# What is the orientation of the vector of friction?

• B
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.

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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.

https://www.physicsforums.com/threa...when-friction-is-involved.986862/post-6322590

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• Lnewqban, etotheipi and 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). 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.

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
Homework Helper
2019 Award
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

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
Homework Helper
2019 Award
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.

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
Homework Helper
2019 Award
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.

• jbriggs444
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)##.

@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
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2019 Award
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.

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: Last edited:
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.

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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.

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.

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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.

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• Lnewqban and sophiecentaur
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).

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A.T.