Buoyant Force Conceptual Question

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Homework Help Overview

The problem involves a rectangular wooden block floating in water, with a specific scenario where masses are added to the block until it is level with the waterline. The context centers around buoyant force and the relationship between the weight of the block and the displaced water.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between buoyant force and weight, with some questioning how the buoyant force changes when additional weight is added to the block. Others discuss Archimedes' principle and the implications of equilibrium in this context.

Discussion Status

The discussion is active, with participants providing guidance on the principles involved, such as the need to consider how buoyant force relates to the weight of the block and any added masses. There is an exploration of how the buoyant force changes with the addition of weight, indicating a productive direction in understanding the problem.

Contextual Notes

Some participants express uncertainty about the initial conditions and the implications of the block being half submerged. There is also mention of assumptions regarding the densities of the block and the fluid remaining constant throughout the discussion.

mantillab
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Homework Statement


A rectangular wooden block of weight W floats with exactly one-half of its volume below the waterline. Masses are stacked on top of the block until the top of the block is level with the waterline. This requires 20g of mass. What is the mass of the wooden block?

Homework Equations



Buoyant Force:
F_b = (p_f)(V_f)(g) = w_o = (p_o)(V_o)(g)

Mass Density:
m = pV

The Attempt at a Solution



Since the block is initially half submerged, its density is one-half of the water:
p_o = 0.50g/cm^3

I'm assuming that the density of the block and the density of the fluid don't change. Without the volume, I don't see how to use the formulas...

My only guess at the answer is 40g because it would require 20g to overcome the initial buoyant force (where the block floats halfway submerged)?
 
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No need to guess. Use Archimedes' principle and equilibrium.

(1) The buoyant force equals the weight of the displaced fluid--the volume of that fluid is just half the volume of the block. BF = weight of block.

(2) After adding weight, what happened to the buoyant force? (How does it compare to the previous buoyant force?) That new buoyant force BF' must equal the total weight of block plus added weights.
 
Sorry, this is all new to me..

If the wood block is floating, then the buoyant force is equal to the weight of the block. But since the wood block with the added weight is now submerged under the water, then the weight of the block must be greater than the buoyant force? Or does the buoyant force change to accommodate the new weight?
 
The buoyant force definitely changes as weight is added--that's why more water is displaced (it sinks lower). In the first case, the buoyant force just has to support the weight of the block; in the second case it must support the block plus the added weight.

Hint: By what factor does the buoyant force increase? (Compare the amount of water displaced.)
 
Thanks for the help! I think I get it now.
 

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