What is the orientation of the vector of friction?

[A.T.]

I'm agree it is a mistake to see that slip,

Your mistake is looking at the motion of the contact location, instead of the motion of the material at the contact location. This is not a math/geometry error, but an error in your understanding/application of the definition of work.

Everything else follows from that. Your needle is just a proxy for the irrelevant motion of the contact location, and thus pointless to discuss.

 

[JrK]

Your mistake is looking at the motion of the contact location, instead of the motion of the material at the contact location.

No, I counted the energy of the length moved by the circle. I count the energy stored inside the elastic. I count all others forces I see. I choose that example where the needle is at the dot of contact, the needle is a free object with constraint by the elastic and the walls (materials) but I count the energy of the material. The needle is an object not a dot. There are 4 objects: the circle, the wall, the needle and the elastic.

 

[Stephen Tashi]

but I count the energy of the material.

If we consider the circle or the wall to be made of a realistic material, we can worry about small deformations storing and releasing energy. However, the problem you originally describe treats the circle and the wall as a "rigid bodies". Rigid Bodies are theoretical idealizations. They do not deform. So there can be no detailed explanation of why there is friction between two rigid bodies. The existence of friction in a problem involving rigid bodies can't be explained by any properties of the bodies. Any forces of friction must be given as abstract assumptions. So the effect of friction between rigid bodies can't be analyzed by a model involving needles or small elastic bands.

If you wish to consider a problem involving realistic materials, I agree that the work done by the force of friction upon a circle made of an elastic material may be different that the work done by friction on a rigid body. However, the work done on an elastic body cannot be analyzed without quantifying the elasticity of the body. You cannot do an analysis only by visualizing needles and elastic bands and get a single result that pertains to all types of elastic bodies.

 

[JrK]

why there is friction between two rigid bodies

I explained it before, at the message #20. The bodies are rigid.

With the needle + elastic I computed all the energies, step by step, and I would like to find my mistake. Maybe a problem with an angle.

 

[Stephen Tashi]

With the needle + elastic I computed all the energies, step by step,

With a rigid body, there is no elastic.

 

[JrK]

Except for the elastic, a perfect elastic, like I explained, just to simplify the calculations.

 

[Stephen Tashi]

and I would like to find my mistake.

Your mistake is assuming the elastic.

 

[JrK]

Your mistake is assuming the elastic.

What do you mean ?

 

[Stephen Tashi]

What do you mean ?

You did an analysis that employs the concept of an elastic. Such an analysis does not apply to rigid bodies.

 

[JrK]

You did an analysis that employs the concept of an elastic. Such an analysis does not apply to rigid bodies.

I wrote except the elastic because it is logic that the elastic is not rigid... but the elastic is an example, just for you, if you want you can place an electrostatic charge on the neddle and on the dot B (different charges). I need a force that pulls, I choose the elastic for example. All the bodies are rigid except the elastic.

Note: or 2 magnets, or 2 electromagnets on the dots A and B, like you want if you think it is a problem the elastic.

 

[A.T.]

I count the energy stored inside the elastic.

If you mean a constant force elastic, that basically just measures the slipping distance relevant for friction work, then the correct method of attaching that elastic is described here:

https://www.physicsforums.com/threa...of-the-vector-of-friction.987415/post-6332661

It doesn't require any needles.

 

[jbriggs444]

We repeatedly see that the circle is described as being pushed by a cylinder and that we are concerned with the work done by the cylinder.

But we are never shown the attachment point of the cylinder, the anchor point for the cylinder nor any accounting for the work associated with the torque from the cylinder as it rotates with respect to the circle.

 

[Stephen Tashi]

I need a force that pulls,

You wish to have a force that impedes the motion of the point of contact.

We are dealing with rigid bodies, there is no property of rigid bodies that says such a force exists. If you want to assume such a force exists then you are adding information to the problem and changing it into a different problem.

You wish to say that work done on the point of contact is the same as work done on the circle.

Calculating work done on a tiny object that stays at the point of contact is not the same as calculating work done on the circle. The circle and the tiny object are not the same object.

 

[JrK]

with the torque from the cylinder as it rotates with respect to the circle.

No, the cylinder doesn't rotate, the circle moves in translation. You can glue the circle with the cylinder. The cylinder will move only in translation.

It doesn't require any needles.

I replied to you before, I used it at #55, that method doesn't use the dot of contact and I would like to use the dot of contact, and I need the needle for that. If I fixe the dot A on the circle I don't have the analogy with my first example with the friction.

You wish to have a force that impedes the motion of the point of contact.

I don't understand, the circle and the wall are controlled in position by external devices: cylinder and motor (for example). I need a force to simulate the force of friction because I see a slip with the friction and I could understand where is my mistake when I use an example that uses the dot of contact. The force of the elastic gives the force that the friction could do in the first example.

Thanks a lot for your patience !

 

[jbriggs444]

No, the cylinder doesn't rotate, the circle moves in translation. You can glue the circle with the cylinder. The cylinder will move only in translation.

Yes the cylinder does rotate. It remains parallel to the wall which is rotating.

If it does not rotate, it must translate. If it translates, then, since it is subject to a torque, that translation involves work.

 

[A.T.]

... I would like to use the dot of contact...

Why?

 

[JrK]

Why?

Because I would like to make the analogy with the example with the friction to understand perfectly the first example.

@jbriggs444: yes the cylinder needs a work ! I counted that work, it is the work to move the circle.

 

[Stephen Tashi]

because I see a slip with the friction and I could understand where is my mistake when I use an example that uses the dot of contact.

To repeat: calculating work done on the dot of contact is a different problem than calculating work done on the circle. If you think of dot of contact as a rigid body, it is not the same rigid body as the circle.


Let's try this: Suppose the point of contact is initially at (2, 15) and there is an atom $A$ of the circle at (2,15). Later atom $A$ has moved to (2+3, 15) = (5,15) and the point of contact has moved to (2+3,27) = (5,27). Atom $A$ has not moved in the y-direction. No other atom of the circle has moved in the y-direction. If you think of the dot of contact as rigid body, its atoms have moved in the y-direction. So the dot of contact and the circle are different rigid bodies.

 

[A.T.]

Because I would like to make the analogy with the example with the friction to understand perfectly the first example.

But in the case with friction the movement of the contact point is irrelevant for the work. So again, why do you insist on focusing on it?

 

[JrK]

So again, why do you insist on focusing on it?

Again, to make the exact analogy with the example with the friction. I know the movement of the dot of contact is irrelevant but I would like to understand my mistake. And, for you: why you don't want to use the needle and the elastic ?

If you think of the dot of contact as rigid body, its atoms have moved in the y-direction.

You think here with the needle, true ? if yes, I'm agree the needle moves in 'y' like my drawings showing. (I supposed x: horizontal and y:vertical).

 

[Stephen Tashi]

You think here with the needle, true ?

Yes, the needle moves in the y-direction. The needle and the circle are not the same object. Hence the work done on the circle and the work done on the needle are different numbers.

 

[A.T.]

I know the movement of the dot of contact is irrelevant but I would like to understand my mistake.

Isn't focusing on something irrelevant the mistake?

 

[JrK]

Isn't focusing on something irrelevant the mistake?

I hope, I could say that only when I will understand the mistake with the needle. The needle + elastic are not complex to study and the advantage: it is geometry, I can measure the difference of potential energy from the length of the elastic.

The needle and the circle are not the same object. Hence the work done on the circle and the work done on the needle are different numbers

Yes, the circle and the needle are not the same object. The work done by the circle is d1*F. There is no work from the needle because the sum of forces on the needle is 0 at each time, the elastic pulls the needle but the walls of the circle and the red wall stop the needle. The other energy is the potential energy stored inside the elastic. And maybe the energy from the forces F3 and F4, I'm not sure about my calculation.

 

[jbriggs444]

Because I would like to make the analogy with the example with the friction to understand perfectly the first example.

@jbriggs444: yes the cylinder needs a work ! I counted that work, it is the work to move the circle.

Then it's not a cylinder. It's an arbitrary external force plus an arbitrary external torque. A cylinder has two attachment points, one at the business end where it attaches to the load and one at the opposite end where it is anchored.

As to your utilization of a needle as a proxy for the contact point, you need to account for the fact that in order to do its job, the needle must exert a non-zero force on the circle parallel to the wall. In order to do this, the needle cannot be exactly at the contact point. It must always be displaced some distance to the side. Further, no matter how small the distance, in order to do its job, the torque produced by the needle on the circle must be a constant.

 

[JrK]

Yes, it is a standard cylinder, fixed on the ground, it pushes the circle:

You can glue the circle to the cylinder (I mean the part of the cylinder that moves to the right).

 

[jbriggs444]

Yes, it is a standard cylinder, fixed on the ground, it pushes the circle:

View attachment 261874

You can glue the circle to the cylinder (I mean the part of the cylinder that moves to the right).

Won't work. You are not accounting for the resulting non-zero net torque on the circle. At least not if we are using friction.

But if we are using the needle and elastic and and some sort of sliding arrangement to clamp the circle to the cylinder and if the flat surface where the cylinder touches the wheel is greased... those are all central forces. No net torque. Yes, that could work.

 

[JrK]

Won't work.

What do you mean ? I can glue the circle on the cylinder (the part that moves). The circle doesn't rotate around itself. I don't want the circle rotate around itself like I specified.

 

[jbriggs444]

What do you mean ? I can glue the circle on the cylinder (the part that moves). The circle doesn't rotate around itself. I don't want the circle rotate around itself like I specified.

If you glue the circle to the cylinder then the other end of the cylinder could ride up and down a vertical greased wall and thereby provide torque to the circle. That could work.

A cylinder anchored to the ground and glued to the cylinder could not work. It would anchor the circle to a fixed horizontal trajectory -- which had not been my understanding of the problem.

 

[JrK]

It would anchor the circle to a fixed horizontal trajectory.

Yes, the circle moves in horizontal translation like I specified.

 

[jbriggs444]

Yes, the circle moves in horizontal translation like I specified.

So the cylinder supplies not only a horizontal force along its axis but also a shear force at right angles to its axis, right?