Archimede's Principle Question?

  • Thread starter Thread starter haflanagan
  • Start date Start date
  • Tags Tags
    Principle
Click For Summary
SUMMARY

The discussion centers on a geologist's method to determine the solid fraction of a hollow rock specimen using Archimede's Principle. The specimen weighs twice as much in air as it does in water, with the solid part's density given as 5.0 x 103 kg/m3. The key equations involved include the buoyant force and the relationship between weight, density, and volume. Participants emphasize the importance of calculating the apparent weight in both air and water to solve for the solid volume fraction.

PREREQUISITES
  • Understanding of Archimede's Principle
  • Knowledge of buoyant force calculations
  • Familiarity with density and volume relationships
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation of Archimede's Principle in fluid mechanics
  • Learn how to calculate buoyant force and apparent weight in different fluids
  • Explore density calculations for composite materials
  • Practice solving problems involving hollow objects and their properties
USEFUL FOR

Students in physics or engineering, educators teaching fluid mechanics, and anyone interested in material properties and buoyancy principles.

haflanagan
Messages
5
Reaction score
0

Homework Statement



To verify her suspicion that a rock specimen is hollow, a geologist weighs the specimen in air and in water. She finds that the specimen weighs twice as much in air as it does in water. The density of the solid part of the specimen is 5.0 x 10^3 kg/m3. What fraction of the specimen’s apparent volume is solid?

Homework Equations



Buoyant force = weight of water displaced
Weight = density * volume * g

The Attempt at a Solution



I don't even know...
Pr=density rock
Pa=density air
Pw=density water
Vs=volume solid
Vh=volume hollow

I thought that maybe:
Weight of rock in air = Pr*vs*g+Pa*vh*g
Therefore, weight of rock in water = (Pr*vs*g+Pa*vh*g)/2

And that's as far as I got. And I don't even think that's right. I'm so lost.
 
Physics news on Phys.org
You are given the density of rock as part of the problem description. So you do know ρr

The density of air is low enough that by comparison to rock or water, it is negligible. So you can safely assume that ρa = 0.

The density of water is a well known constant. One kilogram per liter. So you know ρw

Your formula for the weight of the rock in air is technically correct. But the problem is talking about the apparent weight of the rock in air. That would be the true weight minus buoyancy. Can you write a formula for the apparent weight of the rock in air?

Your formula for the apparent weight of the rock in water is based on the givens of the problem -- it's half of the apparent weight in air. But there is also another formula for the apparent weight of the rock in water...
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
799
  • · Replies 5 ·
Replies
5
Views
2K
Replies
5
Views
4K
Replies
12
Views
2K
  • · Replies 14 ·
Replies
14
Views
7K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
Replies
7
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K