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

[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.

 

[Doc Al]

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.

What determines the buoyant force is not surface area, but volume of fluid displaced.

 

[insightful]

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.

Disregarding surface tension effects, your flat plate will either float (if less dense than water) or sink (if more dense than water) regardless of orientation.

 

[interestedcrl]

interested crl: I don't see dwg. details. But it gets more interesting if the bottom and sides of liquid container matches pretty well the contours of the object placed in it.
Then the liguid level rises surprising high fast, like a person's body in a bathtub, and theoretically a special body could then float on almost no 'weight' of water, and this makes a cute surprise joke for, say, a special solid wood cylinder in a special sized (nearly fitting) cup.