Calculate Raft Buoyancy Depth | Density, Volume & Pressure Formula

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Homework Help Overview

The problem involves calculating the submerged depth of a wooden raft in fresh water, given its density, surface area, and volume. The context is centered around buoyancy and pressure principles in fluid mechanics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the buoyancy principle and the relationship between pressure, density, and submerged depth. There are attempts to derive the submerged depth using different equations and substitutions, with some questioning the correctness of their setups and calculations.

Discussion Status

The discussion is active, with participants checking calculations and questioning the validity of their approaches. Some have provided alternative calculations, while others express confusion about their initial setups. There is no explicit consensus on the correct method yet, but guidance has been offered regarding the equations used.

Contextual Notes

Participants are navigating through the implications of their calculations and the assumptions made about the raft's properties and the fluid's density. There is mention of the need for units in the final answer, indicating a focus on precision in the calculations.

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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
 
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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.
 

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