Jump to content

Lateral force is less than pull force?


Visionary

Recommended Posts

Greetings,

 

I noticed that it takes less force to slide a magnet off a magnetic object(magnet/ferromagnetic material) than directly pulling it off? The force needed to pull the magnet off could be 10LB

while the force applied sliding the magnet off is 1 or less LB, only force countering the applied force is friction not magnetic...
Why is that? I even noticed that only at the edge, there is a stronger force, but not as strong as the direct pull force.

 

 

Link to comment
Share on other sites

Well as you observed the contact force of magnetic attraction is a direct force.

 

That is to say it is a normal force between the magnet and the object.

Newton's 3 rd Law says that this attractive contact force is balanced by a normal reactive force which is equal in magnitude but opposite in direction ie still normal.

 

A lateral force applied to the object is at right angles to this and only faces the force of friction.

The frictional force against sliding is the normal reaction force times the coeficient of friction.

 

Since the coefficient of friction is less than 1 (typically 0.3 to 0.5 for a magnet and a piece of iron strip) the sliding force is less than the normall pull force.

Link to comment
Share on other sites

Good point Studiot,

What about the eddy currents generated to oppose the change in the magnetic field when sliding a magnet off a steel plate?

 

Eddy currents are proportional to the rate at which the field is changing, so they are related to how fast the magnet is sliding.

Link to comment
Share on other sites

 

Eddy currents are proportional to the rate at which the field is changing, so they are related to how fast the magnet is sliding.

 

It's strange how you can have both magnetic attraction and eddy currents at the same time while sliding a magnet around a ferromagnetic surface, I assume the force can never ever be close to the pull force not even 1% of it.

Since it's force is the function of the "change" of the magnetic field.

 

Also, Faraday's law of induction is used to calculate the eddy currents?

Link to comment
Share on other sites

 

It's strange how you can have both magnetic attraction and eddy currents at the same time while sliding a magnet around a ferromagnetic surface, I assume the force can never ever be close to the pull force not even 1% of it.

Since it's force is the function of the "change" of the magnetic field.

 

Also, Faraday's law of induction is used to calculate the eddy currents?

 

Yes, you would use Faraday's law.

 

The induction force can match the external force when you move fast enough with a sufficiently strong field. Drop a strong magnet down a copper tube and it can reach terminal velocity. Constant velocity means the induction force matches gravity.

Link to comment
Share on other sites

Well I was more interesting in knowing the magnitude of the induction force in comparison of the attraction force between a magnet and a steel plate for example, but now... After knowing its possible to reduce such a force by creating air gaps so that the eddy currents would have to pass through small tiny gaps of air, since air has a very large resistance the eddy currents are reduced's substantially regardless of how fast that movement was. So the attraction force now, could be x while the eddy current's induction force is less than x due to the air gaps.

 

I'm trying to reduce the eddy currents between a magnet+ferromagnetic materials without increasing heat as you all know that would cause a huge damp to the magnetic properties of such materials.

Link to comment
Share on other sites

I'd like to make the eddy currents really low, in fact, a value near zero.

So I think by having such a ferromagnetic material shield like so:

figure-24-04-03.jpeg

 

The eddy currents decrease so it would not be a problem. However, the magnetic force decreased due to the change. It could be possibly increased by increasing thickness or the diameter?

Link to comment
Share on other sites

You have been talking about a permanent magnet so the magnetic fields are not varying.

 

Eddy currents are generated in any conductor and flow at right angles to the field and to the motion.

 

You can never fully eliminate them or the machine will not work.

 

However, since the currents increase with the conductance of the conductor, reduction strategies are based on reducing the conductance. A more familiar way of saying this might be to increase the resistance.

 

In order to increase the resistance we can do three things.

 

1) Decrease the cross sectional area. This is not usually an option for mechanical reasons to make the makine work, but it should be kept as compact as practicable.

 

2) Increase the resistivity of the material. We do this by using special steels eg with 5% silicon added.

 

3) Decrease the path length in the direction of (eddy) current flow. We can make this change to the design by using laminated construction.

 

Does this help?

Link to comment
Share on other sites

When sliding a magnet rapidly around a ferromagnetic's surface eddy currents are generated since all ferromagnetic materials are all conductors.

I agree, by increasing the resistance we reduce the eddy currents, My idea is having small air gaps in the ferromagnetic surface. By doing so, it reduces the eddy currents significantly, thus having a low induction force.

Link to comment
Share on other sites

Did you read line 2 of post#10, from where I took the quote?

 

In which direction is the (lateral) motion?

In which direction is the field?

 

Therefore in which direction is the eddy current?

 

post#10 was dedicated to explaining strategies for eddy current reduction strategy 3 shows the geometric segmentation needs to be caried out in the direction of current. It is a three dimensional issue.

Link to comment
Share on other sites

When sliding a magnet rapidly around a ferromagnetic's surface eddy currents are generated since all ferromagnetic materials are all conductors.

 

Only just conductors in some cases. CrO2 is a rather poor conductor.

Anyway, isn't the answer to the original question that the coefficient of friction for smooth metal on smooth metal is fairly low: 0.1 or less.

It's easier to slide a book along a table than it is to lift it.

Link to comment
Share on other sites

Did you read line 2 of post#10, from where I took the quote?

 

In which direction is the (lateral) motion?

In which direction is the field?

 

Therefore in which direction is the eddy current?

 

post#10 was dedicated to explaining strategies for eddy current reduction strategy 3 shows the geometric segmentation needs to be caried out in the direction of current. It is a three dimensional issue.

 

Yes I did read the second line on that post.

I think regardless of the direction of the field, the eddy currents would tend to oppose the change

Look here.

 

The magnet is attracted to the ferromagnetic surface, The black arrow represents the force applied to move it around the surface.

The green arrow represents the eddy currents induced to oppose the change I think?

 

It's like sliding a magnet over a copper plate it you do it quite fast, you'd feel a repulsive force I guess?

 

 

 

Only just conductors in some cases. CrO2 is a rather poor conductor.

Anyway, isn't the answer to the original question that the coefficient of friction for smooth metal on smooth metal is fairly low: 0.1 or less.

It's easier to slide a book along a table than it is to lift it.

 

Well, I knew that. In fact, it could be less when lubricant is applied! Much much more less.

But the issue now is with Eddy currents.

 

Btw,

Thanks for all your help!

Link to comment
Share on other sites

I'm sorry but you don't quite understand eddy currents - they are a bit more complicated.

 

There are two sorts of eddy current (although the loss mechanism is the same in both) and I think you are mixing them up.

 

Firstly there are eddy currents that are generated in transformer cores and motor components.

These are generated when there is a circuit current flowing. These are not the type you are describing.

The energy for these is taken from the electric power of the circuit. It appears as an electrical loss in the circuit.

 

Secondly there are the type that are generated by the mechanically generated motion of a conductor in a magnetic field.

This is the type you are describing.

Since no circuit current is flowing, there is no current to 'oppose'.

Energy is, however, lost to the system and appears as an additional mechanical resistance to the mechanical driver.

 

In both cases the energy is lost to the resistive heating of the conductive material in which the eddy currents are circulating.

 

Do you understand why the eddy currents must be circulatory and some of the conductor must be out of the field for them to occur in the second case?

Edited by studiot
Link to comment
Share on other sites

Well, I knew that but I had the simplest idea about it.

It seems to be more complicated that what I've anticipated...

 

The mechanical resistance isn't it due to the magnetic force that is generated by those "surface" eddy currents that oppose the "lateral" force, since its kind of a "change" to the system.

 

 

 

Do you understand why the eddy currents must be circulatory and some of the conductor must be out of the field for them to occur in the second case?

 

No, I don't. Please do explain.

And in my case there can be a very strong force because while sliding the magnet I'm doing it super fast, so that induced those currents and they oppose the change ultimately.

I'm more concerned about the magnetite of that force.

 

Thanks

Link to comment
Share on other sites

Despite me saying that eddy currents do not oppose things you keep returning to it.

 

Why is this?

 

The effect is nothing like Lenz law where the generated EMF indeed opposes the change to the existing.

 

Rather you should think of it as going into a MacDonalds and ordering a burger.

 

You obtain a burger but there is a cost involved that you have to pay.

 

So it is with any dissipative loss to any system, mechanical, electrical or thermodynamic. The process, like the taxman, exacts a charge for doing it.

 

To oppose, the system would have to provide or generate some force or emf that indeed tends to drive the process in the opposite direction. Dissipative agents can never do this.

A simple mechanical example is a spring. The harder you pull it out the harder it pulls back.

 

Once you have caught this idea I will explain why eddy currents are circulatory in nature.

Edited by studiot
Link to comment
Share on other sites

Despite me saying that eddy currents do not oppose things you keep returning to it.

 

Why is this?

 

The effect is nothing like Lenz law where the generated EMF indeed opposes the change to the existing.

 

Rather you should think of it as going into a MacDonalds and ordering a burger.

 

You obtain a burger but there is a cost involved that you have to pay.

 

So it is with any dissipative loss to any system, mechanical, electrical or thermodynamic. The process, like the taxman, exacts a charge for doing it.

 

To oppose, the system would have to provide or generate some force or emf that indeed tends to drive the process in the opposite direction. Dissipative agents can never do this.

A simple mechanical example is a spring. The harder you pull it out the harder it pulls back.

 

Once you have caught this idea I will explain why eddy currents are circulatory in nature.

Sorry, I didn't understand the concept much, so please bear with me.

I don't know why I'm comparing my case like dropping a magnet through a copper pipe, you noticed the decrease of velocity due to the eddy currents, or moving a piece of copper pendulum through a strong magnetic field, there is a counter force acting on motion. What force is that? Isn't it the eddy currents? And the force is a VERY POWERFUL one. Is that the force we're talking about here?

 

Well, in a conductor there is induced EMF, and with ferromagnets emf is generated, a large conductor surface can have this happen.

Simply before we into detail, lets talk about this issue in detail.

When a magnet slide around a surface of a ferromagnetic material, is there any electromagnetic force at all opposing the change?

 

You say there are losses, please explain the forces involved...

Thanks.

Edited by Visionary
Link to comment
Share on other sites

 

Well, in a conductor there is induced EMF

 

Yes, at last somehting correct.

 

 

You say there are losses, please explain the forces involved...

 

I already have twice. and others have also stated that the losses are the normal I2R losses.

 

 

there is a counter force acting on motion.

 

Is there?

 

Have you heard of

1)Lenz law

2) Flemings left hand and right hand rules

Link to comment
Share on other sites

Studiot,

 

Please explain what will happen if a slide a magnet super fast over a ferromagnetic surface, that also is a good conductor.

What kind of force is acting against my applied initial force(that caused the motion)?

Since I don't understand what's going on, or Eddy currents. I think it's best top propose a system, and ask what would happen... Then understand what is what.

Edited by Visionary
Link to comment
Share on other sites

I shall have to defer further discussion till next week since I shall not be about during the Bank Holiday weekend.

 

However in order to replicate the conditions you are asking about,

 

think of a bar magnet being pulled along across polished iron table, at tight angles to the NS axis.

 

In what direction is the relative motion?

In what direction is the flux or field?

 

So in what direction is the induced EMF?

 

The part of the table where there is no flux because it is not under the bar magnet provides a retun path for circulatory eddy currents.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.