Baffled a bit

[Bluechip]

If I arrange bits as in the pic. ie. Digital kitchen scale, Wixey height gauge and the compression spring, the press 'tare' the scale says zero ...
[ The x bits are two ally discs, otherwise it falls over ...    ]
 
If I put 500g on the top of the spring and measure the height difference the Wixey says 2.18mm.
 
So, a load of 500g will compress the spring by 2.18mm.
 
The kitchen scale says 500g.
 
BUT .... is this the same as a load of 4.905 Newtons ?????
 
ie g/2 ????
 
Dave BC .... out of his depth.... I should KNOW this .... 

[David Jupp]

Yes.  Assuming your local value of g isn't far from 'normal', and that the scale hasn't reached its limit.

[John Rudd]

F= MxA
Where F is Newtons, Mass is in Kilograms and A is acceleration

[Bluechip]

David J. & John R
 
Thanks ... I was getting myself all confused ... again 
 
I am a bit familiar with the F=MA equation, but ...
 
Is it really 500g? Or does it only say that because it is on planet Earth.
 
If someone weighs 98.1kg their mass anywhere in the universe is 10kg, only because they  are on planet Earth do they 'weigh' 98.1kg ???
 
I think that's right, never been anywhere else to find out for sure ... 
 
Dave BC
 
 

[John Rudd]

You are mixed up with force mass and weight.....


A 1kg mass exerts a force of 10 Newtons (rounded up if Gravity is10m/sec/sec)

[raynerd]

If someone has a mass of (common language - weights) 98kg then their mass anywhere in the universe is 98kg

On Earth due to gravity of 9.81 their weight would be 9.81x98 = 961.4N

On the moon due to a gravity of 1.62N their weight would be 1.6x98 = 156.8N

The issue comes in the common use of weight when we actually mean mass - mass being measured in kg and weight a force due to mass x gravity measured in newtons.

[Bluechip]

Aaaaaaaargh ... I feel a saga coming on ....           
 
I think Mr. Jupp answered my question.
 
If he's wrong, say why ...
 
If he's right, it's good enough for my purpose.
 
I have a spring, I know from the calibration data it should compress by 1mm with a force of 5N.
 
I don't happen to have such a force about my person.
 
I do have a 500g test weight.
 
I am satisfied the spring does not qualify as a good spring. [ Not my poxy spring anyway ...    ]
 
We old codgers are usually terminally confused at the best of times. Do try not to make our plight worse ...
 
       
 
Dave BC
 
 
 
 
 
 
 
 

[DaveH]

The kitchen scale says 500g.
 
Dave BC .... out of his depth.... I should KNOW this .... 

I like your "talking scales" 
 
DaveH

[PekkaNF]

I'm trying to follow your thinking, and I think I know where the confusion sets in.

You want spring to generate certain force and you are measuring a distance of deflection with a known weight?

Maybe the scale here is just here to verify the test is succesfull or something is not lost on friction, then photo of the contraption probably would have satisfy our curiosity.

Anyway, the way I would think this experiment is that the said physical weight of 0,5 kg (500g) is used to generate certain force downwards. So this force is F = m * g. Think here "g" only as a fudge factor that will convert your known object "weight in Kg:s to Newtons".

Soo simply put you could think the downward force (in Newtons) of this weight as an object (mass in Kg) is 0,50 kg * 9.81 N/kg = 4,90 N, when this experiment is performed right on this planet. Case closed.

What I think generates this confusion is purely terminological.

Confusion sets in when draw parallel lines with this object that is named "weight" as an shiny brass object and it's physical property "weight", which could mean mass or weight depending who you are talking and when.

Simple way to put this together is to think that mass and energy generally does not dissappear (Conservation of mass). Your mass here is same than in the moon.

Pekka