Mass vs Mass as a Force (Weight)

[rsk]

Are you now denying that “weight” is NOT used to denote mass in some cases and gravitationally induced force in others?

Edit: But to answer your question more directly: 32 pounds-force = 1 pound-mass • 32ft/s^2

I have never ever seen "weight" used to denote "mass" in the context of Newton's laws, no. Not even by middle and high school students.

My students (age 12 upwards) arrive using weight when they mean mass, but most only need to be told once that the everyday use of the word "weight" to mean mass is incorrect in Physics. That said, I have the great fortune to be teaching in Europe and therefore free from your lbs mass and lbs weight nonsense.

 

[jbriggs444]

I have never ever seen "weight" used to denote "mass" in the context of Newton's laws, no. Not even by middle and high school students.

My students (age 12 upwards) arrive using weight when they mean mass, but most only need to be told once that the everyday use of the word "weight" to mean mass is incorrect in Physics. That said, I have the great fortune to be teaching in Europe and therefore free from your lbs mass and lbs weight nonsense.

The terms better used to disambiguate the pound are "pound mass" and "pound force".

The term "pound weight" to denote a unit of measurement is itself ambiguous and would never be used to disambiguate anything. I cannot remember ever having seen that phrase used in that sense.

The term "pound weight" to denote a reference weight with a mass of one pound would be common. e.g. "Can you hand me that (one) pound weight over there?"

The term "weight in pounds" is used fairly commonly, but does not carry with it an indication of whether the intended meaning is force or mass. e.g. "Please enter your weight in pounds".

 

[pbuk]

The term "weight in pounds" is used fairly commonly, but does not carry with it an indication of whether the intended meaning is force or mass. e.g. "Please enter your weight in pounds".

Of course the intended meaning is mass: you don't enter a different number depending on whether you are in Miami or Denver!

 

[jbriggs444]

Of course the intended meaning is mass: you don't enter a different number depending on whether you are in Miami or Denver!

Personally, I agree that mass is intended. But the accuracy with which any given person knows their weight, the size of the noon meal or the elapsed time since the last bathroom break are all probably at least as important as the magnitude of local g. It doesn't matter force or mass is intended -- both are essentially the same number.

 

[gmax137]

Now I am confused, again! Are you telling me that 10 pounds of potatoes have a mass of 10 lbm? I thought I would have to divide the 10 pounds by 32+ to approximate the mass in lbm.

@Quester I've been away, so sorry if this has already been answered. The answer is this: If you push on a one lbm object with a force of 1 lbf, it accelerates at... 32 ft/sec2. You can test this, by hanging the 1 lbm object by a string, anywhere here on earth, and watch what happens when you cut the string.

This is different than the SI system which previous posters have described as "coherent." In SI units, if you push on a 1 kg mass with a force of 1 Newton, the object will accelerate at... 1 m/sec2.

That's really what is at the heart of all the confusion.

 

[sophiecentaur]

That's really what is at the heart of all the confusion.

In the world of the future, in which we easily hop from planet to planet and space station, we would have no confusion. As it is, Earthbound objects of a given mass will all weight the same (unless you have very expensive scales) and we have grown up to be very sloppy about this. That's been the problem.

 

[pbuk]

(unless you have very expensive scales)

NO! You mean "unless you have poorly calibrated scales".

It may be that the mechanism of a weighing scale measures force but that mechinism is calibrated in order to measure mass.

You don't get more flour in a 2lb bag in Denver than you do in Miami.

 

[hutchphd]

NO! You mean "unless you have poorly calibrated scales"

Oh lord we are in the semantic morass once again.

I guess one needs to carefully differentiate a "scale" from a "balance" when in this semantic morass.
A scale usually contains a source of calibrated electrical force for comparison.
A balance compares the unknown mass to a known mass (multiplied by mechanical advantage or not).

I believe that @sophiecentaur was saying that new planetary wide inconsistencies would not arise on Mars (although, on second thought, the mountains are taller and radius is smaller...).

 

[pbuk]

Oh lord we are in the semantic morass once again.

This is only a morass if you are lost in it: the way out should be clear.

Yes a balance compares an unknown mass to a known mass.

If a calibrated scale measures 'electrical force' (whatever that is) then it compares the electrical force produced by an unknown mass to that produced by a known mass.

And no, there won't be any inconsistencies between properly calibrated scales on Mars and on Earth any more than there are between Denver and Miami.

 

[hutchphd]

If a calibrated scale measures 'electrical force' (whatever that is) then it compares the electrical force produced by an unknown mass to that produced by a known mass.

Its easier to use than a nuclear strong force.
I guess I should have used the word "spring". I was attempting to be general when clearly I needed to be pedestrian.

.

 

[JT Smith]

As it is, Earthbound objects of a given mass will all weight the same (unless you have very expensive scales) and we have grown up to be very sloppy about this. That's been the problem.

By very expensive do you mean $25?

The difference in g between Anchorage, Alaska and Bangkok, Thailand is nearly 0.5%.

 

[hutchphd]

They say the cost of living in Bangkok is less than Anchorage...must be true!

 

[PeterO]

When someone asks me "How much do you weigh?", I am tempted to say "I weigh the same as a 105kg mass", but instead I give the short answer they are expecting - namely "I weigh 105 kilograms" They already know that means "I weigh as much as a 105 kg mass".

btw: in this country 1 lb is also a mass, and something that weighs 2.2 lb means something that weighs the same as a 2.2 lb mass.

 

[Quester]

@Quester I've been away, so sorry if this has already been answered. The answer is this: If you push on a one lbm object with a force of 1 lbf, it accelerates at... 32 ft/sec2. You can test this, by hanging the 1 lbm object by a string, anywhere here on earth, and watch what happens when you cut the string.

This is different than the SI system which previous posters have described as "coherent." In SI units, if you push on a 1 kg mass with a force of 1 Newton, the object will accelerate at... 1 m/sec2.

That's really what is at the heart of all the confusion.

Thank you for addressing the question. However, your answer doesn't help me. No matter what the mass of the object, the object will accelerate at a rate of (about) 32 ft/sec² if dropped in a vacuum (to remove the effect of air resistance on very low density objects) .

If f = ma, and in the imperial system, acceleration due to gravity is about 32 ft/sec², then:

f = 32m

so the force required to accelerate any mass at the acceleration of gravity would be equal to 32 times that mass. Ergo, to use the equation f=ma properly, I should divide the weight of an object by 32. Is that correct?

 

[jbriggs444]

Thank you for addressing the question. However, your answer doesn't help me. No matter what the mass of the object, the object will accelerate at a rate of (about) 32 ft/sec² if dropped in a vacuum (to remove the effect of air resistance on very low density objects) .

If f = ma, and in the imperial system, acceleration due to gravity is about 32 ft/sec², then:

f = 32m

so the force required to accelerate any mass at the acceleration of gravity would be equal to 32 times that mass. Ergo, to use the equation f=ma properly, I should divide the weight of an object by 32. Is that correct?

If you are choosing to express force in pounds force, mass in pounds mass and acceleration in feet per second² then yes, you will need a unit conversion factor in $f=ma$ so that the formula reads $f=\frac{1}{32.17}ma$.

If you chose to express force in pounds force, mass in slugs and acceleration in feet per second² then the unit conversion factor becomes 1 and can be ignored.

Similarly, if you choose to express force in poundals, mass in pounds mass and acceleration in feet per second² then the unit conversion factor becomes 1 and can be ignored.

Or you could choose to express force in pounds force, mass in pounds mass and acceleration in gees. Again, the unit conversion factor would become 1 and could be ignored.

If you use a system of units that is not coherent (for the problem you are working) then you will have to insert unit conversion factors into your formulas.

 

[Ibix]

If you use a system of units that is not coherent (for the problem you are working) then you will have to insert unit conversion factors into your formulas.

...and I am grateful you did it so that I didn't have to. Just reading that makes my head hurt...

 

[gmax137]

I should divide the weight of an object by 32. Is that correct?

Yes. But can you see why?

$$f = a~(\frac {ft} {sec^2})~m~(lbm) = a~m~ \frac {lbm ft} {sec^2}$$

so here we have a force f, in units of lbm-ft/sec^2 which is non-standard. if we want the units in pound-force (lbf) then we divide by 32. Why? 32 what? 32 lbm ft/lbf-sec^2 (which is equal to "1")

$$f = a~(\frac {ft} {sec^2})~m~(lbm) = a~m~ \frac {lbm ft} {sec^2}~\frac{lbf~ sec^2}{32~ lbm ~ft}=\frac{a~m}{32}~ lbf$$

They key thing is to write all of the units for each quantity, to help you keep track of what conversion factor you need to use.

Now recall that the acceleration a has a value (here on Earth) of 32 ft/sec^2. So you end up with

$$f = a~(\frac {ft} {sec^2})~m~(lbm) = a~m~ \frac {lbm ft} {sec^2}~\frac{lbf~ sec^2}{32~ lbm ~ft}=\frac{a~m}{32}~ lbf = 32~\frac{m}{32} = m~ lbf$$

Which just shows that a one-pound mass weighs one pound force, here on Earth.

 

[sophiecentaur]

It really is about time to go to SI units, I think. No one could confuse kg with N and the working value of 10 for g makes in-head calculations a doddle.
In UK the only crazy unit we use is the mile. The pint is not metric of course but it’s only used when nobody cares (about anything).

 

[gmax137]

I bet you guys still measure horses in "hands" :).

 

[sophiecentaur]

I bet you guys still measure horses in "hands" :).

The only horse I own is 'about the right height' for cutting logs with my little chain saw.

 

[Ibix]

The pint is not metric of course but it’s only used when nobody cares (about anything).

I don't know - haven't you ever seen a Real Ale enthusiast getting short measure?

 

[Dale]

If you use a system of units that is not coherent (for the problem you are working) then you will have to insert unit conversion factors into your formulas.

Excellent point. The SI is coherent for mechanics problems but not for electromagnetic problems. So if you want to understand what that means you can compare Newton’s laws with Maxwell’s equations with all of its conversion factors in SI.

 

[cmb]

When someone asks me "How much do you weigh?", I am tempted to say "I weigh the same as a 105kg mass"

So, would that mean you'd be tempted to say you weigh the same as a helium balloon with a mass of 105kg?

... discuss ...

 

[jbriggs444]

So, would that mean you'd be tempted to say you weigh the same as a helium balloon with a mass of 105kg?

... discuss ...

Most standards organizations prefer to calibrate scales with brass weights rather than helium balloons.

 

[cmb]

Most standards organizations prefer to calibrate scales with brass weights rather than helium balloons.

So, the comment to be a bit more specific? One may be the same weight as a 105kg mass of brass?

Raises a subtle and interesting question though, actually, I think?

In the case of PeterO's 105kg (let's assume that is precisely his mass), he would only weight precisely the same as a 105kg mass if it also had the same volume as him, else his weight would differ by the additional (or lesser) mass of the air displaced by the difference in volumes, would it not?

 

[Herman Trivilino]

would it not?

No, it would not. The weight of an object does not include any effect due to buoyancy. The comparison is assumed by definition to be made in a vacuum.

 

[cmb]

No, it would not. The weight of an object does not include any effect due to buoyancy. The comparison is assumed by definition to be made in a vacuum.

I would have to disagree with you then, because your reply does not mention what is actually 'weighed'.

If you put an object on scales consisting of a 2kg mass suspended from a 500g 1m^3 helium balloon, then it'd weight about 1 kg.

If we break down why that is, the presence of helium in an enclosed envelope displaces the air above its lower surface. Now consider the total mass of material bearing down from above the lower surface to those atmospheric loads from below.

The balloon has displaced a volume of air that means the atmospheric pressure from above is now less than that from below, and if you measure that on weight scales you'll find it'll be less, or even negative (floats off).

Same with any other mass, but obviously considerably less noticeable for solid masses because the masses involved are a lot more and the displaced air is a lot less.

Let's consider a 1kg 1 litre mass sitting on scales. Let's assume gravity is a perfect 10m^2/s. The mass's force on the scales due to gravity is 10N.

But the volume of the mass has displaced 1 litre of air above its lower surface. Therefore, there is 1 more litre of air bearing on its underside to its top, being 0.01N more from the underside directed upwards.

If we say 1 litre of air is 1 gramme, so a 1 litre 1 kg mass will actually 'weight' 9.99N. A 2 litre 1kg mass will 'weight' 9.98N.. etc..

For a 1000 litre 1kg mass, it'd 'weigh' nothing because it'd then be 1 gramme per litre deducted from the 1kg mass, and it'd be neutrally buoyant.

So, when you get on those bathroom scales, you have to add about one gramme for every litre of your volume, to translate a 'weight' read-out (in kilogrammes) into mass. No adjustment is required for lbs because that sort of 'weight' reading also includes buoyancy.

If you add 1 gramme per litre to the balloon example above you'll get the right answer for the mass (if all those numbers were precise).

 

[Dale]

I am with @Mister T on this. The weight is the force due to gravity. If buoyancy effects are present then a scale will not measure the weight. If your mass is 105 kg then you do weigh the same as a helium balloon with a mass of 105 kg.

 

[cmb]

I am with @Mister T on this. The weight is the force due to gravity. If buoyancy effects are present then a scale will not measure the weight. If your mass is 105 kg then you do weigh the same as a helium balloon with a mass of 105 kg.

Therein lies the whole discussion point of the thread.

I take 'weight' to mean, self-evidently, one's weight measured on scales at the Earth's surface in ambient STP conditions, or

.. it isn't.

If not, please define how and when weight can be measured, because it sounds like it'll be nothing like any lay person understands to be 'weight'.

Medical person; "What is your weight?"
Mister T; "Sorry, I have already explained in public discussion that weight is measured in a vacuum and I have never measured that in a vacuum."
Medical person; "Then I have to record your weight"
Mister T; "No way! I am not going to stand in a vacuum vessel just for that!".

Do you see why I think this is fundamentally part of the thread discussion?

My Chambers dictionary of Science and Technology says;
"Weight; (phys) The gravitational force acting on a body at the Earth's surface"

The ambiguity in that definition is whether it is the gravitational force for 'only' that body, or all sources of gravitational force (that therefore includes the immersive fluid it is in).

I would say the clue there is 'at the Earth's surface' and there is no such point on the Earth's surface that is in a vacuum.

 

[pbuk]

If you put an object on scales consisting of a 2kg mass suspended from a 500g 1m^3 helium balloon, then it'd weight about 1 kg.
...
If we say 1 litre of air is 1 gramme, so a 1 litre 1 kg mass will actually 'weight' 9.99N. A 2 litre 1kg mass will 'weight' 9.98N.. etc..

The word weight in these sentences is not correct; it should be weigh (it'd weigh about 1 kg...will weigh 9.98N).

For a 1000 litre 1kg mass, it'd 'weigh' nothing because it'd then be 1 gramme per litre deducted from the 1kg mass, and it'd be neutrally buoyant.

You got the grammar right this time, but you are confusing three things:

• what a scale is designed to measure (which is mass)
• the mechanism any particular calibrated scale employs to measure mass (which is generally by comparison of the force exerted by a known mass), and any corrections which which may need to be made to the measured mass e.g. correcting for buoyancy using the difference between the densities of the measured mass and the calibrating mass
• the meaning of the weight of an object, which depends on the context e.g. 'an object in orbit is weightless'; 'you weigh less under water', but always refers to a force and is never therefore measured in kg.

So, when you get on those bathroom scales, you have to add about one gramme for every litre of your volume, to translate a 'weight' read-out (in kilogrammes) into mass. No adjustment is required for lbs because that sort of 'weight' reading also includes buoyancy.

How could this possibly be true? Do my bathroom scales magically acquire knowledge of the density of the object that is placed on them when I switch them from kg to lbs?