 What qualifies as work in weightlifting

# Thread: What qualifies as work in weightlifting

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## What qualifies as work in weightlifting

I'm in an introductory to mechanics course(calc based) and my class is in a chapter pertaining to work.

Now,

W=F*(Delta)r*cos(theta)

So work needs not only a force but a displacement

both the book and my professor go on to give examples like this
"Say you picked up a chair with one arm and held it in the air, with your arm completely straight and perpendicular to you body, 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."

I understand the definition and the explanations due to the formula and I know minds much smarter than mine came up with this shit but I just have a hard time believing that when I see weightlifters hold 200kg+ bars over their heads for a few seconds and wait for the signal to drop the bar, that the lifters are doing no work.

Maybe I am missing something, maybe I'm not thinking hard enough on this...idk I just don't like it 2. ## I think you don't like the word "work" as applied to an activity where ATP is being expended even as no motion is occurring. Work is indeed motion-based, and just because something is hard doesn't mean it's "work." The work the lifter did against gravity was the vertical displacement of the bar x the weight of the bar ("weight" being the term I'm using precisely). Holding it there for the down signal is "hard" but not "work."

Tell me about a Farmer's Walk. 3. Member
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##  Originally Posted by MattJ.D. W=F*(Delta)r*cos(theta)

So work needs not only a force but a displacement
...
I just have a hard time believing that when I see weightlifters hold 200kg+ bars over their heads for a few seconds and wait for the signal to drop the bar, that the lifters are doing no work.

Maybe I am missing something, maybe I'm not thinking hard enough on this...idk I just don't like it
If you're talking physics, it doesn't matter if you like it. Outside of that realm though, nobody's going to disagree if you say the lifter is "working" hard even though the bar isn't moving.

Just don't mix up the physics terms with your everyday terms. 4. Member
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## Adding a little more detail. Work is simply the name given to the result of a mathematical formula. Specifically: Both Force and the displacement vector have direction associated with them. The dot product of F*dx "filters" out any products that aren't in either the same, or opposite, direction. This is where the cos(theta) comes from. If there is no displacement (holding a weight overhead), dx= 0 so no work is done. Similarly, if you're carrying something at a constant velocity, all the displacement is entirely perpendicular to the force(Farmer's Walk), our angle is 90, and cos(90) returns zero as well.

So just remember, the next time you do heavy ass farmer walks, you're doing zero work. 5. ## Yeah, that's hard for lots of people to accept. But it also is the basis for our vertical bar path rule, which is very hard to refute. Because of the math. 6. Member
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##  Originally Posted by Mark Rippetoe Yeah, that's hard for lots of people to accept. But it also is the basis for our vertical bar path rule, which is very hard to refute. Because of the math.
Damn math - always gets in the way of bullshit. 7. Member
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## I agree with Rip, PEPCACK and atb5161. Energy can be expended by the lifter even though no work (physics definition) is done. Dare I again raise the question about the eccentric portion of the lift? Related questions: How much work is done and how much energy is spent by the lifter to lower and raise the weight? What is the nature of the stretch reflex?

On the way up, the energy, U1, spent by the lifter can be written
U1 = W + Q1, where W = FD is the work done by the lifter, and Q1 is the heat generated since the human machine isn't perfectly efficient.

On the way down, the energy, U2, spent by the lifter can be written
U2 = - W + Q2, since the weight now does work on the lifter.

The total work done by the lifter going down and up is W - W = 0. The total energy spent, however, is not zero but is equal to U1 + U2 = Q1 + Q2.

Any objections?

I can't imagine how somebody would attempt to calculate Q1 + Q2 for a real lift. As dmworking pointed out in another thread, he and others can measure the energy spent during isometric, concentric, and eccentric contractions of (frog?) muscle, but a squat, for example, is a complicated matter! I remember there was a discussion of measuring the energy using calorimetry, another complicated matter. http://startingstrength.com/resource...589#post700589

Submitted for correction or clarification: During the eccentric portion of the lift, some of the work done on the lifter is converted to heat, but some is temporarily stored in the muscle. Is this temporarily-stored energy the stretch reflex, and is the temporary nature the reason why a paused squat or bench is more difficult than one without the pause? 8. ## 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. If the bar weighs 405, to lower the bar you apply 404 and to raise the bar you apply 406, or something like that. The work calculation must reflect the reality of the situation, and the reality is that the bar moved 2m and muscular force applies to the skeletal levers moved it that distance. The details are impossibly complicated, but this basic solid-body analysis is rather straightforward. 9. Member
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##  Originally Posted by Savs I agree with Rip, PEPCACK and atb5161. Energy can be expended by the lifter even though no work (physics definition) is done. Dare I again raise the question about the eccentric portion of the lift? Related questions: How much work is done and how much energy is spent by the lifter to lower and raise the weight? What is the nature of the stretch reflex?

On the way up, the energy, U1, spent by the lifter can be written
U1 = W + Q1, where W = FD is the work done by the lifter, and Q1 is the heat generated since the human machine isn't perfectly efficient.

On the way down, the energy, U2, spent by the lifter can be written
U2 = - W + Q2, since the weight now does work on the lifter.

The total work done by the lifter going down and up is W - W = 0. The total energy spent, however, is not zero but is equal to U1 + U2 = Q1 + Q2.

Any objections?
From a physics-homework standpoint you are correct. As you define your system, the net work on the barbell ends up being zero.

But as Rip said, it's unproductive and basically jerking off with math. You can define a lot of systems in a way that nets the total work to zero. It's much more convenient and useful to break the system up into two portions, up and down, since we have a convenient tool(barbell) that let's us quickly and easily quantify the "up" portion.

I think quantifying the amount of work done in the eccentric portion of a squat is going to vary much more due to varying bar speeds in the descent. Maybe I'm wrong, but I would guess that bar speeds on the way up for a heavy squat are going to remain relatively stable(slow to really fucking slow). Whereas going down bar speeds may change depending on the individual (controlled to divebomb). The controlled lifter will have a smaller force applied to the barbell over a longer range, while the divebomber will have a larger force applied for a shorter range right near the bottom. I'd actually be curious as to the difference in force for those two situations...

As much as I like all this physics, unless you're doing some kind crossfit style work-capacity comparison, work is pretty useless. I'd rather know the weight on the bar rather than some kind of work estimate. 10. Member
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##  Originally Posted by Mark Rippetoe If the bar weighs 405, to lower the bar you apply 404 and to raise the bar you apply 406, or something like that.
Let's ignore these small changes in force. We agree that they can be any combination to first accelerate the weight (get it moving), then move it at constant velocity for a period of time, then decelerate the weight. Also, let me call the 405 lbs the weight, w = mgh, so I don't mix pounds and meters; otherwise, PEPCAK will be all over my shit. Originally Posted by Mark Rippetoe 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. If the bar weighs 405, to lower the bar you apply 404 and to raise the bar you apply 406, or something like that. The work calculation must reflect the reality of the situation, and the reality is that the bar moved 2m and muscular force applies to the skeletal levers moved it that distance. The details are impossibly complicated, but this basic solid-body analysis is rather straightforward.
I agree that the details are impossibly complicated. I agree that the work calculation must agree with the reality; however, I disagree that the total work done is, or should be, 2mgh. I know this sounds contentious and pedantic as all hell. If we were only talking about the concentric, I'd have no disagreement. In that case, the energy expended by the lifter is work + heat = mgh + heat. If we ask about the lift down and up, then no work is done but energy is expended in the form of heat. Why must we say the work done is 2mgh? If we say that, we're not doing physics anymore.

The reality is the weight was moved against the force of gravity one meter, and then moved along the direction of gravity one meter and brought to rest. Total work = zero. In both cases the force is mgh but in the former movement, the muscle was shortening and in the latter lengthening (for a simplified example, the bicep during a curl). The reality, I think, is it is more difficult to raise a weight than it is to lower it. I think the concentric may be fundamentally different than the eccentric, and therefore I think the respective energies expended may be different.

Maybe you are saying the energy expended by the muscle(s) during the eccentric portion is the same as the energy expended by the muscle(s) during the concentric, and both are equal to mgh? They may be close. I don't know. What I think I've learned so far is that the eccentric and concentric are different. I've also read claims about energy temporarily stored during the eccentric, and maybe that partially explains what is happening when work is done "on the lifter" (or "the lifter does negative work") and maybe it explains the stretch reflex.

Maybe there isn't a desire here to be so pedantic. If so, I'll learn quickly and shut the hell up. #### Posting Permissions

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