# Why is it tiring to push hard against a solid wall even though no work is done?

I've seen this exact question in a book.
(This is not homework)
I've always thought this is because work has nothing to do with effort,
but seeing this question on a book quite shocked me actually, is there any deeper explanation?

Andrew Davies
You are doing work. Your probability wondering how you can do work if your arm and the wall have no displacement. The answer is your muscles fibers are contracting and expanding. The work your doing is spread out throughout your muscle tissue. For example picture a bicycle that is stuck to a brick wall by metal pips welded to the frame. If you get on the bike and peddle you want go anywhere. The wheels will slide along the ground generating heat vie friction. The same thing happens in you muscle tissue. The arm can't move because it can't go though the wall so it does work by have muscle fibers slide around inside your arm, generating heat.

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Mentor
output work/ input energy = efficiency

The human body is very inefficient.

You are doing work. Your probability wondering how you can do work if your arm and the wall have no displacement. The answer is your muscles fibers are contracting and expanding. The work your doing is spread out throughout your muscle tissue. For example picture a bicycle that is stuck to a brick wall by metal pips welded to the frame. If you get on the bike and peddle you want go anywhere. The wheels will slide along the ground generating heat vie friction. The same thing happens in you muscle tissue. The arm can't move because it can't go though the wall so it does work by have muscle fibers slide around inside your arm, generating heat.

O:
is that really so?
should i really except that every time i "get tired" there is work done somewhere?

but that gets really confusing..
because i was just convinced in other topic I've opened that work is frame variant..

hence, there is probably a frame in which there is no work done, and thus, i should not get tired...
not to mention a frame in which I'm doing a negetive work, and thus should get less tired(?)

Now honestly, am I really the only one who gets confused from the energy-work concept?

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Stonebridge
It's tiring to hold a 10kg mass motionless at arm's length even though no "work" is done.
Well, not quite...
If you define work as force times distance moved in the direction of the force (mechanical work) then on the surface, none is done. But this completely misunderstands the way muscles work. In order to hold the mass steady, your muscles have to constantly do small amounts of work as the muscle cells fire and relax. When they fire, the mass is lifted a tiny amount, when they relax, the mass "falls back". It maintains its average position as a result of work being done by the muscles which is not recoverable when the mass falls back.
It's the same thing happening, in principle, when you push against a wall.
To understand the process fully, you need to look at how muscles work.
There is no conflict with the concept of mechanical work if you look carefully at what is going on.

Lsos
You ARE doing work, just not necessarily on the wall.

Imagine a helicopter hovering. It might not be doing any work on itself, since it's not moving, but it's definitely doing ALOT of work on the air.

You have to look at the big picture. Not just the wall, not just the helicopter, but also everything around it. Somewhere, some work is being done.

Gold Member
O:
is that really so?
should i really except that every time i "get tired" there is work done somewhere?

but that gets really confusing..
because i was just convinced in other topic I've opened that work is frame variant..

You have to realize that examples in physics are unbelievably basic. Biological systems are very complex that even include chemical reactions. If you literally just sat on a chair an entire day, you still are burning calories because your body is always 'working'. Energy is always conserved, you just have to look at the complete (and in this case, very complicated) picture.

Mentor
Again, your body doesn't need to be doing any work to consume energy. You consume energy even when sleeping.

JerryClower
The human body is very inefficient.
Yes, we are. We are weaker than our cousins. What is this, survival of the weakest?

Gold Member
I've seen this exact question in a book.
(This is not homework)
I've always thought this is because work has nothing to do with effort,
but seeing this question on a book quite shocked me actually, is there any deeper explanation?

As other replyers have pointed out: your muscle fibers are not like hydraulic pistons.

A hydraulic piston can be pretty much locked in a specific position, by closing the valve through which the hydraulic fluid enters the piston. More generally: any mechanical device can be locked off.

Muscles don't work like that.
Interestingly, the muscles of clam shell organisms are more closely analogous to mechanical devices. Those muscles contract slowly; the layers slide along each other slowly. The layers bond, and this bond releases very slowly.

For animals scurrying around that type of muscle is very unpractical. Animals need muscles that not only contract fast, but equally important: they must also allow fast lengthening. Imagine how stiff you would be if each of your muscels would take minutes to lengthen again.

By necessity the attachments between the layers detach spontaneously, and easily. Conversely, in order to sustain a certain muscle contraction the layers must renew attachments all the time. This continuous breakup and renewal costs energy. That is why merely sustaining a certain muscle tension depletes oxygen and energy resources

O:
Should I really except that every time I "get tired" there is work done somewhere?

But that gets really confusing..
Because I was just convinced in other topic I've opened that work is frame variant..

Hence, there is probably a frame in which there is no work done, and thus, i should not get tired...
Not to mention a frame in which I'm doing a negetive work, and thus should get less tired(?)

can anybody help me understand this?

Mentor
I'm not sure: what part of "your body doesn't need to be doing any work to use energy" didn't you understand? You don't seem to be listening to what people are telling you.

output work/ input energy = efficiency

The human body is very inefficient.

Compared to what? It's my understanding that we utilize 80-90% of the calories we consume: that's significantly higher than any man-made engine.

Muscles range from 25-40% efficiency: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1300991/

F1F0 ATP synthase is close to 100% efficiency: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1692765/

Kinesin is also about 40% efficient. http://www.cellbio.duke.edu/kinesin/KinesinMotility.html [Broken]

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Mentor
Compared to what? It's my understanding that we utilize 80-90% of the calories we consume: that's significantly higher than any man-made engine.
All you are saying there is that 10-20% of the chemical energy injected into a human isn't consumed/burned. Using a similar standard, a good car engine "utilizes" very close to 100% of the fuel injected into it. That doesn't have anything to do with the efficiency of either a car or the human body. It doesn't tell you how much is converted to useful work and how much - in both cases - is simply wasted while idling (for example).
Muscles range from 25-40% efficiency: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1300991/
Not when they aren't moving, they don't. Consider what happens when you hover on a hill with a car, holding your position with the clutch and gas pedals. What's your fuel economy? Output work? Input energy?

The OP is clearly talking about the human body's ability to produce useable output work. Yes, of course the efficiency depends on what you are doing, but unlike many other devices (such as a car with a parking brake or a table holding up a book), the human body can't generate a static force without burning energy (other than directly down...). Output work = 0, input energy = 10 cal (per min) (perhaps). Efficiency: 0/10=0

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Frame Dragger
All you are saying there is that 10-20% of the chemical energy injected into a human isn't consumed/burned. Using a similar standard, a good car engine "utilizes" very close to 100% of the fuel injected into it. That doesn't have anything to do with the efficiency of either a car or the human body. It doesn't tell you how much is converted to useful work and how much - in both cases - is simply wasted while idling (for example). Not when they aren't moving, they don't. Consider what happens when you hover on a hill with a car, holding your position with the clutch and gas pedals. What's your fuel economy? Output work? Input energy?

The OP is clearly talking about the human body's ability to produce useable output work. Yes, of course the efficiency depends on what you are doing, but unlike many other devices (such as a car with a parking brake or a table holding up a book), the human body can't generate a static force without burning energy (other than directly down...). Output work = 0, input energy = 10 cal (per min) (perhaps). Efficiency: 0/10=0

Not to mention that two identical individuals could lift the same weight for the same period of time to the same hight, but the manner in which they did so would cause the METABOLIC activity to vary. There are layers of subtletly to ergonomic efficiency... hence the multi-billion USD/year industry.

@others: To discuss the metabolic activity of an animal in terms of raw newtonian physics is a bit like trying to communicate with a pitchfork. :yuck:

Andrew Davies
can anybody help me understand this?

Yes and No. Anytime there is movement of any kind within your body work is being done. Heart pumping, blood moving, muscles contracting, cells contracting, etc. However Feeling tired is not always the result of physical exertion. For instance if your brain was depleted in some chemical, say you didn't get enough sleep. Then you would still feel tired even though you did no more work then usual

Work is not frame frame variant, if it were the conservation of energy would be violated. However I can think of a few instances (relativity) were it wouldn't be so clear. What form did you get this idea from.

I assume were speaking classically. If I drive past you at 60 m/s. And I see you lifting a break h meters off the ground. I will say you did mgh joules of work. You will too.

Homework Helper
Gold Member
When you say "no work is done" when you push hard on a wall, all it means is that no energy is being transferred to the wall. It's just matter of realizing that 'work' is a technical term in physics, which, in simple terms, means the amount of kinetic energy transferred to a body by a force

Gold Member
Consider what happens when you hover on a hill with a car, holding your position with the clutch and gas pedals.
I was about to make this comparison when I realized Russ already did.

A car hovering on a hill eats up a lot of energy, yet it does not move. By the Physics definition, it is doing no work. But no one doubts that it is consuming fuel.

Think of your muscles like the constantly revving engine with the clutch in - they are always running, whether or not that energy is going into any useful activity.

Frame Dragger
Yes and No. Anytime there is movement of any kind within your body work is being done. Heart pumping, blood moving, muscles contracting, cells contracting, etc. However Feeling tired is not always the result of physical exertion. For instance if your brain was depleted in some chemical, say you didn't get enough sleep. Then you would still feel tired even though you did no more work then usual

Work is not frame frame variant, if it were the conservation of energy would be violated. However I can think of a few instances (relativity) were it wouldn't be so clear. What form did you get this idea from.

I assume were speaking classically. If I drive past you at 60 m/s. And I see you lifting a break h meters off the ground. I will say you did mgh joules of work. You will too.

There is an even bigger give away... What happens when metabolic work is done? What ALWAYS happens? We give off... HEAT. We eat food, break it down into ADP and ATP eventually, then 'burn' it in a chemical process and we give off waste heat. Heat = thermodynamic radiation = proof that we take (more) ordered forms of energy and follow the laws of thermodynamics.

EDIT: Someone mentioned that (human) muscles don't "lock" which is true. Animals which sleep upright however, use a network of tendons and ligaments to "lock" their legs so as to spare their muscles during sleep. Why sleep upright?... well if lions want to eat you, struggling to your feet in a daze probably is a disadvantage. Anyway, as others have said, in the absence of special adaptations metabolic work ends (some short period of time after) when you die.

EDIT2: Davec: Not to mention that to keep that car on an upward grade you either need to engage the brake (generating friction and heat, proof of work), OR you need to have your foot on the gas. I wonder if the term "equilibrium" is just no longer taught to kids these days?

All you are saying there is that 10-20% of the chemical energy injected into a human isn't consumed/burned. Using a similar standard, a good car engine "utilizes" very close to 100% of the fuel injected into it. That doesn't have anything to do with the efficiency of either a car or the human body. It doesn't tell you how much is converted to useful work and how much - in both cases - is simply wasted while idling (for example). Not when they aren't moving, they don't. Consider what happens when you hover on a hill with a car, holding your position with the clutch and gas pedals. What's your fuel economy? Output work? Input energy?

The OP is clearly talking about the human body's ability to produce useable output work. Yes, of course the efficiency depends on what you are doing, but unlike many other devices (such as a car with a parking brake or a table holding up a book), the human body can't generate a static force without burning energy (other than directly down...). Output work = 0, input energy = 10 cal (per min) (perhaps). Efficiency: 0/10=0

But 'work' in the context of mechanics cannot be applied to isometric contraction, much like it cannot be meaningfully applied to your example of holding a fixed position while the engine is running. And an internal combustion engine does not utilize anywhere near to 100% of the available energy: incomplete combustion.

The literature is confusing. The typical claim is that a 2000 kcal/day diet means we are essentially a 100 W light bulb, but that isn't really accurate. It's important to recall that the energy used by the body is not produced by combusting, but is stored via the Gibbs free energy.

I had a hard time finding out how much energy is required to maintain normal body temperature. For example, say I am in a room which is maintained at 98.6 degrees F. What is my basal metabolic rate then, as compared to sitting in a 78 degree room? I'm not sure the data is out there, but that's how to determine how much 'work' (in the Gibbs free energy sense) is required to maintain normal function.

Clearly, sitting in a chair means I am not in equilibrium. Just being alive means I am not in equilibrium- maintaining that disequilibrium requires work, but not mechanical work.

Frame Dragger
But 'work' in the context of mechanics cannot be applied to isometric contraction, much like it cannot be meaningfully applied to your example of holding a fixed position while the engine is running. And an internal combustion engine does not utilize anywhere near to 100% of the available energy: incomplete combustion.

The literature is confusing. The typical claim is that a 2000 kcal/day diet means we are essentially a 100 W light bulb, but that isn't really accurate. It's important to recall that the energy used by the body is not produced by combusting, but is stored via the Gibbs free energy.

I had a hard time finding out how much energy is required to maintain normal body temperature. For example, say I am in a room which is maintained at 98.6 degrees F. What is my basal metabolic rate then, as compared to sitting in a 78 degree room? I'm not sure the data is out there, but that's how to determine how much 'work' (in the Gibbs free energy sense) is required to maintain normal function.

Clearly, sitting in a chair means I am not in equilibrium. Just being alive means I am not in equilibrium- maintaining that disequilibrium requires work, but not mechanical work.

There is a test for what is called the "resting metabolic rate", but it's still an aproximation based on gas measurements to find your idealized "resting" metabolism. The number itself may be +-???, but it WILL let you know roughly how much you burn daily.

And yes... that was a semi-random factoid. End of story.

dacruick
can anybody help me understand this?

I think you are missing the essence of what mechanical work is. If you look at the equation it is F*d sin(theta). that means that it is the force multiplied by the distance in the direction of the force. For you to do work on the wall, you actually have to move the wall. But inside you there is work being done by your triceps to your forearm. and from your forearm to your wrist and so on. and a million other tiny systems where work is being done. When you push the wall it probably moves a little bit, too little for human eyes to see, and when you stop pushing the wall reverts back to its position by the bonds inside the brick doing work on each other. You can bring moving frames of reference into the picture but that seems like working backwards to me. Why would it matter how fast you are driving by if your car is not involved in the system. how much work you do on the wall involves only you and the wall. so if you and the wall are in the same frame of reference then that is all that matters. if you and the wall are not in the same frame of reference, good luck trying to apply a force to it to do work.

Leafhill
So is there any equations anywhere taking into account thermal energy being produced to maintain a helicopter in its hover? If I know the weight of the helicopter and the downward force (gravity = 9.81 m/s^2) I should be able to calculate the amount of watts being consumed. Am I right?

Frame Dragger
So is there any equations anywhere taking into account thermal energy being produced to maintain a helicopter in its hover? If I know the weight of the helicopter and the downward force (gravity = 9.81 m/s^2) I should be able to calculate the amount of watts being consumed. Am I right?

That depends on a lot of external factors, such as the the nature of the main blades: are they designed for heavy lifting (heavy, rigid), maneuvers (piezoelectric flexing), speed (rigid, light). There is also the matter of the efficiency of the helicopter in turning power generated by its gas/electric/etc... fired engine into down-force and therefore life.

Distance from the ground, lateral windage, the shape of the fusalage... and more I'm not thinking of I'm sure.

To be clear, you could read the output on the instruments and figure out how hard the helo's engine is working, and calculate roughly how much of that is ultimately going to end up as heat. Doing that simply by measuring the weight vs gravity however, will tell you not how much energy is consumed, but how much lift you need. How you achieve that lift is a function of many variables that effect efficiency, and therefore avg and peak utilizations of power.

All you are saying there is that 10-20% of the chemical energy injected into a human isn't consumed/burned. Using a similar standard, a good car engine "utilizes" very close to 100% of the fuel injected into it. That doesn't have anything to do with the efficiency of either a car or the human body. It doesn't tell you how much is converted to useful work and how much - in both cases - is simply wasted while idling (for example). Not when they aren't moving, they don't. Consider what happens when you hover on a hill with a car, holding your position with the clutch and gas pedals. What's your fuel economy? Output work? Input energy?

The OP is clearly talking about the human body's ability to produce useable output work. Yes, of course the efficiency depends on what you are doing, but unlike many other devices (such as a car with a parking brake or a table holding up a book), the human body can't generate a static force without burning energy (other than directly down...). Output work = 0, input energy = 10 cal (per min) (perhaps). Efficiency: 0/10=0

I was thinking more about this. The conceptual error here is that in both (idealized) cases of the car moving and staying in place on a hill, if the engine is moving at the same RPM the mechanical efficiency of the *engine* is exactly the same- the crankshaft turns at the same rate. The essential difference here is the power loss due to the *transmission*.

In the body, the situation is qualitatively different. First, the mechanism by which *usable energy* is extracted from food is very different: a Cal of food energy is not really comparable to a Cal of gasoline. For one thing, the body converts the raw food energy into another form of energy and stores it. And I still maintain that biological organisms are highly efficient, much more efficient at doing this, than internal combustion engines.

Now we have the transmission- how that usable energy is converted into mechanical work. Much like a transmission, muscles have a range of operating conditions- constant tension, constant force, constant length, etc. Muscle contraction is significantly more complex than a transmission, but even so, various stages along the energy transduction pathway (the F1F0 ATPase molecule is a 'battery' that charges mitochondria, for example) are again significantly more efficient than engineered motors. By the time we get to the actual motion of a single molecular motor on a filament, the overall efficiency approaches 40% or so, and it's not clear how to incorporate concepts like 'stall'.

The OP already knew that "work != exertion". The fault is the limited definition of 'work'. Surely, there are chemical analogs available- chemical engineers know how to keep a reaction proceeding at a maximal rate, for example.

Frame Dragger
I was thinking more about this. The conceptual error here is that in both (idealized) cases of the car moving and staying in place on a hill, if the engine is moving at the same RPM the mechanical efficiency of the *engine* is exactly the same- the crankshaft turns at the same rate. The essential difference here is the power loss due to the *transmission*.

In the body, the situation is qualitatively different. First, the mechanism by which *usable energy* is extracted from food is very different: a Cal of food energy is not really comparable to a Cal of gasoline. For one thing, the body converts the raw food energy into another form of energy and stores it. And I still maintain that biological organisms are highly efficient, much more efficient at doing this, than internal combustion engines.

Now we have the transmission- how that usable energy is converted into mechanical work. Much like a transmission, muscles have a range of operating conditions- constant tension, constant force, constant length, etc. Muscle contraction is significantly more complex than a transmission, but even so, various stages along the energy transduction pathway (the F1F0 ATPase molecule is a 'battery' that charges mitochondria, for example) are again significantly more efficient than engineered motors. By the time we get to the actual motion of a single molecular motor on a filament, the overall efficiency approaches 40% or so, and it's not clear how to incorporate concepts like 'stall'.

The OP already knew that "work != exertion". The fault is the limited definition of 'work'. Surely, there are chemical analogs available- chemical engineers know how to keep a reaction proceeding at a maximal rate, for example.

I can prove your point (the one I put in bold) with the simplest possible example: plants use photosynthesis which is INCREDIBLY efficient. It's only with the understanding of relatively long-lived coherent superpositioned states in chlorophyl that such effiency can be explained.

If you want less exact, but even more impressive examples I would point to mammalian, reptilian, amphibian hibernation. In terms of computing power vs efficiency the brain is alos pretty astounding. Yes, it takes a LOT of your energy to run the thing (I mean this for all humans, literally), but considering what it's capable of in terms of raw computing and multitasking, memory storage and manipulation, etc...

Mentor
But 'work' in the context of mechanics cannot be applied to isometric contraction, much like it cannot be meaningfully applied to your example of holding a fixed position while the engine is running.
Of course it can:

w=f*d = 10*0=0
And an internal combustion engine does not utilize anywhere near to 100% of the available energy: incomplete combustion.
This is wrong. Think about it: if it were true, there would be substantial amounts of unburned hydrocarbons, nitrous oxides, etc being emitted by cars, in addition to CO2. The introduction of computer controlled combustion and catylitic converters has reduced this to very close to zero. These impurities are measured in an emissions test (in many states) in units of parts per million. http://www.aa1car.com/library/tr1196.htm
Just being alive means I am not in equilibrium- maintaining that disequilibrium requires work, but not mechanical work.
Yes.
The conceptual error here is that in both (idealized) cases of the car moving and staying in place on a hill, if the engine is moving at the same RPM the mechanical efficiency of the *engine* is exactly the same- the crankshaft turns at the same rate. The essential difference here is the power loss due to the *transmission*.
Not only is that not correct (thermodynamic efficiency of the engine is highly variable), you're not applying the analogy correctly. A car engine isn't doing useful work unless it is moving the car and that's what this question is about. Besides which, if a car is idling or hovering on a hill, the energy is all lost due to friction in the drivetrain, so I don't see how you can consider there to be any output work: the purpose of a car engine isn't to generate friction in the gearbox, the purpose is to drive the wheels.
In the body, the situation is qualitatively different. First, the mechanism by which *usable energy* is extracted from food is very different....

.... By the time we get to the actual motion of a single molecular motor on a filament, the overall efficiency approaches 40% or so, and it's not clear how to incorporate concepts like 'stall'. [emphasis added]
All of this simply ignores the OP's question. The OP is asking about pushing against a wall. Period. There is no motion. It actually makes life easy because anything times zero is zero, so you don't really need to calculate the input energy or worry about whether you want to only count the energy being used to run your heart or just what is consumed in the muscles themselves. It doesn't matter: w=f*d=x*0=0
The OP already knew that "work != exertion". The fault is the limited definition of 'work'.
Yes: the whole point of the OP's question was to reconcile the definition of "work" with the fact that the body is using energy even when work is zero.

Mentor
So is there any equations anywhere taking into account thermal energy being produced to maintain a helicopter in its hover? If I know the weight of the helicopter and the downward force (gravity = 9.81 m/s^2) I should be able to calculate the amount of watts being consumed. Am I right?
More or less, yes.

The helicopter is held aloft by a cylindrical column of air being pushed down by its rotor. So the force can be converted to a pressure over the area of the rotor wash. Pressure is related to the motion of the air via Bernoulli's equation. Then velocity, area and density will give you mass flow rate and power.

Alternately, you can do a little less and figure out the change in momentum of a block of air (since force is change in momentum with time). With the cross sectional area, density, then mass flow rate and velocity, you end up with power in a similar way.

<snip>
All of this simply ignores the OP's question. The OP is asking about pushing against a wall. Period. There is no motion. It actually makes life easy because anything times zero is zero, so you don't really need to calculate the input energy or worry about whether you want to only count the energy being used to run your heart or just what is consumed in the muscles themselves. It doesn't matter: w=f*d=x*0=0 Yes: the whole point of the OP's question was to reconcile the definition of "work" with the fact that the body is using energy even when work is zero.

I'm not ignoring the question- all I was doing was pointing out that the situation is more complex than a spherical cow approach, and hopefully pointing out areas where our understanding is incomplete. For example, I can't find a molecular explanation for isometric contraction- since the muscle is not shortening, it's not clear how the actin/myosin picture needs to be modified. Another point, the main end product of muscle exertion is not heat, but lactic acid. So 'useful energy' is not dissipated via heat. For metabolic reactions, the end products are water and CO2- not heat. to a very good approximation, we operate at constant temperature, and the literature is not clear at all about how much heat is actually generated by metabolism.

Perhaps the OP is satisfied with a "shut up and calculate" response, but perhaps not.

<snip> A car engine isn't doing useful work unless it is moving the car and that's what this question is about. <snip>

I just want to make sure I understand what you are saying: If I remove an engine from a car, put it on a test stand, and operate that engine, it now has an efficiency of precisely zero?

Frame Dragger
I just want to make sure I understand what you are saying: If I remove an engine from a car, put it on a test stand, and operate that engine, it now has an efficiency of precisely zero?

Uhhhh... Ok... if that's the definition we're working with, my posts no longer make sense.

Gold Member
I just want to make sure I understand what you are saying: If I remove an engine from a car, put it on a test stand, and operate that engine, it now has an efficiency of precisely zero?

If the engine's operation does not result in a mass displaced by a distance then work done is zero.

Buckleymanor
Why is it tiring to push hard against a solid wall even though no work is done?
Because the wall pushes back with the same amount of force or else it would collapse.

Frame Dragger
If the engine's operation does not result in a mass displaced by a distance then work done is zero.

The mass is displaced... pistons move, belts spin. The only difference is that work isn't going into a transmission so that forward motion of the engine block occurs. Like the human body, an idling engine has many subunits doing work and Work. Maybe you could say the block as a whole isn't doing work if it isn't moving... then again, if you turn on this fictional engine and don't bolt it down it's going to move all over the place.

rewebster
if you push on the wall and nothing happens, it just means that you're not strong enough to break the molecular bonds that hold the wall in that position/structure.

If you were as strong as, say a D8 Cat bulldozer, then the wall may move.

If it was a rock wall with a lot of other rock behind it, it may take more than even a D8.