0
$\begingroup$

In Archimedes principle $F_b = \rho gV$. $\rho$ = density of fluid, $V$ = volume displaced by the fluid, and $g$ = gravity.

If you have an object in the air, like a balloon, how does the volume work? It is challenging to conceptualize how the balloon displaces the air.

Improve this question
$\endgroup$
1
  • $\begingroup$ What is relevant to your query is to displace the column of air in the immediate surroundings of the balloon. That will work well. That is what happens in water too unless it splash out. $\endgroup$
    – Proscionexium
    Commented Nov 14, 2023 at 13:19

2 Answers 2

Reset to default
4
$\begingroup$

It is challenging to conceptualize how the balloon displaces the air.

Think about what happens when you blow up a balloon. In order for the balloon to expand it needs to "push" the surrounding air away from the surface of the balloon. The more you blow it up, the more air that needs to be pushed away. That volume of air pushed away on the outside is the volume of air "displaced" by the volume of the balloon.

Hope this helps.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Nice tutor for understanding basics. However, technically by blowing up balloon we do not help it to displace air, because atmosphere air breathed-in into our lungs is pushed into balloon. So we are actually just re-locating same atmosphere gases. "True" air displacement would be if we would create solid sealed container with perfect vacuum inside. Container total volume would force atmosphere to displace. Same for water. If you would have "almost" sealed box, but just with narrow hole in it - water would sunk-in, and displaced volume would be not container volume, but just it's walls volume. $\endgroup$
    – Agnius Vasiliauskas
    Commented Nov 14, 2023 at 14:17
  • 1
    $\begingroup$ @AgniusVasiliauskas Good point. Perhaps a better example is submerging an inflated balloon in a tank of water and noting that the volume associated with the rise in water level equals the volume of the water displaced by the balloon. $\endgroup$
    – Bob D
    Commented Nov 14, 2023 at 14:44
1
$\begingroup$

Archimedes' principle works the same way in air as it does in liquid. The buoyancy force is again $$F_b=\rho gV,$$ where $\rho$ is the density of the air surrounding the balloon, and $V$ is the volume of the air displaced by the balloon (i.e. the volume of the balloon envelope and the enclosed gas inside).

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.