Is There a Better Model for the Relationship Between Friction and Velocity?

[Baluncore]

I don't think I'm following.

It depends on how the box is placed in contact with the belt. If it is dropped from a great height, or if it slides sideways, to be placed gently onto the belt surface.

If the box falls and bounces, the acceleration is intermittent, but when acceleration occurs it is proportionally greater.

 

[erobz]

If the effects of static friction extend beyond v=0 as the charts in this thread depict, then yes.

Something to the effect of the following?

 

[erobz]

It depends on how the box is placed in contact with the belt. If it is dropped from a great height, or if it slides sideways, to be placed gently onto the belt surface.

I was trying to focus on the latter.

 

[.Scott]

Something to the effect of the following?

View attachment 319422

No. In the "Freshman" model, the coefficient never drops below $\mu_k$.
In the modified model, where we speculate a transition to $\mu_s$, the red line would end by rising to $\mu_s$ as it reached $v=w$.

In the unmodified "Freshman" model, there is no transition - just an abrupt jump from $\mu_k$ to $\mu_s$.

 

[erobz]

No. In the "Freshman" model, the coefficient never drops below $\mu_k$.
In the modified model, where we speculate a transition to $\mu_s$, the red line would end by rising to $\mu_s$ as it reached $v=w$.

In the unmodified "Freshman" model, there is no transition - just an abrupt jump from $\mu_k$ to $\mu_s$.

If the box is has reached the constant belt velocity $w$, then the net force acting on the box must be zero. $\mu_s$ is zero for this problem.

 

[Baluncore]

I was trying to focus on the latter.

And I am saying that it does not matter which, so long as the acceleration time is counted from when the package is released, rather than when it first contacts the belt.

 

[erobz]

And I am saying that it does not matter which, so long as the acceleration time is counted from when the package is released, rather than when it first contacts the belt.

Oh. That's probably interesting on its own, but I'm not so interested at what is happening at the beginning as of yet. I'm trying to figure out what is happening at the end.

 

[jbriggs444]

No. In the "Freshman" model, the coefficient never drops below $\mu_k$.
In the modified model, where we speculate a transition to $\mu_s$, the red line would end by rising to $\mu_s$ as it reached $v=w$.

Naively, I would expect friction to stay at $\mu_k$ until some portion of the block has "caught" on the belt surface. I would then expect momentary excursions of the frictional force between $\mu_s$ and $\mu_k$ as portions of the block "chatter" or "squeak" against the belt until the entirety of the contact surface has caught.

At this point we will have some increasing distortion of the block and/or belt taking place. If this distortion is insufficient to absorb any residual relative kinetic energy of the block, the chattering and squeaking will continue as the surfaces break free before catching again.

At some final moment, the surfaces will catch and the distortion of the block/belt will be sufficient to absorb the residual relative kinetic energy of the block. From this point on we revert to the model of a damped oscillator. We are in a oscillation phase where the displacement is increasing. It will reach a peak which is, by hypothesis, insufficient to cause the surfaces to break loose and is, thus, sure to be less than $\mu_s$ though still greater than $\mu_k$. The oscillation will damp out and we will be left with a frictional force of zero at equilibrium.

Edit: To be clear, the Freshman model is rather simpler than this. I am trying to give some characteristics of a modified model.

 

[erobz]

Naively, I would expect friction to stay at $\mu_k$ until some portion of the block has "caught" on the belt surface. I would then expect momentary excursions of the frictional force between $\mu_s$ and $\mu_k$ as portions of the block "chatter" or "squeak" against the belt until the entirety of the contact surface has caught.

At this point we will have some increasing distortion of the block and/or belt taking place. If this distortion is insufficient to absorb any residual relative kinetic energy of the block, the chattering and squeaking will continue as the surfaces break free before catching again.

At some final moment, the surfaces will catch and the distortion of the block/belt will be sufficient to absorb the residual relative kinetic energy of the block. From this point on we revert to the model of a damped oscillator. We are in a oscillation phase where the displacement is increasing. It will reach a peak which is, by hypothesis, insufficient to cause the surfaces to break loose and is, thus, sure to be less than $\mu_s$ though still greater than $\mu_k$. The oscillation will damp out and we will be left with a frictional force of zero at equilibrium.

Well, that escalated rather quickly. Skirting $\mu_s$ for a moment...The force of friction (net force acting on the box) is going to zero in this problem "when" or "as" $v \to w$ Correct?

 

[.Scott]

If the box is has reached the constant belt velocity $w$, then the net force acting on the box must be zero. $\mu_s$ is zero for this problem.

$\mu_s$ is not the force on the box. It is the static coefficient of friction.
At $w$, the net force is zero. The coefficient of friction is not.
At $w$, the coefficient of friction tells you how much force you would need to get it sliding again.

 

[erobz]

At $w$, the net force is zero. The coefficient of friction is not.

Ok, detour around $\mu_s$, since its climbing (according to what you say). How is the force of friction going to zero? It's clear that it must be(on this we agree), but how?

 

[.Scott]

Ok, detour around $\mu_s$, since its climbing (according to what you say). How is the force of friction going to zero?

It is going to zero because it matches the speed of the belt - and at $w$, the belt is not applying any force to it at all.

When you multiply the coefficient of friction with the normal force, you get the maximum force that can be applied through friction. So if you are standing on the ice and the wind starts to blow, you will remain stationary for as long as your $\mu$ can hold on. Wind speeds beyond that will start you moving. Before you start moving, your $\mu$ will stay at $\mu_s$ while the force of friction rises to a maximum. Then your $\mu$ will suddenly drop to $\mu_k$, and the frictional force drops based on that and you start moving.

 

[erobz]

It is going to zero because it matches the speed of the belt - and at $w$, the belt is not applying any force to it at all.

When you multiply the coefficient of friction with the normal force, you get the maximum force that can be applied through friction. So if you are standing on the ice and the wind starts to blow, you will remain stationary for as long as your $\mu$ can hold on. Wind speeds beyond that will start you moving. Before you start moving, your $\mu$ will stay at $\mu_s$ while the force of friction rises to a maximum. Then your $\mu$ will suddenly drop to $\mu_k$, and the frictional force drops based on that and you start moving.

So you are saying that $\beta (w - v)$ is a correct model for the friction force?

 

[Frabjous]

So you are saying that $\beta (w - v)$ is a correct model for the friction force?

It could be suitable for “fluid“ friction. For ”solid” friction, the force is finite for w-v≠0.

 

[erobz]

It would be suitable for “fluid“ friction. For ”solid” friction, the force is finite for w-v≠0

I don't understand. The force of friction modeled as $\beta ( w - v)$ is finite for $w-v \neq 0$? It is $f_r = 0$ for $v = w$

 

[.Scott]

So you are saying that $\beta (w - v)$ is a correct model for the friction force?

I'm going to answer explicitly because I can interpret your question in a couple of ways.

In the "Freshman" model, the maximum frictional force is the product $\mu N$ where N is the normal force and $\mu$ is the coefficient of friction. For the kinetic case, this will always be $\mu_k N$. In the kinetic case where gravity is holding a box of mass $m$ onto a flat surface, this will be $\mu_k mg$ where $m$ is the mass of the box and $g$ is gravity.

In the situation that you described, only the belt is applying force. So as soon as the friction becomes static, $\mu$ goes to $\mu_s$ and the frictional force goes zero.

 

[Frabjous]

I don't understand. The force of friction modeled as $\beta ( w - v)$ is finite for $w-v \neq 0$? It is $f_r = 0$ for $v = w$

In the limit, (w-v)→0, the force is non-zero. There is a discontinuity at 0.

 

[erobz]

In the situation that you described, only the belt is applying force. So as soon as the friction becomes static, $\mu$ goes to $\mu_s$ and the frictional force goes zero.

The frictional force is the only force acting on the box (in the direction of motion). If $\mu_k$ is tending to some non-zero $\mu_s$ ( as you say) then by some other mechanism the frictional force must be tending to zero as $v \to w$

 

[erobz]

In the limit, (w-v)→0, the force is non-zero. There is a discontinuity at 0.

Well, there is a discontinuity in $f_r = \mu_k m g$ when $v = w$

I can't see the discontinuity in $f_r = \beta( v-w)$?

 

[Frabjous]

Well, there is also a discontinuity in $f_r = \mu_k m g$ when $v = w$

What’s your point? Why is what you said different from what I said?

 

[erobz]

What’s your point? Why is what you said different from what I said?

We are talking about different functions.

 

[Frabjous]

In that case I would say that μ_k is not defined at w=v.

 

[erobz]

In that case I would say that μ_k is not defined at w=v.

In the drag type model, we could have anything happening to $\beta$ as $f_r \to 0$ continuously. In the constant force model $f_r$ is taking a quantum leap to $0$ when $v = w$. Pick your poison.

 

[Frabjous]

In the drag type model, we could have anything happening to $\beta$ as $f_r \to 0$ continuously. In the constant force model $f_r$ is taking a quantum leap to $0$ when $v = w$. Pick your poison.

I would say that drag is a fluid model and is not applicable to simple solid-solid friction. You are using the wrong physics.

 

[erobz]

I would say that drag is a fluid model and is not applicable to simple solid-solid friction. You are using the wrong physics.

Ok, but the constant force seems equally unclear unless we accept quantum jump models in classical mechanics. Clearly there is something happening near $v = w$ that is not commonly understood?

 

[Frabjous]

Ok, but is the constant force seems equally unclear unless we accept quantum jump models in classical mechanics.

Experimentally it fits the data. Eventually one is going slow enough so that distances between asperities matter will begin to matter, but those are not macro features. I hope that when you say “quantum” you are actually mean “discrete”

 

[erobz]

Experimentally it fits the data. Eventually one is going slow enough so that distances between asperities matter will begin to matter, but those are not macro features. I hope that when you say “quantum” you are actually mean “discrete”

Yeah, discrete. I was using quantum to highlight the point about a distinct "jump".

 

[erobz]

I'm not trying to annoy people with these inquiries, I'm just trying to fill some (personal) gaps.

 

[jbriggs444]

Ok, but the constant force seems equally unclear unless we accept quantum jump models in classical mechanics. Clearly there is something happening near $v = w$ that is not commonly understood?

Chattering, creeping, tires squealing, blocks tumbling -- things not contemplated by the freshman model.

In the real world, there are no such things as a rigid bodies or uniform surfaces. As you look closer and closer things look messier and messier. That is an important life lesson which applies to a host of different things.

As H.L. Menken wrote, "Explanations exist; they have existed for all time; there is always a well-known solution to every human problem—neat, plausible, and wrong."

In the freshman model, the decelerating force of friction is given by the normal force multiplied by the coefficient of kinetic friction all the way until relative motion stops. At which point the frictional force drops to zero. For a situation of decelleration to a relative stop, the coefficient of static friction never enters in at all.

In the real world, the block does not have a single velocity. It is not a rigid body. In the real world, portions of the contact surface can be catching on the belt while other parts of the contact surface are still sliding. The parts of the surface that catch lag behind the parts of the surface that slide. Stresses build within the material until the lagging parts catch up and, perhaps, surge ahead while other parts are now catching.

The shear force of friction may not be uniform across the contact surface. The normal pressure may not be uniform either. Certainly, normal pressure will be non-uniform if we account for torques. Vibrations and distortions during chattering may feed back and amplify themselves. The relative velocity between the surfaces may not be constant throughout the contact surface and may not match the velocity of the block's center of mass relative to the belt.

There is plenty of unpredictability and wiggle room so that any simplistic and deterministic model is sure to be inaccurate.

However, all is not lost. If we can measure how stiff the block is, we can calculate how much deflection (strain) would correspond to the difference between static and kinetic friction (stress). We can place reasonable bounds on just how bad the freshman model is likely to be.

Or we can run the experiment.

 

[erobz]

Chattering, creeping, tires squealing, blocks tumbling -- things not contemplated by the freshman model.

In the real world, there are no such things as a rigid bodies or uniform surfaces. As you look closer and closer things look messier and messier. That is an important life lesson which applies to a host of different things.

As H.L. Menken wrote, "Explanations exist; they have existed for all time; there is always a well-known solution to every human problem—neat, plausible, and wrong."

In the freshman model, the decelerating force of friction is given by the normal force multiplied by the coefficient of kinetic friction all the way until relative motion stops. At which point the frictional force drops to zero. For a situation of decelleration to a relative stop, the coefficient of static friction never enters in at all.

In the real world, the block does not have a single velocity. It is not a rigid body. In the real world, portions of the contact surface can be catching on the belt while other parts of the contact surface are still sliding. The parts of the surface that catch lag behind the parts of the surface that slide. Stresses build within the material until the lagging parts catch up and, perhaps, surge ahead while other parts are now catching.

The shear force of friction may not be uniform across the contact surface. The normal pressure may not be uniform either (it certainly will not be if we account for torques). The relative velocity between the surfaces may not be constant throughout the contact surface and may not match the velocity of the block's center of mass relative to the belt.

There is plenty of unpredictability and wiggle room so that any simplistic and deterministic model is sure to be inaccurate.

However, all is not lost. If we can measure how stiff the block is, we can calculate how much deflection (strain) would correspond to the difference between static and kinetic friction (stress). We can place reasonable bounds on just how bad the freshman model is likely to be.

Or we can run the experiment.

After @Frabjous mentioned about the microscopic scale, I was thinking that the rough belt ( a jagged surface) is applying little impulses to the rough box (another jagged surface) as is slide past it. At some speed these impulses are plenty enough to just imperceptibly raise the COM of the package as the interlocking surfaces slide over(under) each other. However, at some point $v$ close to $w$ these impulses can't supply that vertical force ( acting through the "micro normals"), so the surfaces just stay locked relative to each other.

Something like this picture. The impulse delivered to the upper block from lower section of belt would have to be such that the box slides over the microscopic hills and valleys of the belt. Once it fails to deliver sufficient impulse as $v \to w$ they stay locked relative to each other. Basically, the box c.o.m. is oscillating vertically on a microscopic scale as the belt slides underneath it.