press question. Is reduced moment arm between hand and shoulder a good thing?

[Savs]

Hi spacediver! Thanks for your PM, and asking for my opinion (aww, shucks). I haven't read all the posts carefully; they're bouncing between lots of different issues. This issue of torque about a joint is very interesting to me, though. I see the discussion of torques about the knee and hip during the squat has been raised.

I don't have time to get into the nitty gritty discussion with you guys. I think I'm pretty much agreed with Tiedemies here, although I haven't read every sentence of his posts carefully, so I may not agree with everything.

I can summarize my thoughts as follows. When considering a free body diagram, one has to be absolutely certain that all external forces are included, and that's the tricky part here in these discussions about the press (and the squat).

First, an aside. Does anyone else own a copy of Beer and Johnston, Vector Mechanics for Engineers, Statics and Dynamics? I think it's a great textbook. I picked up a copy of the 4th edition back in the early 80s from a roommate (he was a MechE undergrad, and to this day I can't understand why he sold the book). I still flip through it from time to time; I'm doing so right now. I recommend people get a copy if they're interested in learning statics: rigid bodies, reaction forces at joints and supports, trusses, frames, and machines. A second aside! A while back somebody PMed me asking about sources for (re)learning mechanics, and I replied by unloading a long list of textbooks. I think I misunderstood the question, so if you're reading this and you were asking about a quick overview of Lagrangian and Hamiltonian Mechanics, I recommend the first book by Lenny Susskind in his "Theoretical Minimum" series.

Asides aside, spacediver, it might be better to draw your diagram in post #6 differently. Include the reaction forces at point A. Don't draw the torques about points A and B, but instead draw the forces (muscles) pulling on each member AB and BC. Make sure you include the long head of the triceps, because it is an external force acting on the "rigid" body. Sure, if you don't include all the external forces, then a quick examination of the free-body diagram shows that the shoulder muscles don't apply any torque about the shoulder when the weight is directly above the shoulder joint. I think that's an incorrect analysis, though. The same can be said when analyzing the squat, for example.

What else? Yeah... I could be wrong in my statements above. If so, the MechEs can correct me. (Don't correct me about torque vs moment, though. I couldn't care less about MechE jargon.) Oh, about the work done, yes, the muscles producing torque about A do do work. If the muscle is pulling on the limb and the limb rotates through an angle (even if there is no angular acceleration), then work is done. Work is the integral of torque through the angle.

Hope this post helps in some way. I don't check in to any of the forums here anymore, but: Hello, everyone. I hope y'all are doing well.

[Daverin]

Savs replying in 2017

[Tom Campitelli]

If you were to hypothetically get your hands right over your shoulders, the shoulders would need to flex to keep the bar going in a straight line, but the muscles wouldn't be experiencing any resistance from the barbell.

This is incorrect.

The only resistance they'd experience would be due to the weight of the forearm and upper arm.

And the barbell.

Torque generated by the triceps at the elbow joint would also also contribute to shoulder flexion, in the same way that the glutes can straighten out the hips.

The triceps are not shoulder flexors, however.

Torque It's the same in the squat. If the barbell is directly over the knee joint, there is no moment arm at the knee joint, and the quads don't feel any resistance of the barbell. In that situation, the hip extensors bear all the load.

This is also incorrect. Although it would likely be impossible to do so without hanging on to something, if you could squat so that the barbell was directly over a perfectly vertical tibia, the quads still have plenty of work to do. Even without invoking moment arms, the knee still flexes and extends. Knee extension will only happen when the quads contract.

[Tom Campitelli]

It is more or less anatomically impossible to flex the shoulder without introducing moment arms somewhere in the system with a barbell in the hands or to squat without some kind of moment across the knee. It is a neat thought experiment, but the important thing to keep in mind is that moment forces expressed across segments do not have a one to one mapping as far as what muscles do the work. While you can draw moment arms across the thigh, the actions of the muscles on the joints gets blindingly complex very quickly. You can go down a very deep rabbit hole very quickly.

[Tom Campitelli]

And to come back to the original idea of this thread, reducing unnecessary moment arms where possible is, in general, a good idea. Unless there is a specific reason to make things more difficult, you will probably be better served by allowing the load to produce the added resistance instead of adjusting the mechanics of the movement.

[spacediver]

Now consider that point C is the point at which a lifter grips the bar. The friction of the grip makes it impossible for the hand to move horizontally (let us assume a 2-D model here). The force of friction is always assumed to be exactly the same as the horizontal force that would move C to the left or to the right.

Friction would constrain horizontal motion of hand relative to barbell, but not barbell relative to space. When I think of a constraint, I'm thinking of a situation like when you do a "scrape the rack press".

By the way, Spacediver, I do not think you are stubborn. You are wonkish about these matters and I dig that. Some peole get tired of ironing out mechanical models.

And my wonkishness is matched by the unending patience of others, such as yourself

From the above reasoning I conclude that it is perfectly plausible, and even probable, that the optimal training effect -- i.e., the use of as much weight over the maximal range of motion while activating the largest mass of muscles(*) -- is achieved by having the barbell directly above the shoulder joint.

I agree that for maximal range of motion, barbell above shoulders is best. But I'm not convinced (yet) that this is the way to press the most weight, when you don't care about range of motion.

As for the rest of your posts, I'm still not sure I'm groking it all, but see whether my diagram and notes below are on point.

Hi spacediver! Thanks for your PM, and asking for my opinion (aww, shucks). I haven't read all the posts carefully; they're bouncing between lots of different issues. This issue of torque about a joint is very interesting to me, though. I see the discussion of torques about the knee and hip during the squat has been raised.

I swear, I feel like I just summoned batman.

I don't have time to get into the nitty gritty discussion with you guys. I think I'm pretty much agreed with Tiedemies here, although I haven't read every sentence of his posts carefully, so I may not agree with everything.

Fuck you batman! You're helping the bad guy, I'm the victim here

Oh, about the work done, yes, the muscles producing torque about A do do work. If the muscle is pulling on the limb and the limb rotates through an angle (even if there is no angular acceleration), then work is done. Work is the integral of torque through the angle.

My earlier comments were when I assumed the limbs were massless.

First, an aside. Does anyone else own a copy of Beer and Johnston, Vector Mechanics for Engineers, Statics and Dynamics? I think it's a great textbook. I picked up a copy of the 4th edition back in the early 80s from a roommate (he was a MechE undergrad, and to this day I can't understand why he sold the book). I still flip through it from time to time; I'm doing so right now. I recommend people get a copy if they're interested in learning statics: rigid bodies, reaction forces at joints and supports, trusses, frames, and machines. A second aside! A while back somebody PMed me asking about sources for (re)learning mechanics, and I replied by unloading a long list of textbooks. I think I misunderstood the question, so if you're reading this and you were asking about a quick overview of Lagrangian and Hamiltonian Mechanics, I recommend the first book by Lenny Susskind in his "Theoretical Minimum" series.

Thank you for this. I've been hunting for a good text on this stuff, and Beer and Johnston looks excellent. I like susskind also, saw a few lectures of his on relativity. He reminds me (visually) of John Malkovich.

Asides aside, spacediver, it might be better to draw your diagram in post #6 differently. Include the reaction forces at point A. Don't draw the torques about points A and B, but instead draw the forces (muscles) pulling on each member AB and BC. Make sure you include the long head of the triceps, because it is an external force acting on the "rigid" body. Sure, if you don't include all the external forces, then a quick examination of the free-body diagram shows that the shoulder muscles don't apply any torque about the shoulder when the weight is directly above the shoulder joint. I think that's an incorrect analysis, though. The same can be said when analyzing the squat, for example.

Ok, here's my attempt below:



Some notes:

I included some protuberances on the limbs so that I could model the muscles as straight lines. This means keeping track of how forces balance out is much simpler, and I don't think it alters the spirit of the exercise.

The red lines indicate the muscle fibers. I've included shoulder "rotators" (attachment points 1 and 2), the long head of the triceps (5 and 6), and another head of the tricep (3 and 4; yes, I know point 3 should be closer to the shoulder, but didn't want to clutter the figure). We also know that the magnitudes of these pairs of forces are equal (i.e. 1 is equal to 2, etc.).

Assume the blue base, G (roughly equivalent to the scapula), is immovable, and assume limbs are massless. Also assume that the system is in static equilibrium. My earlier contention was that in this situation, the magnitude of force 1 (and 2) is zero. Now I'm not so sure.

The force at point 3 has a horizontal and vertical component. The horizontal component produces a clockwise torque at A, equivalent to F3 Hor * vertical distance between 3 and A. The vertical component produces a counterclockwise torque, equivalent to F3 Vert * horizontal distance between 3 and A. Whether these torques balance out depends upon the angle of force application of the fiber, and how far down the humerus the attachment point 3 is.

We can repeat this analysis for the forces at points 4 and 6. Each will provide a clockwise and counterclockwise torque at A. The key question is whether the sum of all these torques is equivalent to a positive clockwise torque at A. If it does, then the shoulder rotators will need to generate a counterclockwise torque at A, by applying a force at point 2.

Something that may help here is to exploit the law of the conservation of angular momentum. We can do this as follows:

Forces 3 and 4 are all internal forces. Thus, they alone cannot change the angular momentum of the system around any point, whether that be point A, B, C, or some other arbitrary point.

G does exert a force on the system, as it experiences reaction forces that are coupled with the force of the barbell, and points 2, 3, 4, and 6. But importantly, none of these forces can produce a torque at point A, since these forces act at point A, and therefore have no moment arm between the point of force application and the point of rotation.

Based on this, we know that the only forces that can contribute to a net torque around point A are points 2 and 6, as these are truly external forces. Let's ignore 2 for now, because we are only interested in seeing whether the system is balanced without the contribution of the shoulder rotators.

So this leaves point 6 as the only force that can provide a net torque around A. Now in my thinking, this could go one of three ways:

1) the net torque could be a clockwise one. In this case, the shoulder rotators are required.
2) the net torque could be a counterclockwise one. In this case, you'd actually need to use the antagonist shoulder muscles (not drawn here) to prevent the shoulders from rising.
3) the net torque could be zero.

My gut tells me that which of these three cases turns out to be true depends upon the angle of force application at 6, and the precise location of this attachment point.

If I'm correct, then two things are true:

1) If we ignore the long head of the tricep, which crosses two joints, and only restrict ourselves to the medial or lateral heads (which only cross the elbow joint), then, if we were to place the hands directly above the shoulders, the shoulder rotators would not have to produce more force as the barbell got heavier.

2) If we include the long head of the tricep, then the answer to the question depends upon the location and angle of attachment of point 6.

Thoughts?

[spacediver]

And to come back to the original idea of this thread, reducing unnecessary moment arms where possible is, in general, a good idea. Unless there is a specific reason to make things more difficult, you will probably be better served by allowing the load to produce the added resistance instead of adjusting the mechanics of the movement.

I won't be able to respond to your other comments until tomorrow, but I'll quickly say here that I agree that reducing unnecessary moment arms is desirable. But not all moment arms are unnecessary. As a principle, it seems that moment arms between COM and midfoot should always be minimized, if not eliminated. But other sets of moment arms exist in a balance with each other. For example, we could theoretically reduce the moment arm between bar and knee joint, but this would have other consequences, one of which is increasing the moment arm between bar and hip joint.

[Satch12879]

First, an aside. Does anyone else own a copy of Beer and Johnston, Vector Mechanics for Engineers, Statics and Dynamics? I think it's a great textbook.

Yes, these were my Statics and Dynamics textbooks. I still use them as a professional today. To cap off the set you need Timoshenko/Gere's Mechanics of Materials.

[Tiedemies]

I will make this simple; please refer the your model (the latter picture).

This all boils down to very simple mechanics; the need to look at the net resultant force on point C, when force is applied via 1-2, 3-4 and 5-6.

In the absence of any constraint at point C, the net resultant force at C must be vertical, lest the barbell "dip" one way or the other. The force applied via 3-4 will result in a Force at C that is perpendicular to the forearm. This should be obvious. If the barbell is then to move unconstrained by the grip or other factors, then it will start moving in that direction; the moment it is even minimally displaced from it's position, a moment-arm appears at A, and the muscle at 1-2 needs to compensate, to bring point C into balance. A force applied via 5-6 works pretty much the same as 3-4 (and, in your model these are applied at the same time), except that in the presence of a force at C, it also introduces a slight moment arm around A, because the force is not completely parallel with the humerus (If it were, there would not be any torque at A due to the 5-6 force, but there would be the one caused by displacement of C) Henceforth I call the forces of 3-4 and 5-6 "the triceps forces".

Now *both* these forces have to be overcome at 1-2. Ceteris paribus, the force at 1-2 produces a force at C that comes at an angle to the force caused by the triceps forces; this angle is determined by how much the elbow angle (at B) has opened. It is a bit tricky to give the direction of the forces, but one can do it by assuming B is rigid. If the angle is at 90 degrees, for example, the net force at C by 1-2 is in the direction of the forearm, if the elbow is fully open, it is at 90 degrees.

Let us call the triceps forces F1 and the 1-2 forces F2. This force, mechanically speaking, does work only in proportion to the vertical component of the net resultant force. The total amount of work is constrained; it does not change with the other mechanical parameters of the system; if there is a mass m at c and g is the gravitational acceleration, and the point C moves from (approximately) point A to the apex position, then the total work will always be roughly W= mg*(h+f), where h is the length of the humerus and f is the length of the forearm. (roughly because it doesn't really start at the shoulder, it starts a bit off to the side, and h is not equal to f, and so forth)

If you know the positions of the attachment points for the triceps (3,4,5,6), you still would have to model how the triceps forces are distributed between the long-head and the short-head forces. I cannot really tell. This is another constraint of the system, in that it is impossible to only contract one or the other. Perhaps a reasonable model would be such that
one is always some percentage of the other.

Once you know the proportions of these forces, you can select a trajectory x along which the barbell moves, and calculate the forces F1(x) and F2(x) at every point of x. The vertical is easiest to calculate because there the direction of the net resultant force is known at all times. If the trajectory deviates from the vertical, it gets really tricky, because you have to account for speed and sideways acceleration and all that to find out not only the direction, but the magnitude of the net force (I suspect magnitude cancels out over the integral for several solutions, but I haven't solved this so I can't say for certain; the magnitude definitely vanishes for the vertical path); then, the amount of work done by the forces is the integral of the inner product of F1(x) and the gravity field (which luckily is a constant) over x.

There are of course several pathological solutions to this problem, like the one where you first raise the elbow to the apex, then holding it rigid, you simply use the triceps to raise the bar. One solution is where the forearm is fully vertical at all times, and the "canonical" solution is one where the trajectory is fully vertical. All these solutions will have, so I suspect, a different amount of work done by the triceps and the shoulder muscles. In none of the solutions I can think of, including the vertical, is the work done by the shoulder muscles zero, and I very much doubt there is any solution where the work done by either of the muscles is zero. It may even be, that in all but the most devious pathological solutions, the distribution of the work is the same or almost the same, and depends only on the 8 parameters: The attachment points 1,2,3,4,5 and 6, and the lengths of f and h. (The mass only changes the magnitude, not the proportions).

[spacediver]

Tiedemes, thank you for the awesome reply. Thinking of the situation in terms of forces at C is something i hadn't thought about before, and is pretty cool.

This all boils down to very simple mechanics; the need to look at the net resultant force on point C, when force is applied via 1-2, 3-4 and 5-6.

In the absence of any constraint at point C, the net resultant force at C must be vertical, lest the barbell "dip" one way or the other.

Agreed.

The force applied via 3-4 will result in a Force at C that is perpendicular to the forearm. This should be obvious.

I don't think this is true. The force applied at 4 will produce a clockwise torque at B. This in turn will result in a force at C that is perpendicular to the forearm, as you said.

But the force at 4 also has the potential to produce a counterclockwise torque at A (although whether a clockwise or counterclockwise torque is produced depends upon the angle and attachment location). This, in turn, will produce a force at C. The direction of this force, as you point out, is a bit tricky to figure out, but it's conceivable that this force can sum with the tricep-generated force to produce a resultant force in the vertical direction.

You can almost prove this to yourself with a little thought experiment. Suppose that all the muscle fibers are removed, and all that is left is the barbell, limbs, and joints. Now imagine yourself standing to the right of the contraption, and pressing down on the end of the protuberance near point B. It's easy to imagine that if you press down with the right force, and at the correct angle, you could hold the system in balance, without any aid coming from the 1-2 forces.

The key point I'm making here, and which you seem to have not taken into account in your analysis, is that force generated by the 1-2 muscle is not the only way to produce a counterclockwise torque at A.

A force applied via 5-6 works pretty much the same as 3-4 (and, in your model these are applied at the same time), except that in the presence of a force at C, it also introduces a slight moment arm around A, because the force is not completely parallel with the humerus (If it were, there would not be any torque at A due to the 5-6 force, but there would be the one caused by displacement of C)

If the 5-6 force was parallel to the humerus, why would it not produce a torque around A? I can see how this would be true if the point of attachment (6) was spatially coincident with B. But it's not, which means there is a moment arm between 6 and A.

Now *both* these forces have to be overcome at 1-2. Ceteris paribus, the force at 1-2 produces a force at C that comes at an angle to the force caused by the triceps forces; this angle is determined by how much the elbow angle (at B) has opened. It is a bit tricky to give the direction of the forces, but one can do it by assuming B is rigid. If the angle is at 90 degrees, for example, the net force at C by 1-2 is in the direction of the forearm, if the elbow is fully open, it is at 90 degrees.

I'd agree with this, if we ignored the counterclockwise torque produced at A that can potentially be produced by the forces at 4 and 6.

Let us call the triceps forces F1 and the 1-2 forces F2. This force, mechanically speaking, does work only in proportion to the vertical component of the net resultant force. The total amount of work is constrained; it does not change with the other mechanical parameters of the system; if there is a mass m at c and g is the gravitational acceleration, and the point C moves from (approximately) point A to the apex position, then the total work will always be roughly W= mg*(h+f), where h is the length of the humerus and f is the length of the forearm. (roughly because it doesn't really start at the shoulder, it starts a bit off to the side, and h is not equal to f, and so forth)

Agreed, if the barbell moves in a straight line, then the total work done is independent of how efficient the system is.