What is the orientation of the vector of friction?

In summary, the conversation discusses a question about the orientation of the vector of friction between a moving circle and a rotating wall. The participants also discuss the possibility of the normal force being constant and the relation between the friction force and the relative velocities of the objects. More information is needed to determine the velocities and accelerations involved.
  • #36
JrK said:
The circle doesn't rotate around its center, it is only in translation.
If the circle is rigid and non-rotating then there is no distinction between center of mass work and real work. You can consider the external force multiplied by the displacement of the center of mass to get the relevant figure for work done on the circle.
 
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  • #37
jbriggs444 said:
If the circle is rigid and non-rotating
Yes it is rigid and non-rotating. What you call real work ? For you the energy from the friction is measured with d1 not d2 ? I see a 'slip' because the red wall rotates around the circle.
 
  • #38
JrK said:
Yes it is rigid and non-rotating. What you call real work ? For you the energy from the friction is measured with d1 not d2 ? I see a 'slip' because the red wall rotates around the circle.
Real work would be the transfer of energy (other than thermal) to the object on which the force acts. For mechanical work, it is the dot product of the force applied and the displacement of the material in the region where the force acts.

The "slip" is irrelevant to the work done on the target object. But it is relevant to the energy requirements.
 
  • #39
JrK said:
I see a 'slip' because the red wall rotates around the circle.
Again, this motion of the contact point relative to the circle material is irrelevant to work. Only the motion of the material itself matters.
 
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  • #40
JrK said:
the circle is driven by an hydraulic cylinder at a constant velocity. The circle doesn't rotate around its center, it is only in translation.

You haven't explained the arrangement of the hydraulic piston that moves the circle. Is the piston on a fixed axis? Or can piston pivot about one of its ends as it lengthens. If the piston is on a fixed axis, where does the axis point? If the piston can rotate about one of its ends, where is that end?

From the statement "at a constant velocity", one would assume the piston is along a fixed axis and it lengthens at a constant rate.

The spring to have the force of friction between the circle and the red wall doesn't change its length so its potential energy.

Compared to most textbook problems about the frictional forces on objects on inclined planes, this is assumes an unusual situation, but I think there is nothing inherently paradoxical about it.

We could begin with a simplified version of the situation.

The vertices of a triangle at time ##t## are given by ##A(t), B(t), C(t)##

##A(t)## is the constant function ##A(t) = (0,0)##
##C(t)## is the constant function ##C(t) = (x_0, 0)## for some constant ##x_0##.##B(t)## is the function ##B(t) = (B_x(t), B_y(t)) = (k B_x(0), k B_y(0))## for some constant ##k##.

The point ##B(t)## is acted upon by a force of constant magnitude ##F## that points in the direction from ##B(t)## to ##C(t)##.

The simplifed situation does not represent the contact between ##AB## and ##CB## is a circle. If we include the circle, the friction force is not acting along the direction from ##B(t)## to ##C(t)##. Instead it acts along the direction of a tangent to the circle at time ##t## that passes through ##C(t)##.
 
  • #41
@AT: I believe you but I don't understand why. When I use two real physical objects (I tested a lot of time yesterday), I see the distance d2 not d1 for the friction.

@Stephen Tashi : the hydraulic cylinder moves only in translation.

jbriggs444 said:
The "slip" is irrelevant to the work done on the target object.
What that means ? There is the work to move the circle (easy to calculate) and the energy from heating from the friction (depends of the distance d1 or d2) that's all, no ?
 
  • #42
JrK said:
What that means ? There is the work to move the circle (easy to calculate) and the energy from heating from the friction (depends of the distance d1 or d2) that's all, no ?
What do you mean by "the work to move the circle"?

Are you talking about the work done by the hydraulic cylinder? The work done by kinetic friction? The sum of the two?

The slip tells you how much mechanical energy is dissipated into thermal energy.
 
  • #43
jbriggs444 said:
What do you mean by "the work to move the circle"?
The work done by the hydraulic cylinder to move the circle.

jbriggs444 said:
The slip tells you how much mechanical energy is dissipated into thermal energy.
If there are no other energies than the work needed to move the circle and the heating from the friction. I think AT has right : the distance is d1. But I see d2 because there is a slip. If you are on the circle (fixed on the circle) you will see the red wall rotates around the circle, and that rotation gives (for me) a slip, a little difference that gives d2 not d1. And the calculation of that rotation gives the difference between d1 and d2.

In the following example (another position for A0) the rotation is worst:

cye.png
 
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  • #44
JrK said:
@AT: I believe you but I don't understand why.
Two simple examples to see that the motion of the material matters for work, not the motion of the contact point:

1) A Car is diving up an incline at constant speed. The contact point is moving, but the static friction cannot do work on the incline, because the incline is static. Just like when you push against a static wall. So its the (non)motion of the incline material that determines the work done on the incline, not the motion of the contact point.

2) A wheel is spinning in place on a fixed axis, and accelerating a board placed underneath that is free to slide. Here the contact point is static, but the board material is moving. The wheel is obviously doing work on the board which gains kinetic energy. So again, its the motion of the board material that determines the work done on the board, not the (non)motion of the contact point.
 
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  • #45
JrK said:
If you are on the circle (fixed on the circle) you will see the red wall rotates around the circle, and that rotation gives (for me) a slip,
That's not a "slip" but rather a "roll". You have a mix of both here, but that "roll" component is irrelevant for work done by friction.

Imagine a different setup where there is no slip at all, and the wheel is rolling along the wall with some static friction acting. In the frame of the wheel the wall rotating and rolling around the wheel. Here no energy is dissipated by friction, despite the movement of the contact point in both frames.
 
  • #46
I understood your two last examples. Thanks again AT for your patience !

A.T. said:
That's not a "slip" but rather a "roll".
Yes, that I called the rotation of the red wall around the circle for someone fixed on the circle is a roll.

A.T. said:
Imagine a different setup where there is no slip at all, and the wheel is rolling along the wall with some static friction acting. In the frame of the wheel the wall rotating and rolling around the wheel. Here no energy is dissipated by friction, despite the movement of the contact point in both frames.
Yes, it is possible to find the good position of A0 where there is near no energy from friction but the rolling is big so the difference of energy very high. I believe you but it is important for me to understand why and where I'm wrong.

A.T. said:
You have a mix of both here, but that "roll" component is irrelevant for work done by friction.
I would like to understand why. When I manipulate these 2 objects, I can see the friction (I see the combination of the two movements) because the circle moves farther from A0 but the roll in the same time prevents the full distance of friction to have d1. If the radius of the circle is 0, there is no roll, so the energy from friction is well the distance moved by the dot of friction, all is fine with a radius at 0. Why, when the radius is not 0, it is not the movement of the dot of friction ? An isolated roll doesn't give any energy of friction, here the roll is combined with the translation of the circle, so it is odd to say the roll participates in the friction when it is a combination.
 
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  • #47
JrK said:
I believe you but it is important for me to understand why and where I'm wrong.
One way to get intuition on friction, it to imagine geared surfaces instead of smooth surfaces. But the gears are worn off and can skip teeth. The distance relevant for work dissipated by friction is determined by the number of such tooth skips.

JrK said:
An isolated roll doesn't give any energy of friction, here the roll is combined with the translation of the circle, so it is odd to say the roll participates in the friction when it is a combination.
The rolling component doesn't "participate" in the work dissipated by friction, because it doesn't affect the relative movement of the material in contact. It merely affects how the contact location moves, which is irrelevant to work.
 
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  • #48
I try to find a method to understand. I study some different cases to find my mistake. The method: at start, at the contact between the wall and the circle, there are two dots, one on the wall: 'p1' the other on the circle: 'p2'. At final, there are two dots too, one on the wall: 'p3' the other on the circle: 'p4'. With an infinitisimal movement and with friction between the circle and the wall.

1/ The circle is fixed, the wall rolls around the circle: no friction, no energy needed to roll the wall. The distance (p1,p3) is equal to the distance (p2,p4), it is logical there is no energy from friction. All is fine.

2/ The circle is fixed, the wall doesn't roll but I move it in translation along the circle, the energy of friction is the distance moved by the wall. There is a difference of the distance in the calculation: (p1,p3)-(p2,p4), it is the energy of friction. The energy I need to move the wall in translation is equal to the energy of friction.

3/ The circle is fixed, I move the wall in translation along the circle of x millimeters, and in the same time there is a roll, the roll changes the dot of contact of the distance y millimeters. The distance (p1,p3)-(p2,p4) is equal to x, again it is the energy needed to move the wall. The roll is transparent. But the calculation (p1,p3)-(p2,p4) seems correct, at least here.

4/ I come back to my circle in translation and the wall in rotation around A0. I could say the roll is transparent like in the case 3/ but I could also measure the distance (p1,p3)-(p2,p4) it must works too. I have also the dots p1,p2, p3 and p4. The calculation (p1,p3)-(p2,p4) is equal to d2 not d1.

Maybe I can decomposed the movement in another manner:

4.1/ First time, the circle is fixed, the wall is rolled of the angle necessary.
4.2/ Second time, I move the circle with the wall to the distance it must move.
At final, I measure the distance of the wall from A0, it is like translate the wall, I reported to the drawing:

c5.png


's' is the distance of the roll
't' is the distance of the wall from A0 when I translate the circle and the wall after the roll
t=d2, and s+t=d1

The distance of the wall from A0 is d2 not d1.

Is there a physicist method to study that sort of problem ? Or at least, it is possible to discuss about my method (even, it seems very basic).
 
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  • #49
JrK said:
Is there a physicist method to study that sort of problem ?
You compute the velocity of the material that the force is applied to, and then integrate their dot product, to get the work done on the material.
 
  • #50
A.T. said:
You compute the velocity of the material that the force is applied to, and then integrate their dot product, to get the work done on the material.
It is an indirect method, you supposed the direction of the force, there is no other work needed than the circle. I mean a direct physicist method.

What do you think about the difference of the distances (p1,p3)-(p2,p4) moved by the dot on the wall: the distance (p1,p3) and on the circle: the distance (p2,p4) ?
 
  • #51
A.T. said:
You compute the velocity of the material that the force is applied to, and then integrate their dot product, to get the work done on the material.
JrK said:
I mean a direct physicist method.
This is the direct physicist method. See the definition of work as intergral of power:
https://en.wikipedia.org/wiki/Work_(physics)#Mathematical_calculation
 
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  • #52
JrK said:
It is an indirect method, you supposed the direction of the force, there is no other work needed than the circle. I mean a direct physicist method.

What do you think about the difference of the distances (p1,p3)-(p2,p4) moved by the dot on the wall: the distance (p1,p3) and on the circle: the distance (p2,p4) ?
@A.T. has given you the direct method that any physicist would use to define the work done.

Your insistence on using labels like "p1" through "p4" without putting them on your drawings is annoying. I have no idea what you are talking about.
 
  • #53
I didn't know it is the direct method. I thought it was possible to integrate the movement of the dot of contact if I take in account all the movements, I mean the movement of the wall and the movement of the circle. I thought it was a geometric method.

jbriggs444 said:
Your insistence on using labels like "p1" through "p4" without putting them on your drawings is annoying. I have no idea what you are talking about.
I insist because I would like to understand how to think well. At start, at the dot of contact: the dot of contact, p1 and p2 are at the same dot. The dot p1 is fixed on the wall. The dot p2 is fixed on the circle. After a time, the dot p1 is now at a new position I called it p3, and the dot p2 is at a new position I called it p4. I have two vectors: p1->p3 and p2->p4, I measure the difference of these 2 vectors, and each time I use that method with 2 movements I have well the energy needed to move the object equal at that difference. But here, if I use that method I find d2 not d1.
 
  • #54
JrK said:
I didn't know it is the direct method. I thought it was possible to integrate the movement of the dot of contact if I take in account all the movements, I mean the movement of the wall and the movement of the circle. I thought it was a geometric method.I insist because I would like to understand how to think well. At start, at the dot of contact: the dot of contact, p1 and p2 are at the same dot. The dot p1 is fixed on the wall. The dot p2 is fixed on the circle. After a time, the dot p1 is now at a new position I called it p3, and the dot p2 is at a new position I called it p4. I have two vectors: p1->p3 and p2->p4, I measure the difference of these 2 vectors, and each time I use that method with 2 movements I have well the energy needed to move the object equal at that difference. But here, if I use that method I find d2 not d1.
So ##p_2## and ##p_4## are the before and after positions of a dot painted on the circle. The incremental work done on the circle is given by ##\vec{f_{\text{wc}}} \cdot (\vec{p_4}-\vec{p_2})##

Meanwhile, ##p_1## and ##p_3## are the before and after positions of a dot painted on the wall. The incremental work done on the wall is given by ##\vec{f_{\text{cw}}} \cdot (\vec{p_3}-\vec{p_1})##

The sum of the two is guaranteed to be negative and gives the amount of mechanical energy dissipated into thermal energy by the force of kinetic friction.

According to your setup, ##\vec{p_1}## = ##\vec{p_2}##. Both are equal to the initial position of the point of contact.

You do not have a ##p_x## variable for the after position of the point of contact. That is good since that position is irrelevant.
 
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  • #55
I found a method to count the real difference I think. I use for that a theoretical elastic between the circle and the wall. I resumed my method in the following drawing:

jrd.png

I found the distance d2 not d1. Is it correct ?
 
  • #56
If you are going to the trouble of typing in an explanation, please type it into the forums instead of burying it in a graphic in a font that is all but unreadable. The purpose of this is twofold. First, it allows us to read what you have written. Second, it allows us to quote and respond to what you have written.
the graphic said:
An elastic is attached between the dot A and the dot B around the wall of the circle, always around the wall of the circle, I mean the length of the elastic at start is pi/4 * R with R the radius of the circle.

I suppose the force F of the elastic constant when its length changes, at least for the difference of length of that example.

The dot A is fixed on the circle. The dot B is fixed on the red wall. The dot B is always at the same position than the dot of contact between the circle and the red wall.

If the dot B is both fixed to the red wall and continuously at the point of contact, it follows that the wall is moving so as to keep one point always in contact with the circle. It would have been good to have pointed this out explicitly.

the graphic said:
So I need to change the position of dot B so the length of the elastic : the position of the dot becomes B'. Like I change the length of the elastic I recover an energy : the length of B - B'. For the drawing at final I win the energy of d2F
[...]
I think I am finally grasping what you are talking about. The stretching of the elastic measures the mechanical energy that would have been dissipated into thermal energy due to kinetic friction -- assuming that the tension in the elastic is at all times equal to what the force of kinetic friction would have been.

That is nice, but it has little to do with the work done by the wall on the circle.
 
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  • #57
I like the drawings with all information on it but it is true it is better for others people to have the text inside the message.

jbriggs444 said:
That is nice, but it has little to do with the work done by the wall on the circle.
I have the same force on the circle and on the wall than the friction does during the movement. So the energy to move the circle is the same and the energy to rotate the wall is the same (the rotation of the wall doesn't need any energy nor give) with the friction or with the elastic. I need to change the length of the elastic step by step, it is theoretically but I can compare it to the energy recovered by the friction: it is the same. The advantage of the elastic allows to watch the distance d2. And for the friction of for the elastic the energy must the the same than the energy to move the circle.
 
  • #58
JrK said:
(the rotation of the wall doesn't need any energy nor give)
Wrong.
 
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  • #59
JrK said:
The advantage of the elastic allows to watch the distance d2. And for the friction of for the elastic the energy must the the same than the energy to move the circle.
Your elastic would stretch and store energy even if there was no slip, just pure rolling, which dissipates no energy in the sliding friction case. The two are in no way equivalent, and your d2 irrelevant for the energy dissipated by friction.
 
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  • #60
jbriggs444 said:
The stretching of the elastic measures the mechanical energy that would have been dissipated into thermal energy due to kinetic friction -- assuming that the tension in the elastic is at all times equal to what the force of kinetic friction would have been.
I don't think this is true. See my post #59.
 
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  • #61
There cannot be pure rolling given the constraint that a fixed point on the wall (to which one end of the elastic is attached) slides over the circle remaining continuously at the point of contact.
the graphic said:
The dot B is always at the same position than the dot of contact between the circle and the red wall.
 
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  • #62
jbriggs444 said:
There cannot be pure rolling ...
I'm using pure rolling just to show that this elastic band method cannot be used to determine the energy dissipated by friction.

jbriggs444 said:
...given the constraint that a fixed point on the wall (to which one end of the elastic is attached) slides over the circle remaining continuously at the point of contact.
I thought B is merely the contact location, constrained to be somewhere on the wall, but not necessarily fixed to the same physical wall position.

It cannot satisfy both constraints (always contact location & fixed physical wall position) in the OPs scenario either, so this interpretation doesn't make sense to me.
 
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  • #63
A.T. said:
It cannot satisfy both constraints (always contact location & fixed physical wall position) in the OPs scenario either, so this interpretation doesn't make sense to me.
It is the OP's scenario. He specified it.
graphic said:
The dot A is fixed on the circle. The dot B is fixed on the red wall.
 
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  • #64
jbriggs444 said:
It is the OP's scenario. He specified it.
It's either a language problem or it doesn't make any sense. B cannot be the contact location and a fixed physical point of the wall, because the contact location moves along the wall.
 
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  • #65
A.T. said:
It's either a language problem or it doesn't make any sense. B cannot be the contact location and a fixed physical point of the wall, because the contact location moves along the wall.
I agree that it does not make much sense. The only way I saw reconcile it, which indeed may not have been the intent, is to have a fixed point on the wall sliding around the circle and maintaining contact.

This interpretation certainly violates the constraint in the original post that the wall is fixed to a pivot on the ground.
 
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  • #66
Thanks to both of all ! you are very very nice :)

A.T. said:
It's either a language problem or it doesn't make any sense. B cannot be the contact location and a fixed physical point of the wall, because the contact location moves along the wall.
No, it is not a problem of language but I think with a step of integration, I change the position of the dot B at each step of the integration, B is at start (of the step of the integration) at the dot of contact between the circle and the wall and only at that time, after, the dot B is farther because it is fixed on the red wall. The dot B is not in the continuous time at the dot of contact, it is not possible. But I can increase the number of steps.

I know where is my mistake with the elastic: I didn't count the difference of the potential energy stored in the elastic (the potential energy stored from start to end of the study) and there is one ! it is (d1-d2)*F. So, like I drew, the energy lost to move the circle is recover by the elastic.

Remember, I took the idea of the elastic because I would like to find a method to count the energy because I see the friction at d2*F not d1*F. With the dot A fixed on the circle it is not like the friction could do. So, I took an elastic for each step of integration. That elastic at start has a length of 0, the two ends of that elastic is at the dot of contact, one end fixed on the circle, the other end fixed on the red wall. For each step, the elastic will increase its length of d2 not d1, or at least it is like I see that.

I resumed all here:

vffd.png


I think like that I will understand the distance is not d2 but d1.

I copy paste the text in the image because it seems it is not possible to download the image at its full definition:

An elastic is attached between the dot A and the dot B around the wall of the circle, always around the wall of the circle. I suppose the force F of the elastic constant when its length changes, at least for the difference of length of that example. The dot A is fixed on the circle. The dot B is fixed on the red wall. At start the length of the elastic is 0 and at final it is d2 : at start the dots A and B coincide. At each step of the numerical integration, I take a new elastic. The two ends of each new elastic is at the dot of contact between the circle and the red wall, but one end is fixed on the circle the other end is fixed on the red wall. Each elastic increases its length of d2 not d1. The force F1 is the force from the elastic to the red wall at the final of one step. The force F2 is the force from the elastic to the circle at the final of one step. The energy recovered from all the elastics is d2*F. The energy needed to move the circle is d1*F = lg*cos(a)*F. I don’t need nor recover any energy from the rotation of the red wall.
 
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  • #67
JrK said:
The energy recovered from all the elastics is d2*F. The energy needed to move the circle is d1*F.
Then your elastic is not modelling friction correctly, which has to dissipate all the work done moving the circle in the original setup.
 
  • #68
JrK said:
It is very difficult to pose (not resolve) the movement by equations (math) ?

We won't know until we try!

The diagrams you have presented ignore the invention of Cartesian coordinates. They lack representations for crucial dimensions such as the radius of the circle and dimensions that establish the relation of the line of movement of the center of the circle to the point where the wall pivots.

You are attempting to discuss the problem in an abstract geometric style. Archimedes and Newton were able to think about things this way, but the modern approach is to use cartesian coordinates!

For the sake of getting advice about modern methods, you should be specific about cartesian coordindates. For example:

The wall pivots at (0,0). The radius of the circle is r. The center of the circle moves along the line (x,y_1) where y_1 is constant. .

Draw a diagram showing the above information. Try to find the coordinates of the point of contact (x_c, y_c) as functions of x and the constants.

You can get one equation by setting the distance between the center of the circle and (x_c, y_c) equal to r. You can get another equation from the condition that the vector from (x_c, y_c) to the center of the circle is perpendicular to the vector from (0,0) to (x_c,y_c). The complication in getting a solution from these two equations is that there are two possible solutions for (x_c, y_c) since there are two lines from (0,0) tangent to the circle.

I think there are clever people on the forum that can solve the above problem since it is posed in the modern style. Considering the problem in coordinate-free way is only leading to confusion.
 
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  • #69
I like a lot geometry ! Yes, a math method would be better, I tried with my program with a numerical integration (post #24), I used coordinates in the program but I don't know if my calculations are correct. I can explain the program if necessary, the program uses the first drawings: the dots, distances.

To come back to the method with the elastic, I need to place the dot A fixed not on the circle but on a needle, and the needle needs to be between the circle and the red wall. A very thin needle ! Like that the needle is always at the dot I want: the dot of contact between the circle and the red wall. The dot B is fixed on the red wall. I need only one elastic.

I give the details:

gjk.png


fht.png
 
  • #70
@JrK, you're still avoiding cartesian coordinates.

I see no reason for the imaginary needle. You seem to think your coordinate free diagrams are helpful. So far, they haven't helped anybody.
 
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