This might be a ridiculous question but some other lifters and I got into a debate about it.

So, imagine someone is doing Bulgarian Squats. 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 doing the same amount of work/effort compared to the lift being performed with the weight moving with the lifter like the gif above?

  • 1
    I noticed you already asked this on the Physics Stack exchange. I believe Barnaud Juliens answer summarizes it quite well. If you go by the physics definition of 'work', then the amount of work done is the same. If you think about it from a fitness perspective, it does require effort by your leg to even keep the weight in the air. So I think to the average gym rat the answer is: "Yes of course it requires more work/effort even for the legs to keep the weight in the air." Jun 23 '21 at 23:37
  • On a tangent, it reminds me of this question from about 6 months ago that none of us really had a solid answer to. Jun 23 '21 at 23:39
  • Haha yes it was interested to get 3 different answers from the Physics Stack Exchange. To your point i'd say the work is the same to keep it in the air but if the height of the weight stays constant your leg is no long pushing the weight up. Therefore it is doing less work every rep.
    – Dan S.
    Jun 24 '21 at 1:57

The legs don't have to work quite as hard as they would in a regular squat with the same weight, but it's very close, and they certainly have to work harder than in a bodyweight squat.

First of all, let's eliminate the idea that you aren't doing any work because you aren't moving the weight. Yes, you aren't imparting any mechanical work on the weight, but that's not necessary for the muscles to be working, where "muscular work" is defined as the rate at which energy is burned by the muscle to produce force. This is evident from isometric training, where the muscle exert against an immovable object, and in doing so consume energy and fatigue.

Now for how much additional resistance this would add over a regular (split) squat. Holding a dumbbell stationary during the exercise is identical to performing the exercise while pushing upwards on a fixed bar, with a force equal to the weight of the dumbbell. Both the arms are exerting upwards force when they normally wouldn't need to, and this force gets added to what is required of the legs.

As far as the legs are concerned, the big difference between this and a regular dumbbell split squat is that in this case the only additional loading provided by the dumbbell is a constant force equal to the dumbbell's weight, whereas in the regular split squat, the legs must both overcome the external load's weight, and additionally provide enough to accelerate it through the movement. This acceleration component is missing when the dumbbell is held stationary.

But how significant is the acceleration component? Let's model the squat as a vertical movement of 0.5m, with a complete rep following a sinusoidal pattern with period of 2 seconds. So x is the vertical position of the bar, v is its velocity, and a its acceleration.

x = 0.25cos(π*t)
v = -0.25π*sin(π*t)
a = -0.25π²*cos(π*t)

Peak acceleration occurs at the bottom of the movement, at t=1:

a = -0.25π²*cos(π)
  = 2.47m/s²

(Note that m in m/s² is metres, not mass.)

From this we can calculate force required to cause this acceleration, using F=ma, where F is force and m is the mass of the dumbbell:

F = ma
  = 2.47*m

Compare this to the force needed to overcome the weight of the dumbbell, which for any object on Earth is 9.81 times its mass. So with acceleration, peak force would be (9.81 + 2.47)*m, whereas without it, it's just 9.81*m, which is about a 20% reduction.

Also keep in mind that for the vast majority of people, the external load in a Bulgarian split squat is much less than their bodyweight. This means that most of the force they're producing is going towards holding the weight of and accelerating their own body, rather than the external load. So let's say they're a 75kg person using a 25kg dumbbell. The dumbbell only makes up 25% of the weight in that system, so the 20% reduction in peak force due to not needing to acceleration it decreases to a 5% reduction in peak force when bodyweight is accounted for.

  • I like this thought a lot more. I knew my idea had holes in it since the arms are obviously rooted through your feet. The legs not needing to overcome the acceleration makes sense to me.
    – C. Lange
    Jun 26 '21 at 17:15

If we put our plane of reference (x,y) on the feet, the weight is static and the body moves. Theoretically, that means our displacement is 0 and work is 0. By changing the exercise in this manner, the legs don't need to exert any additional effort to move the weight but, how are we keeping the weight static?

If we put our plane of reference on the shoulder, the weight is "static" but the plane of refence is now moving (up and down as the legs move) and therefore our displacement is > 0 and work is some finite number. Logically, the arms need to exert effort in order to keep the weight static relative to the legs.

That's my theory viewpoint. Now practically, I just grabbed a dumbbell and tried doing it myself with 25 lb, and it sucks. Everything about this variation makes this already garbage exercise even worse and I hate it (split squats are a good exercise, I just suck unilaterally). My arms are miles weaker than my legs and I couldn't hold the 25 lb dumbbell static without compromising the exercise. In the end, I had to drop to a weight that I can barely differentiate from bodyweight.

Whoever's right is going to depend entirely on whether you're taking practical effort into account.

  • Yep the lift makes no sense obviously and we also tried it ourselves after the debate started and got multiple opinions on whether it was easier or not. I appreciate the response and giving it a try :)
    – Dan S.
    Jun 24 '21 at 2:02
  • @DanS. -- for sure. It's an interesting thought. Another idea is it can depend on where you're starting your timeframe. Is t = 0 when the weight is already at chest level? In practice, t = 0 should probably be the instant before the weight leaves the rack/floor as it takes effort to lift it into the air initially and it makes more sense as it obviously take effort to keep it there.
    – C. Lange
    Jun 24 '21 at 2:25

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