What is the orientation of the vector of friction?

[JrK]

In my program I used the name of the dot of the first drawing.
The (0,0) is on A0
The dot of contact between the circle and the red arm is 'i'
The center of the circle is the dot 'p'
The intersection between the horizontal line and the red arm is the dot 'q'
The height from A0 to the top of the circle is 'D'
The radius of the circle is 'R'
The length of the distance from A0 to 'i' is 'l' (small L)
I integrate from 'xinf' to 'xsup'For the dots i,p,q, I added a letter for the 'x' or the 'y' and I added an information for take in account the step: the number '1' is the step in calculation and the '2' is the last calculation.

'wf' is the energy to move the circle in translation
'wd' is the energy recovered from the friction
'wp' is to verify the vector of the force of friction is like the orientation of the red wall, must be at 0

What I need to add in the calculations ? Maybe the calculation of wd is not like I calculate because I use the dot 'i' and all people here says it is not the movement of the dot of contact that gives the energy from the friction. Maybe there is a method (an old method used in 16th) but I didn't find on internet, the name of the method could help me to find it and to apply it.

 

[A.T.]

To come back to the method with the elastic,

Here is one way to determine the slipping distance relevant for energy dissipated by friction, using an elastic, that works for your specific case (but not in general!):

Attach one end to the circle at the initial contact point, and the other end to the wall at the initial contact point. The initial length is zero, the final length is the slipping distance.

More general case:

You use two elastics, one for each body, with the ends attached to that body at the initial and final contact points. The slipping distance depends on how the elastics meet at the final contact point:

- If they meet at 180° (straight angle) you add their lengths (special case where you can use a single elastic).
- If they meet at 0° (acute angle) you subtract the shorter from the longer length.

Note that even the more general case works only for monotonic motion, not for back and forth sliding which you would have to do piece-wise. The elastics have to stay in contact with the bodies and cannot take short cuts through air. So for concave surfaces you have put them on the inside of the body, and for changing curvature sign you have to do it piece-wise.

 

[jbriggs444]

Like that the needle is always at the dot I want: the dot of contact between the circle and the red wall.

So this "needle" is not like a physical pin that pierces the elastic at a particular point. It is merely an indicator (like a "needle" on a speedometer) marking the contact point.

Like any other position marker, this indicator has no physical significance. It does not "do" anything. It is merely a convenience for the purposes of accounting.

It seems that you would like to use this marker to separate the elastic into a "circle piece" and a "wall piece". The idea would be that the stretching or shrinking of the "circle piece" corresponds to the work done on the circle by the force of kinetic friction. However, this is not correct.

To gauge the work done on the circle, the relevant place on the elastic is the place (if any) where it is stationary. This is the place where the elastic material does not move past a stationary line at right angles to the contacting surfaces. This place will not, in general, be at the point of contact.

It is worth noting that this point on the elastic changes with your choice of reference frame. Similarly, the work done on the circle changes with your choice of reference frame. Neither energy nor work are invariant quantities.

 

[JrK]

The needle is a real object. It is there to be near the dot of contact, I would like to be at the exact position of the dot of contact but it is possible only in theory. I suppose all the objects rigid and without any friction. Remember I done that because I don't know another method to calculate that example with the friction between the objects. I cannot place the image in the message, a bug ?

 

[A.T.]

Remember I done that because I don't know another method to calculate that example with the friction between the objects.

I gave you a method using integrals and I gave you a method using elastics. The needles seem needless here.

 

[JrK]

You changed your message, I didn't see it at start. Ok, I will try with your method.

The needle is useful, I can suppose the 2 bodies rigid and a small diameter for the needle, for example some atoms, like that the needle is near the dot of contact. I have well the distance I though with the friction. I have only one elastic and in straight shape (not curved). I calculate with my program the energy from F3 and F4 and it is the same:

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

int main()
{
long double pi=M_PI;

long int N=1000000;
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,dlrc=0,dlpc=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);
   dlpc+=(px2-px1)*sinl(a1)*1000.;
   dlrc+=l*(a2-a1)*1000.;

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

Edit: @A.T. : the first method you gave for my specific case is the method I used before and I realized it is not like the friction. At least, I don't see it like the friction because for me the force of friction moves only of d2.

@jbriggs444: no, the needle is not an indicator, it is a real needle (without friction), but with a theoretical diameter very small, some atoms, to be very near to the dot of contact. Maybe in theory, I can take the diameter of the needle of 1 atom. I used the needle to block the force of the elastic with the two lateral walls (circle and red wall) like my images show in the message #74. I drew the forces F2 from the elastic and the forces F3 and F4 to the lateral walls. And I calculate the energy from F3 and F4, they cancel themselves. So, at final there is only the energy stored in the elastic and the energy to move the circle. And the distance moved by the needle is well d2 like I could compare with my first message.

 

[A.T.]

Edit: @A.T. : the first method you gave for my specific case is the method I used before and I realized it is not like the friction.

It gives you the distance of slippage, relevant for energy dissipated by friction

At least, I don't see it like the friction because for me the force of friction moves only of d2.

Wrong.

You should look at your problem from the rest frame of the wall. It is easier to understand relative motion, if one object is static. In the frame of the wall the circle is doing back-spinning , like the wheel of a car sliding forward on ice, while spinning the wheels in reverse gear. This increases the distance of slippage, which here is the sum of the contact point movements relative to each body.

In other cases, if the wheels were spinning forwards you would have to take the difference. In pure rolling that difference becomes zero.

 

[jbriggs444]

@jbriggs444: no, the needle is not an indicator, it is a real needle (without friction), but with a theoretical diameter very small, some atoms, to be very near to the dot of contact

But what does the needle do?

Does it fasten the elastic to the wall?
Does it fasten the elastic to the circle?
Does it pierce all three and thereby fasten the wall to the circle?

If it does not do any of the three, it might as well not exist. If it does one of the three, just say so. [You cannot fasten the needle to the point of contact because the point of contact is not a physical entity]

 

[JrK]

But what does the needle do?

I didn't says but if I use the needle I need to control in position the red arm. I use an hydraulic cylinder to control the position of the circle and I count the energy that the hydraulic cylinder consumes. I have an electric motor that controls the position of the red wall, so its rotation, like the hydraulic cylinder I count all the energy needed/recovered from it.

The elastic pulls the needle closer to the dot B.

The needle is between the two walls and it cannot move closer to the dot B because the distance between the two walls reduces more and more at the dot of contact. At the dot of contact the distance between the two walls is 0, so with a diameter of the needle greater than 0 the needle is blocked. The elastic pulls the needle but the two walls prevent it to move, because I control the position of the circle and the position of the red wall. The needle (in theory) is at the dot of contact.

 

[jbriggs444]

I didn't says but if I use the needle I need to control in position the red arm. I use an hydraulic cylinder to control the position of the circle and I count the energy that the hydraulic cylinder consumes. I have an electric motor that controls the position of the red wall, so its rotation, like the hydraulic cylinder I count all the energy needed/recovered from it.

The elastic pulls the needle closer to the dot B.

The needle is between the two walls and it cannot move closer to the dot B because the distance between the two walls reduces more and more at the dot of contact. At the dot of contact the distance between the two walls is 0, so with a diameter of the needle greater than 0 the needle is blocked. The elastic pulls the needle but the two walls prevent it to move, because I control the position of the circle and the position of the red wall. The needle (in theory) is at the dot of contact.

So if I understand it properly, the needle is oriented so that it is perpendicular to the sheet of paper on which the wall and the circle are portrayed. It is acting like a watermelon seed, being squeezed away from the point of contact. Is that correct?

 

[JrK]

Is that correct?

Yes.

 

[jbriggs444]

Yes.

Let me restate the scenario in my own words and you can tell me whether you agree.

This means that you must have a very large squeezing force on the needle and must have a very large force holding the circle against the wall. That is fine. One could imagine arrangements to accomplish this without any use of energy.

The elastic band is attached at one end to the needle and at the other end to the wall at the starting contact point of circle with wall. This point of attachment of elastic to wall is point B. Point B is fixed to the wall. We might consider a label "B" painted on the wall here.

The initial contact point of circle with wall is labelled point A. Point A moves with the circle. We might consider a label "A" painted onto the circle here. Nothing is attached at point A.

The wall has a stationary and frictionless pivot point down and to the right where it is anchored to the ground. The wall is massless and infinitely thin. The wall is rotating clockwise, restrained by a motor or generator that controls the rate of rotation.

The circle is driven up and to the right by a hydraulic cylinder that pushes on its center and which somehow braces the circle against rotation. The anchor mechanism for this hydraulic cylinder has not been specified. The details of the anchoring mechanism are important if one is examining the work done on or by the hydraulic cylinder. But for now we can consider it as an arbitrary external force applied at the center of the circle and directed up and to the right and as an arbitrary external clockwise torque on the circle.

I want to present a simple schematic picture of the arrangement after the circle has been displaced some distance up and to the right and after the wall has rotated through some angle clockwise just for the purposes of clarifying the scenario. Not for the purposes of analyzing it.

Do we agree that this is an accurate description of the scenario currently under discussion?

 

[JrK]

1/ The dot A is the dot where the needle is. Why do you want to add a dot fixed on the circle ? Note: in the program I used the dot 'i' (intersection).

2/ I don't understand your force F on the drawing you drew. The elastic pulls. The circle moves in HORIZONTAL translation to the right, not up and right.

3/ Yes, I didn't speak a lot about the hydraulic cylinder. The cylinder is fixed on the ground. The circle is glue to the cylinder: the circle cannot rotate around itself. Is the dot of attachment of the cylinder important ? Yes, it could be an arbitrary force applies on the center of the circle.

I added a pdf file with all the drawings. It is better than an image. Please, look at the drawing at left, I drew two positions of the circle and the wall.

Note: in the program lowercase 'a' is the angle of the red wall relatively to the horizontal.

 

[Stephen Tashi]

I added a pdf file with all the drawings.

Those sorts of drawings have not been helpful.
I repeat my suggestion that you use Cartesian coordinates.

Let's write the path of the point of contact $(x_c,y_c)$ between the circle and the "wall" as a function of the length $x$ of the piston.

The distance from $(0,0)$ to $(x_c,y_c)$ is $d = \sqrt{ x^2 + y_h^2 - r^2}$

$\angle \alpha = \arctan{ \frac{r}{d}}$
$\angle \theta = \arctan{ \frac{y_h}{x}}$
$\angle \phi = \angle \theta - \angle \alpha$

$x_c = d \cos{ \phi}$
$y_c = d \sin { \phi}$

Apparently, we are to assume that a force $F = (F_x, F_y)$ of constant magnitude $|F|$ acts on the circle at $(x_c, y_c)$ in the direction from that point toward the origin. The $x$-component $F_x$ of that force acts to oppose the piston.
$F_x = -|F| \cos{\phi}$

The work done by $F$ on the piston as it lengthens from $x = a$ to $x = b$ is computed by doing the line integral representing a force $F_x(x)$ acting on the path $(x_c(x), y_c(x))$ from $x = a$ to $x = b$.

A numerical approximation for the answer should be a numerical approximation to that line integral.

 

[JrK]

Notice this is for the example with the friction not the needle (for others users that could read too).

I wrote the numerical integration:

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

int main()
{
 int i,N=1000000;
 long double xc=0,yc=0,x_last=0,d=0,alpha=0,theta=0,phi=0,F=1.,Fx=0,r=0.1,yh=0.9,w=0,x=0,pi=M_PI;
 long double a=0.758579;
 long double b=0.788021;
 
 w=0;
 x=a;
 x_last=a;
 for(i=0;i<N;i++)
 {
  x=a+(b-a)/N*i;
  d=sqrtl(x*x+yh*yh-r*r);
  alpha=atanl(r/d);
  theta=atanl(yh/x);
  phi=theta-alpha;
  xc=d*cosl(phi);
  yc=d*sinl(phi);
  Fx=F*cosl(phi);
  w+=Fx*(x_last-x);
  x_last=x;
 }
    
 printf("\nw=%Lf",w);   
  
}

It gives the same work I found for the wall (with the friction). You didn't give the formula for the energy from the friction, do you have it ?

 

[Stephen Tashi]

You didn't give the formula for the energy from the friction, do you have it ?

How do you define "the energy from friction"? Isn't it the same as the work done by friction on the circle?

 

[JrK]

How do you define "the energy from friction"? Isn't it the same as the work done by friction on the circle?

I see less friction than the energy from the displacement of the circle and I would like to have calculations to understand where is my mistake. My program computes the energy from friction but it uses the dot of contact. So, I found the method with the needle + the elastic. In that method, I have the same result I thought with the friction: the elastic won d2*F but the energy needed to move the circle is d1*F.

 

[Stephen Tashi]

I see less friction than the energy from the displacement of the circle

How do you define "the energy from the displacement of the circle"?

In order to displace the circle, the piston exerts a force against the circle that is equal and opposite the force of friction on the circle - provided we are allowed to move the circle without accelerating it. So why do you think that the "energy from the displacement of the circle" is different than the work done by friction upon the circle?

 

[A.T.]

I see less friction than the energy from the displacement of the circle and I would like to have calculations to understand where is my mistake.

Your mistake is right here:

My program computes the energy from friction but it uses the dot of contact.

Why this approach is wrong is demonstrated here:
https://www.physicsforums.com/threa...of-the-vector-of-friction.987415/post-6330146

 

[JrK]

I'm agree it is a mistake to see that slip, even I see it. I replied by private message at Stephen to not overflow that thread. I found the method with the needle + elastic to have the SAME dot compared to the friction. So, I would like to understand the mistake in the example with the needle + elastic BUT WITHOUT FRICTION because the length won by the elastic is well d2 not d1, so the energy won by the elastic is d2*F and the energy needed to move the circle is d1*F. I calculated the energy needed/given by the forces F3 and F4 and the sum is at 0. So maybe a problem with an angle ?

The example, the circle moves in horizontal translation, the red wall rotates around A0, I drew the start and final position, the circle and the red wall are controlled in position with a cylinder and a motor for example. There are 4 objects: the circle and the wall, the elastic and the needle:

Enlargement of the previous image:

Enlargement of the area where there is the elastic, I drew the start and final position, the elastic is taken by the dots A and B, the dot B is fixed on the red wall, the dot A is fixed on the needle, the needle is near the dot of contact between the circle and the red wall:

To watch the forces of the elastic, I drew the start and final position, at start the elastic has a length at 0, and at final its length is d2:

To watch the needle (the needle is perpendicularly to the screen), I drew the final position, the needle is pulled by the elastic but the lateral walls prevent it to move closer the dot B:

Enlargement, I drew the final position, to watch the forces from the needle to the lateral walls (circle and red wall):

The program :

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

int main()
{
long double pi=M_PI;

long int N=100000000L;
long double R=.1;
long double D=1.;
long double xinf=1.;
long double xsup=xinf/tanl(44.99/180.0*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,ix2=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,dlrc=0,dlpc=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;
  if(i==0 || i==N-1) printf("\npx1=%Lf",px1);
  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-ix2)-fabsl(px1-px2))),ix1,iy1,180/pi*a1);
  suma+=a1;

  if(i>0)
  {
   sumdx+=(fabsl(px1-px2)-fabsl(ix1-ix2))*cosl(a1);

   l=(sqrtl(ix1*ix1+iy1*iy1)+sqrtl(ix2*ix2+yi2*yi2))/2.;
   xf=ix2+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);
   dlpc+=(px2-px1)*sinl(a1);
   dlrc+=l*fabsl(a2-a1);

  }
  px2=px1;  py2=py1;  qx2=qx1;  qy2=qy1;  ix2=ix1;  yi2=iy1;  xl2=xl1;  yl2=yl1;  a2=a1;

}
printf("\nN=%ld , R=%Lf , D=%Lf , xinf=%Lf , xsup=%Lf",N,R,D,xinf,xsup);
printf("\ndlp=%Lf , dli=%Lf, dlpc=%.18Lf , dlrc=%.18Lf, diff=%.18Lf",px1-xsav,ix1-xi1sav,dlpc,dlrc,(dlpc-dlrc));
printf("\nsuma=%Lf , cos=%Lf , wf=%Lf , wd=%Lf , wp=%Lf , diff=%Lf , eff=%Lf , ddx=%Lf , sumdx=%Lf\n",(asav-a1)*180/pi,cosl(suma/N),wf,wd,wp,wf-wd-wp,(wd+wp)/wf, fabsl(px1-xsav)-fabsl(ix1-xi1sav),sumdx);
return 0;
}

 

[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.