TBR Physics Buoyancy Question

[water123]

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I was reading through TBR Physics Book 2, the fluids and solids chapter and I am confused about a buoyancy question

Example 3.7B

The question is: Which of the following are TRUE regarding three balls of equal size that are fully submerged in water, if ball A sinks, ball B is stationary and Ball C rises?

I. The density of ball C is less than that of ball B.
II. The buoyant force on ball C is greater than that on ball A.
III. The density of ball A is greater than that of water.

The answer is both I and III.

I understand why I and III are right but I don't understand why II isn't.

I think I am missing some important concepts here. So is the buoyant force on C and A the same (because they have the same size and therefore the same volume (for Archimedes principle)? Or is the buoyant force greater for A and why?

And also, since A has a greater density than C, doesn't that mean it has a greater buoyant force than C?

I am really confused. Any help is appreciated. Thanks!

 

[Eye Cue]

Buoyant force = density * volume * gravity

Therefore, buoyant force is directly proportional to density, since volume (and obviously gravity) are constant.

Ball A sinks, ball C rises. Ball A, therefore, has a greater density than ball C, and a greater buoyant force.

This is how I would have approached it at least. Hope this helped!

 

[Obi Wan]

How I understood it:

Buoyant Force B = (density of fluid) * (Volume of object) * g

Since all three objects have the same size, by extension they have the same volume. Since they are in the same fluid, both density of fluid and g are constant; so all 3 have the same buoyant force.

Pg. 97 in TBR has a really awesome derivation of net force on an object, which is

a = g(density of water - density of object)/ (density of object)

if a is a positive value, the object accelerates upwards; if a is negative, the object accelerates downward.

Since Object A sinks, density of A must be greater than density of water. But they all have the same buoyant force.

 

[Jeff85]

How I understood it:

Buoyant Force B = (density of fluid) * (Volume of object) * g

Since all three objects have the same size, by extension they have the same volume. Since they are in the same fluid, both density of fluid and g are constant; so all 3 have the same buoyant force.

Pg. 97 in TBR has a really awesome derivation of net force on an object, which is

a = g(density of water - density of object)/ (density of object)

if a is a positive value, the object accelerates upwards; if a is negative, the object accelerates downward.

Since Object A sinks, density of A must be greater than density of water. But they all have the same buoyant force.

Hopefully someone can confirm, but I believe that volume aspect to buoyancy refers to the volume of water displaced by the object, not the object volume persay. When something sinks, volume water displaced and volume of object happens to be the same, but when something is floating, there is a volume of the object that is underwater.

 

[loveoforganic]

Jeff is right. Buoyant force is directly proportional to the volume of water displaced and has nothing to do with the density of the object. All the objects are the same size and all begin fully submerged, so all have the same buoyant force.

 

[aurevoir0711]

I get why part I and III are correct, just not sure why II is wrong.

Here is my approach. Since the balls have same size, but they all act differently, their weight in the water is different due to their density difference.

Ball A sinks, so it has the highest density, where p ball A * V * g (weight of ball A) > p water * V * g (buoyant force). I will refer to balls' volumes as V as they all have same volume.

Ball B remains stationary, so it has the second highest density, where p ball B * V * g (weight of ball B) = p water * V * g (buoyant force).

However, for Ball C, since it starts rising, it has the lowest density and its density is lower than that of water. This results in force balance equation, p ball C * V * g (weight of ball B) < p water * Vdisplaced * g (buoyant force).

Because the object rises to the surface until W (weight) = B (buoyant force), some portion of the ball will not be submerged in the water when the ball reaches equilibrium. This means that volume of water displaced by ball C would be smaller than its volume. Hence, I would think that the buoyant force is smaller for ball C compared to other two options because less fluid is displaced. It would still get the right answer, but the explanation says that the buoyant force is same for all three balls since they all have equal volume. I see their points if all three are fully submerged, but ball C can't be fully submerged based on my understanding... I think I am missing an important concept.

Also, generally speaking, if two objects have different masses and same volumes but both of them float, doesn't the object with higher mass have more buoyant force underneath because more volume would be submerged?

I am a little confused on this topic. Any thoughts on this? I would sincerely appreciate your response.

 

[aldol16]

Because the object rises to the surface until W (weight) = B (buoyant force), some portion of the ball will not be submerged in the water when the ball reaches equilibrium. This means that volume of water displaced by ball C would be smaller than its volume. Hence, I would think that the buoyant force is smaller for ball C compared to other two options because less fluid is displaced. It would still get the right answer, but the explanation says that the buoyant force is same for all three balls since they all have equal volume. I see their points if all three are fully submerged, but ball C can't be fully submerged based on my understanding... I think I am missing an important concept.

Also, generally speaking, if two objects have different masses and same volumes but both of them float, doesn't the object with higher mass have more buoyant force underneath because more volume would be submerged?

Well, as you point out, the buoyant force is calculated by the displaced volume, not the actual volume of the object, unless that object is fully submerged. So if you're forcibly holding the three balls underwater at t = 1, then they will all have the same buoyant force acting upon them. Then you release the balls. Now they're no longer at equilibrium because the force you applied to hold them in place is suddenly removed. So now, there will be two forces acting on the balls - the buoyant force and gravity. For the least dense one, the buoyant force exceeds that of gravity and it rises. For the most dense one, the buoyant force is less than that of gravity and it sinks. Eventually, equilibrium will be reached and the buoyant force will again be balanced by gravity. What's changed? Well, the V(displaced), of course. For the least dense one, the buoyant force is going to go down as the object rises out of the water until the V(displaced) is small enough to reduce the buoyant force to equal that of gravity. So the buoyant forces at the end should be different because the V(displaced) will be different in each case, after the balls have come to equilibrium. And since the least dense ball will now have the least volume displaced, it's going to have the least buoyant force acting on it.

 

[aurevoir0711]

Well, as you point out, the buoyant force is calculated by the displaced volume, not the actual volume of the object, unless that object is fully submerged. So if you're forcibly holding the three balls underwater at t = 1, then they will all have the same buoyant force acting upon them. Then you release the balls. Now they're no longer at equilibrium because the force you applied to hold them in place is suddenly removed. So now, there will be two forces acting on the balls - the buoyant force and gravity. For the least dense one, the buoyant force exceeds that of gravity and it rises. For the most dense one, the buoyant force is less than that of gravity and it sinks. Eventually, equilibrium will be reached and the buoyant force will again be balanced by gravity. What's changed? Well, the V(displaced), of course. For the least dense one, the buoyant force is going to go down as the object rises out of the water until the V(displaced) is small enough to reduce the buoyant force to equal that of gravity. So the buoyant forces at the end should be different because the V(displaced) will be different in each case, after the balls have come to equilibrium. And since the least dense ball will now have the least volume displaced, it's going to have the least buoyant force acting on it.

That was my thought process exactly. I think I just overcomplicated the problem by thinking that the answer wanted Ball C at equilibrium position. Thank you for your answer.