- #1
Ethan Godden
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Homework Statement
A beach ball is made of thin plastic. It has been inflated with air but the plastic is not stretched. BY swimming with fins on, you manage to take the ball from the surface of a pool to the bottom. Once the ball is completley submerged, what happens to the buoyant force exerted on the beach ball? (a) Increases (b) Remains constant (c) decreases (d) Impossible to determine
Homework Equations
Archimedes's Principle: The magnitude of the buyant force on an object always equals the weight of the fluid displaced by the object.
B=ρfluidgVdisplaced
The Attempt at a Solution
I think the key words in this question is that the plastic is not stretched. The answer is, apparently, the buoyant force decreases. I think this reason for this is that the ball is compressed more as the depth increases meaning, by Archimedes principle, the buoyant force decreases. But from a previous question on here, doesn't that also mean the water is compressed with depth meaning more water is displaced with depth? I am assuming that the amount the water is compressed is negligible compared to how much the beach ball is compressed. Am I correct with this?
Also, the question never mentions the ball of water being compressible. I am wondering if this answer is still possible assuming everything is incompressible. If so how? I thought by Archimedes principal that the buoyant force remains constant as depth increases assuming the volume displaced and density of water remains constant with depth.
Thank you,
Ethan