# How can mass be measured independently of gravity?

In summary: The formula Mgh is commonly accepted as the work done by raising a mass by a distance h, where M is defined as the mass of the object raised.However, is this really the mass, or the weight, simply obtained by weighing the object? If it's the weight, then doesn't the equation effectively contain the acceleration due to gravity twice?The scales I'm using are for weighing items to go into the post. They are giving a reading in kgm.But surely if I used the scales on the Moon, then they would give a reading of 1/6 what it was on Earth, whereas the mass of the object would stay the same.This is a problem that will have
TL;DR Summary
Does the formula Mgh contain the acceleration due to gravity twice?
Hello

The formula Mgh is commonly accepted as the work done by raising a mass by a distance h, where M is defined as the mass of the object raised.

However, is this really the mass, or the weight, simply obtained by weighing the object? If it's the weight, then doesn't the equation effectively contain the acceleration due to gravity twice?

best regards ... Stef

It's the mass of the object, not the weight.

russ_watters
Many thanks.

So to be clear, I would have to weight the object and then divide this by 9.8 to get the mass?

That's right.

russ_watters
Most balances are calibrated in kg, so they are indicating mass. A true weight meter would give a calibration in Newtons. In such a case it is correct to divide by 9.8, but otherwise not.

russ_watters
Ah OK, that's interesting, and confusing.

The scales I'm using are for weighing items to go into the post. They are giving a reading in kgm.

But surely if I used the scales on the Moon, then they would give a reading of 1/6 what it was on Earth, whereas the mass of the object would stay the same.

anorlunda
Ah OK, that's interesting, and confusing.

The scales I'm using are for weighing items to go into the post. They are giving a reading in kgm.

But surely if I used the scales on the Moon, then they would give a reading of 1/6 what it was on Earth, whereas the mass of the object would stay the same.
That's a problem that will have to be solved when there is a postal service to and from the Moon.

tech99, russ_watters and berkeman
Very witty.

It avoids the question of whether the scales are measuring mass or weight.

It avoids the question of whether the scales are measuring mass or weight.
Since simple scales use compressing springs and displacement to measure the object, they are measuring force directly, no? So that would be a weight measurement (F=ma=kΔx). If the scale is used on the surface of the Earth, you can correlate that weight measurement with the associated mass and display whichever you want.

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... The scales I'm using are for weighing items to go into the post. They are giving a reading in kgm.
...
Your scale may show the kg symbol, rather than kgm.
That symbol means kilogram-force.
https://en.m.wikipedia.org/wiki/Kilogram-force

As far as I know, [kgm] is an old days way of writing [kg], because [lbf] (pound-force) can be confused with [lbm] (pound-mass) if one writes just [lb], due to the way "weight" and "mass" are mixed up in the English language. So scales would be marked with [lbm] to indicate it's already discounting gravity from the measurement, thus it's mass, and when converting to [kg], it was customary to write [kgm] (kilogram-mass, not kilogram-meter) to match the same thought process.

It's not normally used outside English speaking countries, as most of the world uses SI units, where [kg] is never spelled as [kgm], as that's an American-Anglo-Saxon thing. When SI users see [kg], they never think [kgf], thus [kgm] only muddle things up due to the default interpretation of "m" as meaning "meter", not "mass", whilst in Imperial units, "m" is mass, as "meter" is replaced by either [yd], [ft] or [in].

PeroK
Very witty.

It avoids the question of whether the scales are measuring mass or weight.
Witty, but correct. The direct answer is that they are calibrated for use on Earth's surface and if they are sent to the moon or even a mountain top they need to be recalibrated.

PeroK
Summary: Does the formula Mgh contain the acceleration due to gravity twice?

However, is this really the mass, or the weight, simply obtained by weighing the object? If it's the weight, then doesn't the equation effectively contain the acceleration due to gravity twice?
You might enjoy a mental exercise. How would you measure mass independent of gravity? Swinging a mass in a circle at the end of a string
comes to mind. What other ways can you think of?

## What is work done in raising a weight?

The work done in raising a weight is the amount of energy required to lift an object against the force of gravity. It is measured in joules (J) and is calculated by multiplying the weight of the object (in newtons) by the height it is lifted (in meters).

## How is work done in raising a weight related to potential energy?

Work done in raising a weight is directly related to potential energy. As the weight is lifted, it gains potential energy due to its increased height. The amount of work done is equal to the change in potential energy of the object.

## What factors affect the amount of work done in raising a weight?

The amount of work done in raising a weight is affected by the weight of the object, the height it is lifted, and the force of gravity. The greater the weight and height, the more work is required. Additionally, the force of gravity can vary depending on the location on Earth.

## Can work be done on a weight while it is being raised?

Yes, work can be done on a weight while it is being raised. This work is done by the force applied to the object to lift it against gravity. As the weight is lifted, the force applied is doing work on the object, increasing its potential energy.

## How is work done in raising a weight different from work done in lowering a weight?

The work done in raising a weight is positive, as energy is being added to the object. On the other hand, the work done in lowering a weight is negative, as energy is being taken away from the object. This is due to the direction of the force applied and the change in potential energy of the object.

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