Challenging buoyancy!

A metal cube is placed at the bottom of an empty vessel and water is poured in. When the water level just reaches the top of the cube, how does the force on the bottom of the vessel in contact with the cube compare to the force before the water started to pour in?

Details

  • Note, the cube forms an airtight seal with the bottom of the vessel.
It is greater than It is less than It is the same

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2 solutions

Trung Vu
Sep 22, 2018

It is known that metal has a higher density than that of water. Denoting them ρ m e t a l \rho_{metal} and ρ w a t e r \rho_{water} , then: ρ m e t a l > ρ w a t e r \rho_{metal} \gt \rho_{water} Thus: ρ m e t a l V s u b m e r g e d g > ρ w a t e r V s u b m e r g e d g \rho_{metal} V_{submerged} g \gt \rho_{water} V_{submerged} g W c u b e > F b u o y a n c y W_{cube} \gt F_{buoyancy} The weight of the cube will always be greater than the buoyant force, meaning that the cube will always sink. The amount of water around the cube has no role in the contact force, because the buoyancy force is always F B = ρ w a t e r V s u b m e r g e d g F_B = \rho_{water} V_{submerged} g , and that is independent in terms of how much water is poured in. Therefore, the contact force before and after pouring in water are the same.

But when cube is under water, water weight will be added.

Sergei Gontsarenko - 12 months ago
Auper Bhadra
Mar 5, 2019

Metalic cube will always be touched with vessel. Thus there should be no water between them.And no normal force. And so forces always remain same. So simple.

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