Find the volume of the object sticking out of the water

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Homework Help Overview

The problem involves determining the volume of an object that is partially submerged in water, specifically focusing on the portion that remains above the water's surface. The context includes concepts of density and buoyancy, with the object having a specified density and mass, and the surrounding seawater density provided for calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Archimedes' Principle and the relationship between buoyant force and the weight of the displaced fluid. There are questions about how to set up the equations correctly and whether to use the weight of the object or the buoyant force in calculations.

Discussion Status

Some participants have provided guidance on how to apply the principles involved, while others are exploring different interpretations of the equations. There is an ongoing exchange about the correct setup of the equations and the implications of the results, with no clear consensus reached on the final answer.

Contextual Notes

Participants are navigating potential confusion regarding the use of gravitational acceleration in their calculations and the interpretation of the results, including the volume of the object and the portion above water. There is also mention of the relationship between the volume of the object and its density, which is critical to the problem.

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Homework Statement


One of the problems with debris in the ocean is that it is often difficult to see, because much of the object is under the surface of the water. An object with a density of 801 kg/m3 and a mass of 1039 kg is thrown into the ocean. Find the volume of the object sticking out of the water (use ρsea water = 1024 kg/m3).


Homework Equations


p = m/v

The Attempt at a Solution


I am not sure how to go about solving this problem?
 
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What is p=m/v here?

Remember Archimedes' Principle? Use it. State it if you're not sure how to use it.
 
Fb = Pf * Vf * g
Not sure how to use it though.
Pf = 1000kg/m^3
 
It means that the upward buoyant force on an object is equal to the weight of fluid it displaces. So, if d is the density of the fluid, V is the vol of the object that is under the fluid, then the upward force is d*V*g. Can you apply it now?
 
Ok so I understand the concept but when I solve it do I use 0 for Fb and then solve for Vf? Or am I supposed to use the weight of the object there?
 
Let me illustrate by using the data given in this prob. Suppose V is the vol of the object that is submerged. Then Fb = weight of water with vol V = V*1024.

This upward force balances the total weight of the body, since the body is floating. So,
V*1024 = 1039*g.

So, you know V. You also know that the total vol of the object = mass/density = 1039/801.

Now, by subtracting, you can tell how much is sticking out.
 
So is Fb always going to equal the density * V?
okay I get an answer of 8.65 m^3
it says incorrect I did 9.94 - 1.29
 
I didn't mean for you to just copy me blindly. By mistake I have written, "V*1024 = 1039*g", where it should be V*1024*g, because I'm equating weights. Try it now.
 
Sorry I didn't know..I am still trying to figure this out...
V * 1024 * g = 1039 * g
V = (1039 * g) / (1024 * g) = 1.01

1039/801 = 1.29

So I subtract those volumes? I get 0.28...very small?
 
  • #10
Right answer. What's the problem?

The whole volume of the object is 1.29 cubicm. Sticking out portion is 1/5th of the whole. In an iceberg, only 1/12 of the whole sticks out.
 

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