Work done by a normal force (or rather, work NOT done)

In summary: I don't really know how to put this. Yeah, another one: A box being carried on an accelerating flat bed truck. Noone would dispute static friction is doing work in that case either. Maybe its this whole "reference frame" idea you bring up that can be used to fix...yeah, I don't really know how to put this.
  • #1
Ebby
41
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Homework Statement
What force does positive work such that the woman's kinetic energy remains constant?
Relevant Equations
W = F * d
Screenshot 2023-05-06 085222.png


This question states that the normal force of the stairs on the woman does NO work. I do not understand how this can be. I would reason like this:

The woman propels herself up the stairs using her legs. Her legs push down against the stairs, and the consequent normal force pushes upwards on her legs.

Therefore it is the normal force that's directly responsible for her progress. Since she moves at a constant speed, this normal force must be equal and opposite to her weight.

The normal force does positive work.

Why am I wrong?
 
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  • #2
It's certainly true that the normal force is needed for her to make progress up the stairs, but that doesn't mean that force is doing the work. For a force to do work on an object, there must be displacement at the point of contact.

Another example: Imagine you were lying on the floor and then stood up. Does the normal force do work in that case?

Another hint: What's the source of the energy needed for her to climb the stairs (or for you to stand up)?
 
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  • #3
Suppose a force is applied to point P on an object.

The force only does work on the object when P moves in the direction of the force.

(Or, to be a bit more rigorous, work is done on the object by the force when P has a non-zero component of displacement in the direction of F. The work done is the dot product of the force and displacement: ##W = \vec F \cdot \vec s## if you have met the dot product.)

The normal force of a stair on a foot does no work because there is no relative movement between the foot and the stair while they are in contact.

[Minor edit.]
 
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  • #4
Further to the other replies, suppose you were to stand on a stack of blocks with legs bent then straighten them. All the blocks exert a greater normal force as you accelerate. Which is doing the work?
 
  • #5
Steve4Physics said:
The normal force of a stair on a foot does no work because there is no relative movement between the foot and the stair while they are in contact.

And yet, static friction - a force with no relative displacement between what its applied to, can do work.

You might think I'm just trying to cause a derailment, but in just a few threads down

https://www.physicsforums.com/threads/work-done-by-a-force-down-a-ramp.1052212/

We have static friction doing work.

Perhaps, if students bring this up to their professors it can be resolved.
 
  • #6
erobz said:
And yet, static friction - a force with no relative displacement between what its applied to, can do work.

You might think I'm just trying to cause a derailment, but in just a few threads down

https://www.physicsforums.com/threads/work-done-by-a-force-down-a-ramp.1052212/

We have static friction doing work.

Perhaps, if students bring this up to their professors it can be resolved.
Good point.
How about "the normal force from the stair does no work because the contact point is not displaced in the direction in which the force acts"?
 
  • #7
haruspex said:
Good point.
How about "the normal force from the stair does no work because the contact point is not displaced in the direction in which the force acts"?
I don't have a resolution, maybe that's all it needs. But it certainly needs some explaining. I'm not qualified to do that. Obviously, this fact has been brought up before, with several examples of a "glitch" in the standard definition in my opinion, I'm just noting that no wonder students can become confused, if they aren't, they aren't really paying attention.
 
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  • #8
erobz said:
I don't have a resolution, maybe that's all it needs. But it certainly needs some explaining. I'm not qualified to do that. Obviously, this fact has been brought up before, with several examples of a "glitch" in the standard definition in my opinion.
The version I offered has the merit that it says ##\vec F.d\vec s=0##, as required.

Note that whether a force does work depends on the reference frame. To a person riding an up escalator, neither the normal force from the stair nor gravity do work. To the bystander at rest relative to the Earth, gravity and escalator are doing equal and opposite nonzero work.
 
  • #9
haruspex said:
The version I offered has the merit that it says ##\vec F.d\vec s=0##, as required.

Note that whether a force does work depends on the reference frame. To a person riding an up escalator, neither the normal force from the stair nor gravity do work. To the bystander at rest relative to the Earth, gravity and escalator are doing equal and opposite nonzero work.
Yeah, another one: A box being carried on an accelerating flat bed truck. Noone would dispute static friction is doing work in that case either. Maybe its this whole "reference frame" idea you bring up that can be used to fix it?
 
  • #10
erobz said:
Yeah, another one: A box being carried on an accelerating flat bed truck. Noone would dispute static friction is doing work in that case either. Maybe its this whole "reference frame" idea you bring up that can be used to fix it?
It depends on what conundrum you are trying to get fixed up.

Neither energy nor work are "invariant". Both depend on a choice of reference frame. As this post by @sophiecentaur from earlier in the week points out.

The formulation by @haruspex takes this into account perfectly well: ##W = \vec{F} \cdot d \vec{S}##. Pick a frame where the point of application of a force is motionless and ##d \vec{S} = 0##. Pick a frame where the point of application is moving (like an escalator or a pickup bed from the Earth frame) and ##d \vec{S} \ne 0##

The intent of the problem setter for the OP hear appears to be that the rest frame of the Earth is adopted. We can see that because of the claim that "the normal force does no work". This means that we have adopted a frame where the stairs is motionless.

The OP makes another important point: "her legs are not rigid". It would appear that the textbook where this problem appears is making an effort to distinguish between the "real work" calculated based on the displacement of the point of contact and the "net work" calculated based on the displacement of a body's center of mass. This book appears to use the unqualified term "work" to refer to the former.

If we multiply the normal force by the displacement of the woman's feet, do we get a zero result?

If we multiply the normal force by the displacement of the woman's center of mass, do we get a zero result?

What is the effective point of application of the gravitational force on the woman?

What is the point of application of the normal force on the woman?
 
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  • #11
jbriggs444 said:
It depends on what conundrum you are trying to get fixed up.

Neither energy nor work are "invariant". Both depend on a choice of reference frame. As this post by @sophiecentaur from earlier in the week points out.

The formulation by @haruspex takes this into account perfectly well: ##W = \vec{F} \cdot d \vec{S}##. Pick a frame where the point of application of a force is motionless and ##d \vec{S} = 0##. Pick a frame where the point of application is moving (like an escalator or a pickup bed from the Earth frame) and ##d \vec{S} \ne 0##
In the case of the rolling wheel on the motorcycle in the frame of the ground the point of contact is at rest ##v = 0##, and yet the point of contact is in motion. Which is it?
 
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  • #12
erobz said:
In the case of the rolling wheel on the motorcycle in the frame of the ground the point of contact is at rest ##v = 0##, and yet the point of contact is in motion. Which is it?
It takes a lot of words to clarify that one. So many that I usually let it slide.

Technically, it is not the displacement of the point of contact that matters. It is the displacement of the material of the acted-on object at the point of contact that is relevant.

That makes the relevant displacement for a the force of road on tire (or tire on road) zero for this purpose.

It also means the the work done on a knife being sharpened can be zero and the work done on the grinding wheel can be non-zero despite the point of contact being stationary. [You can focus in on that interaction further and make it messy -- everything is messy when you focus in tightly]
 
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  • #13
jbriggs444 said:
It takes a lot of words to clarify that one. So many that I usually let it slide.
Maybe an Insight is justified? I find It hard to believe that if it's that nuanced, no one will find it interesting.
 
  • #14
erobz said:
Maybe an Insight is justified? I find It hard to believe that if it's that nuanced, no one will find it interesting.
The last time I tried to preemptively get that one covered a number of years back, I caught backlash with several posters asking what I'd meant by the clumsy phrasing. Which is part of why I now "let it slide". Kinetic friction (alluded to briefly in #12) is the other common case where points of contact can move relative to the contacted body. That one caught pushback as well because the details of kinetic friction get messy when you focus in tightly.

So yeah, maybe someone else can do a better job than I at putting together a useful Insight.
 
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  • #15
erobz said:
Maybe an Insight is justified? I find It hard to believe that if it's that nuanced, no one will find it interesting.
I would say that the word 'nuanced' is not appropriate here. There is nothing subtle or nuanced about applying the appropriate basic laws. What is wrong here is the misconception about agency; something 'has to' do something to something else to satisfy some people. That 'something' is Energy and some of the energy transfer can be calculated in the form of work and, very often, the net useful work can be zero with the Energy all being transferred as heating things up.

jbriggs444 said:
The last time I tried to preemptively get that one covered a number of years back, I caught backlash with several posters asking what I'd meant by the clumsy phrasing. Which is part of why I now "let it slide". Kinetic friction (alluded to briefly in #12) is the other common case where points of contact can move relative to the contacted body. That one caught pushback as well because the details of kinetic friction get messy when you focus in tightly.
I can see how you decided to let it slide. Any successful Insight on this topic would need to put a lot of misconceptions to bed - the main one is that anthropomorphism and over-classification creep in. I have a feeling that any Insight would be followed by a string of Yebbuts. Strangely, the same difficulty doesn't seem to arise in Electrical Circuit theory with Amplifiers where no one insists on Resistance having a direction.
 
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  • #16
haruspex said:
Further to the other replies, suppose you were to stand on a stack of blocks with legs bent then straighten them. All the blocks exert a greater normal force as you accelerate. Which is doing the work?
I believe the answer to be none of them. I see the point you're making. Thanks.
 
  • #17
Thank you all so much for your answers. I am mulling them over. One thing that occurs to me regarding the displacement is which one of the following is better:

There must be displacement AT the point of contact in the chosen reference frame.

or

There must be displacement OF the point of contact in the chosen reference frame.
 
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  • #18
haruspex said:
Further to the other replies, suppose you were to stand on a stack of blocks with legs bent then straighten them. All the blocks exert a greater normal force as you accelerate. Which is doing the work?
We are assuming that the blocks are rigid(?). So which part of which bock moves relative to the reference frame? How is Work defined and what work could the blocks possibly be doing? I think @haruspex may just be stirring the pot here.

(Beaten to it by 'a moment')
 
  • #19
sophiecentaur said:
We are assuming that the blocks are rigid(?). So which part of which bock moves relative to the reference frame? How is Work defined and what work could the blocks possibly be doing? I think @haruspex may just be stirring the pot here.

(Beaten to it by 'a moment')
I was just trying to show that the blocks cannot be doing the work.
 
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  • #20
Ebby said:
Thank you all so much for your answers. I am mulling them over. One thing that occurs to me regarding the displacement is which one of the following is better:

There must be displacement AT the point of contact in the chosen reference frame.

or

There must be displacement OF the point of contact in the chosen reference frame.
Neither really captures it. This gets close:
There must be a displacement of the contact (i.e. both points together while making contact), and the direction of that displacement must have some component in the direction of the force.

Examples:
  • Riding a skateboard, constant velocity on level ground. The feet maintain contact with the board, but the normal force is perpendicular to the movement, so no work done.
  • Car driving up hill, no skidding. At the instantaneous point of contact, there is no movement. The static friction is not doing work. The work is done by the forces from the engine on the axle. This is similar to the walking up stairs example.
  • Lifting a cup by gripping the vertical sides. The surfaces in contact move in the direction of the force the hand applies to the cup, so it does do work.
But frames of reference can lead to different views. If I walk downhill past a parked car, in my frame of reference the static friction from the road is doing work on the car! Gravity, fortunately, is doing equal and opposite work on it.
 
  • #21
sophiecentaur said:
I have a feeling that any Insight would be followed by a string of Yebbuts.
And? If one has a theory, It must contend with the "Yebbuts"... if its easily punctured, it's probably not good. I don't see why this should be an issue.
 
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  • #22
haruspex said:
  • Car driving up hill, no skidding. At the instantaneous point of contact, there is no movement. The static friction is not doing work. The work is done by the forces from the engine on the axle. This is similar to the walking up stairs example.

But you yourself don't fully believe this ( quoted from https://www.physicsforums.com/threads/work-done-by-a-force-down-a-ramp.1052212/ )?

haruspex said:
"Average force" is defined as a time average, ##\frac{\int\vec F.dt}{\int dt}##, but let's accept we are here asked for an average over distance. Since the force (static friction) is necessarily in the direction of travel, ##dW=\vec F.\vec{ds}=F.ds##. But force is a vector, so the average force (over distance) is ##\frac{\int\vec F.ds}{\int ds}##.

The only possible external force acting on the car propelling the car up the hill is the force of static friction. I have found this problem in physics texts. Static Friction is the external force, and it is doing Work in the problems. The methodology clearly works, even though it is also clearly inconsistent.
 
  • #23
erobz said:
But you yourself don't fully believe this ( quoted from https://www.physicsforums.com/threads/work-done-by-a-force-down-a-ramp.1052212/ )?
The only possible external force acting on the car propelling the car up the hill is the force of static friction. I have found this problem in physics texts. Static Friction is the external force, and it is doing Work in the problems. The methodology clearly works, even though it is also clearly inconsistent.
It is often convenient to "zoom out", ignoring the details of an interaction and look at the bulk effects. We are back to the distinction between "real work" and "center of mass work"

Real work looks at the displacement of the material of the acted-on object at the point of contact under a particular force.

Center of mass work looks at the displacement of the center of mass of the acted-on object under a particular force. Or under the vector sum of all forces. With only one displacement to worry about, you are free to apply the distributive law and sum the forces before multiplying by displacement. ##\sum_i \vec{F_i} \cdot \Delta \vec{s} = \Delta \vec{s} \cdot \sum_i \vec{F_i}##

Zoom in (real work)...

If we are looking at the interface between tire and road, no work is being done. The two surfaces have zero relative motion (so no energy is being dissipated into heat) and, in addition, the surfaces have zero motion in our selected frame of reference (so no energy is being transferred as work between the mating surfaces).

Zoom out (center of mass work)...

But if we are looking at an abstract interface between motorbike and road, work is being done. There is relative motion between the bike and road. So energy can either be dissipated or injected at the interface. As it turns out in the case at hand, energy is injected. The bike's engine produces the energy and the drive train transmits it into the interface (the rotating and driven wheel). The bike as a whole is in motion. We are zoomed out. We ignore the details of the engine and drive train and see an abstract force from the road on a moving bike. Non-zero work is done on the bike. Zero work is done on the road.

If you are treating the acted-on object as a either point-like or as a rigid, non-rotating blob effectively located at the center of mass then you are adopting the zoomed out point of view.

If you are paying attention to the inner details of the acted-on object such as rotation, vibration, waves, eddies and moving parts then you are adopting the zoomed in point of view.
 
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  • #24
Leg extension applies the force and elevates the lady to the next step, while the alternate leg contracts forward to the next step. This repeats to create steady kinetic energy against the gravitational potential energy.
 
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  • #25
TonyStewart said:
Leg extension applies the force and elevates the lady to the next step, while the alternate leg contracts forward to the next step. This repeats to create steady kinetic energy against the gravitational potential energy.
Yes. Here we have zoomed in to consider the leg as an abstract interface between moving body and stationary stairs. The original problem statement urged us to do so.

If we wanted to, we could zoom in further and look at the gluteus maximus, thigh and calf muscles and see that those are the actual energy sources while the tarsals, metatarsals, tibia, fibula, femur and the associated ligaments and tendons do zero work.
 
  • #26
The macro view should do of cyclic process of leg extension and flexion. Keep it simple, but no simpler.
 
  • #27
jbriggs444 said:
It is often convenient to "zoom out", ignoring the details of an interaction and look at the bulk effects. We are back to the distinction between "real work" and "center of mass work"

Real work looks at the displacement of the material of the acted-on object at the point of contact under a particular force.

Center of mass work looks at the displacement of the center of mass of the acted-on object under a particular force. Or under the vector sum of all forces. With only one displacement to worry about, you are free to apply the distributive law and sum the forces before multiplying by displacement. ##\sum_i \vec{F_i} \cdot \Delta \vec{s} = \Delta \vec{s} \cdot \sum_i \vec{F_i}##

Zoom in (real work)...

If we are looking at the interface between tire and road, no work is being done. The two surfaces have zero relative motion (so no energy is being dissipated into heat) and, in addition, the surfaces have zero motion in our selected frame of reference (so no energy is being transferred as work between the mating surfaces).

Zoom out (center of mass work)...

But if we are looking at an abstract interface between motorbike and road, work is being done. There is relative motion between the bike and road. So energy can either be dissipated or injected at the interface. As it turns out in the case at hand, energy is injected. The bike's engine produces the energy and the drive train transmits it into the interface (the rotating and driven wheel). The bike as a whole is in motion. We are zoomed out. We ignore the details of the engine and drive train and see an abstract force from the road on a moving bike. Non-zero work is done on the bike. Zero work is done on the road.

If you are treating the acted-on object as a either point-like or as a rigid, non-rotating blob effectively located at the center of mass then you are adopting the zoomed out point of view.

If you are paying attention to the inner details of the acted-on object such as rotation, vibration, waves, eddies and moving parts then you are adopting the zoomed in point of view.
I guess I was under the impression that people are implying static friction is not responsible for the center of mass work when they say "static friction does no work" in such a problem... , in context of Newtons Laws which apply to the center of mass of a system.
 
  • #28
erobz said:
I guess I was under the impression that people are implying static friction is not responsible for the center of mass work when they say "static friction does no work" in such a problem... , in context of Newtons Laws which apply to the center of mass of a system.
Newton's Laws work at all scales of course.

In particular, the second law applies to the center of mass of a system: ##\sum \vec{F} = m_\text{tot} \vec{a}_\text{com}##

It also applies to the pieces of the system. Also to the pieces of the pieces. Also to the infinitesimal mass elements which could be integrated into the pieces of pieces.

Sometimes it is a struggle to find an appropriate level of abstraction to describe a system. Or to reach an agreement about what level of abstraction is intended in a particular description.

But I get where you are coming from. If we are adopting the notion of center of mass work then static friction (a force at the low level interface level) is indeed responsible for the change in momentum which we see for the body as a whole (a body viewed at the high level summary level).

If we are doing an accounting for total momentum using Newton's second law then this description is apt.

On the other hand, if we are adopting the notion of real work, then static friction (again, a force at the low level interface level) is not responsible for the change in energy of the wheel, tire and motorcycle. The energy source is the engine. Or the pistons in the engine. Or the expanding high pressure combustion gasses.

If we are doing an accounting for energy on a part by part basis then a more detailed description is appropriate.
 
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  • #29
jbriggs444 said:
Newton's Laws work at all scales of course.

In particular, the second law applies to the center of mass of a system: ##\sum \vec{F} = m_\text{tot} \vec{a}_\text{com}##

It also applies to the pieces of the system. Also to the pieces of the pieces. Also to the infinitesimal mass elements which could be integrated into the pieces of pieces.

Sometimes it is a struggle to find an appropriate level of abstraction to describe a system. Or reach an agreement about what level of abstraction is intended in a particular description.
I agree with this, but I also feel it muddies the whole "zooming in argument". We drawn a system boundary... fine. We can we break the system up, sure. But all rivers must merge somewhere.
 
  • #30
erobz said:
I agree with this, but I also feel it muddies the whole "zooming in argument". We drawn a system boundary... fine. We can we break the system up, sure. But all rivers must merge somewhere.
And we are back at "yebbut" and the motivation to "let it slide".
 
  • #31
jbriggs444 said:
And we are back at "yebbut".
I'm just saying there has to be a connection. Fundamentally, static friction is just a model, not a law. There is no guarantee its consistent over all scales. It's almost certainly not.

I'm not denigrating anything you've stated. Quite the opposite. I'm fine with the bulk model. It's just that even very talented people are flip flopping all over the place. "It does work, It doesn't do work, it still, but its translating, real work- zoomed in, center of mass work - zoomed out, Newtons laws hold over all scales, ect... All valid "Yebuts" if you ask me. To me that sounds more like the results of applying a not so robust (yet paradoxically works quite well in most cases) model to a complicated interaction.
 
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  • #32
jbriggs444 said:
And we are back at "yebbut" and the motivation to "let it slide".
Little joke here, the problem with letting static friction slide is that "it sticks".
 
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  • #33
I am probably a little more confused now than when I asked the original question :)

When you guys are talking about "relative motion", do you mean relative motion between the point of application of the force that's acting and the point on the material that's being acted upon, or do you mean the motion of the contact point (as a whole) in the chosen reference frame?

By the way, my book (Physics by Ohanian) has an example of the "accelerating car" problem (which I show below for reference). In the Earth reference frame, I find it hard to convince myself that the point of contact of the tyre on the road is instantaneously at rest. I imagine the car at a time t, and then I imagine the car at a time a fraction of a second later, and this point (to me at least) seems to have moved...

Capture.PNG
 
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  • #34
Ebby said:
In the Earth reference frame, I find it hard to convince myself that the point of contact of the tyre on the road is instantaneously at rest. I imagine the car at a time t, and then I imagine the car at a time a fraction of a second later, and this point (to me at least) seems to have moved...
Something can be instantaneously at rest but still be accelerating. What if you tossed a ball straight up? Would you agree that at the top of its motion its velocity is zero? And yet it will move.
 
  • #35
Ebby said:
I imagine the car at a time t, and then I imagine the car at a time a fraction of a second later, and this point (to me at least) seems to have moved...
If you look at the trajectory of a point on the periphery of a rolling disc, it follows a cycloid. As it approaches the ground, its motion approaches vertical. At the instant it touches ground, it has no forward velocity. What is critical here is that the forward motion approaches zero faster than its distance from the ground does.
 
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