What qualifies as work in weightlifting

[Giri Kotte]

I really think it is unproductive to consider the net change in position as being zero from start to finish of a squat when calculating the work. If the eccentric ROM was 1m and the concentric ROM was 1m, it seems rather apparent that the net change in position was 2m and that muscular force produced 2m of motion.

Work/energy is a scalar so work isn't zero even if the net displacement is zero. Work done will be zero only when the universe does the same work on the system as the system does on the universe.
Work is done in both eccentric and concentric motions as in both the cases, barbell is not being allowed by the lifter to fall freely at the rate of 9.8 m/s^2. Obviously the lifter doesn't work as much in the eccentric but, he's still doing work.
As you said initially, ATP energy expended is the correct measure of work done by a lifter for all practical purposes.

[strideknight]

In pure mechanical sense, "W" is the work done. But, work done in the terms of human intuition/parlance is always "Q", i.e., change in internal energy + mechanical work done.
P.S.: this the best picture of first law of TD I could find. So please interpret it in terms of fundamental physics.

In terms of barbell movements, this is probably the easiest way to think about the work being done. We can ignore heat exchange for all practical purposes, and the bar starts from a dead stop and ends in a dead stop (if you're doing it right) so the change in Kinetic Energy between the two positions is 0. So what happens is that all the work being done is either Potential Energy being added to the barbell, or removed from the barbell, as we move it through the gravitational field. This makes sense... You drop a barbell from a foot and it has a certain amount of energy when it hits the ground - you drop it from 1m higher and it has alot more energy when it hits because gravity has acted on it over a further distance.

[hagrid's dad]

A bit off topic, but does anyone know of a good reference for what's going on at the actin-myosin level during the eccentric portion of a lift?

I thought I had a good intro-text-level picture of actin and myosin, but that picture falls apart when I try to think of them consuming ATP to produce muscle tension while the muscle fibers are lengthening.

[rageshrub]

A bit off topic, but does anyone know of a good reference for what's going on at the actin-myosin level during the eccentric portion of a lift?

I thought I had a good intro-text-level picture of actin and myosin, but that picture falls apart when I try to think of them consuming ATP to produce muscle tension while the muscle fibers are lengthening.

I would say that not ALL fibers are relaxing during the eccentric phase.

Force is still being produced and exerted on the object, but that force happens to be less than the force of gravity acting on that object, and therefore it heads towards the ground.

So while the overall next effect is a lengthening contraction, some muscle fibers are still very much contracting to produce force in order to control the decent. Otherwise the object would just drop straight to the ground.

[Manofglass]

Holding it there for the down signal is "hard" but not "work."

It is hard but also work. Bcharles already mentioned the key aspect:

On the way up some muscled fibers contract and shorten.

For 'work' to be done, be it in on a schoolbook point-particle or in a complex multi-body system like a lifter, a displacement is needed in the direction of the force. Since muscle fibres always contract in the direction in which force is delivered they always do 'work'.

Lacking any (secret) parts that move in the direction of the force applied that is why a a floor can carry a heavy barbell 'without effort' (as long as the barbell is not so heavy as to compromise the structural integrity of the floor) and a lifter might be 'trembling' (=those contracting fibres at work) with exhaustion to hold it in lockout for a couple of seconds.

But the more we put ourselves in a situation where we let the weight rest on our (bone) structures instead of carrying them with our muscles the less work we do and the easier it is to maintain those. Which is why certain positions like the rack after a clean and the overhead lockout are so much easier to maintain than other bar-positions during the lift ('Press sticking point' I'm looking at you).

Back to Matt's teacher/textbook:

for some time x; you are performing no work because while you are you applying force for some time x on the chair, the displacement of the chair in some time x is zero, so the work done is zero."

Let's assume the purely hypothetical case where the chair doesn't move at all vertically (which it will): the force keeping the chair up is the 'normal force'. Which is the floor pushing back against the feet of the lifter+barbell. For the normal force to keep the chair up it needs to be transferred through the body of the lifter to the barbell, but for that to be done successfully the (stabilizing) muscles in the lifter's body must contract really hard to keep the lifter upright, and that means they are doing work.

Work done on the chair: zero
Work done by the muscle fibres on the body: nonzero

[MattJ.D.]

Its funny that everyones going to the internal work of the body now because my professor went to that in another course, he did a body example like you guys are doing but then went to a harrier jet...the ones that can hover, and used that too

[hagrid's dad]

I would say that not ALL fibers are relaxing during the eccentric phase.

Force is still being produced and exerted on the object, but that force happens to be less than the force of gravity acting on that object, and therefore it heads towards the ground.

So while the overall next effect is a lengthening contraction, some muscle fibers are still very much contracting to produce force in order to control the decent. Otherwise the object would just drop straight to the ground.

The fibers aren't "relaxing" but they are all lengthening, on average, over the course of a full ROM movement, aren't they?

The force *can't* just be produced by contracting fibers... because any path along the muscle from attachment to insertion has to be getting longer -- so at least some of the fibers in the path have to be lengthening, but still maintaining tension.

I was guessing that the resistive force would be caused by something like friction, rather than active "strokes" of the myosin heads... with the friction turned on and off by calcium binding to troponin to open up binding sites...

But my google-fu appears to be weak, since I can't find any good references.

[hagrid's dad]

Since muscle fibres always contract in the direction in which force is delivered...

That can't be right, can it?

The muscle *does* get longer in an eccentric movement. So there have to be muscle fibers that are simultaneously lengthening and exerting force.

[Giri Kotte]

In terms of barbell movements, this is probably the easiest way to think about the work being done. We can ignore heat exchange for all practical purposes, and the bar starts from a dead stop and ends in a dead stop (if you're doing it right) so the change in Kinetic Energy between the two positions is 0. So what happens is that all the work being done is either Potential Energy being added to the barbell, or removed from the barbell, as we move it through the gravitational field. This makes sense... You drop a barbell from a foot and it has a certain amount of energy when it hits the ground - you drop it from 1m higher and it has alot more energy when it hits because gravity has acted on it over a further distance.

Actually, there is an error in the formula. "Q" should be "delta Q".
With respect to the universe (that doesn't include the lifter), the lifter adds potential energy and then removes it.
But for the lifter as a system, he expends energy lifting the bar (increasing it's PE), holding the bar (by making sure his body imparts same force on the barbell as gravity does) and dropping the bar (making sure it doesn't fall freely under gravity).
One can also say that zero work is done. But, that's from the view point of a universe that includes the lifter (but, that system is not the one mentioned in the OP). Even in such a case, the entropy (randomness) of the universe increases even if the increase in the temperature is ignored!

[strideknight]

Actually, there is an error in the formula. "Q" should be "delta Q".
With respect to the universe (that doesn't include the lifter), the lifter adds potential energy and then removes it.
But for the lifter as a system, he expends energy lifting the bar (increasing it's PE), holding the bar (by making sure his body imparts same force on the barbell as gravity does) and dropping the bar (making sure it doesn't fall freely under gravity).

My point was to discuss the barbell system itself, adding in the lifter complicates things unnecessarily because all we are is basically electrochemical machines that pump out heat.

If we care to look at the barbell/universe model including the effects of the lifter but not the lifter, then there is technically an increase in internal energy because the lifter transfers enthalpy - you lose heat and impart kinetic energy to the bar. Any "PE removed" isn't really removed, it's been added as heat to the system. If we look at the barbell+universe+lifter system, then conservation of energy kicks and there's no change during the movement.

One can also say that zero work is done. But, that's from the view point of a universe that includes the lifter (but, that system is not the one mentioned in the OP). Even in such a case, the entropy (randomness) of the universe increases even if the increase in the temperature is ignored!

No. In none of these scenarios is no work done, and I'm not sure why people don't get this. Also, you realize that Entropy is defined using temperature right? You really can't disentangle the two and act like one increases without the other changing.