If someone is doing a weighted bulgarian squat and keeps the height of the weight static is their leg doing the same amount of work?

Question

So imagine someone is doing this lifting exercise. However instead of the dumbbell being held at his chest, the lifter uses his arms to keep the height of the dumbbell static to the ground. So at the bottom of the rep the dumbbell would maybe be a little above his head and at the top of the rep the dumbbell would be back at his chest.

My question is if you performed the exercise with the weight's height remaining static would your leg still be using the same amount of force to get to the top of the rep compared to the lift being performed with the weight moving with the lifter like the gif above.

Answer 1

Ok, so here is my answer, I will divide it in two parts, the first one probably  being actually irrelevant to the actual core of the question, but simply included for the sake of discussion  

1. On a theoretical point of view (physics definition of work)

I disagree with joseph’s assertions. I have to mention I am not actually sure! Maybe Joseph is right and I’m wrong, but I include my arguments here for the sake of discussion. The picture below is a very schematic depiction of a “squat up” in what you are describing.

In the regular version, both the center of mass (COM) of the person and the weight are going up. Let’s call the $\Delta h$ the distance travelled by the weight and you indeed get that the work that has been performed on the weight during its travel to raise it up has been $W_{person->weight}=-mg\Delta h$ where the minus sign is just a thermodynamics convention to indicate that the weight received this energy. By conservation of energy on the (weight+person) system,

$W_{ weight-> person }=- W_{person->weight}=mg\Delta h$

where the positive sign indicates that the person used up this amount of energy to raise the weight. Now you see where I’m going. In the second situation, the height of the weight is constant. It means that the total work $W_{person->weight}$ performed during the movement is zero. Reciprocally, the work performed by the weight on the guy is also zero. The total amount of physical work due to the weight in that case is null (though the work needed to move that butt aka COM remains unchanged)

EDIT: Although I am struggling to formulate it properly, the total null work can be split up in equal and opposite work from the legs and arms. On a squat down, the legs are receiving work from the weight (weight is pushing in the direction of the legs movements), while the arms exert work on the weight. On a squat up, it is the contrary (arms receive work while legs exert work). Therefore this would lead to the conclusion that both legs and arms are exerting mechanical work on the weight in an alternating fashion.  

2. On a sane person's point of view

Well actually, the definition of “work” is not the same for a physicist than for a fitness guy. Who thought physicists were not big on fitness? Indeed, let’s say you just lift a bar above your head and keep it there for ten minutes. A physicist passing by will tell you with a smirk that during those 10 minutes, no work has been done on this stationary system. You will beg to differ while you sweat and grunt and try your best not to get physical while the physicist keepsdoing physics, since you very much feel the work in your arms. This is because it does take energy to contract your muscles and keep them contracted to maintain the weight in place although no “mechanical work” is actually performed.

(what’s worse, if you do a squat down, the mechanical work kind of implies that the weight is helping that movement, when clearly it is harder to do with a weight, because you have to control the descent. Just like it can be very tiring in practice to climb down a very steep slope although gravity is theoretically helping you on your way down).

In short, the mechanical definition of work is almost garbage when it comes to estimating how much your muscles will actually “work”. I suspect that having the weight will indeed make the exercise more difficult in the second case, since the muscles in your arms and legs do have to sustain the weight, but I can not give a satisfying answer as to how this effort is distributed between arms and legs. I guess trying it and seeing how it feels compared to the regular version is not the worst option ;)

Answer 2

If I understand what you are saying, then yes, if the distance his arms move the weight is the same as the distance his legs move it.

In the second case, his arms are instead doing the up-and-down motion. The work done is now being shifted to his arms, when initially, that work was done exclusively by his legs*.

In both cases, the total work done $$W=mgy$$ where $y$ is the distance that the weight is moved, whether it is done by his arms or legs, and the force required to do so, given by $$W=F\cdot y=mg\cdot y$$ is also the same in both cases. It is also important to note that in both cases work is being done first against gravity and then by gravity, so that after each rep, the net work done adds to zero. How this translates to how much calirometric energy is being used, is probably a question for a biophysicist.

* This is by no means rigorous and in the second case there may be work also being done by the legs, though a great majority, I think is done by the arm-weight system. Once again, someone proficient in biomechanics can provide a more rigorous explanation. Where are you? :)

Answer 3

Here's my attempt. The total work done in the second case is the same as in the first case less the work allowed to be done by gravity by the moving arms, $m_{weight}gh$, but the work done by the leg is the same. This is because the leg doesn't care what the arms do, it still exerts the same amount of work on the body-dumbbell system, it's just that now the arms allow for an external force to lower the weight relative to the person.

You could also hold the dumbbell was as high as possible at the start of a rep and then lower it as low as possible when the leg is fully extended. In that case, the work gravity does is higher, in modulus, than $m_{weight}gh$, and so the total work done on the weight would be negative, but I don't see why that would change the amount of work done by the leg or the "effort" required by the exercise.