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This is an undergrad physics problem, which puzzled me because of the little background provided; see also the comment below by @RC_23.

Problem. Compute the force exercised by the air to an object whose volume is $V=1000$ dm$^3$.

Solution. Now $V=1000$ dm$^3$ = 1 m$^3$. My reasoning is that the object is pressing the air causing a shift in its mass. This shift must be equal to the hydrostatic force, which, assuming the object is in equilibrium, must be equal to the gravity force.

Thus, given the density of air $d = 1.225$ kg/m$^3$ and the gravity $g=9.81$ m/s$^2$, the force exercised by air over the object should be $$F_p = mg = (dV) g = 12.0\,\mathrm{N}.$$

                                                     [solution: 12.7 N]

Do you agree with my reasoning? If not, how would you attack the problem?

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    $\begingroup$ I don't know the context, but at 1 atm the density of air is about 1.2 kg/m³ $\endgroup$
    – RC_23
    Commented Aug 14, 2022 at 18:21
  • $\begingroup$ that's all the context provided in the book. $\endgroup$
    – utobi
    Commented Aug 14, 2022 at 18:24
  • $\begingroup$ sorry, there was a typo in d, you are obviously right. $\endgroup$
    – utobi
    Commented Aug 14, 2022 at 18:26
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    $\begingroup$ What have you tried, do you have a specific concept you have a problem with? Please elaborate these points, otherwise the question will probably be closed. $\endgroup$
    – Kuhlambo
    Commented Aug 14, 2022 at 18:27
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    $\begingroup$ Sounds reasonable, it does not even ask for net force or anything like that, so that seems to be all. It's just the buoyancy force it seems. $\endgroup$
    – Kuhlambo
    Commented Aug 18, 2022 at 7:59

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I would call it the buoyancy force of the air , but your calculation is right, if you take the density of air at 0°C you get the given solution

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  • $\begingroup$ that makes sense indeed. For the buoyancy force to be in equilibrium, the object must be far from the Earth's surface (assuming we are on earth), where the temperature is typically lower. Right? $\endgroup$
    – utobi
    Commented Aug 18, 2022 at 16:35
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    $\begingroup$ This has nothing to do with being far from earth's surface! you can have 0° on many places on or near the surface, and the buoyancy force goes down with air pressure. There was no question of "equilibrium" That you calculate was the buoyancy force you did it only for another air density $\endgroup$
    – trula
    Commented Aug 18, 2022 at 16:50

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