How much deeper does the barge lower?

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SUMMARY

The discussion focuses on calculating how much deeper a rectangular barge sinks when a 400 kg block is added. Using Archimedes' principle, the barge, measuring 5 m in length and 2 m in width, must displace an additional volume of 0.4 m³ to accommodate the weight of the block. The key to solving the problem lies in determining the relationship between the displaced volume and the area of the barge's base. The solution requires calculating the depth increase based on the area of the barge and the volume displaced.

PREREQUISITES
  • Understanding of Archimedes' principle
  • Basic knowledge of volume and area calculations
  • Familiarity with the concept of buoyancy
  • Ability to perform unit conversions (e.g., kg to m³)
NEXT STEPS
  • Calculate the depth increase using the formula: Depth = Volume Displaced / Area of Base
  • Explore examples of buoyancy problems involving rectangular objects
  • Study the implications of Archimedes' principle in real-world applications
  • Investigate the effects of different materials on buoyancy and displacement
USEFUL FOR

Students in physics or engineering courses, educators teaching buoyancy concepts, and anyone interested in practical applications of Archimedes' principle.

keithcuda
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Homework Statement


  1. A rectangular barge floats in freshwater. When a 400 kg block is loaded on the 5 m long by 2 m wide barge, the barge sinks a few centimeters deeper. How much deeper does the barge lower?

Homework Equations


Archimedes' principle
This is where I am stuck. My textbook doesn't talk much about this at all.
barge.jpg


The Attempt at a Solution

 
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You've done all the hard work. How much deeper does the barge have to "sink" to displace the extra 0.4 m3?
 
Bystander said:
You've done all the hard work. How much deeper does the barge have to "sink" to displace the extra 0.4 m3?
barge.jpg
 
keithcuda said:
Yes, but you could have got there a bit faster from knowing the volume displaced and the area.
 
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