Buoyant Force Conceptual Question

Click For Summary
SUMMARY

The discussion centers on a physics problem involving buoyant force and the equilibrium of a wooden block floating in water. The block, initially half-submerged, requires an additional 20g of mass to reach the waterline, indicating that the mass of the block is 40g. The key equations discussed include the buoyant force formula, F_b = (p_f)(V_f)(g), and the mass density equation, m = pV. The application of Archimedes' principle clarifies that the buoyant force equals the weight of the displaced fluid, which changes as additional weight is added to the block.

PREREQUISITES
  • Understanding of buoyant force and Archimedes' principle
  • Familiarity with mass density calculations
  • Knowledge of equilibrium in fluid mechanics
  • Basic algebra for solving equations
NEXT STEPS
  • Study Archimedes' principle in detail
  • Explore the relationship between buoyant force and submerged volume
  • Learn about density variations in different materials
  • Investigate real-world applications of buoyant force in engineering
USEFUL FOR

Students studying physics, particularly those focusing on fluid mechanics, as well as educators looking for practical examples of buoyant force and equilibrium concepts.

mantillab
Messages
27
Reaction score
0

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)?
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
36
Views
4K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 18 ·
Replies
18
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K