What is the orientation of the vector of friction?

  • #1
JrK
139
1

Summary:

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

Main Question or Discussion Point

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:

vtg5.png

Have a good day guys !
 

Answers and Replies

  • #2
A.T.
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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.
 
  • #3
JrK
139
1
Why? Is the normal force constant?
Just to simplify the thinking and yes I can suppose the normal force constant. Can I ?
 
  • #4
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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: $$\vec v - \frac{\vec v \cdot \vec A}{\vec A \cdot \vec A}\vec A$$
 
  • #5
A.T.
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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?
 
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  • #6
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.
 
  • #7
JrK
139
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@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 ?
 
  • #8
Stephen Tashi
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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.
 
  • #9
JrK
139
1
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.
 
  • #10
A.T.
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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).
 
  • #11
JrK
139
1
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.
 
Last edited:
  • #12
A.T.
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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.
 
  • #13
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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 :cool:

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.
 
  • #14
JrK
139
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@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:


cjf2.png



fhs2.png


lg is the distance the circle need to move, and it is higher than d2/cos(a).
 
  • #15
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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.
 
  • #16
JrK
139
1
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 ?
 
  • #17
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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 ?
:rolleyes: ??? :rolleyes:

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).

:cool:
 
  • #18
JrK
139
1
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.
 
Last edited:
  • #19
A.T.
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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.
 
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  • #20
JrK
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If the net normal force (spring + contact) is zero
Yes, I drew the forces from the spring:

fg9e.png


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 ?
 
  • #21
A.T.
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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.
 
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  • #22
JrK
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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
 
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  • #23
A.T.
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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.
 
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  • #24
JrK
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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:

Code:
#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.
 
Last edited:
  • #25
A.T.
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....the energy from the friction ...
How did you calculate that? If using your d2, then it is wrong. See my previous post.
 

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