Calculate Raft Buoyancy Depth | Density, Volume & Pressure Formula

  • Thread starter Thread starter Jbreezy
  • Start date Start date
  • Tags Tags
    Buoyancy
Click For Summary
SUMMARY

The discussion focuses on calculating the submerged depth of a wooden raft in fresh water using the principles of buoyancy. The raft has a density of 625 kg/m³, a surface area of 5.5 m², and a volume of 1.5 m³. The correct submerged depth, h, is determined to be 0.9375 meters, derived from the equation h = (density of wood x volume of wood) / (surface area x density of water). Participants clarified the correct application of buoyancy equations and identified errors in initial calculations.

PREREQUISITES
  • Understanding of buoyancy principles and Archimedes' principle
  • Familiarity with pressure equations: P = ρgh
  • Knowledge of density calculations for different materials
  • Ability to manipulate algebraic equations for problem-solving
NEXT STEPS
  • Study the derivation and application of Archimedes' principle in fluid mechanics
  • Learn about pressure calculations in different fluid contexts
  • Explore the relationship between density, volume, and mass in physics
  • Practice solving buoyancy problems with varying materials and conditions
USEFUL FOR

Students in physics or engineering, educators teaching fluid mechanics, and anyone interested in understanding buoyancy and related calculations.

Jbreezy
Messages
582
Reaction score
0

Homework Statement



A raft is constructed of wood having a density of 625 kg/m^3. It's surface area is 5.5 m^2, and it's volume is 1.5 m^3. When the raft is placed in fresh water (denistiy = 1000 kg/m^3), to what depth, h, is the raft submerged under the water?

Homework Equations



P = ρgh
Where P = pressure, ρ = density

The Attempt at a Solution



P = ρgh
Sub, F /A = P
F/A = ρgh

F = mg (g will cancel when we rearrange to solve for h)
m here is the mass of wood which is the density of wood times it's volume.
So, V x ρ_wood = M_wood.
Putting this all together.
And solving for h

M_wood / ( A(ρ_h20) = h
With values I got.

.1704 meters.

What do you think? Right wrong please confirm my process.
Thanks
 
Physics news on Phys.org
Check your calculation.
625x5.5x1.5x9.8 = 5.5xhx1000x9.8
Find h.
 
I didn't get the same answer when I substituted h back into 625x5.5x1.5x9.8 = 5.5xhx1000x9.8. The equations were not equivalent. Where did I mess up?Thanks
 
No I didn't . Is it the right height though?
 
Last edited:
Calculate h from my post#2. What is the value of h?
 
625x5.5x1.5x9.8 = 5.5xhx1000x9.8

625 x 5.5 x 1.5 x 9.8/5.5 x 1000 x 9.8 = h
h = I got .9375?
 
Put the unit. That is the correct answer.
 
What? I know what you mean put the unit but what did I do wrong?
 
h = 0.9375 m.
 
  • #10
I know the unit. I was just asking from my original set up what did I do wrong? I don't understand why my setup is wrong?
 
  • #11
Show your calculation. How did you get your answer?
 
  • #12
(1.5 m^3 x 625 kg/ m^3 ) / (5.5m^2 x 1000 kg/m^3) = .1704m

? This is what I'm asking why is this the wrong approach to the problem?
Mass of wood / area times density of water
Thanks
 
  • #13
Mass of the wood = density of the wood x area of the raft x height of the raft. In your calculation, you have left the area.
 
  • #14
I think I just used the wrong equation well sort of.
I have F/A = ρgh
Should be m_wg = ρgh
m_w = mass of wood.
Because if I do F/A then I have an extra multiplication by 1/A. In your equation that cancelled.

Yeah thanks
 
  • #15
You are welcome.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
695
Replies
10
Views
11K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K