Calculating Buoyancy: Solving for the Height of a Boat in Salt Water

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Homework Help Overview

The discussion revolves around calculating the height a boat will float in salt water compared to fresh water, given its weight and surface area. The problem involves concepts of buoyancy and the relationship between displaced volume and water density.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between the boat's surface area, the height it floats, and the volume of water displaced. Questions arise regarding how to relate the submerged volume to the height above water and the implications of the boat's shape.

Discussion Status

Some participants have provided calculations for the displaced volumes in both fresh and salt water, while others are seeking clarification on how to interpret the surface area and its role in determining the height the boat will float. There is ongoing exploration of the implications of these calculations.

Contextual Notes

Participants note the importance of understanding the surface area as it relates to the water line and the distinction between the submerged and above-water portions of the boat. There is also mention of the need for clarity regarding the boat's shape to facilitate understanding.

BrainMan
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Homework Statement


A small boat weighing 1000 N has a surface area of 3 m^2. It floats only 5 cm above the water level when in a fresh-water lake. How high out of the water will it ride in a salt-water lake? Assume the surface area of the boat does not change as it rises (salt water has a density of about 1.03 x 10^3 kg/m^3).

Homework Equations


B = pVg

The Attempt at a Solution


I can use the above equation to solve for the volume of the water displaced which will be the same as the volume of the boat. Unfortunately I am not sure were to go from there. I am not sure how to find the relationship between the height the boat rises in the water and it's volume. I also am not sure what to do with the information it gave me on the boats surface area.
 
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The "surface area" of the boat should be understood as the area enclosed by the boat's outline at the water line. If this area is 3 m^2 and the boat rises by 5 cm, by how much will the submerged volume of the boat be reduced?
 
I'm sure it would be much clearer if they'd said the boat is in the shape of an upright rectangular box!
 
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jbriggs444 said:
The "surface area" of the boat should be understood as the area enclosed by the boat's outline at the water line. If this area is 3 m^2 and the boat rises by 5 cm, by how much will the submerged volume of the boat be reduced?
So it's not the surface area of the entire boat just the part of the boat under the water?
 
What is the displaced volume of salt water required to support the weight of the boat? What is the displaced volume of fresh water required to support the weight of the boat? What is the difference in displaced volume between salt water and fresh water?

Chet
 
Chestermiller said:
What is the displaced volume of salt water required to support the weight of the boat? What is the displaced volume of fresh water required to support the weight of the boat? What is the difference in displaced volume between salt water and fresh water?

Chet
Volume of water displaced for fresh water = .102 m^3
volume of water displaced for salt water = .0991 m^3
difference in water displaced = .0029 m^3
 
If the boat cross sectional area at the water line is 3 m^2, for the volume difference you calculated, how much higher will the boat float in salt water?

Chet
 
Chestermiller said:
If the boat cross sectional area at the water line is 3 m^2, for the volume difference you calculated, how much higher will the boat float in salt water?

Chet
I'm not sure that's the part I'm having trouble with
 
BrainMan said:
I'm not sure that's the part I'm having trouble with
But it's the only step you have left to do.
 
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