# Confusing case of work done by friction

The work done by the frictional force on a surface in drawing a circle of radius r on the surface by a pencil of negligible mass with a normal pressing force N(co-efficient of friction μ) is ?
my textbook says it is zero but i don't understand it why.

AlephZero
Homework Helper
I don't understand why either.

To a good approximation the work done against friction would be $2 \pi r N \mu$.

Unless I think "drawing a circle" means something different from what your textbook means. I don't see why the mass of the pencil is relevant either.

OK, this could be a sort of trick question, and the book is saying all the work is done in plastic deformation of the pencil lead, which isn't "friction"...

I don't understand why either.

To a good approximation the work done against friction would be $2 \pi r N \mu$.

Unless I think "drawing a circle" means something different from what your textbook means. I don't see why the mass of the pencil is relevant either.

OK, this could be a sort of trick question, and the book is saying all the work is done in plastic deformation of the pencil lead, which isn't "friction"...

now two things arise in my mind
1. if the mass of the pencil is negligible hence there will be no normal reaction and hence no friction.
2. work is defined in terms of displacement and displacement is 0 so work done by friction is 0.
please let me know if my reasoning is correct.

AlephZero
Homework Helper
1. if the mass of the pencil is negligible hence there will be no normal reaction and hence no friction.

No. the force on the pencil is not the same as its weight.

2. work is defined in terms of displacement and displacement is 0 so work done by friction is 0.

No. If drive my car 100 miles and finish at the same place where I started, does that mean I don't need any fuel for the trip?

Doc Al
Mentor
my textbook says it is zero but i don't understand it why.
What textbook are you using?

rcgldr
Homework Helper
Assume the surface is a flat plate resting on a frictionless plane. Some of the work done on the plate would be in the form of torque times angular displacement and result in the plate rotating with some amount of angular energy corresponding to the net work done on the surface.

256bits
Gold Member
The work done by the frictional force on a surface in drawing a circle of radius r on the surface by a pencil of negligible mass with a normal pressing force N(co-efficient of friction μ) is ?
my textbook says it is zero but i don't understand it why.

Because the pencil leaves a trace on thw paper due to static friction and static friction plays not a part in work - it has a force but no displacement - the graphite particules are being left behind stationary on the surface.
Where the work comes from is the breaking chemical bonds between the graphite molecules at the tip of the pencil. Why they mention drawing a circle in this problem is for the textbook authors to explain as there would be no work when drawing a line or any other figure due to the static friction causing the graphite to remain behind.

That could be the authors' reasoning.

It's NOT static friction. Both in static and in kinetic friction there's deposition of material from one surface to another, with stationary particles left behind (wheel marks in a road are a good example of stationary particles left behind by a car slipping over the road - kinetic friction). That's not a valid reason for calling the friction in this case "static".

The question, as stated, should have the answer AlephZero gave.

Someone should scroll down to equation 2 in the wiki article on work and see if there's a clue there. I read the sentence: "Equation (2) explains how a non-zero force can do zero work. The simplest case is where the force is always perpendicular to the direction of motion, making the integrand always zero. This is what happens during circular motion. However, even if the integrand sometimes takes nonzero values, it can still integrate to zero if it is sometimes negative and sometimes positive."

It's in the section "Force and Displacement"

http://en.wikipedia.org/wiki/Work_done

It seems it might apply to the circular motion of a compass drawing a line. I'm not sure.

Someone should scroll down to equation 2 in the wiki article on work and see if there's a clue there. I read the sentence: "Equation (2) explains how a non-zero force can do zero work. The simplest case is where the force is always perpendicular to the direction of motion, making the integrand always zero. This is what happens during circular motion. However, even if the integrand sometimes takes nonzero values, it can still integrate to zero if it is sometimes negative and sometimes positive."

It's in the section "Force and Displacement"

http://en.wikipedia.org/wiki/Work_done

It seems it might apply to the circular motion of a compass drawing a line. I'm not sure.

Wikipedia is right, but not in this case. The friction force will be parallel to the movement of the tip of the pencil, so there's work being done.

AlephZero
Homework Helper
Assuming the OP hasn't misunderstood what the textbook said (and we don't have either the name of the book or an exact quote from it), in a sense this is just a question about language.

It is obvious (by experiment) that work is being done. It is obvious the work is approximately proportional to the normal force between the pencil and the paper. So you can either describe the situation as (the Coulomb model of) friction, or you can describe it in some other more complicated way.

But how you describe it doesn't change the amount of work done.

Assuming the OP hasn't misunderstood what the textbook said (and we don't have either the name of the book or an exact quote from it), in a sense this is just a question about language.

It is obvious (by experiment) that work is being done. It is obvious the work is approximately proportional to the normal force between the pencil and the paper. So you can either describe the situation as (the Coulomb model of) friction, or you can describe it in some other more complicated way.

But how you describe it doesn't change the amount of work done.

but displacement is zero and so work done by friction(not any other force) is zero.if not what is wrong with the definition of work?

Doc Al
Mentor
but displacement is zero and so work done by friction(not any other force) is zero.if not what is wrong with the definition of work?
Why do you think the displacement is zero?

Please tell us the name of the textbook you are using.

olivermsun
You can see in this thread there are different ideas of what "work" is being quantified. The integral $\int \mathbf{F} \cdot \mathrm{d}\mathbf{x}$ evaluated along the path is indeed nonzero, so I would say the book, which appears to have multiplied constant F times the integral $(F \int \mathrm{d}\mathbf{x} = F \cdot 0 = 0)$, seems to be wrong in this case.

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256bits
Gold Member
It's NOT static friction. Both in static and in kinetic friction there's deposition of material from one surface to another, with stationary particles left behind (wheel marks in a road are a good example of stationary particles left behind by a car slipping over the road - kinetic friction). That's not a valid reason for calling the friction in this case "static".

The question, as stated, should have the answer AlephZero gave.

Certainly kinetic friction between the paper and pencil lead does not rub off the particles of graphite from the lead in this case. The work a person does in moving the pencil across the paper results from the breakage of chemical bonds.

Unroll a carpet across the floor. Is the static friction that holds the unrolled part of the carpet doing in work? Where is the kinetic friction in this case?

Certainly kinetic friction between the paper and pencil lead does not rub off the particles of graphite from the lead in this case. The work a person does in moving the pencil across the paper results from the breakage of chemical bonds.

Unroll a carpet across the floor. Is the static friction that holds the unrolled part of the carpet doing in work? Where is the kinetic friction in this case?

I'm afraid I have to disagree.

When I use a pencil to draw a picture, I'm sliding two surfaces (the pencil over the paper). That would be, by definition, kinetic friction.

If it were static friction, there would be no sliding (the tip of the pencil and the sheet of paper would move together). The maximum one could draw doing it is a dot. But the problem asks the work done in drawing a circumference, radius R. That would imply in sliding between surfaces, and, thus, kinetic friction.

Notice that you may break chemical bonds in both cases.

256bits
Gold Member
I'm afraid I have to disagree.

When I use a pencil to draw a picture, I'm sliding two surfaces (the pencil over the paper). That would be, by definition, kinetic friction.

If it were static friction, there would be no sliding (the tip of the pencil and the sheet of paper would move together). The maximum one could draw doing it is a dot. But the problem asks the work done in drawing a circumference, radius R. That would imply in sliding between surfaces, and, thus, kinetic friction.

Notice that you may break chemical bonds in both cases.

The two surfaces - ie the graphite layer at the tip and the paper - do not slide against one another.
A layer of graphite is sheared off from the pencil tip and left on the paper. This layer that touches the paper remains in place while the pencil moves forward.

The graphite layer that touches the paper does not slide but stays in place is acted on by a force and that would be the static friction between the paper and that graphite layer.

The force between the graphite layer staying on the paper and the graphite layers still on the pencil are acted upon by a force and that is the chemical bond.

The static friction overcomes the chemical bonding and we have wear of the pencil lead.

Normal kinetic friction also has wear between surfaces but mostly only atoms at a time

You are saying that as the pencil is moving across the paper, since there is movement between mating surfaces, the friction force should be labelled a kinetic friction, and one applies a force to the pencil against a kinetic friction, which I can comprehend from a statics and dynamics viewpoint.

I am saying that if one looks down closely at what is happening at the pencil tip, there is only static friction in play and breakage of chemical bonds, and as the pencil moves across the paper the force one applies is that to ovecome and break the chemical bonds between graphite layers.

That is the distinction and perhaps what the book explanation, is also.
Who knows, the book in question could be a chemistry testbook, with a section on intermolecular forces, and not a statics and dynamics textbook, which would be that assumed from the question and from most of the postings.

rcgldr
Homework Helper
> static ... kinetic ... friction

I don't think this matters. What matters is that a force parallel to the surface the circle is drawn on is applied over a distance along that surface. As mentioned previously, if this surface was a plate resting on a frictionless plane, the plate would end up with angular energy due to the force being applied in a circular path (which would eventually create a torque). It the plate's movement was restricted so that it could only rotate, and the drawn circle shared a common center with the plate's rotation, then all of the work done would end up as angular energy. Since the plate would rotate while the circle was being drawn, the actual distance would be greater than 2 π r.

The two surfaces - ie the graphite layer at the tip and the paper - do not slide against one another.
A layer of graphite is sheared off from the pencil tip and left on the paper. This layer that touches the paper remains in place while the pencil moves forward.

The graphite layer that touches the paper does not slide but stays in place is acted on by a force and that would be the static friction between the paper and that graphite layer.

(...)

That is the distinction and perhaps what the book explanation, is also.
Who knows, the book in question could be a chemistry testbook, with a section on intermolecular forces, and not a statics and dynamics textbook, which would be that assumed from the question and from most of the postings.

@ 256 bits: That's a rather interesting description! I hadn't though that before. Could you give me any references?

(I'm sorry, I think I was too harsh on you in my previous posts...)

@rcgldr: It makes all the difference. Static friction would do no work in this case. The textbook's answer would be totally valid, and we would have the answer to the question the OP posted. The textbook would be right.

rcgldr
Homework Helper
@rcgldr: It makes all the difference. Static friction would do no work in this case.
That assumes no movment of either object. A car can accelerate using just a static friction force applied to the pavement, and work is done by the cars engine. If the surface itself moves (which it does in real life, since no object is unmovable), then the static friction force times distance the surface is moved represents work done.

I think this is a case over semantics though. Using the thought experiment where the surface being drawn on is a plate free (no friction) to rotate but not translate, it seems that drawing a circle on the plate will cause the plate to experience angular acceleration, regardless of what the friction is. Newton's third law applies here, if it takes a force parallel to the surface (torque) to draw the circle, then the plate is responding with an equal and opposing reaction force (torque) (due to angular acceleration).

256bits
Gold Member
@ 256 bits: That's a rather interesting description! I hadn't though that before. Could you give me any references?

(I'm sorry, I think I was too harsh on you in my previous posts...)

Harsh - not at all. :)

References.
Still looking - found some blogs and Q&A, but not something with a name attached to it.

I think this is a case over semantics though.

That is exactly what Acut and I were discussing, though we did not express it as semantics per se.

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if the amount of friction produced by the surface and the pencil are the same they naturally get cancelled. so it becomes 0.

256bits
Gold Member
if the amount of friction produced by the surface and the pencil are the same they naturally get cancelled. so it becomes 0.

Friction always comes in a pair of forces of equal magnitude but opposing sense:
One force acts on the surface in the direction motion of the object.
Second is on the object but in the opposite direction of motion of the object.

NascentOxygen
Staff Emeritus
The two surfaces - ie the graphite layer at the tip and the paper - do not slide against one another.
A layer of graphite is sheared off from the pencil tip and left on the paper. This layer that touches the paper remains in place while the pencil moves forward.

The graphite layer that touches the paper does not slide but stays in place
When using an artist's 6B pencil, this may be true. But when using an ordinary HB or 2H there is friction. That scraping sound I can hear, it's definitely more than the muted protest of graphite sheets uncleaving! It's kinetic friction from kaolin particles in the 'lead'.

And if I twirl the compasses multiple times to obtain a darker score, it's possible to wear a hole in the paper. That's doing more than just leaving a trace of graphite powder.

When using an artist's 6B pencil, this may be true. But when using an ordinary HB or 2H there is friction. That scraping sound I can hear, it's definitely more than the muted protest of graphite sheets uncleaving! It's kinetic friction from kaolin particles in the 'lead'.

And if I twirl the compasses multiple times to obtain a darker score, it's possible to wear a hole in the paper. That's doing more than just leaving a trace of graphite powder.
You're using a pencil that's too hard for the value you want. H pencils are designed to be used for light values. Ideally, the situation should be as close as possible to what 256bits describes. If the paper isn't sheering graphite off the pencil, you're less drawing than incising.

NascentOxygen
Staff Emeritus
You're using a pencil that's too hard for the value you want.
One always does, in practice. Choice of pencil is a compromise. Otherwise, there's a constant need to be sharpening a rapidly bluntening pencil. That's why when erasing a mistake, students always an indent left in the paper. FRICTION!
H pencils are designed to be used for light values.
Mostly, the use of compasses in maths and drawing requires light marks, as most are construction marks, and not meant to stand out starkly. (Quite different to an artist's needs.)

In my opinion, today's HB pencils are not what they used to be. These days HB pencils seem to be like H or even 2H pencils were a few decades back.

One always does, in practice. Choice of pencil is a compromise. Otherwise, there's a constant need to be sharpening a rapidly bluntening pencil.
That's just what has to be done with dark values: constant resharpening. The sharper the pencil the darker the line you can draw with that pencil. The blunter the pencil the more brute force that's required.

Mostly, the use of compasses in maths and drawing requires light marks, as most are construction marks, and not meant to stand out starkly. (Quite different to an artist's needs.)
Technical drawing is all done on computer nowadays, but back in the day the pencil work was generally preparation for inking, "construction lines" as you say, meant to be erased once the inking was done. Anyone doing serious pencil work had, or has, a set ranging in value. They go from 9B to 9H now.

Artists, though, almost always draw light construction lines as well.

In my opinion, today's HB pencils are not what they used to be. These days HB pencils seem to be like H or even 2H pencils were a few decades back.
I agree. They've gotten harder and I suspect that's because clay is probably cheaper than graphite. If you want the old HB softness you probably have to go to an art store and buy a B.

In any event, the situation you described is not the proper use of a pencil and doesn't change what 256bits was saying. You're not supposed to be tearing or scraping the paper.