Reconcile poundforce w/slug and lbmass

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The discussion revolves around the confusion between pound-force (lbf), pound mass (lbm), and their relationship through acceleration due to gravity. A key point is the incorrect assertion that 1 lbm multiplied by 32 ft/s² equals 32 lbf, which contradicts established definitions. The correct relationship involves recognizing that 1 lbf is equivalent to 1/32 slug multiplied by 32 ft/s², leading to the conclusion that 32 lbf equals 32 lbm times 32 ft/s². The conversation also touches on the complexities of unit conversions and the historical context of measurement systems, emphasizing the importance of coherent units in scientific calculations. Ultimately, the discussion highlights the necessity of understanding these conversions to avoid errors in engineering and physics applications.
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Homework Statement
Def of lbf=32lbm*1ft/sec^2=1lbm*32f/s^2 So, 1lbf=1/32slug*32f/s^2. But, 1lbm*32f/s^2=32lbf.
Relevant Equations
def of lbf=32lbm*1ft/sec^2
Where's the discrepancy?
 
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Roady said:
But, 1lbm*32f/s^2=32lbf.
No it isn't. That contradicts the previous definition.
 
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precisely where is the contradiction? but the product means something, 1lbm*32f/s^2=32lbm*ft/s^2. That is the unit of force! Explain yourself. This is not one pound of force.
 
Roady said:
But, 1lbm*32f/s^2=32lbf.
No. See if you can spot your mistake.
 
I've pulled this formula inside out. There are only 3 terms. F=ma, a=32.17xxx, i simplified. Even the base definition of lbf=32units of mass/a^2, but is reported as 1lbf. So, i'm not seeing something. If I had, I wouldn't of asked. Take ur time. This is a 45yr problem. I probably don't see a lot of things. Can we straighten this out before proceeding to next issue?
 
Roady said:
But, 1lbm*32f/s^2=32lbf.
That is wrong. The above equation (which I'll call equation A) doesn't follow from the other equations.

Consider your (correct) equation:
1 lbf = (1/32)slug * 32 ft/s^2

Multiply both sides by 32 which gives:
32 lbf = 1 slug * 32 ft/s^2

Since 1 slug = 32 lbm this gives:
32 lbf = 32 lbm * 32 ft/s^2
which is the corrected version of equation A.

BTW, the usual symbol for a 'pound mass' is simply 'lb', not 'lbm'.

Edit - typo'.
 
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The numerical value of rt does not =left. 1lbf= 1/32slug*32f/s^2. Cancel, not multiply 32. Yields,
1slug*1f/s^2 units which is def of pound force, Steve. You just didn't simplify. Nice try.
 
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Roady said:
I've pulled this formula inside out. There are only 3 terms. F=ma
As I pointed out in your other thread, the correct formula for systems of units that are not coherent is ##F=kma## where k is a unit conversion factor. For coherent systems, ##k=1##.

Using pounds force, pounds mass and feet per second squared, ##k \approx \frac{1}{32.17}##

There are three commonly known systems of units involving the pound.

1. Pounds mass, pounds force, feet per second squared

##F=\frac{1}{32.17}ma##

2. Pounds mass, poundals, feet per second squared. One poundal is about 1/32.17 of a pound force.

##F=ma##

3. Slugs, pounds force, feet per second squared. One slug is about 32.17 pounds mass.

##F=ma##
 
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Thank you. I had completely forgot about this constant. I have thermo txt enroute which will supply plenty problems of lb steam into horsepower. This should help me acclimate to lb mass problems. I had resolved to convert lb to slug since i got numerical equity out of that eq. Which is essentially the k. Forgot where that come from. Long distant coursework. Realized i really didn't understand it then and i'd like to fill that hole. This makes me want to get a FORTRAN compiler and relive the glory day's of freshman engineering. The modular nature of C++ is very versitile, but, FORTRAN, you had to have some programming finesse to efficiently 'talk' to the computer. Like expanding a determinant using the program.
 
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  • #10
Roady said:
The numerical value of rt does not =left. 1lbf= 1/32slug*32f/s^2. Cancel, not multiply 32. Yields,
1slug*1f/s^2 units which is def of pound force, Steve. You just didn't simplify. Nice try.
Hmm. If ##x = \frac 1{32}a \times 32b## then:
i) what is ##32x## ?
ii) what is ##\frac x{32}##?

EDIT. Since @jbriggs444 has resolved your problem, feel free to ignore this!
 
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  • #11
Roady said:
Thank you. I had completely forgot about this constant.
I love it when a discussion ends in agreement.

One reason that many of us prefer to work with SI (the metric system) is to avoid unit conversion difficulties like this. In some fields, one goes even further to a geometrized unit system.
 
  • #12
Incoherent systems: it's lost pieces of information like that this and law of tangents wouldn't of happened; amongst other things. Hilarious wracking my brain for want of the lost piece. Most text dont address British Engineer Sys, or Eng Engr Sys. But, ton of chill water used for hvac and manipulating those quantities should be thoroughly understood. Hydrologic work done in English units Corps of Engr's. No reason to cease addressing it. Power generation analysis probably EES. So, it's kind of like local MoS&T not requiring thermo for civil engineers. They'll never convert steam to hp, but mass flows act on civil systems and need to be understood. Pouring of Hoover dam concrete had to be modified for thermodynamic reasons. If this hadn't accounted for the accumulation of heat in too large mass of concrete, the dam would of failed. Fortunately this was detected. But it wouldn't be by a MoS&T grad. I experienced a new approach to statics/dynamics evaluations; it was called conservation principles and taught to id every force in sys, draw fbd and apply forces, tabulate all the forces +/-. Then solve for unknowns. Whereas, the Hoover dam generation knew sum of forces, x,y=0 for equilibrium. Cause dynamics was a mechanical engineering problem. But, c.e. taught dyn force resolution these days, least @ TxA&M. It is simply an work/energy term to be resolved, rather than force. Been looking into Bernoulli eq too to model energy in propeller stream @ full power. I did basic xperiment. I physically measured take off roll and computed avg acceleration @ liftoff. I compared this to acceleration derived from given engine horsepower and the discrepancy was depressingly large. I realize the propeller was incredibly inefficient xfer of power and if I had been around i'd of questioned the efficiency of the system as too wasteful to be useful. Especially considering the internal energy of the fuel. You're not getting much of the ttl qty of internal energy of the gas. Round and round of jet such a better xfer, but seem to be quite consumptive of fuel as well. But, much higher efficiency of intE than piston. The electrical system consumption on jet a/c kilowatts easy. That's a lot B to turn a coil against. I've cranked a home telephone unit. Even that wear you out after awhile. This is funny. I asked a mechanical engineer how a falling body generates 2000lbf? Structure tied off to had to withstand this by OSHA. He told me to go ask a c.e. I was shocked. I could resolve that now w/work energy relation to torque on structure.
 
  • #13
Roady said:
Homework Statement: Def of lbf=32lbm*1ft/sec^2
No. There is no officially-sanctioned definition of the pound-force. The definition of the pound Avoirdupois is 0.453 592 37 kg.

The closest definition of the pound-force makes use of the formerly officially-sanctioned definition of the kilogram-force that depended on this so-called standard value of the free fall acceleration g = 9.806 65 m/s2. Thus the pound-force would equal (0.453 592 37 kg)(9.806 65 m/s2).

This is, at least, how NASA defines it.

I've never seen units like the slug or the poundal used anywhere other than a physics classroom.
 
  • #14
Mister T said:
No. There is no officially-sanctioned definition of the pound-force. The definition of the pound Avoirdupois is 0.453 592 37 kg.

The closest definition of the pound-force makes use of the formerly officially-sanctioned definition of the kilogram-force that depended on this so-called standard value of the free fall acceleration g = 9.806 65 m/s2. Thus the pound-force would equal (0.453 592 37 kg)(9.806 65 m/s2).

This is, at least, how NASA defines it.

I've never seen units like the slug or the poundal used anywhere other than a physics classroom.
I would say it sent man to the moon. FORTRAN sent man to the moon. A guess; I wasn't directly involved. 60yrs ago, mks was hardly taught in Am universities.
 
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  • #15
Roady said:
I would say it sent man to the moon. FORTRAN sent man to the moon. A guess; I wasn't directly involved. 60yrs ago, mks was hardly taught in Am universities.
By the time I went to high school seven or eight years later the science classes were teaching nothing but MKS (or cgs). Like @Mister T, I've never seen a slug or a poundal in the wild.

As recently as 1998, NASA lost a Mars mission due to measurement units. Quoting from the article:

"The primary cause of this discrepancy was that one piece of ground software supplied by Lockheed Martin produced results in a United States customary unit, contrary to its Software Interface Specification (SIS), while a second system, supplied by NASA, expected those results to be in SI units, in accordance with the SIS. Specifically, software that calculated the total impulse produced by thruster firings produced results in pound-force seconds. The trajectory calculation software then used these results – expected to be in newton-seconds (incorrect by a factor of 4.45)"
 
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  • #16
Roady said:
I would say it sent man to the moon.
The definition I gave based on SI units is indeed the definition of the pound-force used by NASA's manned lunar missions in the late 1960's and early 1970's. I fail to see your point.

NASA's Mars Climate Orbiter was lost due to a unit conversion error in 1999. The spacecraft's navigation software used SI units (meters and kilograms), while the ground-based software used English units (pounds and feet).

Roady said:
60 yrs ago, mks was hardly taught in Am universities.

That makes sense. That would have been 1965. In 1960 the GCPM established the International System of Units (SI), the modern version of the metric system, setting the meter, kilogram, and second as three of the base units. It takes years for universities to catch on. And even then many of the professors, steeped in the old cgs system, went along with reluctance.

By the time I got to college in 1973, we were taught what was still called the MKS system, with a nod to the "good ol' days" of cgs.

Again, though, I fail to see your point. It's 2025. In the 2019 meeting of the GCPM the SI underwent a significant revision, redefining four of the seven base units (kilogram, ampere, kelvin, and mole) in terms of fundamental physical constants rather than artifacts or experimental observations. Despite any nostalgia some folks might have for former glory days, modern science and technology demands more precise definitions of the units of measure used to manufacture modern devices and conduct meaningful research.

We need the new unit definitions and conventions if we are to continue to advance our civilization.
 
  • #17
Units of BTU, ton's of chillwater, lb mass of steam, are in common use today. I doubt they changed units on Apollo missions to ease calculations, because current crop of engineers were trained in them.

This is funny. 40's something female in my engineering class make humorous state, something about this unit conversion made G sub c(gravitational constant)=1. I never got the chance to explain it is a constant and nothing is going to change it. She left the engineering field for Journalism. Ya'll can rest easy.

Above error. Testing should of revealed units of instrumentation. That this wasn't noticed is like when flying in simulator squadron and flt leader call out for fuel levels and you don't answer right away cause you don't know where fuel gauge is. I would be quite disappointed if PhD in engineering couldn't explain this to me.

Too bad the theroetical survey didn't pan out in 1700's. The guy never considered earth was ellipsoid, not sphere and his 1/1,000,000 meridan measurement never fit. The krypton wavelength std couldn't be achieved until when? English units served the world pretty well. It wasn't deficiant just inconvenient and arbitrary.
 
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  • #18
Roady said:
This is funny. 40's something female in my engineering class make humorous state, something about this unit conversion made G sub c(gravitational constant)=1. I never got the chance to explain it is a constant and nothing is going to change it. She left the engineering field for Journalism. Ya'll can rest easy. Above error.
You may want to read about geometrized units. Indeed, in this system Newton's universal gravitational constant ##G = 1##.

I'd mentioned this in post #11 in this thread.

More generally, almost any physical constant we choose can have any numeric value we choose if we pick a set of units that makes it so. The constants for which this cannot be done are "dimensionless physical constants". They are pure numbers. The same in every system of units.

For instance, the fine structure constant ##\alpha## is dimensionless and is approximately ##\frac{1}{137}##.
 
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  • #19
Roady said:
Units of BTU, ton's of chillwater, lb mass of steam, are in common use today. I doubt they changed units on Apollo missions to ease calculations, because current crop of engineers were trained in them.

You seem to be connecting the choice of units used to the validity of the science and engineering done with those units.

There is no such connection!

One can do perfectly valid science and engineering using any system of units.

Metrology, on the other hand, requires precision in its definitions of those units. Once the metrologists have made their choices scientists and engineers are free to do their calculations using any units they choose. Precision, however, requires that whatever units one uses, they must have their definitions based on those choices made by the metrologists.

This is nothing new. It's been in use since the Treaty of the Meter was signed back in the 1870's. For example, the United States, having signed the treaty, accepted their copy of the International Prototype of the Kilogram (IPK) in a ceremony at the White House presided over by then-president Grover Cleveland. Ever since then the pound Avoirdupois has been defined in terms of the kilogram, as required by that treaty.

The definition was refined a few times over the decades, but in 1959 the definition we now use was adopted as 0.453 592 37 kg.

None of this has anything whatever to do with the validity of scientific calculations, or which branch of science is required to get an engineering calculation correct. It simply establishes standards of precision.
 
  • #20
1 ##lb_f## is the force needed to give a mass of ##1\ lb_m## an acceleration of 32.2 ft/sec^2; it is also equal to the force needed to give a mass of 1 slug an acceleration of 1 ft/sec^2

$$1 slug = 32.2 lb_m$$

So, in terms of slugs and ##lb_f##, $$F =ma$$ but in terms of ##lb_f## and ##lb_m##, $$F=\frac{ma}{g_c}$$ where $$g_c=32.2\ \frac{lb_m}{lb_f}\frac{ft}{sec^2}$$
 
  • #21
Chestermiller said:
1 lbf is the force needed to give a mass of 1 lbm an acceleration of 32.2 ft/sec^2

What if you're near the equator where the free fall acceleration is less than 32.1 ft/s2?

If you set an object with a mass of one pound on a spring scale, don't you want it to exert one pound of force on the scale?
 
  • #22
If the spring scale was calibrated with the factor 32.2 in mind, it would read less than 1 pound of force
 
  • #23
Gordianus said:
If the spring scale was calibrated with the factor 32.2 in mind, it would read less than 1 pound of force
Which is illegal if you're using the scale for commerce. Buyers would benefit, but sellers would lose money. An unsatisfactory arrangement.

This is why things are not bought and sold using a unit of measure that changes with location. The phrase "sold by weight" refers to a quantity that does not change with location. We call it mass.

This is required by law.

The famous motto of the Toledo Scale Company is "Honest Weight, No Springs".
 
  • #24
Herman Trivilino said:
What if you're near the equator where the free fall acceleration is less than 32.1 ft/s2?

If you set an object with a mass of one pound on a spring scale, don't you want it to exert one pound of force on the scale?
No. The 32.2 is fixed. And the measured force at the equator would be < 1 lb.
 
  • #25
Herman Trivilino said:
What if you're near the equator where the free fall acceleration is less than 32.1 ft/s2?

If you set an object with a mass of one pound on a spring scale, don't you want it to exert one pound of force on the scale?
The standard way to calibrate a spring scale (including scales that use load cells) involves standard reference masses.

e.g. You put a 1 pound mass on the scale at the place where it will be used and calibrate it to display 1 pound.

From an operational point of view, this means that a properly calibrated spring scale accurately reports mass but does not necessarily accurately report force.
 
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  • #26
Chestermiller said:
No. The 32.2 is fixed. And the measured force at the equator would be < 1 lb.
Any reputable company would calibrate the scale so that it reads one pound. They would not agree with you that the 32.2 is fixed.

Moreover, people engaged in commerce who insist that the 32.2 is fixed would be in violation of the law. So they would not agree with you, either. Neither would the people who write or enforce those laws.


Sorry for the above incorrect statements that I had to cross out. If you define the pound-force in the way you have, which by the way is not the only definition out there, then you are correct.

So long as it's understood, as @jbriggs444 says, a properly-calibrated scale is not measuring force, it's measuring mass. So in this case it would read one pound, even though the force may be different from one pound, depending on how you define the pound-force.

The kilogram-force used to be an officially-sanctioned unit of force whose definition was based on a fixed value of the free fall acceleration equal to 9.806 65 m/s2. And the same problem existed. Which is probably why the GCPM BIPM abandoned it.
 
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