What Happens to the Buoyant Force on a Beach Ball When Submerged?

[Ethan Godden]

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=ρ_fluidgV_displaced

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

 

[Rx7man]

Liquids compress VERY little in comparison to any gases... I think you have the right idea, as the ball is submerged deeper, the gas inside it is compressed at a far faster rate than the surrounding water, thus volume decreases, and subsequently buoyancy as well.

 

[Ethan Godden]

Liquids compress VERY little in comparison to any gases... I think you have the right idea, as the ball is submerged deeper, the gas inside it is compressed at a far faster rate than the surrounding water, thus volume decreases, and subsequently buoyancy as well.

Thank you,

I am assuming by your answer that it's not possible for the buoyant force to decrease if everything is incompressible?

 

[Rx7man]

I think that would be right...

If you filled your beach ball with water, it would have 0 buoyancy at any depth since both are compressing at the same rate.

 

[Chestermiller]

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=ρ_fluidgV_displaced

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?

Definitely.

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.

This is correct. And you already correctly judged that the ball (filled with air) would compress much more than the water. So your assessment was totally correct.