Buoyant force on a Beach Ball

In summary, when a beach ball made of thin plastic is fully submerged in water, the buoyant force exerted on it decreases due to the compression of the air inside the ball. This is because the gas compresses at a faster rate than the surrounding water, leading to a decrease in volume and subsequently, buoyancy. Even if everything is incompressible, the buoyant force would still decrease as the depth increases due to the constant volume and density of water.
  • #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
 
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  • #2
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.
 
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  • #3
Rx7man said:
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?
 
  • #4
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.
 
  • #5
Ethan Godden said:

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?

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.
 
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Related to Buoyant force on a Beach Ball

What is buoyant force?

Buoyant force is the upward force exerted by a fluid on an object placed in it. This force is equal to the weight of the fluid that the object displaces.

How does buoyant force affect a beach ball?

A beach ball experiences buoyant force when it is submerged in water. The amount of buoyant force acting on the beach ball is equal to the weight of the water that is displaced by the ball. This force helps the beach ball to float on the surface of the water.

What factors affect the buoyant force on a beach ball?

The buoyant force on a beach ball is affected by the density of the ball, the density of the fluid it is placed in, and the volume of the fluid displaced by the ball. The larger the volume of the ball or the less dense the fluid, the greater the buoyant force will be.

Why does a beach ball float on water?

A beach ball floats on water because the buoyant force acting on it is greater than its weight. The air inside the beach ball makes it less dense than the water, causing it to float on the surface.

Can the buoyant force on a beach ball change?

Yes, the buoyant force on a beach ball can change if any of the factors that affect it change. For example, if the beach ball is deflated and becomes more dense, the buoyant force will decrease and the ball may sink. Additionally, if the water it is placed in becomes more or less dense, the buoyant force on the beach ball will also change.

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