.

 

[TonyCross]

You are correct. A fraction of the mechanical energy in the form of kinetic energy is converted to heat, sound, etc. which means that when the ball rebounds, its mechanical energy will be reduced.

Thanks

 

[A.T.]

Newtons 3rd every action has an equal and opposite REACTION.

Newtons 3rd Law just says that forces come in equal but opposite pairs. The naming is irrelevant.

 

[TonyCross]

Newtons 3rd Law just says that forces come in equal but opposite pairs. The naming is irrelevant.

Yes in the case of my example a brake disk and a rotor (a pair) each acts in opposition so reactive force is
important as this is the force seen by the disk.

 

[vela]

The one that makes the can fall over? It's the force exerted on the can by the ball. There's also a force that the can exerts on the ball, which slows the ball down. The two form an action-reaction pair, but which one is called the action and which one is called the reaction is arbitrary.

Here, you're only talking about the force exerted by the pads on the disk. You're not concerned about the force the disk exerts on the pad, so referring to a reactive force is pointless.

 

[A.T.]

... they are most definitely linked ...

Everything is linked somehow. That doesn't mean that some energy can be divided into two components, one of which is a force.

 

[A.T.]

...reactive force is important ...

But calling it "reactive" is not.

 

[TonyCross]

But calling it "reactive" is not.

what should I call it if not a reactive force

 

[A.T.]

what should I call it if not a reactive force

force by ... on ...

 

[TonyCross]

Take a hammer and nail, I move the hammer at a constant speed with a certain mass this is kinetic energy or potential energy, it hits the nail with a Force (Newtons) the action of the nail has to be the reactive force causing
the nail to move into the wood with the transferred kinetic energy, or am I wrong.

 

[TonyCross]

The one that makes the can fall over? It's the force exerted on the can by the ball. There's also a force that the can exerts on the ball, which slows the ball down. The two form an action-reaction pair, but which one is called the action and which one is called the reaction is arbitrary.

Here, you're only talking about the force exerted by the pads on the disk. You're not concerned about the force the disk exerts on the pad, so referring to a reactive force is pointless.

The friction between the pad and the rotor causes the pad to try and move forward, if I place a strain gauge on
the pad I will see a force trying to move the pad forward, the caliper stops this happening, this is not the same as the rotating rotor, this must be classed as a reactive force acting on the pad.

 

[vela]

The problem is that you have a vague understanding of forces, energy, work, etc., and you're trying to draw a specific conclusion based on just suggestive connections. Until your knowledge of basic physics improves, it's highly unlikely you'll be able to reason correctly and reach a valid conclusion.

 

[TonyCross]

The problem is that you have a vague understanding of forces, energy, work, etc., and you're trying to draw a specific conclusion based on just suggestive connections. Until your knowledge of basic physics improves, it's highly unlikely you'll be able to reason correctly and reach a valid conclusion.

Thanks for your condescending reply. I was not looking for insults I simply had what I thought was a valid question.

 

[russ_watters]

Take a hammer and nail, I move the hammer at a constant speed with a certain mass this is kinetic energy or potential energy, it hits the nail with a Force (Newtons) the action of the nail has to be the reactive force causing
the nail to move into the wood with the transferred kinetic energy, or am I wrong.

You can call it that, but it isn't necessary to do so. In any case, I think this is a digression and not helping answer your original question.

The hammer applies a force (variable) over a distance to a nail, doing work/transferring energy to it. Where does that energy reside/what form is it in after that?

 

[TonyCross]

You can call it that, but it isn't necessary to do so. In any case, I think this is a digression and not helping answer your original question.

The hammer applies a force (variable) over a distance to a nail, doing work/transferring energy to it. Where does that energy reside/what form is it in after that?

The energy is work done (nail in the wood) and heat caused by the friction between the wood and the nail.

 

[Dale]

Does this heat reduce the reactive force experienced on the disks? If there were no heat would there be an increase of lateral force on the brake pads?

No, heat does not reduce force in general.

There was a mention of brake fade, but that is not heat reducing force that is heat altering the material properties in an undesirable way. In other words, brake fade is due to a specific technology not a general physical principle.

In particular, it seems like you are thinking that the “action” force will be larger than the “reaction” force because heat will reduce the “reaction” force. This is incorrect. The “action” and “reaction” forces are equal and opposite.

 

[russ_watters]

The energy is work done (nail in the wood) and heat caused by the friction between the wood and the nail.

No, "work done on the nail" is the process, not the result. It is being done only while the nail is in motion. Once finished, then what?

 

[vela]

Thanks for your condescending reply. I was not looking for insults I simply had what I thought was a valid question.

Sorry if it came off as condescending. It wasn't intended as an insult. No one is born with a knowledge of physics, and for many people, it takes years to learn (correct) physics. There's no shame in not knowing basic physics. (Indeed I'd say most of my friends and family don't.) But if you really do want to understand a basic physical explanation to your question, your reasoning does have to be more sophisticated than "energy and force are related, so if one goes up, the other must as well."

 

[kuruman]

Take a hammer and nail, I move the hammer at a constant speed with a certain mass this is kinetic energy or potential energy, it hits the nail with a Force (Newtons) the action of the nail has to be the reactive force causing
the nail to move into the wood with the transferred kinetic energy, or am I wrong.

The hammer has some speed when it hits the nail. As a result of the collision, the hammer transfers some its kinetic energy and momentum to the nail during the time interval $\Delta t$ they are in contact. There is a force acting on the nail during that interval which causes the nail to accelerate. It is that force that drives the nail into the wood and varies with time. However you can calculate an average force exerted by the hammer on the nail which is the momentum change of the hammer divided by the time interval $\Delta t$. As for the kinetic energy of the nail, it very quickly dissipates into heat.

That the full blown description. There is no no need for a "reactive" force anywhere.

 

[TonyCross]

No, heat does not reduce force in general.

There was a mention of brake fade, but that is not heat reducing force that is heat altering the material properties in an undesirable way. In other words, brake fade is due to a specific technology not a general physical principle.

In particular, it seems like you are thinking that the “action” force will be larger than the “reaction” force because heat will reduce the “reaction” force. This is incorrect. The “action” and “reaction” forces are equal and opposite.

Thanks for your reply, I didn't think there would be a difference in the action/reaction force, I also see that material properties in brakes system may cause brake fade. Are you saying that any heat in the brakes has no relation to the loss of kinetic energy of the vehicle while slowing down?

 

[TonyCross]

No, "work done on the nail" is the process, not the result. It is being done only while the nail is in motion. Once finished, then what?

equilibrium assuming the hammer doesn't rebound.

 

[TonyCross]

The hammer has some speed when it hits the nail. As a result of the collision, the hammer transfers some its kinetic energy and momentum to the nail during the time interval $\Delta t$ they are in contact. There is a force acting on the nail during that interval which causes the nail to accelerate. It is that force that drives the nail into the wood and varies with time. However you can calculate an average force exerted by the hammer on the nail which is the momentum change of the hammer divided by the time interval $\Delta t$. As for the kinetic energy of the nail, it very quickly dissipates into heat.

That the full blown description. There is no no need for a "reactive" force anywhere.

Sorry I have to disagree, your description is great however:
I quote you (There is a force acting on the nail during that interval) for EVERY force there is an opposing force.
when you walk the ground forces back at you, if you fire a canon there is a opposing force, are hammers exempt?

 

[Dale]

Are you saying that any heat in the brakes has no relation to the loss of kinetic energy of the vehicle while slowing down?

No, the heat is approximately equal to the loss of kinetic energy. It does not cause a loss of force.

 

[TonyCross]

No, the heat is approximately equal to the loss of kinetic energy. It does not cause a loss of force.

So the heat would cause the vehicle to lose some kinetic energy in the transfer process.

 

[russ_watters]

equilibrium assuming the hammer doesn't rebound.

Yes, but the question we're trying to answer is: where is the energy now?

So the heat would cause the vehicle to lose some kinetic energy in the transfer process.

That's a scrambled order of events. It's kinetic energy -> work -> heat.

 

[TonyCross]

Yes, but the question we're trying to answer is: where is the energy now?

The heat created will dissipate into the environment, not sure how you can quantify how tight the nail is in the wood though.

 

[russ_watters]

The heat created will dissipate into the environment, not sure how you can quantify how tight the nail is in the wood though.

Right. There is some temporary deformation of the wood that absorbs some of the energy, but it generally isn't much.

 

[TonyCross]

Right. There is some temporary deformation of the wood that absorbs some of the energy, but it generally isn't much.

Because you cannot create or destroy energy I guess the nail would need similar force to remove it.

 

[russ_watters]

Because you cannot create or destroy energy I guess the nail would need similar force to remove it.

Yes, and you'll notice the wood doesn't completely return to its prior condition/shape, so a lot of the deformation is permanent.

 

[Dale]

So the heat would cause the vehicle to lose some kinetic energy in the transfer process.

Yes, heat is a type of energy transfer. In this case energy is being transferred from kinetic energy to thermal energy. So in this case heat reduces kinetic energy. It does not reduce force.

 

[TonyCross]

Yes, heat is a type of energy transfer. In this case energy is being transferred from kinetic energy to thermal energy. So in this case heat reduces kinetic energy. It does not reduce force.

Thanks, that is the answer to my initial question, however I may have worded it in the wrong way.

 

[kuruman]

Sorry I have to disagree, your description is great however:
I quote you (There is a force acting on the nail during that interval) for EVERY force there is an opposing force.
when you walk the ground forces back at you, if you fire a canon there is a opposing force, are hammers exempt?

Of course hammers are not exempt. However my description of what happens concentrated on the nail. You wish me to talk about what happens to the hammer, no problem. The Newton third law counterpart of the force exerted by the hammer on the nail is an equal and opposite force exerted by the nail on the hammer. That force acts in a direction opposite to the velocity of the hammer, which means there is an acceleration opposite to the velocity of the hammer the end result being that the hammer decelerates rapidly and moves at the speed of the nail.

When you talk about forces you have to be clear about what your system is, that is on what entity these forces act. If the system is the nail, it is acted upon by the hammer, the surface into which it is driven and gravity. If the system is the hammer, it is acted upon by the hand that holds it, the nail that it strikes and gravity. Once you sort out (a) what object (system) you are talking about and (b) what forces act on the system, Newton's second law takes over and shows you what the acceleration of the system will be under the action of these forces. Needless to say that somewhere out there, outside the system you will find the Newton's 3rd law counterparts of all the forces that act on your system but they have nothing to do with the acceleration of your chosen system.

You will achieve clarity of thought if you decide exactly what your system is before you consider the forces acting on it. That's a first step. The second step is to consider all the pieces of the universe that can act on your system and count them. Then you look at the forces, one per piece of universe, that act on this system.

 

[jbriggs444]

Because you cannot create or destroy energy I guess the nail would need similar force to remove it.

No. That does not follow.

Energy conservation says that the energy present in the hammer will still be there (as heat, stress, vibrations, residual motion or whatever). It does not say that the energy required to insert a nail is the same as the energy required to extract it.

In the typical case of a nail sunk into fresh wood and withdrawn shortly thereafter, the squeezing force of wood on nail will be similar on the way in and on the way out. The friction of wood on nail will then be similar on the way in and on the way out. So one would expect the required work done both ways to be similar.

But if you've ever pulled a nail from a piece of rotten wood, you'll know that the work done on extraction does not have to be as great as the work done on insertion. And if you've ever tried to extract a bolt from a tapped hole after it has rusted in, the work done to extract can be greater than the work done on insertion.

 

[Dale]

Then you look at the forces, one per piece of universe, that act on this system.

Sometimes an external body can interact with the system through multiple forces. Like the normal force and the gravitational force from the Earth acting on a box on the ground.

 

[kuruman]

Sometimes an external body can interact with the system through multiple forces. Like the normal force and the gravitational force from the Earth acting on a box on the ground.

That's a tricky one. I would call the ground "Surface" as I would call the surface of a table that is firmly attached to the Earth. To me "Earth" gives rise to $\frac{GMm}{r^2}$ (or $mg$). Strictly speaking you are, of course, correct.

 

[A.T.]

Take a hammer and nail, I move the hammer at a constant speed with a certain mass this is kinetic energy or potential energy, it hits the nail with a Force (Newtons) the action of the nail has to be the reactive force causing
the nail to move into the wood with the transferred kinetic energy, or am I wrong.

Now you have three bodies and two 3rd law pairs. Your action and reaction labels make no sense, but they are meaningless physically anyway.

 

[TonyCross]

Of course hammers are not exempt. However my description of what happens concentrated on the nail. You wish me to talk about what happens to the hammer, no problem. The Newton third law counterpart of the force exerted by the hammer on the nail is an equal and opposite force exerted by the nail on the hammer. That force acts in a direction opposite to the velocity of the hammer, which means there is an acceleration opposite to the velocity of the hammer the end result being that the hammer decelerates rapidly and moves at the speed of the nail.

When you talk about forces you have to be clear about what your system is, that is on what entity these forces act. If the system is the nail, it is acted upon by the hammer, the surface into which it is driven and gravity. If the system is the hammer, it is acted upon by the hand that holds it, the nail that it strikes and gravity. Once you sort out (a) what object (system) you are talking about and (b) what forces act on the system, Newton's second law takes over and shows you what the acceleration of the system will be under the action of these forces. Needless to say that somewhere out there, outside the system you will find the Newton's 3rd law counterparts of all the forces that act on your system but they have nothing to do with the acceleration of your chosen system.

You will achieve clarity of thought if you decide exactly what your system is before you consider the forces acting on it. That's a first step. The second step is to consider all the pieces of the universe that can act on your system and count them. Then you look at the forces, one per piece of universe, that act on this system.

Thanks for you detailed clarification, much appreciated.