What is the orientation of the vector of friction?

[JrK]

Hi,

It is my first message :) I hope you are all fine and safe in these difficult days !

I cannot find the good orientation of the vector of friction. A circle moves in translation to the right and in the same time the wall rotates around A0. A0 is fixed to the ground. There is always the contact between the circle and the wall. The circle doesn't rotate around itself, it is just a translation. There is a friction between the circle and the wall, I supposed the friction constant in value, not in orientation, for a small angle of rotation of the wall. At start, I thought the vector of friction has the same orientation than the wall but if I supposed that : the work from the translation of the circle is higher than the work from the friction. So, I think the orientation of the vector of friction is not like I think, how can I draw the good orientation of the vector of friction ? When I draw the length that the force moves along the wall I find a distance lower than I thought because there is a "slip" due to the modification of the angle of the wall and so the position of the dot of contact between the circle and the wall. So, is there a method to construct by drawings the orientation of the vector of friction ?

I drew 3 positions of the device with a small angle of rotation of the wall. And I drew an enlargement of the dot of contact:



Have a good day guys !

 

[A.T.]

I supposed the friction constant in value

Why? Is the normal force constant?

I thought the vector of friction has the same orientation than the wall

It's always parallel to the wall, per definition, and opposes the relative sliding motion.

 

[JrK]

Why? Is the normal force constant?

Just to simplify the thinking and yes I can suppose the normal force constant. Can I ?

 

[Dale]

Summary:: It is a cinematic question of a basic movement of two shapes.

I cannot find the good orientation of the vector of friction.

If you have the normal vector to the plane $\vec A$ and the velocity of the ball $\vec v$ then the component of $\vec v$ in the plane is given by: $$\vec v - \frac{\vec v \cdot \vec A}{\vec A \cdot \vec A}\vec A$$

 

[A.T.]

Just to simplify the thinking and yes I can suppose the normal force constant. Can I ?

I don't think so, if the incline changes. Also, why assume no rotation of the circle if there is friction?

 

[wrobel]

Let a particle of mass $m$ moves on a surface with nonzero velocity $\boldsymbol v$ and reaction force acting on the particle from the surface is $\boldsymbol R=\boldsymbol N+\boldsymbol T$.
Here vector $\boldsymbol T$ is tangent to the surface and $\boldsymbol N$ is perpendicular to the surface.

By definition the Coulomb friction force is $$\boldsymbol T=-\gamma|\boldsymbol N|\frac{\boldsymbol v}{|\boldsymbol v|};$$
here $\gamma>0$ is a friction coefficient. If $\boldsymbol v=0$ then $\boldsymbol R$ is defined from static equations.

 

[JrK]

@wrobel and @Dale: I try to understand your equation and I reply later.

Also, why assume no rotation of the circle if there is friction?

Because the circle cannot rotate, it can only move in translation.

I don't think so, if the incline changes.

The value (not the vector) of the force of friction cannot be constant ? I mean 1 N for example.

The normal vector is always perpendicularly to the wall ?

 

[Stephen Tashi]

I cannot find the good orientation of the vector of friction.

It is an interesting question.

First we should settle the elementary question of whether you ask for the vector of friction that describes the force of the wall (inclined plane) on the circular object? - or are you asking for the vector of friction that describes the force of the circular object upon the wall? Which object are you considering as a free body? - the circular object? - or the wall?

Imagine the circular object falling and the wall turning downward at such a rate that it maintains contact with the circle but exerts no force on it. This shows that the magnitude and direction of the frictional force depends on the relative velocities of the wall and the circular object. Drawing a vector representing friction is a representation of that force at a given time. In a dynamic situation, forces are function of accelerations, so we must worry about accelerations also.

The information you give isn't sufficient to determine velocities and accelerations. If you specify how the angle $a$ varies as function of time then we might make progress.

 

[JrK]

First we should settle the elementary question of whether you ask for the vector of friction that describes the force of the wall (inclined plane) on the circular object? - or are you asking for the vector of friction that describes the force of the circular object upon the wall? Which object are you considering as a free body? - the circular object? - or the wall?

The circle is driven by an hydraulic cylinder with a constant velocity (at least during the study). The hydraulic cylinder moves to the right the circle and the circle cannot rotate around itself, it is a mechanical constraint. I suppose the mass of the wall very low because I don't want to think with the mass nor the acceleration because it is difficult enough for me like that (but after why not). A spring (for example) pinces the 2 walls at the dot of the contact between the circle and the wall. The dot of contact between the circle and the wall changes its position for the circle and for the wall, so I suppose the "spring" pinces the 2 walls at the dot of contact just to have the force of friction between the circle and the wall.

 

[A.T.]

The circle is driven by an hydraulic cylinder with a constant velocity (at least during the study). The hydraulic cylinder moves to the right the circle and the circle cannot rotate around itself, it is a mechanical constraint. I suppose the mass of the wall very low because I don't want to think with the mass nor the acceleration because it is difficult enough for me like that (but after why not). A spring (for example) pinces the 2 walls at the dot of the contact between the circle and the wall. The dot of contact between the circle and the wall changes its position for the circle and for the wall, so I suppose the "spring" pinces the 2 walls at the dot of contact just to have the force of friction between the circle and the wall.

This is the description you should have posted in right away. But all the answers were eventually given already: You have a constant friction force that opposes the relative motion of the contact patches (parallel to the wall).

 

[JrK]

You have a constant friction force that opposes the relative motion of the contact patches (parallel to the wall).

If the force of friction is parallel of the wall, the rotation of the wall around A0 doesn't give any energy. Ok, so what work I forgotten ? The energy needed to move the circle is greater than the energy recovered from the friction because there is a slip. I measured the distances d1 and d2 and for a small of rotation 'a' it is easy to compare these energies.

 

[A.T.]

If the force of friction is parallel of the wall, the rotation of the wall around A0 doesn't give any energy. Ok, so what work I forgotten ?

The friction is not doing any work on the wall here. If the wall is mass-less and has no resistance to rotation, the normal forces don't do work on it either, because they sum to zero. Otherwise the normal contact force and the normal pinching force from the spring are not equal and opposite, and the remainder is doing work on the wall.

The energy needed to move the circle is greater than the energy recovered from the friction because there is a slip. I measured the distances d1 and d2 and for a small of rotation 'a' it is easy to compare these energies.

There is no energy recovered by the wall from the friction here. The energy recovered from the normal forces is still smaller then the input to the circle, because kinetic friction dissipates energy.

 

[Lnewqban]

For a small angle of rotation around 'a', the energy of friction is $d2*F$ and the energy needed to move the circle is $d1*F$, I measured a difference so I need to count another energy if it is not from the rotation of the red wall, where it could be ?

Welcome JrK

The difference is that your hydraulic cylinder will need to use more energy to overcome friction between the circle moving to the right and the wall rotating clock-wise.

That happens because the hydraulic cylinder moves with a constant velocity, but the wall rotates with an angular velocity that progressively decreases.
The sliding relative velocity between circle and wall progressively increases, which means more sliding distance as time goes by.

Another important factor to consider is that according to the geometry of your pressing spring, the normal force (always perpendicular to the wall) that induces friction (always parallel to the wall) may remain constant with rotation or not.

 

[JrK]

@Lnewqban: sorry, I was editing the message in the same time :/

With F the force of friction (constant), the energy from friction is d2*F and the energy needed to move the circle is d1*F, I measure a difference:

lg is the distance the circle need to move, and it is higher than d2/cos(a).

 

[Lnewqban]

It is a geometrical problem first, then the friction problem follows.

Those distances will be equal only when the wall becomes horizontal.
For the condition of wall being vertical, that distance would be zero.
Therefore, friction distance will grow with right/clock-wise movement.

Whether cylinder work to overcome friction resistance grows or not, depends also on how the normal force changes as rotation proceeds.

 

[JrK]

Whether cylinder work to overcome friction resistance grows or not, depends also on how the normal force changes as rotation proceeds.

I consider the normal force is constant to have a constant force of friction.

Those distances will be equal only when the wall becomes horizontal.
For the condition of wall being vertical, that distance would be zero.
Therefore, friction distance will grow with right/clock-wise movement.

Why I cannot thinking like I do ? The energy from friction is well d2*F ? And the energy to move the circle is not d1*F ? why ?

 

[Lnewqban]

I consider the normal force is constant to have a constant force of friction.

What mechanism do you use to achieve that?

Why I cannot thinking like I do ?

?

The energy from friction is well d2*F ?

There is no energy from friction, energy for needed work comes from the hydraulic cylinder (which is probably much higher as needed for other productive uses of the mechanism than the useless work consumed by friction).

And the energy to move the circle is not d1*F ? why ?

The energy to move the circle varies with rotation.
Extreme cases considering only discussed friction:
- Wall in vertical position: Consumed energy (supplied by cylinder) is normal force times horizontal distance (no work needs to be used for friction).

- Wall in horizontal position: Consumed energy (supplied by cylinder) is friction force times horizontal distance (no work needs to be used for normal force).

- Wall in intermediate position: Consumed energy (supplied by cylinder) is horizontal resultant of normal plus friction forces times horizontal distance (work needs to be used for both normal force and friction).

 

[JrK]

What mechanism do you use to achieve that?

I press the two walls (the red wall and the circle) with, for example, a spring. The force of the spring is always perpendicularly to the 2 walls and passes by the dot of contact of the walls. The spring needs to move, to follow the dot of contact, and I think (I'm not sure) I don't need any energy to move it, and the length of the spring is always constant (in theory the length of the spring is 0).

?

I would like to say, where is my mistake in my thinking ?

There is no energy from friction

There is a friction between the red wall and the circle and like there is a relative movement (the distance d2), there is an energy from friction: d2*F. Why do you think there is no energy from friction ?

The energy to move the circle varies with rotation.

Yes, but I take the position I drew and I move a little the circle (and a small rotation for the red wall too). I think around the angle: a=45° . For example, like I drew, from 45° to 44°.

Edit: maybe we are not agree in the "normal force", for me the red wall has a very low mass (I consider 0 in theory), the normal force for me is the force I apply with the spring but that force cancel itself with the walls. The sum of forces on the red wall due to the presence of the spring is 0 because the circle cancel the force of the spring. And the sum of forces on the circle due to the presence of the spring is 0 because the red wall cancel the force of the spring.

 

[A.T.]

Edit: maybe we are not agree in the "normal force", for me the red wall has a very low mass (I consider 0 in theory), the normal force for me is the force I apply with the spring but that force cancel itself with the walls. The sum of forces on the red wall due to the presence of the spring is 0 because the circle cancel the force of the spring. And the sum of forces on the circle due to the presence of the spring is 0 because the red wall cancel the force of the spring.

If the net normal force (spring + contact) is zero, then only friction can do work:
- The work by friction on the circle is the integral of F_friction dot circle_velocity (should turn out negative)
- The work by friction on the wall is zero, because the F_friction creates no torque around the pivot.

 

[JrK]

If the net normal force (spring + contact) is zero

Yes, I drew the forces from the spring:

For the integral, I can think for a small angle of rotation (smaller than I drew). And could I think with the distance moved by the spring instead of the velocity ?

 

[A.T.]

Yes, I drew the forces from the spring:

OK, then as I said, only the horizontal component of the friction does negative work on the circle, according to the horizontal displacement of the circle.

 

[JrK]

OK, then as I said, only the horizontal component of the friction does negative work on the circle

Yes, I'm agree. I applied the cosine function to have the horizontal component of the force. When I measured the energy to move the circle I need more energy than I have from the friction, it is easy to measure the distance and I take the mean of the cosine function.

For 1° from a=45°

d2=0.17
lg=0.27

I must have cos(44.5°) or near that value but I have 0.63 far from 0.72

 

[A.T.]

Yes, I'm agree. I applied the cosine function to have the horizontal component of the force. When I measured the energy to move the circle I need more energy than I have from the friction, it is easy to measure the distance and I take the mean of the cosine function.

For 1° from a=45°

d2=0.17
lg=0.27

I must have cos(44.5°) or near that value but I have 0.63 far from 0.72

Your d2 is irrelevant. Work is determined by the motion of the physical material to which the force is applied, not but the motion of the contact point (which moves along the material). Your lg is the infinitesimal displacement, but since the force direction changes you have use integration for larger displacements.

 

[JrK]

Your d2 is irrelevant. Work is determined by the motion of the physical material to which the force is applied, not but the motion of the contact point (which moves along the material). Your lg is the infinitesimal displacement, but since the force direction changes you have use integration for larger displacements.

I built the drawing with a big step for the grid so I redrew it with a thin grid to be sure. I found a difference even I divided by 2 or even by 4 the angle 1° (around 45°) and the ratio of the energy from the friction divided by the energy needed to move the circle is always the same . But ok, it could be a mistake in the construction so I wrote a program to have the numerical integral, and like that I verified the angle of the force of friction:

#include <stdio.h>
#include <math.h>

int main()
{
long double pi=M_PI;

long int N=10000;
long double R=.1;
long double D=1.;
long double xinf=1.;
long double xsup=xinf/tanl(44/180.0*pi);

long double px1=0,py1=0,a1=0,px2=0,py2=0,a2=0,dl=0,wd=0,wf=0,ix1=0;
long double iy1=0,xi2=0,yi2=0,qx1=0,qy1=0,s=0,xl1=0,yl1=0,xl2=0,yl2=0;
long double sx=0,sy=0,qx2=0,qy2=0,xi1sav=0,yi1sav=0,xsav=0,suma=0,l=0,asav=0;
long double xf=0,yf=0,wp=0,af=0,sumdx=0;

long int i;
qy1=D;

for(i=0;i<N;i++)
{
  //a1=(long double)(asup/180.0*pi-ainf/180.0*pi)/N*i;
  qx1=xinf+(xsup-xinf)/N*i;
  a1=atanl(qy1/qx1); if(i==0) a2=a1;
  s=R/tanl(a1/2.);
  px1=qx1-s;
  py1=qy1-R;
  ix1=px1+R*sinl(a1);
  iy1=py1-R*cosl(a1);
  if(i<1) {xi1sav=ix1;yi1sav=iy1;xsav=px1;asav=a1;}
  printf("\npx1=%Lf , py1=%Lf , a=%Lf , ix1=%Lf , iy1=%Lf , a1=%Lf",px1,py1,180/pi*atanl((fabsl(iy1-yi2)-fabsl(py1-py2))/(fabsl(ix1-xi2)-fabsl(px1-px2))),ix1,iy1,180/pi*a1);
  suma+=a1;

  if(i>0)
  {
   sumdx+=(fabsl(px1-px2)-fabsl(ix1-xi2))*cosl(a1);

   l=(sqrtl(ix1*ix1+iy1*iy1)+sqrtl(xi2*xi2+yi2*yi2))/2.;
   xf=xi2+l*sinl(a1)*fabsl(a2-a1);
   yf=yi2-l*cosl(a1)*fabsl(a2-a1);
   af=fabsl(atanl((iy1-yf)/(ix1-xf)));
   wf+=fabsl(px2-px1)*cosl(af);
   wd+=sqrtl((ix1-xf)*(ix1-xf)+(iy1-yf)*(iy1-yf))*cosl(a1-af);
   printf("\nl=%Lf , xf=%Lf , yf=%Lf , af=%Lf , diffa=%Lf",l,xf,yf,180/pi*af,180/pi*(a1-af));
   wp+=fabsl(sinl(a1-af))*l*fabsl(a2-a1);
  }
  px2=px1;  py2=py1;  qx2=qx1;  qy2=qy1;  xi2=ix1;  yi2=iy1;  xl2=xl1;  yl2=yl1;  a2=a1;

}

printf("\nN=%ld , R=%Lf , D=%Lf , xinf=%Lf , xsup=%Lf \ndlp=%Lf , dli=%Lf , suma=%Lf , l=%Lf , cos=%Lf , wf=%Lf , wd=%Lf , wp=%Lf , diff=%Lf , eff=%Lf , ddx=%Lf , sumdx=%Lf\n",N,R,D,xinf,xsup,px1-xsav,ix1-xi1sav,(asav-a1)*180/pi,l,cosl(suma/N),wf,wd,wp,wf-wd-wp,(wd+wp)/wf, fabsl(px1-xsav)-fabsl(ix1-xi1sav),sumdx);
return 0;
}

I have always a difference. I can take an angle of 0.0001° (around 45°) the ratio is always the same. And now, the program give near the same ratio than the drawing. I found a ratio of 0.917 with the program and 0.92 with the drawing and the accuracy of the drawing is +/-0.5% and take in mind I use the mean of the cosine function and it is not exactly that. I'm not able to have the math integral but the program doesn't change the result when I increase the number of steps. So, maybe I'm wrong in the drawing and in the program or there is a force that gives another small energy somewhere.

 

[A.T.]

...the energy from the friction ...

How did you calculate that? If using your d2, then it is wrong. See my previous post.

 

[JrK]

How did you calculate that? If using your d2, then it is wrong.

Yes, I used d2. I don't understand why the distance d2 can't give the energy from friction. I take a dot fixed on the wall at start (45°), and I measured the distance between that dot and the new dot of contact (at 46° or less). Could you explain more please ? and how can I draw it on the drawing ?

Work is determined by the motion of the physical material to which the force is applied

I suppose you mean "work" from friction because I used d2 for that. I used the physical materials: the wall and the circle. When the circle moves, the angle between the wall and the circle changes (when I'm on the circle, I see the wall rotates) and the friction has less distance to move because there is a slip due to the modification of that angle.

 

[A.T.]

Yes, I used d2. I don't understand why the distance d2 can't give the energy from friction. I take a dot fixed on the wall at start (45°), and I measured the distance between that dot and the new dot of contact (at 46° or less). Could you explain more please ? and how can I draw it on the drawing ?

I explained this in post #23.

See also this recent thread where this misconception is discussed:
https://www.physicsforums.com/threa...when-friction-is-involved.986862/post-6322590

 

[JrK]

In the drawing, I measured the distance 'd2' but in the program I calculate the distance (at each step) relatively to a fixed dot on the red wall. Maybe it is not possible to draw the correct distance to find the energy of the friction but in the program I can calculate it, how can I do ?

 

[A.T.]

Maybe it is not possible to draw the correct distance to find the energy of the friction ...

The correct displacement is your d1 (the component of lg parallel to the force of friction).

 

[JrK]

The correct displacement is your d1

Thanks for your patience AT ! I read your link yesterday. Here, for my example, when the circle moves in translation, the movement (circle in translation and the rotation of the red wall around A0) is like the red wall rotates around the circle in the same time without any friction, I mean the rotation of the red wall rotates around the circle without adding a friction. It is for that I see a slip in the distance. I imagine well the circle moving in translation, and it increases its distance from A0 but like the red wall rotate around A0, the dot of contact, so the dot where the friction comes from, is closer to A0. Maybe the calculation could help me to understand.

 

[A.T.]

...the dot of contact, so the dot where the friction comes from...

How that dot moves is irrelevant to work done. Only how the material moves matters.

 

[JrK]

Only how the material moves matters.

Yes, and I try to understand the movement of the matters:

1/ The circle moves in translation
2/ The red wall rotates around A0
3/ The translation of the circle and the rotation around A0 of the red wall create a rotation of the red wall around the circle, I can't forget that movement

If I take in account 1/ and 2/ I'm agree with you and all is fine. But 3/ exists in the same time. It is very difficult to pose (not resolve) the movement by equations (math) ?

 

[A.T.]

Yes, and I try to understand the movement of the matters:

1/ The circle moves in translation
2/ The red wall rotates around A0
3/ The translation of the circle and the rotation around A0 of the red wall create a rotation of the red wall around the circle, I can't forget that movement

If I take in account 1/ and 2/ I'm agree with you and all is fine. But 3/ exists in the same time. It is very difficult to pose (not resolve) the movement by equations (math) ?

1/ matters for the work done on the circle
2/ matters for the work done on the wall
3/ is irrelevant and redundant, because you have already defined the motion of the wall in /2. Pick one kinematic description and stick to it, don't mix them up in a single analysis.

 

[jbriggs444]

Yes, and I try to understand the movement of the matters:

1/ The circle moves in translation
2/ The red wall rotates around A0
3/ The translation of the circle and the rotation around A0 of the red wall create a rotation of the red wall around the circle, I can't forget that movement

Let us try to be systematic about this.

We can describe the motion of the circle as the translation of its center combined with rigid rotation about that center. There are other ways of describing the system. But this is probably the most intuitively appealing. Having decided on this description, we need to be consistent about then using it.

The motion of the wall is trickier. However, the original post gives us the answer. We are invited to view the motion of the wall as pure rotation around the fixed point A0 at its lower left.

We can adopt the ground frame of reference in which point A0 is at rest for purposes of both descriptions. This puts us in a position to meaningfully talk about energy and work. Let us first talk about energy.

The translational kinetic energy of the circle is given by $\frac{1}{2}m_c v_c^2$ where $m_c$ is the mass of the circle and $v_c$ is the velocity of its center of mass.

The rotational kinetic energy of the circle is given by $\frac{1}{2}I_c \omega_c^2$ where I is the moment of inertia of the circle and $\omega$ is its angular rotation rate about its center. This could be rewritten in a variety of ways using the mass of the circle, its radius or the rotational velocity of the rim.

The total kinetic energy of the circle is the sum of the two.

The translational kinetic energy of the wall is zero. We have decided on a description where the wall is rotating about a fixed point at A0 in a frame of reference where point A0 is not moving.

The rotational kinetic energy of the wall is given by $\frac{1}{2}I_w \omega_w^2$. Since the translational kinetic energy of the wall is zero, this is also the total kinetic energy of the wall.

Now let us talk about work. We are specifically concerned about the work done by the force of kinetic friction at the point of contact between circle and wall. We want a careful accounting so that we do not over-count or under-count.

What figure for "work done by kinetic friction" would figure into the translational kinetic energy of the circle?

This would be the vector dot product of the force of kinetic friction (wall on circle, parallel to the wall) times the incremental displacement of the center of mass of the circle. That matches your "1\" quoted above

The vector dot product of the normal force (wall on circle, perpendicular to the wall) times the incremental displacement of the center of mass of the circle would also figure in, but we are only talking about kinetic friction here.

What figure for "work done by kinetic friction" would figure into the rotational kinetic energy of the circle?

This would be the vector dot product of the force of kinetic friction (wall on circle, parallel to the wall) times the incremental displacement of the material at the rim of the circle relative to the center of the circle. This incremental displacement might be easily found using the rotation rate of the circle, its radius and the time increment.

What figure for "work done by kinetic friction" would figure into the total kinetic energy of the circle?

This would be the vector dot product of the force of kinetic friction (wall on circle, parallel to the wall) times the incremental displacement of the material at the rim of the circle relative to our chosen rest frame.

What figure for "work done by kinetic friction" would figure into the translational kinetic energy of the wall?

Zero. Given our choice of description for the wall, it is not translating, only rotating. The dot product of the force of kinetic friction times zero is zero.

What figure for "work done by kinetic friction" would figure into the rotational kinetic energy of the wall?

Zero. The force of kinetic friction on the wall and the motion of the wall material at the point of contact relative to the wall's reference point are at right angles. Their dot product is zero.

 

[JrK]

@jbriggs444 : thanks for your message, for me it is more difficult, too to be honest, because I just want to study the comparaison of the energy from friction (between the circle and the red wall) and the energy to move the circle. I can consider there is no acceleration/decelation, near no mass, 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. The spring to have the force of friction between the circle and the red wall doesn't change its length so its potential energy. And I don't need nor give any energy to move the spring. So the study is only the direction of the force of friction, all people are agree it is parallel to the red wall, at start I thought the vector of friction was not exactly parallel to the wall. The energy to move the circle is calculated with the program. So, my last question concerned the length moved by the force of friction to have the energy of friction. I thought it was d2 but AT said it is d1 and I don't understand why. If the radius of the circle is 0 (a dot), the distance d2 is equal to d1 and the energy to move the dot is equal to the energy of friction. It is well the distance moved by the dot of contact between the 2 objects. Why it is not the same with a radius different of 0 ? With a radius different of 0 for the circle, for me I see a slip between the two objects. My program calculates the small movement between the 2 objects, step by step and I have near the same result than the measure in the drawing.

 

[jbriggs444]

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.

 

[JrK]

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.

 

[jbriggs444]

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.

 

[A.T.]

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.

 

[Stephen Tashi]

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 $C(t)$.

 

[JrK]

@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.

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 ?

 

[jbriggs444]

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.

 

[JrK]

What do you mean by "the work to move the circle"?

The work done by the hydraulic cylinder to move the circle.

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:

 

[A.T.]

@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.

 

[A.T.]

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.

 

[JrK]

I understood your two last examples. Thanks again AT for your patience !

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.

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.

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.

 

[A.T.]

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.

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.

 

[JrK]

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:

'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).

 

[A.T.]

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.

 

[JrK]

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) ?