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ldn
08-17-2010, 11:31 AM
Hi Everyone,

When you have two flat plates that slide against each other, is it better to have a smaller or larger area of contact?

With a larger area, the force is more spread out, with a smaller area, there's less opportunity for friction. I know clutch disks and brakes like to have a large area so I'm thinking smaller is better, but those surfaces aren't lubricated.

This is for the slide on my milling attachment. The slide plate is about 2 inches wide and 4 inches long. I'm thinking about filing out the middle of the plate so it just makes contact on the outer 1/2". That would also make it easier to scrape should I choose to do so.

What do you think?

rkepler
08-17-2010, 11:38 AM
I've only seen a few slides with a hollowed out center, and that's when they were 24" long and the center section was about 6". I've never seen a cross slide or compound on a lathe deliberately hollowed out in the middle. That's not to say that there aren't any, just that I haven't seen one.

DannyW
08-17-2010, 11:51 AM
More trouble than that it's worth, to hollow them out.
The "gain" will be negligible.
Furthermore, you create pockets that swarf gets in, resulting in all kinds of mischief!

Danny

Carld
08-17-2010, 01:37 PM
The more square inches of surface are in contact the higher the friction will be. If you machine the two plates flat and grease them they should move fairly easy.

As to lathe cross slides and compounds the contact area is very small relative to the size of them because the contact area is on the dovetails on each side and the drag is adjusted with the gibs that take up the slop. They do not touch together in the center area between the dovetails.

How are going to guide them? what will keep them running straight? How will you adjust the side movement?

Fasttrack
08-17-2010, 01:39 PM
Theoretically, it makes no difference. The force of friction is equal to the normal force times the coefficient of friction.

In practice, it becomes a little bit more complicated. In the case of bearing slides, I always opt for more surface area. This is because you want the surface to slide on a film of oil, not on the other surface. If the surface area is too small, the pressure increases (same weight but distributed over a smaller area). This pressure will cause the part to rupture the oil film and now you will be sliding on the steel surface.

ldn
08-17-2010, 01:42 PM
More trouble than that it's worth, to hollow them out.
The "gain" will be negligible.
Furthermore, you create pockets that swarf gets in, resulting in all kinds of mischief!

Danny

I'm still curious about the original question in the abstract sense though.

Carld
08-17-2010, 01:44 PM
Hmm, I guess you didn't read my explanation in my post #4 did you.

ldn
08-17-2010, 01:56 PM
How are going to guide them? what will keep them running straight? How will you adjust the side movement?

I am building it from a plan I found online. See this link and scroll all the way to the bottom for an exploded diagram:

http://i.imgur.com/OkGaO.jpg

It runs between two rails with a brass gib to take up the sideways slack. It is held in place by two wider rails that contain it -- and there is no provision for adjusting wiggle in that dimension. I'm going to have to just shim it by trial and error.

It actually looks like a pretty poor design and I hadn't realized that at the outset. However it does have the advantage of being able to be constructed out of flat stock without any fancy cutting.

Gib Material

While I'm asking questions, I also wanted to ask what is a good type of brass to use for the gib. I ordered 385 brass, but ended up with 360 brass because the 385 was out of stock.

The instructions specified "hard brass".

Thanks.

ldn
08-17-2010, 02:01 PM
Hmm, I guess you didn't read my explanation in my post #4 did you.

Carl, before you posted I clicked to reply. Then I got a phone call, came back and wrote my message. In the meantime you had posted yours and I hadn't seen it yet.

darn, another call, be back in a while with an edit.

Black_Moons
08-17-2010, 02:09 PM
Id think a smaller area would wear much faster too.. Less metal to wear before it wears down. Course, that only matters if this is a sliding deal

Of course, if your milling attachment is designed to clamp down and not move, you still want a larger surface area way.. Better grip, less chance of damageing ways under more pressure (At least, changeing the rather small, poorly cast clamp on my tailstock for a huge one (So big it actualy sticks out behind the tailstock 3", Enough space for ANOTHER clamp to be mounted ontop if I ever find the need) made outta CRS sure did improve the holding force!.. but could also be because I perposefuly made the slide way with a 'poor' finish with lots of machine marks for extra grab/place for oil to go)

Carld
08-17-2010, 02:13 PM
I don't think you will have much trouble with it as long as you can devise a way to remove any wiggle or side play. Having looked at the plans there is a gib to take up side movement but nothing to adjust upward movement so that may be the issue to solve.

philbur
08-17-2010, 02:19 PM
Carld, in the abstract sense your statement:

"The more square inches of surface are in contact the higher the friction will be."

Is not correct. In it's simplist form:

F = mu x N

Where F = friction force or drag. mu = coefficient of friction for the two contact materials and N is the applied force at 90 degrees to the sliding surfaces. No area involved.

I have to support Fastrack's view with the qualification that if the surface area becomes to large then oil viscosity may come into play. So the real answer is that there is no standard answer as the friction in any system is dependent on many parameters that are inter-related.

Phil:)

Hmm, I guess you didn't read my explanation in my post #4 did you.

ldn
08-17-2010, 02:22 PM
I don't think you will have much trouble with it as long as you can devise a way to remove any wiggle or side play. Having looked at the plans there is a gib to take up side movement but nothing to adjust upward movement so that may be the issue to solve.

OK, back from the phone call. As I said, our posts crossed in the mail and I did carefully read yours afterwards. At first blush it seems to conflict with Fasttrack's post but I think that the "film of oil" thing is the key difference.

Edit: I've got a plan to deal with the unaccounted-for play by using shims between the two stacked side rails.

ldn
08-17-2010, 02:25 PM
F = mu x N

Where F = friction force or drag. mu = coefficient of friction for the two contact materials and N is the applied force at 90 degrees to the sliding surfaces. No area involved.

OK, I'm not going to argue, but I have to ask, why is more area better when talking about a brake pad or clutch?

Heat distribution? Wear resistance?

MuellerNick
08-17-2010, 02:26 PM
because the contact area is on the dovetails on each side and the drag is adjusted with the gibs that take up the slop. They do not touch together in the center area between the dovetails.

That certainly is not true.
Hollowing out the contact area can only be done, if the short guide always covers the long guide, like the saddle is shorter than the bed.
This is not true for many cross slides and most of the compounds.

Edit:
And the reason why this is done is not to reduce friction, but to compensate for wear.

Nick

gary350
08-17-2010, 02:36 PM
One of the examples in my physics book shows a 2"x4"x8" brick being pulled by a set of scales.

The weight of a brick = 8 lbs.

If the brick is standing on its 2x4 side force is 1 lb per square inch.

If the brick is standing on its 2x8 side force is 1/2 lbs per square inch.

If the brick is laying on its 4x8 side force is 1/4 lb per sq inch.

The surface area of the 2x8 side is double that of the 2x4 side.

The surface area of the 4x8 side is double that of the 2x8 side and 4 times that of the 2x4 side.

1 lb / sq in x friction for 8 sq inches = 8

1/2 lb / sq in x friction for 16 sq inches = 8

1/4 lb / sq in x friction for 32 sq inches = 8

Friction is based on surface area and weight. If surface area doubles and lbs per sq inch is reduced by 1/2 and weight stays the same friction always stays the same.

lazlo
08-17-2010, 02:38 PM
Theoretically, it makes no difference. The force of friction is equal to the normal force times the coefficient of friction.

Dang it Tom, quit muddying the discussion with the correct answer -- you don't need that fancy edumacation :)

http://images.wikia.com/uncyclopedia/images/1/18/Edumacation.jpg

philbur
08-17-2010, 02:53 PM
Exactly. More material available for wear and better heat dissipation from the larger surface area.

Phil:)

OK, I'm not going to argue, but I have to ask, why is more area better when talking about a brake pad or clutch?

Heat distribution? Wear resistance?

strokersix
08-17-2010, 04:11 PM
The Physics 101 assumption of constant friction coefficient is incorrect in practice. In practice, friction coefficient will vary not only by unit pressure but also by sliding velocity.

Familiar example: Lower air pressure in drag race tires results in greater tractive force available without increased vehicle weight.

Another example: Motor vehicle brakes and clutches have lower friction coefficient when the sliding velocity is high.

Think of the difference between static and dynamic coefficient presented in Physics 101 textbooks. The transition between static and dynamic isnt a step function, it's continous.

winchman
08-17-2010, 04:27 PM
How can it not be a step function? The static coefficient of friction can only apply at one speed...zero. The dynamic CoF applies at all other speeds.

Fasttrack
08-17-2010, 05:53 PM
How can it not be a step function? The static coefficient of friction can only apply at one speed...zero. The dynamic CoF applies at all other speeds.

That's correct in "theory", again. ;) Like Strokersix mentioned, there is a transition region between kinetic and static friction. We know this as "stick-slip", only it occurs on a much smaller scale. To make matters worse, the kinetic friction is not constant. Friction (on the molecular level) is actually a very complicated problem.

That's why I said, in theory, it doesn't matter. When it comes to two machine components sliding at low speed, the surface area is not a concern except for the danger of rupturing the oil film on which you'd like your components to slide. When it comes to "hard" materials, the coefficient of friction's dependence upon pressure is completely negligible until you get to very high pressures (where you have already ruptured the oil film).

Apparently Carld didn't read my post :rolleyes: ;)

Thanks Robert :D

philbur
08-17-2010, 06:02 PM
Yes but:

1) The improved grip is due to the deflated tyre better conforming to the track surface irregularities, it's nothing to do with the classical definition of coefficient of friction. Winter tyres do better for a similar reason. Also in the same manner that longer studs on your shoes gives you better grip on particular types of surface. The principle however has nothing to do with the coefficient of friction.

2) The clutch/brake example could possibly be due to temperature differences. At least the area has not changed so the relevance is not clear.

3) Stick slip is a step function, both values for which are valid in the classical equation.

If you look up the coefficient of friction for most materials you will find only one or two values, neither of which defines area as being an influencing factor. It is however accepted by all mechanical engineers that the classical definition of F = mu x N is only the starting point for real life applications.

Phil:)

Familiar example: Lower air pressure in drag race tires results in greater tractive force available without increased vehicle weight.

Another example: Motor vehicle brakes and clutches have lower friction coefficient when the sliding velocity is high.

Think of the difference between static and dynamic coefficient presented in Physics 101 textbooks. The transition between static and dynamic isnt a step function, it's continous.

Fasttrack
08-17-2010, 06:05 PM
3) Stick slip is a step function, both values for which are valid in the classical equation.

Friction is not a purely classical phenomenon.

It is however accepted by all mechanical engineers that the classical definition of F = mu x N is only the starting point for real life applications.

Agreed. :)

topct
08-17-2010, 06:26 PM
Gib Material

While I'm asking questions, I also wanted to ask what is a good type of brass to use for the gib. I ordered 385 brass, but ended up with 360 brass because the 385 was out of stock.

The instructions specified "hard brass".

Thanks.

Brass would not be a very good gib material. Especially 360.

One of the bronzes will slide much better.

philbur
08-17-2010, 06:56 PM
Taking into account the OP and the discussion that followed it's clear to me that the issue is not about rubbing friction and surface areas but about quickly re-establishing an oil film when you start to crank the handle. So the issue is about a good oil distribution system, i.e., distribution channels and scraping. The coefficient of friction and surface area is then effectively irrelevant.

Phil:)

Carld
08-17-2010, 07:30 PM
Philbur, I can't agree with that. The more square inches of contact the greater the friction dry or oiled or greased. Of course it would be relative to only dry, only oil or only grease. One square inch of contact will have less friction than two square inches of contact. We can't use the brick application here because we are talking about a flat plate that is a given thickness but more or less square inch area. In this case more square inches=more friction.

You guys sure do like to convolute what a poster says to establish your own answer.

Nick's post,
Quote:
because the contact area is on the dovetails on each side and the drag is adjusted with the gibs that take up the slop. They do not touch together in the center area between the dovetails.

That certainly is not true.
Hollowing out the contact area can only be done, if the short guide always covers the long guide, like the saddle is shorter than the bed.
This is not true for many cross slides and most of the compounds.

Edit:
And the reason why this is done is not to reduce friction, but to compensate for wear.

Nick, please tell me of any lathe that has full contact of the cross slide or compound between the dovetails on each side. I have never seen one but you may know of one so tell me. I didn't say anything about hollowing out the center, the center just does not set on the other surface. Only the dovetails make contact and the bottom area under the outside of the dovetail.

A.K. Boomer
08-17-2010, 07:47 PM
Carld --- more square inches does not always equate to greater friction if the unit pressure surpasses the lubrication film --- then the more area involved is only lubrication/viscosity friction VS the smaller area that's got metal to metal contact (I.E. great seizures ghost:p )

gary350
08-17-2010, 07:54 PM
The Physics 101 assumption of constant friction coefficient is incorrect in practice. In practice, friction coefficient will vary not only by unit pressure but also by sliding velocity.

Familiar example: Lower air pressure in drag race tires results in greater tractive force available without increased vehicle weight.

Another example: Motor vehicle brakes and clutches have lower friction coefficient when the sliding velocity is high.

Think of the difference between static and dynamic coefficient presented in Physics 101 textbooks. The transition between static and dynamic isnt a step function, it's continous.

A tire has a certain radius and only a certain amount of the tire makes contact with the road but when you lower the pressure it causes the tire to become flatter where it touches the road therefore it has larger surface area and more friction. But if you lower the tire pressure too much the tire cups up and wrinkles in the middle and the tire does not touch the road where it is cupped this reduces the friction.

When I use to race we use to experement with tire pressure 1 psi makes a world of difference under certain conditions. At the start of the race the tire is cold then after a few laps the tire is warmed up and friction increases several times. But after several more laps expecially on a hot summer day the tire gets too hot and friction drop off way less than the friction of a cold tire. If your running a 30 lap race the last 20 laps tire will be extremely hot so tire pressure is more important then because the last 20 laps is 2/3 of the whole 30 lap race. On a 15 lap race mid range tire pressure might be the most important expecially if it is not hot outside that day. There is a lot of science to tires and you should buy tires for the street vehicle you drive according to your daily driving conditions.

Carld
08-17-2010, 07:54 PM
This is what I was talking about Nick.

http://i82.photobucket.com/albums/j276/yeathatshim/contactareas.jpg

Maybe I didn't explain it well enough.

A.K. Boomer
08-17-2010, 07:59 PM
Hey - is that some babes phone # at the bottom of the drawing -- well --- Is it???

philbur
08-17-2010, 08:46 PM
Carld the accepted equation for two surfaces rubbing together is F = mu x N. This is not my opinion it is an accepted engineering fact. Area of contact doesn't figure in the equation. Your misconception comes up in every high school class on friction and gets corrected immediately. The common reaction is to go with intuition but this is a mistake. Look up the cof value for say steel and then look for anything that says it is area dependent.

Phil:)

Philbur, I can't agree with that. The more square inches of contact the greater the friction dry or oiled or greased. Of course it would be relative to only dry, only oil or only grease. One square inch of contact will have less friction than two square inches of contact. We can't use the brick application here because we are talking about a flat plate that is a given thickness but more or less square inch area. In this case more square inches=more friction.

You guys sure do like to convolute what a poster says to establish your own answer.

Lew Hartswick
08-18-2010, 12:25 AM
Carld the accepted equation for two surfaces rubbing together is F = mu x N. This is not my opinion it is an accepted engineering fact. Area of contact doesn't figure in the equation. Your misconception comes up in every high school class on friction and gets corrected immediately. The common reaction is to go with intuition but this is a mistake. Look up the cof value for say steel and then look for anything that says it is area dependent.

Phil:)
Yea Phil, I wonder how we can ever get rid of that seemingly universal
idea. Any first physics class demonstrates the falsity of the area
dependance on frictional forces. But you just can't tell some people,
It's one of those, " are you asking or telling me " things. :-)
...lew...

Carld
08-18-2010, 12:43 AM
So your telling me if I have a flat steel floor and I drag a 12"x12"x1" plate over it and measure the pull it takes to move it and then drag a 24"x24"x1" plate over the same floor it will take the same force as the 12"x12" plate.

Your saying no matter how big the plate is it will never take more force to move it. If that is true then if I drag a 20'x20'x1" plate it will take the same force as dragging a 12"x12"x1" plate.

Mcgyver
08-18-2010, 12:47 AM
So your telling me if I have a flat steel floor and I drag a 12"x12"x1" plate over it and measure the pull it takes to move it and then drag a 24"x24"x1" plate over the same floor it will take the same force as the 12"x12" plate.

Carl they're not saying that, I think the part you're missing is the friction is a product of the force and cooefficient of friction.... in this case the force pressing down on your floor is by the weight of the plate and the coefficient of friction is some number for steel and concrete - that determines friction. Your 24x24 plate has 4x the force of 12x12 so will have 4x the friction....BUT.... if you had a 12x12x4 block, it would have the same friction as 24 x24 x1 plate....as both exert the same for downward force

Fasttrack
08-18-2010, 12:53 AM
One square inch of contact will have less friction than two square inches of contact. We can't use the brick application here because we are talking about a flat plate that is a given thickness but more or less square inch area. In this case more square inches=more friction.

You guys sure do like to convolute what a poster says to establish your own answer.

As Philbur says, this is simply NOT the case. As long as the mass of the machine component is fixed, increasing the area will NOT increase the friction. What it WILL do is prevent the component from rupturing the oil film and thus increase life and decrease friction.

BUT if we compare two dry plates of steel or similar material moving at modest speed, each with equal mass, then the friction is COMPLETELY unaffected by the surface area in contact.

I don't see why we can't use the brick. The brick is the same mass (just like the machine component) and all we do is change the surface area that is sliding.

This is not a matter of opinion. Period. :)

It is worth pointing out that F = mu * N is an empirical formula. It does not hold true for some materials. That is why the "misconception" remains. A perfect example is a soft material in contact with a hard material. The actual contact area (on the molecular level) does not scale linearly with pressure for these materials, so more surface area can result in more friction (i.e. tires on a road, etc). :)

EDIT: Mcgyver typed faster.

whitis
08-18-2010, 01:00 AM
To first order (and in ideal physics), surface area does not affect friction, all other things being equal.

Now, if you compare a 1 square foot 2" plate vs a 2 square foot 2" plate on a flat horizontal surface in normal gravity, friction will be double but only because weight has doubled.

In the situation the OP is talking about, the total force is the same and whether it is distributed over a small area or a large one doesn't change the friction. Force per unit area goes down as area goes up, so total friction comes out the same. If the surfaces are ultra polished, then molecular forces will increase the friction with surface area. Hydrodynamic losses may increase with surface area.

Tires are a different situation - tires run on other than ideal surfaces. The asphalt is covered in dust, rain, snow, etc. The gravel or concrete itself could break loose if you put too much force in one place. Worse, the rubber has low shear strength, as demonstrated every time you leave skid marks. Concentrate too much force in a small area, and something will give. Thus you need to be conerned not just about friction/stiction but also the integrity of the surfaces. Plus once you break free, you are in sliding friction, not stiction. Heat also complicates things.

Friction vs stiction is a step function. The actual friction doesn't change at low velocities. However, you may spend part of your time in stiction and part in friction regimes such that a time weighted average resembles an intermediate coefficient of friction. Lubrication introduces some velocity dependent effects.

Too small a contact area and you concentrate wear. Too large and you may have problems with kinematics (rocking) and with keeping the surfaces flat. Concave surfaces on a flat are stable, convex ones are not, so you may deliberately remove some material in the middle to simulate a concave surface - i.e. bearing at the edges rather than the middle. Heat, loads, and other forces may transform an initially flat surface into a convex one.
Since there are likely to be some serious tilting forces, having your bearing surfaces as far apart as possible improves stability.

The effect of your geometry on your lubricating films should be considered.
A slightly beveled leading edge allows more lubricant on the ways to enter the sliding area, rather than being swept away, and pressurizes it (but may also allow more grit to get in). Grooves can help. Oil injection ports.

Rigidity may be more of an issue.

ldn
08-18-2010, 01:07 AM
Thanks everyone, I think I get it now. I appreciate the detailed descriptions.

The Artful Bodger
08-18-2010, 01:08 AM
Your saying no matter how big the plate is it will never take more force to move it. If that is true then if I drag a 20'x20'x1" plate it will take the same force as dragging a 12"x12"x1" plate.

I think that would be right if you loaded the 12"x12" plate to weigh the same as the 20'x20' plate

Carld
08-18-2010, 01:26 AM
Mcgyver, that is EXACTLY what I am talking about. The OP was asking if he reduced the area of contact on a given size plate would the friction be lower and it should be. Then things got convoluted and I used the example of one thickness and different size plates and for some reason no one understood what I was saying. Perhaps I should have made a 1000 word description with drawings so that I would be understood somewhat like Tiffie does.

perhaps the last post explained better what I said. There is no way in hell that doubling the square inch area of a 1" plate would not make it harder to move that plate. I don't understand why anyone is arguing that point.

Art, I said NOTHING about changing the thickness of the plate, just the square inch area. Your trying to change what I said. Just stick with the 1" thick plate and change the square inch area of the contact and tell me why the friction won't change.

I still don't quite understand how the friction doesn't change in Whitis's explanation below

In the situation the OP is talking about, the total force is the same and whether it is distributed over a small area or a large one doesn't change the friction. Force per unit area goes down as area goes up, so total friction comes out the same. If the surfaces are ultra polished, then molecular forces will increase the friction with surface area. Hydrodynamic losses may increase with surface area.

------------------------------------------------------------------------------------------------------------------------

A few posts into the thread someone asked if the friction would increase if the plate size was increased and I assume he meant the thickness would remain the same and that is where this all got started.

Fasttrack, in your example the mass stayed the same but in what I was saying the mass varied. That is where everyone is making their own interpretation of what I said.

Let me repeat it, if you take a plate of the same thickness and change the square inch area of contact the friction will change.

QUIT CHANGING WHAT I AM SAYING TO WHAT YOUR THINKING.

MuellerNick
08-18-2010, 02:44 AM
didn't say anything about hollowing out the center, the center just does not set on the other surface.

This statement is from the trivia department.

Nick

Paul Alciatore
08-18-2010, 03:44 AM
The F = mu * N formula is indeed given in the low level physics courses as a way of predicting frictional forces. But there are assumptions that go along with it. It assumes dry and fairly flat surfaces. Lubrication will change things. An oil or grease film will both decrease the value of mu and also add a new element that does depend on area. Just try putting grease between two plates that are only 1/4" square and comparing the force needed to overcome the grease friction with that of two greased plates that are 12" square. Do it vertically to eliminate the normal force between them so you are just comparing the friction from the grease. I guarantee the 12" X 12" plates will take a lot more force to slide apart than the 1/4" square ones.

Another point here, and this was discussed in a more advanced physics class I had, is that although the generally given formula for friction says it depends only on the normal force and not at all on the area, exactly the opposite is actually true. The area of contact is what determines the actual frictional force because it is molecular attraction that causes the frictional force and the more molicules that are in contact, the greater the frictional force will be.

So why does the normal force formula work? Well, any surface, even a highly polished one, even an optically "perfect" one will have some irregularities at the molecular level. These irregularities determine the actual AREA that is in contact as only the high points are actually touching the other surface. But the more normal force you apply, the more these high points are smashed down flat and the greater the area that is actually in contact. So the normal force increases the area in contact and it is actually the area in contact that determines the frictional force. Taking this to an extrems, melting the two surfaces together (A.K.A. welding them) brings the contact area to a maximum and keeps it that way permanently, even with no normal force and this produces the maximum amount of "friction". And the larger the area of the weld, the stronger it is.

In reality, friction is indeed determined only by the actual area in contact - molecular contact, that is.

And larger scale irregularities can create mechanical interference, kind of like small gear teeth or velcro. This can also add an element of "frictional force".

A.K. Boomer
08-18-2010, 04:49 AM
There's all kinds of factors in calculating friction between surfaces and its not just surface area and lubrication viscosity/quality but the speed in which the action is taking place and the material hardness temperatures and on and on...

Back in the day when automobiles were still in kind of an experimental era the typical volkswagon and porsche engines shared many similarities - both aircooled horizontally opposed 4 cylinders ----- but the porsche's produced more horsepower and rapped out at higher RPM's,

early porsche's ate a few connecting rod bearings due to the demand of increased power, Like I stated there are many factors involved but one of the biggest contributing issue's was the massive big end's of the connecting rods and therefore bearings/crank journals very similar to VW's - great surface area - but this amount of radius is coupled to something that is counterproductive - bearing speed between the crank journal and the rod bearings, The low RPM volkswagons seemed to survive better (well - they had their own problems like only holding 2.7 quarts of oil and no filter) but for the porsche's the added RPM's was one of the culprits in pushing the oil too far and creating to much "stir friction"

So what's the solution -- they needed to keep their surface area to handle the extra unit pressures of the extra ponies but they can't go any bigger,,,

The answer was to make the crank shaft journals much smaller in diameter yet widen them out to allow for similar sized contact patch - a total Win-Win...

J Tiers
08-18-2010, 09:47 AM
The area independence is true, but filled with assumptions.

In the real world, lots of other things come into play. hardness of materials, type of surface, etc, etc, etc.

one kg on the area of the point of a pin will give a great deal different net real-world apparent sliding resistance on a piece of brass, or lead, than one kg on an area of 100 cm^2.

On the small area, the surface will be indented and "lock" the "pin" against sliding. Not a frictional effect, truly, but a real world effect even so.

Then again, brakes..... large area means more heat dissipation area.... and out friend surface indentation again if you get silly about area.

So there is an optimal area for nearly any situation. that will have to do with lube, or lack of it, surface hardness and type, and what you want.

Automobile brakes need to actually fit in an auto wheel assy. so even if they might be best if larger, they are what they are for another reason. They would not be smaller, because the wear, heating, and hardness of available materials suggests they be larger, while the unsprung weight, etc suggests they be smaller. any design is a balance of these.

A slideway would perhaps be best with a low interface pressure from a wear point of view, but also must fit the machine, and not contribute too much friction of the lube.....

MuellerNick
08-18-2010, 04:18 PM
he F = mu * N formula is indeed given in the low level physics courses as a way of predicting frictional forces. But there are assumptions that go along with it. It assumes dry and fairly flat surfaces.

After 3 pages, pseudo-sience is taking over!
The µ_r states the conditions. That means, it is not one value for all, but many values for material match and lubrication (water, grease, oil, nothing, ...).

Nick

beanbag
08-19-2010, 01:47 AM
Carld is right in that surface contact area strongly affects friction. That's because most of the time, the friction vs force is sub-linear. If you push down on something twice as hard, it doesn't makes twice as much friction. By the same token, if you press down half as hard, it makes more than half the friction. So if you spread the load over twice the area, you'll end up with a little more friction than before (>1/2 X 2= >1). This shows up all the time in automotive stuff relating to tires, and is why people try to equalize the weight distribution between inside and outside tire for best grip.

Some other times, friction vs force is super-linear, like lubricated sliding surfaces. If you push down too hard, you break thru the lubrication layer and create galling.

MuellerNick
08-19-2010, 04:57 AM
This shows up all the time in automotive stuff relating to tires, and is why people try to equalize the weight distribution between inside and outside tire for best grip.

Useless example.
a) That is stick friction between road and tire, not slip friction
b) A tire is an elastic body and compared to metal-metal contact a liquid. In this case, the two surfaces mesh into each other. Viscous friction has other laws, the two surfaces adhere.

Nick

strokersix
08-19-2010, 07:45 AM
Tire friction a useless example? I don't agree. I think it is one illustration of where the simple formula does not work which is the point of this discussion. The simple formula should be used as a starting point, then improved by testing of the actual friction couple of interest which has been said several times.

The argument that it's one kind of friction or another kind of friction only serves to reinforce the inadequacy of the simple formula.

Surprised noone has mentioned gage blocks wrung together.

jugs
08-19-2010, 08:04 AM
Tire friction a useless example? I don't agree. I think it is one illustration of where the simple formula does not work which is the point of this discussion. The simple formula should be used as a starting point, then improved by testing of the actual friction couple of interest which has been said several times.

The argument that it's one kind of friction or another kind of friction only serves to reinforce the inadequacy of the simple formula.

Surprised noone has mentioned gage blocks wrung together.

Thats air pressure !

MuellerNick
08-19-2010, 09:04 AM
think it is one illustration of where the simple formula does not work which is the point of this discussion.

Did the OP ask about the physics of tire friction? No! He asked about friction between two solid materials. And that's where the simple model works quite well.

The argument that it's one kind of friction or another kind of friction only serves to reinforce the inadequacy of the simple formula.

It helps in picking the right model. The friction between a solid and an elastic material is the wrong application for the right model. But it can help to confuse and lead to a useless and unrelated discussion.

Nick

J Tiers
08-19-2010, 09:19 AM
Agreed.

the simple model was described as "inadequate", but trying to apply a complete model and accompanying formulas to the simple case is unreasonably complicated and does not help.

You do not need a textbook on economics to pay for your bus fare, or buy some bananas. You use a very simple model for those actions.

NO

What it means is that the model is adequate for some things but not for others. That is the way of models, they apply in their proper conditions, and not otherwise. To argue they are useless, or inadequate because of that is to be uselessly academic.

Paul Alciatore
08-19-2010, 10:58 AM
Tire friction a useless example? I don't agree. I think it is one illustration of where the simple formula does not work which is the point of this discussion. The simple formula should be used as a starting point, then improved by testing of the actual friction couple of interest which has been said several times.

The argument that it's one kind of friction or another kind of friction only serves to reinforce the inadequacy of the simple formula.

Surprised noone has mentioned gage blocks wrung together.

Yes, friction is a very complicated subject. The simple formula is very limited. I did think of gauge blocks. They are a good example of increasing the area in actual contact without adding any normal force to accomplish it and as such, provide a perfect counter example against the simple formula.

In designing a slide, the OP needs to reach a compromize between a lot of factors. I suspect that a range of sizes would work quite well. This is not a really critical application like the automotive engine examples given above and a design that "looks right" will probably be OK. A good choice of lubrication will probably make more difference than any small adjustments in the dimensions. The choice of materials would also be important.

MuellerNick
08-19-2010, 11:16 AM
They are a good example of increasing the area in actual contact without adding any normal force to accomplish it and as such, provide a perfect counter example against the simple formula.

They are a good example of the van-derWaals forces and how a gecko can walk on the ceiling. That is not friction.

Nick

jugs
08-19-2010, 02:43 PM
Originally Posted by strokersix
Surprised noone has mentioned gage blocks wrung together.

Thats air pressure !

They are a good example of the van-derWaals forces and how a gecko can walk on the ceiling. That is not friction.

Nick

Take some larger gauge blocks wrung together, (sm ones could be held on by surface tension from oil or moisture)
clamp one block on small stand so other blocks are suspended,
cover with bell jar & remove air with vac pump,
watch in amazement as blocks fall off :eek:
Thats air pressure ! NOT friction or 'Special Forces' :D .

john

:)

MuellerNick
08-19-2010, 02:59 PM
watch in amazement as blocks fall off

No wonder. A bit of air and oil still is trapped in the gap. Oil evaporates, air expands. **plopp**

Now wring them together under vacuum or explain the gecko that he can't walk on the ceiling.

Or, try to explain the effect with a 100 mm Jo-block that still sticks to an other block but is way too heavy for air pressure to be held in place.

What page are we at? Nonsense-o-meter is almost at its limit!

Nick

jugs
08-19-2010, 03:42 PM
[quote=MuellerNick]No wonder. A bit of air and oil still is trapped in the gap. Oil evaporates, air expands. **plopp**

Now wring them together under vacuum How ??

or explain the gecko that he can't walk on the ceiling. gecko's not @ issue (different technology)

Or, try to explain the effect with a 100 mm Jo-block that still sticks to an other block but is way too heavy for air pressure to be held in place. Whats the surface area of your joblock X 14 psi

What page are we at? Nonsense-o-meter is almost at its limit! If you want nonsense ask Evan to join in :eek:

john

MuellerNick
08-19-2010, 04:27 PM
or explain the gecko that he can't walk on the ceiling. gecko's not @ issue (different technology)

Do yourself a favor and read what Van der Waals forces are.

Nick

Paul Alciatore
08-20-2010, 02:51 AM
They are a good example of the van-derWaals forces and how a gecko can walk on the ceiling. That is not friction.

Nick

Perhaps I should have been clearer. I was not talking about the force that tends to hold the blocks together against an anti-normal force. I was talking about the resistance to sliding or twisting them apart in the manner they are usually assembled and disassembled. That resistance is friction.

MuellerNick
08-20-2010, 03:49 AM
I was not talking about the force that tends to hold the blocks together against an anti-normal force. I was talking about the resistance to sliding or twisting them apart

Ah, I didn't read it that way. Sorry.
Yes, that is friction. But not of the simple kind. Because of the really flat surface there are many many contact points at the (almost) atomic level. And here, the simple model fails.

Physics always (?) work with models. Sometimes they work very well, sometimes they are crude and only work for specific applications with a lot of exemtions. Tribology is one of the less completely understood fields

Nick