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

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SUMMARY

The discussion focuses on calculating the height a boat will float in salt water compared to fresh water. A boat weighing 1000 N with a surface area of 3 m² floats 5 cm above the water in fresh water. Using the buoyancy equation B = pVg, the volume of water displaced is calculated as 0.102 m³ for fresh water and 0.0991 m³ for salt water, resulting in a volume difference of 0.0029 m³. This difference directly correlates to the height the boat will rise in salt water, which requires further calculation based on the boat's cross-sectional area.

PREREQUISITES
  • Understanding of buoyancy principles and Archimedes' principle
  • Familiarity with the buoyancy equation B = pVg
  • Knowledge of density values for fresh water and salt water
  • Basic geometry related to volume and surface area calculations
NEXT STEPS
  • Calculate the height the boat will float in salt water using the volume difference and the boat's surface area.
  • Research the effects of varying densities of different types of salt water on buoyancy.
  • Explore the implications of boat design on buoyancy and stability in different water types.
  • Learn about the relationship between weight, volume, and density in fluid mechanics.
USEFUL FOR

Students studying physics, marine engineers, and anyone interested in fluid mechanics and buoyancy calculations.

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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