Is friction necessary for a Tractrix Curve?

Question

Is friction necessary for a Tractrix Curve? (ANSWERED)

1.If friction is necessary, *what curve will the particle trace if friction is not present?* (NOT ANSWERED)

2.If friction is not necessary, *what curve will the particle trace if friction is present?*(ANSWERED)

3.Does the setup have a string or a rigid rod attached to the tractor? (ANSWERED)

I have asked this on a Math forum because here friction is to be viewed Mathematically, and not Physically.

NOTE This came when a question was asked on a WhatsApp group on physics I was part of. My level of physics is nowhere near to solve this problem, but the person who solved on the group did it by assuming friction was not present. But Wikipedia said friction is present. I was confused, hence this question.

Answer 1

The mathematical description of a tractrix does not even mention friction.

The tractrix is defined as a curve with the property that the distance from any point on the curve to a fixed straight line measured along the tangent of the curve is constant.

This can be interpreted in classical mechanics as the path of an object that is pulled by a string in certain circumstances -- and for this interpretation one probably needs to assume that the intertia of the object is negligible compared to the friction, such that its direction of movement at any point is exactly the direction of the string that pulls it. However, this is purely a matter of interpretation and not part of the definition of the curve, which is entirely independent of such physical concepts.

Answer 2

The only book I came across friction in relation to the pseudosphere revolution surface from tractrix is E.H. Lockwood's A Book of curves, Cambridge 1961 that too on a single line. may be there are other references.

It has the property of constant tangent (red line) length $a$ defined as

$$ \frac {\sin \phi}{r} = \frac{1}{a} $$

Unless you draw the force/vector diagram you are not connecting physics and maths. So try to draw one. Perhaps it is without friction, ( iirc) the shape does not change with wear out of grinding stone.

A rudimentary situation could be depicted thus: If $F_N$ is normal force at interface we have friction force

$$ \tan \phi = \frac{F_{horiz}}{F_{vert}}$$ then the horizontal grinding force that is frictional force creating wear on a stone of given width $w$ approximately could be $ \dfrac{F \,\sin \phi}{2 \pi r w} $ constant on projected area for grinding stone shaped as a tractrix.

EDIT1:

I am trying to figure out another context of how OP may have imagined tractrix generation. It has to do with the orthogonal trajectories of circles whose rolling/dragging on a line which he appears to assume creates a tractrix locus. The tractrix is produced as tangent envelope but not as a locus of a point on a rolling circle in any manner.

This is a drag curve without rolling. There is no rolling at all, but only dragging , so circles slip severely everywhere except at at cusp when circle contacts x-axis, a situation locally like the cycloid with zero x-axis speed, no drag. Otherwise friction is unnecessary and no reference need be made for this scenario.

The variable drag as a parameter $h$ is valid for the pushed circles ( please note, not tractices!) :

$$ (x-h)^2 +(y-a)^2 = a^2 \tag1$$

with parametric equations

$$ x= (h - a \cos \phi),\, y= a ( 1 -\sin \phi) \tag2$$

They produce an orthogonal trajectory of tractrices I sketched roughly here. The envelope of cranks (green line) $PC = a $ touches the tractrix at $P$. The tractrix is one in a set of orthogonal trajectories of above displaced/sliding circles, along lines parallel to x-axis.

Parametric equations of orthogonal trajectories tally with standard tractrix as:

$$ y = a (1-\sin \phi) ,\, -x = a ( \cos \phi + log(tan (\phi/2))) ,\, \tag3 $$

where slope $ \phi$ is an arbitrary constant (single parameter) associating to constant length $a$.

EDIT 2:

Derivation of orthogonal trajectories

Differentiate (1)

$$ 2 (x-h) + 2 ( y-a) y^{\prime} =0 \tag4 $$

For Orthogonal Trajectories take negative reciprocal slope

$$ \frac{dy}{y-a}= \frac{dx}{x-h} \tag5 $$

Next integrate using $\phi$ for arbitrary constant slope as generating parameter

$$ \frac{y-a}{x-h}= \tan \phi \tag6 $$

or this is same thing as

$$ \frac{(y-a)}{\sqrt{(x-h)^2 + (y-a)^2}}=\frac{(y-a)}{a}=\sin \phi \tag7 $$

which is the ODE of standard tractrix (upto$\pm$ sign of radical) for this chosen axes direction resulting in (3).

EDIT3:

Drawing a tractrix is not so difficult. No need of modelling with friction and all, need not complicate more than necessary.

Assuming we have a dragged object weight $W$ on a flat table of coefficient of friction $\mu$ then force or tension in weightless string should be more than $\mu W$ for motion to ensue at all.

Dragging a magnet on a vertical plane I include a simple demo ( magnet force instead of gravity, magnet pulled by a length of paper).

Tractrix followed but without trace

Errors in Wiki of reference. The arc length should be $ a \log \dfrac {y_2}{y_1}$.