Archimedes' Buoyancy: Which Ball Will Experience a Greater Buoyant Force?

[Jmiz]

Question: Balls A and B of equal mass are floating in a swimming pool, as shown below. Which will produce a greater buoyant force? (Image shows two circles with circle A larger than circle B)

A. Ball A
B. Ball B
C. The forces will be equal
D It is impossible to know without knowing the volume of each ball

I think this question is poorly written. Based on the givens, I deduced that volume A > volume B, and thus density A has to be less than density of B in order for the two balls to have equal mass. So, the volume submerged in order for A to float has to be less than the volume submerged for B to float. Thus, it doesn't necessarily have to be ball A that will produce a greater buoyant force. Thoughts? Thanks.

 

[Buzz Bloom]

Thoughts?

Hi Jmiz:

Your analysis seems completely correct to me, up to

So, the volume submerged in order for A to float has to be less than the volume submerged for B to float.

although you didn't say explicitly your choice of A, B, C, D.

I don't understand you comment:

Thus, it doesn't necessarily have to be ball A that will produce a greater buoyant force.

Did you pick "A"? If so I am surprised. I think the greater the submerged volume then the greater the buoyant force.

ADDED
I goofed. The buoyant force = the gravity force. Therefore "C".
https://en.wikipedia.org/wiki/Buoyancy

Regards,
Buzz

 

[Merlin3189]

What does it mean to float?
Apart from any other properties or deductions, what is the essential of floating?

 

[Jmiz]

Hi Jmiz:

Your analysis seems completely correct to me, up to

although you didn't say explicitly your choice of A, B, C, D.

I don't understand you comment:

Did you pick "A"? If so I am surprised. I think the greater the submerged volume then the greater the buoyant force.

ADDED
I goofed. The buoyant force = the gravity force. Therefore "C".
https://en.wikipedia.org/wiki/Buoyancy

Regards,
Buzz

Yes the answer was ball A. Kaplan answer key probably assumed that the volume submerged will be greater for ball A because it was drawn as larger...

 

[Jmiz]

What does it mean to float?
Apart from any other properties or deductions, what is the essential of floating?

The essential of floating is when the buoyant force equals the weight.

 

[Merlin3189]

And you are told the balls have equal mass - what does that say about their weight?

Then you are asked about about the buoyant forces.

(PS I just noticed your other post where you say, you looked up the answer (in Kaplan?) and ball A has the greater buoyant force. IS that what you said? Or have I misread it?)

 

[Jmiz]

And you are told the balls have equal mass - what does that say about their weight?

Then you are asked about about the buoyant forces.

(PS I just noticed your other post where you say, you looked up the answer (in Kaplan?) and ball A has the greater buoyant force. IS that what you said? Or have I misread it?)

The answer key stated choice A. I picked choice C because it was the only one that made some what sense to me, but if there was a choice that said "it is impossible to know without knowing the density of each ball" I would have selected that.

Given consideration of your information, the balls will have equal weight. So the buoyant forces resulting from each ball will be equal, which makes logical sense to me. However, Kaplan's answer key disagrees:

"The buoyant force is equal to the weight of water displaced, which is quantitatively expressed as F = m (fluid displaced) g = p (fluid) V (fluid displaced) g
The volume of displaced fluid is equal to the volume of the ball. The density of the fluid remains constant. Therefore, because ball A has a larger volume, it will displace more water and experience a larger buoyant force."

The situation that their answer applies for I believe is not for this question's situation but for a setting that has both balls completely submerged in water, which the question stem does not explicitly state.

 

[Doc Al]

The situation that their answer applies for I believe is not for this question's situation but for a setting that has both balls completely submerged in water, which the question stem does not explicitly state.

Just curious, does the image show the balls floating on the surface (as the text implies) or submerged (as the answer key implies)?

 

[Jmiz]

Just curious, does the image show the balls floating on the surface (as the text implies) or submerged (as the answer key implies)?

The image is just as I described: 2 circles nothing else one circle, labeled A, is larger than the second, labeled B. No bodies of water in picture.

 

[Doc Al]

The image is just as I described: 2 circles nothing else one circle, labeled A, is larger than the second, labeled B. No bodies of water in picture.

Strange. As you said, a poorly crafted question. If you ignore the diagram, which is meaningless if it doesn't show the water level, then the answer is clear. (And not what the answer key gives.)

 

[Merlin3189]

Well, I don't know what to make of that book answer. As far as i can see, you would be absolutely correct with C. If they have equal weight and are floating, then the buoyant force must be the same and the volume displaced must be the same. All other factors are irrelevant.

The only possibility that I can see is, they are asking, "what WOULD the buoyant forces be, IF the balls were pushed under until completely submerged" (rather than left floating.)

The comment that, "The volume of displaced fluid is equal to the volume of the ball." is nonsense. It is one of the common mistakes that students make, to assume that the volume displaced is always equal to the volume of the object. This is only true when the object is completely submerged and this is normally when the object is sunk or barely floating.
When an object is "floating" (even barely floating), the only thing you can say for sure, is that the weight of the object is equal to the weight of fluid displaced. Thence the weight of fluid of known density allows you to calculate the volume of fluid displaced and this must be the submerged volume of the object (, which must be less than or equal to the total volume of the object, else it would not float.)

Your comment, "if there was a choice that said "it is impossible to know without knowing the density of each ball" I would have selected that." is effectively D. If you knew the volume of each ball, you would know the relative density and if you knew the density of each ball, you would know the relative volumes of the balls ( & NB, the total volumes of the balls, not the submerged volumes.) But neither would matter one jot! It is the weight (or mass) and the fact that they are floating that answers the question. The only volumes that would help you answer the question are the submerged volumes and neither D, nor your alternative formulation of knowing the density, will tell you that.

 

[Doc Al]

As far as i can see, you would be absolutely correct with C. If they have equal weight and are floating, then the buoyant force must be the same and the volume displaced must be the same. All other factors are irrelevant.

Exactly.

 

[russ_watters]

No bodies of water in picture.

That's odd because the way you worded the question states that the body of water is in the picture ("floating in a swimming pool, as shown below"). You sure you didn't miss the surface and think it was a border of some kind?

 

[Jmiz]

Well, I don't know what to make of that book answer. As far as i can see, you would be absolutely correct with C. If they have equal weight and are floating, then the buoyant force must be the same and the volume displaced must be the same. All other factors are irrelevant.

The only possibility that I can see is, they are asking, "what WOULD the buoyant forces be, IF the balls were pushed under until completely submerged" (rather than left floating.)

The comment that, "The volume of displaced fluid is equal to the volume of the ball." is nonsense. It is one of the common mistakes that students make, to assume that the volume displaced is always equal to the volume of the object. This is only true when the object is completely submerged and this is normally when the object is sunk or barely floating.
When an object is "floating" (even barely floating), the only thing you can say for sure, is that the weight of the object is equal to the weight of fluid displaced. Thence the weight of fluid of known density allows you to calculate the volume of fluid displaced and this must be the submerged volume of the object (, which must be less than or equal to the total volume of the object, else it would not float.)

Your comment, "if there was a choice that said "it is impossible to know without knowing the density of each ball" I would have selected that." is effectively D. If you knew the volume of each ball, you would know the relative density and if you knew the density of each ball, you would know the relative volumes of the balls ( & NB, the total volumes of the balls, not the submerged volumes.) But neither would matter one jot! It is the weight (or mass) and the fact that they are floating that answers the question. The only volumes that would help you answer the question are the submerged volumes and neither D, nor your alternative formulation of knowing the density, will tell you that.

Thanks for your very helpful explanation!

 

[Jmiz]

Thanks a lot to everyone who contributed! It seems clear to me that this was a mistake by Kaplan.

 

[nasu]

@russ_watters
You are right, of course, but looking for the problem in their MCAT book, the OP is also right. :)https://books.google.ca/books?id=aY-9AwAAQBAJ&pg=PA143&lpg=PA143&dq=Balls+A+and+B+of+equal+mass+are+floating+in+a+swimming+pool,+as+shown+below&source=bl&ots=zpB6VIsTxC&sig=3wYreTW6sHvRlzkrbR0uM6ZnLZw&hl=en&sa=X&ved=0ahUKEwjQy-amzZPKAhUH7R4KHWEcChsQ6AEIKjAC#v=onepage&q=Balls%20A%20and%20B%20of%20equal%20mass%20are%20floating%20in%20a%20swimming%20pool%2C%20as%20shown%20below&f=false

 

[Buzz Bloom]

Thanks a lot to everyone who contributed! It seems clear to me that this was a mistake by Kaplan.

Hi Jmiz:

If both balls have a density greater than water, then both would sink. Since A is larger, it would displace more water, have greater buoyancy, and therefore sink slower

Regards,
Buzz

 

[Dan Allred]

I would think that the "bouyancy force" for two objects of identical mass are the same--even if one of the two is not floating--provided both are at rest.

 

[jbriggs444]

I would think that the "bouyancy force" for two objects of identical mass are the same--even if one of the two is not floating--provided both are at rest.

Ignoring the corner case of neutral buoyancy, a mass that is not floating is either sinking or resting on the bottom. If it is sinking, it is not at rest. If it is resting on the bottom, it is not displacing its own weight.

 

[Merlin3189]

I would think that the "bouyancy force" for two objects of identical mass are the same--even if one of the two is not floating--provided both are at rest.

I don't know what one can say about this. It appears to be a completely different concept of buoyancy from that I know (and since the OP asked about two FLOATING balls, it seems irrelevant here.)
For me, buoyancy equals the weight of fluid displaced, which in turn equals g x the density of the liquid x the volume of liquid displaced.
Nowhere does the mass of the object ever enter into consideration.
For floating objects THEIR volume is also irrelevant.

As for the 'at rest' bit, I prefer to think that the buoyant force is unaffected by movement and to consider dynamic effects separately, but that is a matter of choice.

 

[Jmiz]

Hi Jmiz:

If both balls have a density greater than water, then both would sink. Since A is larger, it would displace more water, have greater buoyancy, and therefore sink slower

Regards,
Buzz

Buzz, I believe based on the stem of the question, one cannot conclude A would sink slower based on your logic because though it has a larger volume it would have a lower density because both A and B needs to have the same mass. Secondly, stating that both balls have a density greater than water is out of the question, which stated both balls are floating on water.

 

[Doc Al]

I wonder if you have an older version of the Kaplan book. I just found the following on Amazon:

Looks like they cleaned up the wording in this edition. Link: https://www.amazon.com/dp/1625231180/?tag=pfamazon01-20

 

[Merlin3189]

I wouldn't say they have cleaned up the question. It is a completely different question. In the first they are floating and in the second they are submerged.

I find it more confusing than the original question.
First they say the masses are equal. This is irrelevant and can only be a distractor.
Then they offer option D. Since you do need to know something about the volumes of the balls, this is a correct answer. So you are left wondering whether you really can draw any conclusion about their volumes from the diagram. If you are supposed to take the diagram as evidence rather than just illustration, then you already know their volumes, because you can measure their diameters and you can conclude A is more buoyant, answer A.
The only argument one could put against D being correct, is to say you don't need to know the exact volumes, only which is the bigger volume.

 

[Doc Al]

I wouldn't say they have cleaned up the question. It is a completely different question. In the first they are floating and in the second they are submerged.

Just changing "floating" to "submerged" makes a big difference. I'd say the revised question is pretty clear. (And agrees with the answer provided.)

First they say the masses are equal. This is irrelevant and can only be a distractor.

Nothing wrong with distractors. ;)

Then they offer option D. Since you do need to know something about the volumes of the balls, this is a correct answer. So you are left wondering whether you really can draw any conclusion about their volumes from the diagram. If you are supposed to take the diagram as evidence rather than just illustration, then you already know their volumes, because you can measure their diameters and you can conclude A is more buoyant, answer A.
The only argument one could put against D being correct, is to say you don't need to know the exact volumes, only which is the bigger volume.

I see your point, somewhat. Nonetheless, I believe one can reasonably assume that the diagram is relevant and accurate enough. I would have picked A without hesitation. You don't need the volume, only the relative volume, which the diagram provides.

 

[Merlin3189]

Fair enough.

 

[nasu]

The link in post 14 is to the older version, where it says "float in the pool".
I suppose it was easier to change a few words than to change the image, when they realized the error.
They should have changed option D as well to avoid confusion. Even for the floating case.
Maybe in the next edition, if they read PF.:)

 

[Quark_abc]

For the 'floating' question, if we take the atmospheric pressure into account, then force due to atmospheric pressure (downward) will be more on the bigger ball (greater surface area). The job of the buoyant force is to balance the weight and atmospheric force. Weights of A and B being equal, as atmospheric force is more on bigger ball A, hence buoyant force will be more on A, the bigger ball.

 

[Merlin3189]

That's an interesting thought. But why do you think the net downward atmospheric force on the ball A is greater than on B?

If the sizes were such that B were exactly half submerged, then A, the larger, would be less than half submerged. So the atmosphere is bearing only downwards on B, but partially upward on A. In fact A needs less buoyant force from the water than B.

Putting aside this atmopheric pushing, because we normally simplify the calculations using buoyancy notions,
If we had the two balls floating on the water, but a vacuum above, then the weight of water displaced would be the same for both balls of equal mass.
Both balls would have the same volume submerged, so A would have greater volume in the vacuum above the liquid.
So when we now fill the vacuum with another fluid (air in the question, but we could also think of a light hydrophobic oil) both balls now displace some of that fluid and gain some upward buoyant force from that. A gets more buoyant force from this new fluid than B.
Since Weight of ball = Buoyant force from water + Buoyant force from new fluid
if A gets more lift from the new fluid than B, then A must get less lift from the water than B.

 

[Quark_abc]

Very well explained merlin3189. From above what I can gather is:
1. If balls are in vacuum, buoyant force will be same on both A and B (since weights are same)
2. If another lighter fluid is above water, then buoyant force due to the other fluid will be more on A as compared to B because more of A's volume is above water. This means buoyant force due to water will be less on A than B.
3. I have a doubt here with respect to atmosphere as the other fluid above. As in the reply above, Merlin 3189 talks of a 'downward' atmospheric bearing on B in case it is half (or more than half submerged). Why is it that when we treat the atmospheric effect by buoyancy concept, the force is upward, but if seen as Force= pressure x protected area, the push is downwards. What's the discrepancy here?

 

[jbriggs444]

3. I have a doubt here with respect to atmosphere as the other fluid above. As in the reply above, Merlin 3189 talks of a 'downward' atmospheric bearing on B in case it is half (or more than half submerged). Why is it that when we treat the atmospheric effect by buoyancy concept, the force is upward, but if seen as Force= pressure x protected area, the push is downwards. What's the discrepancy here?

Atmospheric pressure does not go away at the surface of the water. The water pressure at depth h is the sum of atmospheric pressure plus $\rho g h$. So the presence of atmospheric pressure above increases the water pressure below and the net is almost zero.

The total buoyant force on the ball is still given by the weight of the displaced fluid. It is the sum of the weight of the displaced water [by the portion of the ball that is below the surface] plus the weight of the displaced air [by the portion of the ball that is above the surface]. We normally ignore the air because its contribution is so small.

 

[Dr_Zinj]

Question: Balls A and B of equal mass are floating in a swimming pool, as shown below. Which will produce a greater buoyant force? (Image shows two circles with circle A larger than circle B)

A. Ball A
B. Ball B
C. The forces will be equal
D It is impossible to know without knowing the volume of each ball

I think this question is poorly written. Based on the givens, I deduced that volume A > volume B, and thus density A has to be less than density of B in order for the two balls to have equal mass. So, the volume submerged in order for A to float has to be less than the volume submerged for B to float. Thus, it doesn't necessarily have to be ball A that will produce a greater buoyant force. Thoughts? Thanks.

The buoyant force on the balls will be the same, as the balls have equal mass. Which means as long as they are floating, they'll displace the same volume of water. A will float higher in the water than B, but will displace water over a larger area of the surface. The buoyant force will only change if you use force to submerge the balls, in which case A, displacing more water, will require greater force to push down than B.

 

[Jim Kadel]

Back to the original question: not poorly worded, IMHO, but...
..it would seem simply a case of the answer key being incorrect: A instead of C for "the floating situation".
[It happens, altho students claim it much more frequently :>]

 

[Rerry]

Hi Jmiz:

Your analysis seems completely correct to me, up to

although you didn't say explicitly your choice of A, B, C, D.

I don't understand you comment:

Did you pick "A"? If so I am surprised. I think the greater the submerged volume then the greater the buoyant force.

ADDED
I goofed. The buoyant force = the gravity force. Therefore "C".
https://en.wikipedia.org/wiki/Buoyancy

Regards,
Buzz

I was just wondering, if ball A has the same mass as B, but A being larger than B, would A not cover more surface area than B, so there for would have more volume coming in contact with water, creating a greater force on A, If you have two balls, one having A larger surface area, if you try to push these two balls underneath the water,the bigger ball you get more resistance and it's harder to push under, The smaller ball B isn't as hard because it does not have the same surface area pushing back. A being larger has to be more buoyant. Explain to me why C is correct.

 

[Doc Al]

Back to the original question: not poorly worded, IMHO, but...
..it would seem simply a case of the answer key being incorrect: A instead of C for "the floating situation".

Since the answer key, detailed solution, and diagram all seem to be consistent with the word "floating" changed to "submerged" in the problem statement, I suspect that was the error. That is further supported by a later version of the book showing just that correction.

 

[Aleena786]

Ball A. Phrased wrong but imagine a flat plate. With larger surface area displaced when it's lying flat, it floats upwards. When tilted vertically it would sink.