Fluid Statics (multiple choice question)

AI Thread Summary
The discussion centers on buoyant force as described by Archimedes' Principle, which states that the buoyant force on a submerged object equals the weight of the fluid it displaces. For two blocks of the same size, one lead and one Styrofoam, the buoyant force is the same since it depends solely on the volume of fluid displaced, not the object's density. In the case of a boat floating in salt water versus fresh water, the buoyant force is greater in salt water due to its higher density, allowing the boat to displace less volume while still floating. The key takeaway is that buoyant force is determined by the displaced fluid's weight, not the density of the object itself. Understanding these principles clarifies the relationship between buoyancy and fluid density.
Ariel P
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Homework Statement



#1 If you hold a block of lead and an identical size block of Styrofoam under water, the buoyant force will
a) be greater on the lead block.
b) be greater on the Styrofoam block.
c) be the same on both blocks.
d) depend on the depth that each block is held. #2 Salt water has greater density than fresh water. A boat floats in both fresh and in salt water. Where is the buoyant force greater on the boat?
a) in salt water
b) in fresh water
c) buoyant force is the same in both
d) not enough information given to determine the answer

Homework Equations



I think this principle is relevant,
Archimedes' Principle:
A fluid exerts an upward buoyant force on a submerged object equal in magnitude to the weight of the volume of fluid displaced by the object.

The Attempt at a Solution



#1 Well, what I can guess from that principle is that the buoyant force will be the same beacuse it depends on the "weight of the volume of fluid displaced" and both the lead block and styrofoam block are the same size.

#2 I got that the answer is (C) from somewhere but I'm unable to reason for that answer. From my experience, I assume the boat will float "more" in salt water than fresh water, which would differ the amount of volume of water displaced. Then, the answer cannot be (C). I'm not so sure,..thanks!
 
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Your reasoning for the first part is right. For the second part use what you know,

"weight of the volume of fluid displaced"

For the second part the boat weighs the same so the "weight of the volume of fluid displaced" is the same but in one case the fluid is water and in the other it is more dense salt water. So the "weight of the volume of fluid displaced" is,

W = ρ_water*volume_water displaced =
ρ_salt water*volume_salt water displaced

Because the salt water is more dense the boat does not have to "displace" as much water.

Hope this was not too confusing.
 
Spinnor said:
Your reasoning for the first part is right. For the second part use what you know,

"weight of the volume of fluid displaced"

For the second part the boat weighs the same so the "weight of the volume of fluid displaced" is the same but in one case the fluid is water and in the other it is more dense salt water. So the "weight of the volume of fluid displaced" is,

W = ρ_water*volume_water displaced =
ρ_salt water*volume_salt water displaced

Because the salt water is more dense the boat does not have to "displace" as much water.

Hope this was not too confusing.

I don't see how the buoyant force on the Styrofoam would be the same as the lead since they may be the same size, but their densities are far different

and the equation is

ρsubfluid = msub/mfluid

ρst<<ρPb

and for c) he is right because the boat is not being held at the same depth in salt water and water unlike the first question.
 
Liquidxlax said:
I don't see how the buoyant force on the Styrofoam would be the same as the lead since they may be the same size, but their densities are far different

and the equation is

ρsubfluid = msub/mfluid

ρst<<ρPb

and for c) he is right because the boat is not being held at the same depth in salt water and water unlike the first question.

If the following is correct,

"Archimedes' Principle:
A fluid exerts an upward buoyant force on a submerged object equal in magnitude to the weight of the volume of fluid displaced by the object."

then I think I correctly translated the above into,

W = ρ_water*volume_water displaced =
ρ_salt water*volume_salt water displaced

But I make mistakes all the time.
 
Liquidxlax said:
I don't see how the buoyant force on the Styrofoam would be the same as the lead since they may be the same size, but their densities are far different

The buoyant force is a function of how much liquid is displaced and the density of the liquid. Any two objects of the same volume will have the same buoyant force if they are held entirely under the liquid's surface. Density of the object is not a factor.

Now if you were under water and you wanted to weigh 1 cubic foot of aluminum and one cubic foot of steel, naturally the steel would weigh more. But the buoyant force on each is the same.
 
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