Don't have a handy enginerd to talk to face to face, so this will have to suffice. Plenty of smart folks to interface with...

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Okay, velocity and acceleration are easy, and I think I said I already understand those, but I guess you included them just for completeness?

Velocity is speed PLUS direction. IE, moving 9.8 m/s DOWNWARD. Acceleration is how velocity changes, or how much velocity is added per second. That I get, pretty simple algebra, methinks.

However:

"Velocity is in units of m/s and its velocity at any time during the fall is given by v=at where a is the acceleration of gravity (9.8 m/s^2) and t is the number of seconds. See how the s terms cancel out?"

Actually, no, I don't see ANYTHING cancelled out of the equation... Probably 'cause you didn't show your work. Methinks you'd get a c- from any algebra teacher for that. Hehe...

But, I see what you MEANT to say. velocity = acceleration times time, where the velocity at any point in time can be obtained by knowing the rate of acceleration and the time an object has been accelerated:

V = A * T
or
M/S = M/(S^2) * S

OR

A = V * 1/T
or
M/(S^2) = M/S * 1/S

"Multiplying a mass by a length and dividing the result by a time seems incomprehensible ..."

Yup, sounds about right... However...

"As the jug falls, its mass stays the same, but it's velocity is increasing so we describe the jug with a new quantity called the momentum."

This part I wonder about, particularly, the mass staying the same... If it's being accelerated, isn't energy being imparted from the gravitational field. IE, work is being done on the jug to increase or change its velocity, which consumes energy (even if only nominal amounts). Doesn't conservation mean that the energy has to go somewhere, IE into the jug in the form of, what, kinetic energy? And Energy = Mass * C^2. So, it seems to me that as the jug is accelerated (imparted energy) it would increase in mass (even if only a nominal amount), yes?

Wasn't this one of the effects of the E = M * C^2 equation? Hence as velocity -> light speed, mass -> infinity (been a while since I worked with 'limits')? (The whole Mass->infinity thing doesn't really make a whole ton of sense, but hey, we work with what we've got, for the moment, despite its flaws).

"Newton's law defines a force as a mass times an accelleration, and since we know the mass is 4kg and the accelleration of gravity is 9.8 m/s^2 we know the force of gravity on that jug is a constant 39.2 m/s^2, or 39.2 Newtons. Unlike velocity, accelleration and momentum, the unit of force has a name."

Well, the fact that it has a name, is neither here nor there. As far as I'm concerned, naming is of no relevance. We could call Kilogram-meters "Bobs" or "Susans" and it still wouldn't make them make any more sense... The naming has nothing to do with what's actually being described. One of those language-y things that has no relevance to the actual maths, as you so nicely pointed out. Language is imperfect. Language is an approximation. Even math is an approximation to the best of our ability.

Technically, if one wanted to get silly one could even say that since math is a language, science-fiction can be written in math. IE, complete gibberish can be written in math that doesn't accurately decribe OUR reality, but only some hypothetical universe (frankly, I think that divorcing physics from observation and saying that only pure math can describe the universe is silly; any language can be written well or written poorly based on the thought processes that go into it, math is no better or worse than any other language that way, and since it's often very abstract it may end up describing nothing useful even if both sides of the equals sign match up).

And again, isn't the above example/equation you mentioned a problematic statement (exemplary of what NOT to do and how silly errors lead to bad data)? I think you cancelled one too many terms? 'Cause we're multiplying a mass times an acceleration. Where did the mass go in the final number you proferred? Mass doesn't cancel with meters/second^2... Shouldn't that be 39.2 Kg-m/s^2? Congratualtions. You've just written some good science fiction in "math."

IE, 4Kg * 9.8m/(s^2) =/= 39.2 m/(s^2)

Rather:

4Kg * 9.8m/(s^2) = 39.2 Kg-m/(s^2)

If that's not right, how does the "mass" cancel out when there is no other mass to cancel with from the acceleration equation? Are you sure YOU learned basic algebra? *wink* j/k, I'm sure it was just a typo. Ri-iight?

On an aside, we still seem to not really understand what a mass-meter is (just like in momentum; except this time we have a unit for it; whooppee! We named something we don't understand! Can't we call kilogram-meters Bobs? C'mon. Pretty please?). And this "term" still seems slightly incomprehensible. I mean what does mass times distance really MEAN? And divided by time? Heck, even mass over time...

Well, mass over time I could maybe understand, if we're saying the mass CHANGES over time (kind of like distance; instead of changing position, we change massor energy in the system)? Kg/s I suppose we could understand if we knew how the mass was changing over time. But we'd need to know at what rate the mass changes over time. But Mass * Distance? That just seems weird. I mean what does it MEAN?

Now, if we look at the parts, divided out:
Force = Kg * D/(S^2)
vs.
Force = Kg/S * D/S

Individually these make sense, I think? Change in mass over time, and change in distance over time. But, again what does it MEAN to multiply them together? I guess we'd have to know how both change over time. We've said that the distance over time is, what, 9.8 m/s, so, what would the change in mass be? I'm wondering, can we use the first equation above in the example you used? Then rearrange the terms?

IE, you said 4Kg dropped, accelerating 9.8m/(S^2).

So could we fairly say:
Force = 4Kg * 9.8m/(S^2) {?}

If so, can we rearrange the equation as above to make more sense of it:

Force = 4Kg * 9.8m/S * (1/S)
Force = 4Kg/S * 9.8m/S {?}

Obviously, the final answer if we mash it all together still comes out to 39.2Kg-m/(S^2), but we get slightly more meaningful pieces with which to work...

So, what does it mean to multiply a velocity (change in position) of 9.8m/S times a change in mass of 4Kg/S?

Well, we've already stated that mass is a resistance to acceleration, or conversely (when in motion) a resistance to deceleration. And Force is what, momentum over time? (Mass * (Distance/time)) * 1/time.

So, what is all this telling us? Well, to quote a famous phrase momentum has been very generally decribed in terms of "Objects in motion tend to stay in motion, unless acted upon by an outside force."

IE, the mass will continue at the same velocity until acted upon by something to change its velocity (speed or direction).

Don't know if that shed more light on the whole change in mass * change in distance bit or not. It's still difficult to see how distance influences or relates to mass and how we can even consider multiplying them. Division I could understand, that would be a measure of, what, weight, I guess, or mass-by-volume. But what does multiplication mean in terms of mass and distance?

I'm wondering if we can modify the equation again, slightly (this may be a dead end, but let's go with it). So, we had:

Force = Kg * D/(S^2)
vs.
Force = Kg/S * D/S

So, I was wondering what does multiplication mean in terms of distance and mass. But then I remembered the rule that division and multiplication can be interchanged by flipping the thing we're working with. IE, rather than multiplying by D, we could divide by 1/D. And we know that division means taking whatever we're dividing and dividing it over a specific area. Then it becomes almost more like an area or volume measurement. IE, some unit, per some other unit... I mean, heck, if we're playing with math, why not, right? Again, I don't know what it means exactly, but we could certainly rephrase the equations:

Force = Kg * D/(S^2)
Force = Kg * D * 1/(S^2)
Force = Kg * 1/(1/D) * 1/(S^2)
Force = Kg/(1/D) * 1/(S^2)

So, what does this actually MEAN? Not quite sure. As distance goes up, inverse of distance goes down, Kg over inverse goes up. Basically, it's just like multiplying... *Grumble* If one said Y=1/X, as X -> 0 Y -> infinity and as X-> infinity, Y -> 0. Odd function. Probably not especially useful...? Ahh well, back to the drawing board.

Still not entirely clear on what i means to multiply a distance and a mass... Since they're NOT like units.

I mean it's nice that in some manner they cancel out of equations at eventually (or just get hidden so we don't notice them anymore). I'm just not clear on how they interact... I mean what we're saying when we multiply a kilogram by a meter. Somehow, that just seems gibberish-y to me.

But, somehow it seems to me to make more sense if we're saying something akin to, the longer it drops, the more energy is imparted, the more momentum changes, and the more mass it gains due to work performed on it, and conservation laws. Is that a fair assessment? Maybe I'm just having trouble with the math of it. But that seems to me to be what's going on.

IE, the more energy something has, the more mass it gains. By doing work on something, energy (thus mass) is imparted increasing velocity (even if only nominally). So, an object accelerated in a gravitational field gains mass. Yes?

Now, if an object were imparted energy in such a way as to give it spin, would it still gain mass? IE, if work is done on it, to cause it to rotate faster and faster, does it gain mass?

As I'm thinking about this, I'd want to say that as it spins it will be forced apart by centrifugal forces, reducing the mass per volume slightly. If we were to assume that it didn't expand, but remained as say a sphere in the same positino and with the same radius, but it was moving faster and more energetically, we could rightly say that it had more mass, yes?

Big if, of course. But I was also wondering, does that also make mass relative? 'Cause what if you were in the frame of reference of the spinning object such that you thought you were stationary and the universe around you was moving faster and faster (but in the opposite direction)? IE, an equivalence of motion problem from different frames of reference... If I were on the inside of the sphere, thinking I was stationary and the universe was moving. But someone on the outside in the universe saw the sphere, thought they were stationary, and the sphere was rotating, who's right? And how would yo ucalculate mass? IE, from the sphere's point of view, or from the universe's point of view? Obviously there's a huge perspective/scale difference, yeah? If one thinks that it is the sphere that is stationary, and the universe that has increased in speed, that would seemingly throw off all the calculations, wouldn't it? 'Cause it would take far more energy to cause the whole universe to move than the sphere... So, is there any easy way to resolve this seeming perspectives paradox? If you're on the side of the sphere, it seems like the universe just got some massive magical kick of speed requiring untold amounts of energy. Whereas if you're the universe, the energy to cause a tiny sphere to rotate is miniscule...

This seems... Weird. And I've gotten a bit off track here. *Sigh*

So many issues, so little time.

So, to recap. velocity equal direction and speed (meters/second). Fine.

Acceleration/deceleration = change in direction/speed

Force = mass * acceleration or mass/time * distance/time (depending how you arrange the equation).

But like momentum, force seems weird 'cause how do you multiply disparate units and get any meaningful result? What *IS* a kilogram-meter? For that matter, what is a kilogram? What is the physical definition of a kilogram? Do we even know what "mass" actually is, or only its effects? The answer becomes rather obvious that we only know its effects and not what actually underlies it.

Time to inject a bit of ironic wikipedia humor...

<<<The kilogram was originally defined as the mass of one litre of pure water at standard atmospheric pressure and at the temperature at which water has its maximum density (3.98 degrees Celsius). This definition was hard to realize accurately, partially because the density of water depends slightly on the pressure, and pressure units include mass as a factor, introducing a circular dependency in the definition.

To avoid these problems, the kilogram was redefined as precisely the mass of a particular standard mass created to approximate the original definition. Since 1889, the SI system defines the unit to be equal to the mass of the international prototype of the kilogram, which is made from an alloy of platinum and iridium of 39 mm height and diameter, and is kept at the Bureau International des Poids et Mesures (International Bureau of Weights and Measures). Official copies of the prototype kilogram are made available as national prototypes, which are compared to the Paris prototype ("Le Grand Kilo") roughly every 10 years. The international prototype kilogram was made in the 1880s.

By definition, the error in the repeatability of the current definition is exactly zero; however, in the usual sense of the word, it can be regarded as of the order of 2 micrograms. This is found by comparing the official standard with its official copies, which are made of roughly the same materials and kept under the same conditions. There is no reason to believe that the official standard is any more or less stable than its official copies, thus giving a way to estimate its stability. This procedure is performed roughly once every forty years.

The international prototype of the kilogram seems to have lost about 50 micrograms in the last 100 years, and the reason for the loss is still unknown (reported in Der Spiegel, 2003 #26). The observed variation in the prototype has intensified the search for a new definition of the kilogram. It is accurate to state that any object in the universe (other than the reference metal in France) that had a mass of 1 kilogram 100 years ago, and has not changed since then, now has a mass of 1.000 000 050 kg. This perspective is paradoxical and defeats the purpose of a standard unit of mass, since the standard should not change arbitrarily over time. The philosopher Saul Kripke elaborated on the philosophical implications of this kind of problem, referring, however, to the then-current definition of the metre in terms of an artifact, a choice which was later dropped.>>>

So, basically what they're saying is that the "international standard" of "mass" has gotten lighter over the last 100 years. Heh. That's awesome. It just shows that even our most precise calculations and models are error prone and relative. So, really we don't know what mass is, if we can't even decide on a standard for it that DOESN'T change over time. Aside from which, the original definition apparently included a circular dependency on itself (via pressure, which apparently includes mass as a variable?)... Heh. Wonderful! I love irony.

No wonder we can't actually express anything meaningful. Our standards are relative and sometimes rely on circular logic to prove themselves. I don't mean to indict science, but seriously, one has to wonder just slightly...

Still, the rearranged equation with mass over time multiplied by distance over time seems to be interesting. Though the brain seems at a slight loss as to wrapping completely around the idea as to what it means, specifically.

One might say that if distance covered is zero, then velocity is zero. And the change in mass is therefore zero as well, 'cause anything times zero is zero, yeah? Well, really, if either mass or velocity hits zero, the whole equation hits zero.

So, if it has mass, but no velocity, there's no force, yeah? And if it has velocity, but no mass (is that even possible?) there's no force? And if there's no mass and no velocity (again, is no mass even possible?), it probably doesn't even exist.

If it has a little velocity/acceleration and a lot of mass, it still seems to exert a lot of force. And if it has little mass, but a lot of velocity/acceleration, it also exerts a lot of force...

So for example, it seems like something with 5 Kg of mass increasing in speed 10,000 meters/(second^2) would wield the same force as a 10,000 Kg mass accelerated at 5 meters/(second^2). Yeah? 'Cause when all the maths are done you still get 50,000 Kg-m/(s^2)... Weird, but okay. ;o)

Anywho, I'll have to ponder a bit more to figure out what the heck is going on...