EXACTLY. So if you were carrying YOUR weight on your back you'd naturally lean forward.
This is
absolutely the wrong way to think about the problem and why such problems must be solved by appropriately trained and licensed people when the public safety is at stake. You would kill people if you tried to do this work. This "ratio" business you are stuck on is irrelevant. I've explained several times how it is irrelevant. Expressing the weight of the backpack as "your weight" or some ratio of "your weight" does not properly formulate the problem.
Imagine a person just standing there with no backpack. If that person weighs 180 lbs on Earth, that person would weigh 30 lbs on the Moon. But since that weight is directly over his feet in both cases, it's a simple case of axial loading. There's no part of the load that can be considered eccentric, therefore no moment.
Now wave a magic wand and suddenly a backpack appears on the person's back. That doesn't change the axial loading part of the problem. So the weight that causes that axial loading doesn't factor into any balance equation. Therefore it doesn't matter whether it's Earth weight or Moon weight. We simply disregard the man's own weight. It's not part of the equation.
Since we started doing this in EES we'll have to continue. The moment arm for the backpack is about 1 ft. That's factoring in the average thickness of the human torso and the known thickness of the PLSS. If the PLSS weighs 150 lbs on Earth, that's a moment (or torque, if you prefer) of 150 pounds-feet (lbf⋅ft). It literally doesn't matter what fraction that might be of the person's weight. The person's weight is being properly borne axially and therefore it does not contribute to any tipping moment.
That same PLSS would weigh only 25 lbs on the Moon. Again, it doesn't matter in the slightest what percentage that might be of the axial-only loading in lunar gravity. Only the eccentric load contributes to the moment. The moment here is 25 lbf⋅ft.
That is absolute.
How that moment applies depends on how you model the response of the man. People are squishy, so you can model it as a bending moment acting at the man's shoulders. The man's response would be plastic deformation backwards.
But you are specifically concerned with a tipping moment, which is the balance problem. We therefore consider the astronaut to be a vertical rigid body in gravity attached to a pivot at the feet. Now we need to consider the problem as a triangle. You have an eccentric gravity load applied at a certain shoulder height—say 5 feet. The 1-ft moment arm doesn't change with gravity. The resulting hypotenuse—5.1 ft—doesn't change with gravity. All these are
geometric properties of the problem that maintain their relationships irrespective of gravity.
But we have to decompose the moment (or torque) along that hypotenuse to find out how much of the eccentric gravity load acts perpendicular to the hypotenuse and therefore the tipping moment. Again, the astronaut's axial load (his own body weight)
is not part of this equation. Gravity always acts down, but the part of it acting along the hypotenuse changes as the topple happens and the hypotenuse becomes more horizontal. Since that's a more advanced topic, we start with the initial moment.
The hypotenuse is 11.3º off vertical, so the portion of the eccentric load W acting perpendicular to it, F, is
F = (sin 11.3º) W = 0.196 W
Solving for W = 150 lbs yields 29.4 lbs. Acting over a 5.1 ft moment arm gives 149.94 lbf⋅ft.
Solving for W = 25 lbs yields. 4.9 lbs. Acting over a 5.1 ft moment arm gives 24.99 lbf⋅ft.
24.99 is much less than 149.94, therefore the tipping moment is much less.
But of course as an astronaut begins to topple, that axial load of his own weight becomes an eccentric load as soon as he is far enough off vertical. That moment arm is 3.3 ft long (to his body center of gravity), and the value acting over it is the astronaut's weight. The problem is set up the same way, for some given angle of astronaut lean. If the astronaut has tipped 10º off vertical, the eccentric load of the backpack would be computed from 11.3º (the initial eccentricity) + 10º total system lean. The newly eccentric load of the astronaut would use just the 10º lean, the astronaut's weight alone
in lunar gravity, and the moment arm of 3.3 ft. Then you combine the two. The entire problem literally scales by the 1/6 gravity ratio and gives you 1/6 as much tipping moment no matter what values you use for the astronaut mass and backpack mass.
Now the engineer will soon realize that if everything is a rigid body, you can reformulate the whole problem by recomputing the combined system's total new center of gravity and just do one moment arm calculation. That's more involved for irregularly shaped bodies, but there are straightforward (though tedious) ways to accomplish it. When you get into it, you realize that the mass-at-a-distance computations start to look a lot like what we did above. The same math that lets us decompose different moments according to system components gets folded into the total system mass properties calculation. The latter is how we do it in practice.
Done. You're wrong.
