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

[erobz]

Let say a package is entering a conveyor belt (velocity $w$) at zero initial velocity $v$ in the direction of the conveyor.

Initially the package slips as its being accelerated (from rest) to conveyor belt speed. A linear drag force seems reasonable to me for a model:

$$ \beta( \mu, N ) ( w - v ) = m \frac{dv}{dt} $$

$\beta$ has dependency on $\mu$ (which has dependency on the relative velocity), and the normal force $N$, such that $\beta$ is dimensionally consistent. However, if we ignore that for a moment( assuming $\beta$ is a constant) you get the following solution for $v(t)$:

$$ v(t) = w \left( 1 - e^{-\frac{\beta}{m} t } \right)$$

The problem, with $\beta$ as some constant (if its a problem?) is in this simplification, the package never quite reaches belt speed... it just becomes arbitrarily close to $w$ as $t \to \infty$:

If we address the fact that $\beta$ is not constant by examining a plot of $\mu$ vs $(w-v)$: it should look something like this:

The question I have is if you were to formulate that function for $\beta$ using the characteristics of $\mu$ above would it fix the issue of the package reaching belt speed in finite time, because it at least seems to do so in reality?

 

[Frabjous]

I think the issue is w-v which is always going to zero.

 

[erobz]

I think the issue is w-v which is always going to zero.

What would the function for $\beta$ need to be such that in the limit as $( w - v) \to 0, \dot v \to 0$. It seems like some higher order inverse function in $(w-v)$ could work, but it wouldn't be what was anticipated for $\mu$? or is it that it couldn't work period?

 

[Baluncore]

... β has dependency on μ (which has dependency on the relative velocity), ...

You have not defined β or μ.
If μ was the friction coefficient, then the force accelerating the package, would be constant, not a function of velocity difference.

 

[Frabjous]

What would the function for $\beta$ need to be such that in the limit as $( w - v) \to 0, \dot v \to 0$. It seems like some higher order inverse function in $(w-v)$ could work, but it wouldn't be what was anticipated for $\mu$? or is it that it couldn't work period?

You could have an increasing $\beta$, but that does not seem physical. I also do not like the high speed limit ($\infty$). In my experience, friction decreases with velocity. Given that both limits are questionable, I would avoid the functional form. My first attempt would be a constant frictional force, $\mu mg$.

 

[kuruman]

Suppose you are an observer sitting on a box that is already at rest relative to the conveyor's inertial frame. You see a box coming down with zero horizontal velocity. As soon as it touches the belt, it receives an impulse that gets it sliding away from you until it stops. The modeling is the same as that of a block on a horizontal surface being given initial velocity $v_0$ and moving across until it stops. In that case we teach that the force is constant until the very end when it becomes velocity-dependent and drops sharply.

I agree that assuming a constant force of kinetic friction is only an approximation, but you don't need the conveyor belt to devise a more realistic model. Just model kinetic friction differently. I think what you want is the red line below. The vertical axis is frictional force divided by $mg$. Please excuse the sloppy plot.

 

[erobz]

Thanks, for your replies everyone.

I think I was getting a bit confused. For some reason in my mind, I saw $\mu$ increasing from $\mu_k$ to $\mu_s$ as if the belt itself were accelerating. I realize now, that in the case of a constant belt velocity the net force acting on the box goes to zero as $v \to w$, so $\mu$ should drop to zero as @kuruman suggests.

 

[Frabjous]

Thanks, for your replies everyone.

I think I was getting a bit confused. For some reason in my mind, I saw $\mu$ increasing from $\mu_k$ to $\mu_s$ as if the belt itself were accelerating. I realize now, that in the case of a constant belt velocity the net force acting on the box goes to zero as $v \to w$, so $\mu$ should drop to zero as @kuruman suggests.

Given that the force is dropping to zero, I am not sure it will solve your asymptote problem.

 

[erobz]

Given that the force is dropping to zero, I am not sure it will solve your asymptote problem.

I was just going to express my concern for that.

 

[Frabjous]

I was just going to express my concern for that.

Have you tried a constant frictional force for w≠v; 0 for w=v?

 

[erobz]

Have you tried a constant frictional force for w≠v; 0 for w=v?

No, I haven't yet. I haven't solved any ODEs with the step function for a while! Time brush up on the Laplace Transforms.

 

[Frabjous]

No need to integrate through zero.

 

[erobz]

No need to integrate through zero.

Looking back, I don't believe I ever knew how to do what you are talking about...

how do you write a unit step function in terms of the dependent variable?

$$\mathscr{U}( v - w)$$

in terms of the time I think I'd have:

$$ m \frac{dv}{dt} = \beta ( 1 - \mathscr{U}( t - a )) $$

How do I get $a$? I feel like I need the solution first!

 

[erobz]

Maybe I'm overcomplicating it.

If you are proposing I just do:

$$ m \frac{dv}{dt} = \mu m g $$

The $v$ is just linear

 

[Frabjous]

Maybe I'm overcomplicating it.

If you are proposing I just do:

$$ m \frac{dv}{dt} = \mu m g $$

The $v$ is just linear

You caught me in mid-composition and saved me the LateX of a formal solution.
Yes.
Since we do not have real data, keep it as simple as possible.

 

[erobz]

You caught me in mid-composition and saved me the LateX of a formal solution.
Yes.
Since we do not have real data, keep it as simple as possible.

Yeah, I guess any dynamics that are going to happen with $\mu$ are going to happen very near $v = w$ regardless.

 

[Frabjous]

It is not clear to me whether $\mu$ increases or decreases at small velocity differences. I would argue that at some point it begins to increase to static friction values. I do not know if there is a region before that where it decreases.

 

[erobz]

It is not clear to me whether $\mu$ increases or decreases at small velocity differences. I would argue that at some point it begins to increase to static friction values. I do not know if there is a region before that where it decreases.

Yeah, it seems to do some funny things.

if the belt is moving at constant velocity, then as $v \to w$ the net force acting on the box has to go to zero. However, if the belt was accelerating at $a$ the then net force acting on the box as $v \to w$ has go to $ma$, indicating $\mu$ climbs to $\mu_s$...

I guess there is a likelihood that if the belt were accelerating "it never catches up" to consider.

 

[Frabjous]

Yeah, it seems to do some funny things.

if the belt is moving at constant velocity, then as $v \to w$ the net force acting on the box has to go to zero. However, if the belt was accelerating at $a$ the then net force acting on the box as $v \to w$ has go to $ma$, indicating $\mu$ climbs to $\mu_s$...

I guess there is the possibility that if the belt were accelerating "it never catches up" to consider.

It is more than a changing $\mu$, it is changing friction mechanisms. Also, as you point out, slip can occur.

 

[erobz]

It is more than a changing $\mu$, it is changing friction mechanisms. Also, as you point out, slip can occur.

I don't know if it's considered interesting, but it's not surprising to me anymore how one can find gaps in knowledge in seemingly mundane problems like this. I find something almost everywhere I look.

 

[Lnewqban]

I don't know if it's considered interesting, but it's not surprising to me anymore how one can find gaps in knowledge in seemingly mundane problems like this. I find something almost everywhere I look.

I would see it as a similar mysterious to me problem with sliding motorcycle tires that suddenly grab the pavement again.
While such slide is happening, many things quickly change around the contact patch: undulations of the road, vibrations of the chassis and transmission, response of the suspension, very small changes in temperature and roughness of the surfaces, etc.

For the case of the conveyor I would consider similar things being induced by the response of the flexible belt, rollers and frame to the impact of the packages falling on it.

 

[erobz]

I would see it as a similar mysterious to me problem with sliding motorcycle tires that suddenly grab the pavement again.
While such slide is happening, many things quickly change around the contact patch: undulations of the road, vibrations of the chassis and transmission, response of the suspension, very small changes in temperature and roughness of the surfaces, etc.

For the case of the conveyor I would consider similar things being induced by the response of the flexible belt, rollers and frame to the impact of the packages falling on it.

Everything is probably at least an order of magnitude more difficult to nail down than we care to pretend!

 

[Frabjous]

All of the great physicists that I have known have been fascinated by the entire world, not just the subject area that they have specialized in.

 

[jrmichler]

Let say a package is entering a conveyor belt (velocity w) at zero initial velocity v in the direction of the conveyor.

Initially the package slips as its being accelerated (from rest) to conveyor belt speed.

I have considerable real world experience with exactly this problem. Stacks of napkins and sheets of corrugated instead of boxes, but the analysis is identical. The analysis is as follows:

1) Assume/determine initial velocities and dynamic friction coefficient.
2) Calculate time and sliding distance to accelerate to the speed of the belt or skid to a stop.
3) Then it moves with the belt or stays stopped.

Assuming constant dynamic coefficient of friction makes the calculation easy, and the results agree very well with observations. These calculations are done by hand because they are so simple. Real world calculations of this type are done using the simplest method that gets the job done because the engineer has two goals:

1) Solve the problem.
and
2) Effectively communicate the results to other engineers and to management, so we do not mention differential equations.

This figure from a patent application, is a direct copy of the figure produced by the engineer of a complex version of the problem in the OP. The complexity is only from the challenges involved in keeping track of numerous single box problems, or single napkin stacks in this case.

 

[erobz]

I have considerable real world experience with exactly this problem. Stacks of napkins and sheets of corrugated instead of boxes, but the analysis is identical. The analysis is as follows:

1) Assume/determine initial velocities and dynamic friction coefficient.
2) Calculate time and sliding distance to accelerate to the speed of the belt or skid to a stop.
3) Then it moves with the belt or stays stopped.

Assuming constant dynamic coefficient of friction makes the calculation easy, and the results agree very well with observations. These calculations are done by hand because they are so simple. Real world calculations of this type are done using the simplest method that gets the job done because the engineer has two goals:

1) Solve the problem.
and
2) Effectively communicate the results to other engineers and to management, so we do not mention differential equations.

This figure from a patent application, is a direct copy of the figure produced by the engineer of a complex version of the problem in the OP. The complexity is only from the challenges involved in keeping track of numerous single box problems, or single napkin stacks in this case.

View attachment 319232

When I was actively engineering (for a short while a few years ago) I was bad at engineering goal 2! The running joke among my boss\coworkers started was don't believe anything I say until I came back with my third revision! It wasn't unjustified. I was always came out equations blaring , too quick to believe my analysis. I'd spill the beans... then I'd find a mistake, and repeat, and repeat...eventually the process leads to a solution, but people tend to recall how many times I change my own position.

This, however, was just intellectual (if you could call it that) wandering spurred by someone else's problem on here, but thank you for confirming that a constant friction force is a reasonable fit for the real-world application.

 

[Baluncore]

If the box falls onto the belt from above, then the force of vertical impact with the belt, should through friction during impact, provide a similar horizontal acceleration to a second box that was pushed horizontally onto the belt at the same time the first box was released to fall. The concept may also apply to a bouncing box.

The time for; v to approach w; appears to begin with release of the box, not on impact with the belt.

 

[.Scott]

$$ m \frac{dv}{dt} = \mu m g $$

The $v$ is just linear

... and the $m$'s cancel out.

The normal Freshman Physics model is:

Given:
1) $\mu_k$, the kinetic coefficient of friction;
2) $v_b=(w-v)$, the velocity relative to the belt;
3) $g$, gravitational acceleration; and
4) Ignoring static friction then:

This model is a fairly easy one to validate. All you need is a long flat tiltable surface and a box. Tilt the surface until you get your $\mu_k$, start the box sliding, then check its speed as it slides down. If the model was perfect, the speed would be constant. Or if the tilt was inexact, the acceleration would be constant.

So using this for the OP problem, the package will accelerate linearly at $g\mu_k$ until $v_b$ reaches zero.

The force does not drop as the relative velocity ($v_b$) drops. In fact, if we allowed the effects of static friction to come into play before $v_b$ reached zero, the acceleration would momentarily increase.

 

[erobz]

This model is a fairly easy one to validate. All you need is a long flat tiltable surface and a box. Tilt the surface until you get your $\mu_k$, start the box sliding, then check its speed as it slides down. If the model was perfect, the speed would be constant.

Thats wouldn't seem to be able to verify the points of interest though $v \to w$ . I was trying to think about what happens in the transition for the box velocity $v$ to a constant conveyor belt velocity $w$ as slipping ceases?

The force does not drop as the relative velocity ($v_b$) drops. In fact, if we allowed the effects of static friction to come into play before $v_b$ reached zero, the acceleration would momentarily increase.

You mean, it would momentarily increase before decreasing to zero. Correct?

 

[erobz]

If the box falls onto the belt from above, then the force of vertical impact with the belt, should through friction during impact, provide a similar horizontal acceleration to a second box that was pushed horizontally onto the belt at the same time the first box was released to fall. The concept may also apply to a bouncing box.

The time for; v to approach w; appears to begin with release of the box, not on impact with the belt.

I don't think I'm following.

 

[.Scott]

You mean, it would momentarily increase before decreasing to zero. Correct?

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

 

[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.