# Friction/heat kinetic energy - reactive force

TonyCross
Hi,
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

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.

• sophiecentaur
TonyCross
Maybe I am but lets 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.

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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.

• Lnewqban
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?

Mentor
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??

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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. Science Advisor 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.

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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)##.

• Digcoal
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

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.

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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.

• Digcoal and nasu
... 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.

...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

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.

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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.

Mentor
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.

Mentor
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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.

Mentor
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?

Staff Emeritus
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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."

• Digcoal and russ_watters
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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.

• Digcoal, nasu and russ_watters
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?