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Consider a weigh machine and an object. Both are inside a case. Weigh the object. Then remove all the air in the case. Weigh the object again. If the weight of the object with air is M kg. What will be the weight without air?

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    $\begingroup$ possible duplicate of physics.stackexchange.com/q/449433 $\endgroup$ – niels nielsen Jan 13 at 7:16
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    $\begingroup$ Hmm what a spring balance measures is the tension force .... So in the case with air the bouyant force is there but without air it is not there $\endgroup$ – Aditya Garg Jan 13 at 7:48
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Yes in principle buoyancy force is in action. It will make no difference only with an arm balance if it happens that your sample and the reference masses have the same effective volume.

The effect is anyway negligible and not considered even in gravimetric analysis (precision down to tenth/hundredth of a mg).

However, for very high precision and to account for propagation of errors, or to define standards in weighing, compensation methods and balances must be considered. Otherwise the operator determines an apparent weight.

More can be found here: buoyancy force in weight measurements

https://is.gd/WrEA0R

a pdf which I've briefly checked for just its relevance to the question.

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Assuming you have kept the case on solid ground, then the weight measured will be equal to the normal reaction force (or the restoring force $F=-kx$ as it really is a spring that is compressed because of the mass).

If your case is in free fall, then as all masses are in free fall, no mass exerts any force on each other apart from their mutual gravity (but there is no visible acceleration).

Your weighing machine would give approximately the same result (unless you want to consider buoyant forces acting on the system).

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  • $\begingroup$ Can the downvoter provide a reason why it has been done so? $\endgroup$ – KV18 Jan 13 at 16:18
  • $\begingroup$ Because first paragraph assumes a spring scale and simply explain it. The second one deals with measuring weight on free falling scales which is quite, say, unusual. The third one came to the point but very quickly and put into brackets what OP was presumably looking for. $\endgroup$ – Alchimista Jan 14 at 19:21

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