Physics Question - ma = mg

[IanB]

To divert a bit, I know (or believe to be true) that if I lift a box fast or slow, the same work is still done. If W = F * h, and F = ma, why would picking up the box faster not use more Work?

Assuming no other forces except Earth gravity and the Force of the box, if I pick up the box 1m in 1s, my acceleration is faster than if I pick up the box 1m in 10s; thus making the Work larger.

Picking up the box faster does use more work. You are (1) doing work to overcome gravity, and (2) doing work to accelerate the box.

Usually the work to accelerate the box cannot be recovered and goes to waste (like, for example, heat in the brakes of your car).

The ideal minimum work to overcome gravity usually only occurs if you pick up the box infinitely slowly. In thermodynamics this is often called a "reversible" process. The box is never moving at any measurable speed, so you never do any excess work to accelerate it.

Conceptually, you could get the work of acceleration back again (like regenerative braking), but this would never be 100% efficient and you always lose some energy in such a process.

[Nominal Animal]

If W = F * h, and F = ma, why would picking up the box faster not use more Work?

Because regardless of how long it takes, lifting a kilogram of weight in standard gravity of g=9.80m/s² upwards one meter requires one Joule of energy.  (1 J = 1 kg m s⁻².)

Remember, the a in W=mah is not the acceleration of movement, but the acceleration due to gravity against which you are doing the work.

In physics, you cannot do the old scam of paving a driveway without agreeing about it with the owner, and then charging them for it.  W does not describe *how* you did the work, only the *amount* of work.  The forces you choose to apply to do work against gravity is your choice; they only affect whether a specific end result can be achieved, and how long that will take, not the work needed to achieve that specific end result.

Consider, for example, that instead of lifting the 100 kg box we discussed in previous posts by hand, you use a mechanical hoist.  What is, exactly such a hoist?  Why, it is nothing but a mechanical device that acts as a force multiplier.  There are many other such devices; hydraulic cylinders being among my favourites.  Just by dint of their geometry, a force applied to a control surface, yields a much larger force on the output surface!  And other than friction and mechanical wear, they spend no energy to do so!  What magic is this?

Well, no magic.  A force in itself is not something we can exploit in any way.  We can let the force do work, and convert or collect almost all of the energy released; or we can spend energy to cause a force to do useful work for us.  A force in itself is not useful; it isn't even real, tangible, in the human sense.

A force has the same utterly annoying feature potential energy has: to use them, a physical change of some sort must occur.  It is not the force or the potential energy itself that we can exploit; only the change can be exploited.  (This is much more fundamental than one might think, and is at the core of why the misguided dreamers dreaming of zero point energy or free energy will never see their dreams fulfilled: they dream of exploiting something that is, rather than something that happens.)

Picking up the box faster does use more work.

Bullshit.  That makes the asinine assumption that the energy spent to accelerate the box is not recovered.  There is no reason for such an assumption; that energy is not "lost", nor does its amount affect the height lifted.

Basically, you just made the claim that because you can do other work at the same time, more work is spent in the original task.

Recovering just about 100% from acceleration is not hard, either.  Converting it into useful energy in a specific form (say, electricity instead of say heat) can be hard; but there is nothing physically preventing such conversions from reaching essentially 100%.  ("Essentially 100%" in this context means that in the human scale, we can approach infinitely close to the limit, even if certain (but not all) such energy conversions can never reach 100% mathematically.)

If you, IanB, amend the claim to a human will spend more energy to lift a box faster, then I obviously agree.  For a human, because of the antagonistic makeup of our musculature –– we have separate muscles for gross movements, and smaller, weaker muscles for fine movements –– the optimum "speed" (with respect to energy expenditure) is not zero, and depends on the physiology of the individual.  We also have basically no facilities for recovering any of the work our musculature has done for later use.

It is rather surprising, given that human is the one land animal species that given enough time, can run down every single other known land animal.  There is no other land animal in existence that does distance running better than humans do.  (Yes, a healthy trained human can run down a horse.  There just aren't that many humans around that can still run the 200 miles or 300 kilometers necessary.)  There have been several publicity stunts, "competitions" between a man and a horse, in the last century or so; but the human runners in these competitions are laughably bad compared to e.g. message runners in ancient Greece or Rome.

(Consider the Mongol Derby, the longest horse (riding) race in the world.  In ten days, you need to ride 1000 kilometers; you do get to switch horses very often, though, as we don't want to kill the horses.  However, the record for a human running 1000 km is under six days for men, and under eight days for women.  No horse can do that.  Look up "ultramarathon"; you'll be surprised at what humans are actually capable of.)

Humans cannot really recover any of the energy their musculature spends, other than a little bit of the waste heat, and with mechanical implements.  Passive exoskeletons – collections of braces and springs – are already in use, that can double the range a fully laden infantryman can travel under their own power in any given day without compromising their fighting ability.  They basically recover the work done by human musculature into springs et cetera, and use that energy to augment future movement.  However, a bicycle is even more efficient (as it reduces the work needed in the first place), if there is a road one can use.

[David Hess]

Go back to the beginning and figure out what is actually being measured.  Scales measure *force* and not *mass*.  Weight is force.

Units of pounds are force.  The Imperial unit for mass is the slug, although in the industry it is more useful to talk about pounds force and pounds mass.

[TimFox]

If you apply more vertical force F to the mass M than the weight W of that mass, then when you stop applying that force, after moving through H vertical distance,
the mass is still moving upwards, now being slowed by the acceleration of gravity g.  The net force on the mass while you are applying it is (F - W), which accelerates it vertically so it may rise.
If F = W, the mass does not accelerate, but may rise at a constant velocity (assuming you apply enough extra to overcome air resistance and other factors not included in this discussion).
The kinetic energy of the mass when the force stops will be K = (F - W) x H, and the mass will continue to rise for a further vertical distance
D = K / W, whereupon it stops moving up and begins to fall, accelerated by the acceleration of gravity g.

[SiliconWizard]

Sorry if it has been said, I didn't read quite all.

Your "weight", per se, doesn't just depend on gravity, but also on the fact there is a reaction force (equal but opposed to gravity when you are at equilibrium.)
This is this reaction force (to illustrate: that just prevents you from "falling" down to the center of the Earth) that effectively gives you a "weight". Your weight is not an inherent characteristic that just depends on mass and gravity. It "expresses" itself through a reaction force. Without a reaction force that opposes gravity, you effectively have no "weight".

The most (normally) well known case is freefall. When in freefall, any massive object has virtually no weight.

[TimFox]

You have merely re-stated Newton's Third Law.
Weight is defined as the force of gravity on a mass, which in a uniform gravitational field (i.e., over a region much smaller than the distance to the heavy source of local gravitation, such as within a meter of the Earth's surface) is equal to the product W = m x g, where g is the acceleration of gravity that is essentially uniform over this small region.  On the moon, a similar region will have a different value for g, and therefore the mass will have a different weight.
In freshman physics labs, in order to demonstrate the acceleration of gravity, a useful teaching method is the "Atwood Machine"  https://en.wikipedia.org/wiki/Atwood_machine , where two masses with a known relatively small difference are connected by strings through pulleys, so that the net force is the difference in their weights. 
Yes, if the mass is sitting on a table, the table resists the force of gravity and the mass does not accelerate or move (if the surface is spongy, the mass will sink into the surface until the spring force of the foam rubber balances the weight).
However, if you drop the mass, and it falls freely with nothing resisting it until it hits your foot, it accelerates according to Newton's Second Law, where the accelerating force equals the weight (defined above) and the mass is the mass.  The velocity when it hits your foot is left as an exercise for the reader.

[bostonman]

Many anomalies exist while picking up a box fast versus slow. I'm trying to leave out the human factor in all this, air resistance (which I think has been ignored in the last replies), etc...

Also, my physics background is only two or three courses (physics 1, 2, and 3), so I'm limited in knowledge, but obviously these discussions (and reading) have helped over the years.

Leaving out recouping energy or methods of picking up a box as in holding the box out in front of me. If say a 100kg box is on the ground, I hover over it, pick it up exactly vertical 0.5m, and drop it. The Force on the box is 980N (100kg * 9.8m/s^2), and the Work is 490J (0.5m * 980N).

Now if I picked up the box the same height in 1s versus 10s, the acceleration is still only 9.8m/s^2 regardless? So the fact I've accelerated the box faster or slower has no bearing on how many Joules my body used (using a conversion from Google: 117 calories).

It makes sense to a degree from the chasing the horse analogy. The horse may outrun a human, but it burned the candle at both ends and runs out of energy 100 miles down the road after only 1hr (let's assume the animal and human is unable to take a "breather" and start running again). Whereas the human took 50hrs to run 100miles. In the long run, both have burned the same calories based on whatever factors would go into calculating Joules of a human/animal running - something I'm not asking to be calculated.

I know from years of going to the gym that some people think running sprints periodically for short duration burns more calories than those who run a constant regular speed. But I imagine it would be the same thing. The person next to me on the treadmill will have run 10miles in a short time, but they didn't (in theoretical physics definitions) haven't burned anymore calories than me who took five times longer to run the same 10miles.

Getting back to the box analogy, as someone said, if I now hold the box (maybe say hugging it against my chest), no Work is done. How could this not be true? My arms would get tired, and I'm assuming (again, leaving out the human factor) that my body is burning calories in order to overcome gravity. By the formula Work = Fd, the Work is 0J, and, if it were a motor holding something, the heat lost would be due to wire resistance but no Work done by the motor, but it seems my body would need energy to hold that box still.

[Nominal Animal]

Now if I picked up the box the same height in 1s versus 10s, the acceleration is still only 9.8m/s^2 regardless? So the fact I've accelerated the box faster or slower has no bearing on how many Joules my body used (using a conversion from Google: 117 calories).

You are confusing work and the energy spent by your body; the same mistake IanB made above.

To quote from Wikipedia:
    Work is the energy transferred to or from an object via the application of force along a displacement.

Perhaps the error is my fault, though.  I should have been careful to always use the word only, as in "the work in only lifting a box upwards one meter in constant gravity"; but I just didn't think of properly excluding adding (unrelated/unnecessary) kinetic energy to the object.  (Kinetic energy is not just linear motion, it includes rotation also.)

If we have a box massing ten kilograms, standing still on the floor, with a table top nearby with its top one meter off the floor, and you move the box from the floor to the table, from standstill to standstill, the work done – the energy transferred to the object – is always 10 joules (10 J = 10 kg m² s⁻²), regardless of whether you do so in a fraction of a second, or take a year to do it.  Or even if you do it twice, moving the box back to the floor in between.

This amount, 10 joules, is also the minimum amount of energy you will need to spend to achieve this result.  There is no upper limit, but the difference in energy will not be transferred to the object; it goes somewhere else.  Often waste heat.

If the box starts from standstill, but does not end at standstill, we can increase the work without limit by transferring extra energy as kinetic energy (classically, 0.5 m v², where m is the mass, and v is the change in velocity, for a non-rotating object).

If the box does not start at standstill, but does end at standstill, then the work depends on the direction and speed (and rotation, if rotating) of the box.  We could even end up with "negative" work, ending up gaining more energy from the box by stopping it than we need to transfer to it to move it a meter higher.  A perfect example is using a trebuchet or similar device, and consider the situation where the box has just been "lobbed" as the starting point.  Given a suitable trajectory, the box will end up on the table without any energy transfer, any work at all; the box just converts some of its kinetic energy to potential energy (by moving upwards in the local gravity potential) without any further external assistance.

Basically, if we do not assume standstill at both starting and ending points, we can stuff as much kinetic energy in the non-standstill case as we want, and increase the work (energy transferred to or from the object) without limit.

I so wish I could explain this stuff better!


For simplicity, let's assume that we apply a constant lifting force for as long as possible, then drop the lifting force to zero at the precise moment, so that the box ends up with zero speed at the one meter elevation point.  We ignore friction and buoyancy due to air, and both initial and final positions have zero velocity.  No rotation either; the box has the same kinetic energy at the end that it had at the start.

During the lifting phase, the acceleration (positive upwards) of the box is a-g > 0, where g is the (downwards) acceleration due to gravity, 9.80 m/s², and a = F/m, F being the lifting force applied and m being the mass of the box (10 kg).  During the coasting part, the acceleration is -g.

When a constant acceleration a is applied for duration t, the change in velocity is at, and change in position is vt+0.5at², where v is the initial velocity when the acceleration started.

When a force acts on an object, the integral of the force (vector) over the interval, describes the linear momentum imparted to the object by that force during that interval.  This is called the impulse.  Its units are Newton seconds (1 N s = 1 kg m s⁻1).

(Remember that the box standing still on the floor has two major forces acting on it: one is gravity, and the other is the static force of the floor resisting deformation.  Their impulses cancel out, whenever the box stays in the same location.  So, if one does not consider all forces acting on an object, the impulses due to only some of the forces acting on it aren't really relevant in the real world: it is like measuring the velocity of a car with respect to an aeroplane flying overhead.  By picking a suitable subset of forces, you can make the impulses whatever you want, just like you can make said car velocity whatever you want by picking a suitable aeroplane as the reference.)

I suspect, but have no proof, that the chemical energy spent in human musculature is, as a first approximation, roughly linearly dependent on the impulse the forces generated by those muscles, for a relatively wide (but not full) range of human muscle power.  In the absense of a better model, I'll use this.

So, to minimise the human energy spent, we don't try to minimise the force, but the integral of the lifting force over time; i.e, we minimize the impulse due to the forces the muscles manage to generate.

If you work out the math, and define the upwards lifting force F as F=kmg, i.e. as k times the force due to gravity (and k>1), then the lifting impulse is sqrt(2gm²k/(k-1)); the force kmg is applied for duration sqrt(2/(gk(k-1)).  For k>1, this impulse is a monotonically decreasing function, so in this model, the harder the constant force you can apply, the less chemical energy your muscles will burn.

For k=10, and g=9.80m/s², the duration is just 50 milliseconds (0.048 seconds); for k=5, 100 milliseconds (0.101 seconds); for k=2, 320 milliseconds (0.319 seconds). (The k scale factor is nice, because then the mass m drops out from the duration.  The units do not seem to match, but that's just because the formulae has the fixed height to be lifted upwards, one meter, "hidden" in the constants.  If you write it out using proper dimensions, it does work out correctly.)

It does make sort-of intuitive sense – at least, if you've ever compared the effort needed to do say pull-ups slowly versus fast.  I can do a dozen fast pullups (not the swingy ones – those are cheating via angular momentum –, nor using your legs to kick in the air to get some extra help; just fast but proper ones), but ask me to stay stationary midway, and I'll drop in a few seconds, much faster than it would take to do those pullups.  I think.  Need to verify at the gym (on separate days).

However, this result is not because of work done to the box, i.e. energy transferred to the box, but because we assume the energy spent in the muscles is directly proportional to the impulse applied (by the forces generated by those muscles, as measured where those limbs grasp something).

[CatalinaWOW]

This longish thread has shown how much meat there is in even simple problems.  And how important it is to frame the question properly.  Things which seem obvious to those with long exposure can be either quite confusing or devilishly simple depending on presentation and/or point of view. 

Pertinent topics which have been touched on are the relation of weight and mass, the inability to distinguish (at non relativistic speeds) the difference between a fixed acceleration and being in a static gravitational field and the differences between work and power and the difference between power stored in a system through changes in it's gravitational potential and work elsewhere (muscles and the like).

Just think of all the complexity a hard problem would bring to the table.

[Nominal Animal]

Classical physics is easy and straightforward.   Until you try to truly understand it, or to apply your understanding to solve a problem.  It's only easy if you've already worked it all out before.

Quantum physics is where you need to discard whatever vestiges of intuition you have left, because intuition and its odd evolved rules it operates by becomes a severe hindrance for proper understanding here.

String theory is just mathematicians having a very long running joke on physicists.

You can construct similar sequences for any topic you choose.  If you disagree with my three choices above, just replace them with something more appropriate in your experience; the difference is just due to the differences in our personal experiences, and sequences that apply to me probably do not apply to you and vice versa.

Learning is so darned subjective.

[RJSV]

THANK YOU, DAVID HESS, for mentioning the British units 'SLUG', as mass, (with the corresponding British unit of force I.E. 'POUNDS').
   Now my concerns, today, are to merely point out how this thread is bit wild, maybe, for my taste. I mean, you have one person mixing up the two main standard systems, (English and Metric standards).
You've got someone stating that speed is same a mass, I think he got mixed up, thinking inertia increase is same as mass increase, plus some other (yahoo) stating or mis-stating that relativistic effects are major in EVERY situation. So, yeah, mass increases with speed, but...arrghph not practically speaking.
  Like I've tried to teach, I think historically, that 'g' came into the picture, in f=mg as 'no suprise' rather than a 'direct' acceleration! The units match, and heck, it's another 'gravity related' deal.
   On that subject, probably Newton got to f=ma through some observations / measurements, but also by way of the universal force equation, that is mass x mass divided by separation distance (squared).
   
   So...anyone with a history context would help clarify these questions, as I am merely speculating, on Newton's exploration.

But what I DON'T see is little 'tendrils' of acceleration IE 'g' shooting up past that stationary weight. 'g' is just in the force-related EQUATION, dude.
Word salads. AND then there is the fellow that stated "words don't matter anyway, they are just there to describe stuff."
Uh, ok, so if I owe you money, $100, let's just be loose, and call it '100 cents'.

[JohnnyMalaria]

Go back to the beginning and figure out what is actually being measured.  Scales measure *force* and not *mass*.  Weight is force.

Units of pounds are force.  The Imperial unit for mass is the slug, although in the industry it is more useful to talk about pounds force and pounds mass.

This is not true. It depends on which system of measurements you are using. e.g., the British Gravitational system does define pound as a force but the FPS system defines it as a mass. The UK legally defines a pound as 0.454 kilogram and, therefore, is defined as mass.

I was educated in the UK in the 70s and 80s. Pounds were always units of mass and pounds-force (lbf) specifically used to indicate force.

[Nominal Animal]

I use either SI units, Planck units, Hartree atomic units, or bananas.

The last one covers all those slugs and pounds and pints and stones and wheelbarrows and such quite nicely, thank you.

I even use decimal degrees instead of arcminutes and arcseconds, to the exasperation of some of my astronomist friends.  (I also like to "accidentally" use the word "logy" instead of "nomy", just to see if they're still listening.)

If you use inch ounce-force to measure torque, does the scale depend on whether it is a fluid (fl. oz) or not (notfl. oz)? And what does australia (Oz) have to do with this, anyway?

[T3sl4co1l]

What, you don't use bananas for angle too!? 0/10 disappointed.

Tim

[TimFox]

To avoid ambiguity, use “oz av” for small forces and “fl oz” (a pronounceable acronym) for volume.  1.5 fl oz is roughly “two fingers”, except here in Chicago where we measure with vertical digits.

[SiliconWizard]

You have merely re-stated Newton's Third Law.

Yes, it all comes from Newton's laws. Nothing special. It's just that the notion of weight is something that is actually not all that intuitive.

However, if you drop the mass, and it falls freely with nothing resisting it until it hits your foot, it accelerates according to Newton's Second Law, where the accelerating force equals the weight (defined above) and the mass is the mass.  The velocity when it hits your foot is left as an exercise for the reader.

I'm not sure what an "accelerating force" is. There is force, and there is acceleration.

And actually, in freefall, the weight has absolutely no effect. Acceleration is the gravity (if you simplify things of course and suppose there are no friction forces due to air - hence why true experiments must be done in vacuum), and it doesn't depend on mass whatsoever. In turn, the speed doesn't depend on mass either. That's something not really intuitive and that many people, even educated ones, have a hard time with.

[TimFox]

An accelerating force is that in Newton's First Law, usually stated now as "an object at rest will stay at rest, and an object in motion will stay in motion unless acted on by a net external force".  In the original, "Corpus omne perseverare in statu suo quiescendi vel movendi uniformiter in directum, nisi quatenus a viribus impressis cogitur statum illum mutare".  See "Newton's Principia for the Common Reader" by S Chandrasekhar, available at https://oiipdf.com/newtons-principia-for-the-common-reader
The Second Law gives the result of such an external force, usually stated now as "F = m a," where a is the resulting acceleration from that force.  Actually, Newton's original statement translates as "F = dP/dt", where P is now called the "momentum" (original "quantity of motion"), and this version works better in Special Relativity.  A discussion of this in https://blogs.bu.edu/ggarber/archive/bua-py-25/newtons-second-axiom/  clarifies this:

"Lex II: Mutationem motus proportionalem esse vi motrici impressae,
et fieri secundum lineam rectam qua vis illa imprimitur.
In translation (by Andrew Motte) this becomes:
LAW II.
The alteration of motion is ever proportional to the motive force impressed;
and is made in the direction of the right line in  which that force is impressed.
Newton then goes on to clarify:
If any force generates a motion, a double force will generate double the
motion, a triple force triple the motion, whether that force be impressed
altogether and at once, or gradually and successively. And this motion
(being always directed the same way with the generating force), if the body
moved before, is added to or subducted from the former motion, according
as they directly conspire with or are directly contrary to each other ; or
obliquely joined, when they are oblique, so as to produce a new motion
compounded from the determination of both."

I believe that Newton assumed that the inertial mass and the gravitational mass are equal, and this became a cornerstone of Einstein's theory.
Once again, I suggest considering the Atwood Machine, where gravitational acceleration and inertial masses form an interesting demonstration.

[RJSV]

Sorry TIM, I disagree with the use of the term 'MOTION': That's too loose, why not try use the direct terms, Changes in position= velocity (or called speed or rate) and if accelerating/decelerating then you have another 'rate', of change, that of velocity change.
To drive a point home I would ask: "...So then, what are the units of this 'MOTION' thingy that you mentioned ?"
  Are you going to reply: "...Why.. it's feet per second
..of course...".
But that's called VELOCITY.!!!  Don't misunderstand, this is a friendly, and voluntary activity,JUST BUSINESS
and so I gotta speak out. The sarcasm is only all in fun. (Refer to Nominal animal previous post)
   I'm going to re-name VELOCITY: How about 'Wiggles'? ... Or BANNANAs per sector!
Wordsaladwordsalad. (thanks for your patience)

[TimFox]

In terms of Newton's work:  "quantity of motion" is what is now called "momentum".  Terminology changes through the centuries.  In non-relativistic mechanics, P = m v , where the vectors P and v are, respectively, the momentum and the velocity, while the scalar m is the mass.  In Principia, if I remember correctly, Newton defined the mass as the product of the volume and density, but modern usage defines the density as the mass divided by the volume.  There are a lot of circular definitions in the history of science.

[Nominal Animal]

What, you don't use bananas for angle too!? 0/10 disappointed.

Next time it comes up, I swear I'll say "it's two bananas short of ninety degrees".  Let the mayhem commence.

[RJSV]

'MAYHEM'?, 'BANANNAs'? I'm stealing all my jokes, from NOMINAL ANIMAL!

   O.K. : BANANNAs it is:
   (You) are pretty sure that 'Banannas per second' is a ridiculous term, but substituting 'loose' descriptors for actual precisely defined terms of PHYSICS is maybe OK. No, no,no:
   'MOTION' is just a general umbrella term that could mean either, (or both) SPEED or ACCELERATION.
I think what got my attention, was a guy, back a ways, who posted that a parameter of ACCELERATION, well then that had to be a 'velocity' simply because he spotted the equation term: "PER SECOND". Must be a speed, right? ...  No one will ever know!
Heck, even the term 'Specific Meaning' has no weight here.
SOOOO, For this reasons, I vote
       NOMINAL  ANIMAL for PRESIDENT

[TimFox]

Try reading what I actually wrote.  I did quote Newton’s original statements for historical purposes, but I carefully used the terms acceleration, momentum, and velocity when writing in my own voice.  Modern versions of Newton’s theories are easier to understand and, where they are relevant (non-quantum and non-relativistic) are internally consistent and an accurate description of motion with its various attributes.  For further discussion, please see any textbook on freshman physics.

[RJSV]

Yes, Tim, I was a bit out of line by posting in generalities (like some people seem to 'smoosh' together a bunch of tech jargon, without accurate definitions.) But your latest posts, I like, just maybe a bit of time lag, as I comment addressed back a bit ( posts from couple weeks ago.)
   But, still, not you but one recent post was directly confusing MOMENTUM (increasing) and saying that was 'WEIGHT' increasing (that being, again, momentum as mass X speed.) That's a pretty simplistic and not only wrong, but mis-leading (due to the fact that 'mass' is in the mix.
   That (other) person's posts appear to me like an ignorant summary, saying 'No one will ever know' IE. the real use of equations and concrete/specific definitions.
Heck, I still see that (helpless) expressed, in the discussion of 'fluid ounces' and how that somehow complicates things, impossibly. Just: "no one will ever know".  Yeah and 'Einstein'?!!! Gotcha there.
  I think I will include 'Einstein'+ in everything I post.
Gives it an air of legitimacy.
Einstein !  Ha ha, gotcha again.
AND Thanks, I mean no insults here.

[CatalinaWOW]

The disparagement of traditional systems leaves out poundals and a few other variants.  These mass and force quantities are fine if kept straight, but the one real advantage of the metric system for anyone of average or above intelligence is that the metric community has done a better (not perfect) job of keeping them straight.  Dimensional analysis is always useful.  It was nearly mandatory  in the traditional units world and still when referencing older texts.  Even metric folk can miss out when using an older text that nonchalantly uses physical units expressed in cgs or slightly obscure units like wavenumber.

[David Hess]

Go back to the beginning and figure out what is actually being measured.  Scales measure *force* and not *mass*.  Weight is force.

Units of pounds are force.  The Imperial unit for mass is the slug, although in the industry it is more useful to talk about pounds force and pounds mass.

This is not true. It depends on which system of measurements you are using. e.g., the British Gravitational system does define pound as a force but the FPS system defines it as a mass. The UK legally defines a pound as 0.454 kilogram and, therefore, is defined as mass.

I was educated in the UK in the 70s and 80s. Pounds were always units of mass and pounds-force (lbf) specifically used to indicate force.

My point is that it is very easy to get into trouble with the unit pound because it can mean force or mass depending on the system, and literature will not always be explicit.