Buoyancy of Spheres: max mass added without sinking

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SUMMARY

The maximum mass that can be added to a 40-cm-diameter Styrofoam sphere without sinking in water is calculated using the formula for buoyant force (FB = ρoVg) and the volume of the sphere (V = 4/3 π r³). The density of Styrofoam is 300 kg/m³, leading to a calculated maximum mass of approximately 1005 kg. However, the total mass required to sink the sphere, including the mass of the Styrofoam, is found to be 2345 kg, indicating that the mass needed to submerge the sphere is significantly higher than typical expectations.

PREREQUISITES
  • Understanding of buoyancy principles and Archimedes' principle
  • Familiarity with density calculations and units (kg/m³)
  • Knowledge of volume calculations for spheres
  • Basic algebra for solving equations
NEXT STEPS
  • Review buoyancy calculations in fluid mechanics
  • Learn about the properties of different materials and their densities
  • Explore the implications of Archimedes' principle in real-world applications
  • Investigate how to calculate the volume of irregular shapes
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Students studying physics, particularly those focusing on fluid mechanics, engineers working with buoyant materials, and educators teaching principles of density and buoyancy.

BennyQuixote
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Homework Statement



Styrofoam has a density of 300 kg/m3. What is the maximum mass that can hang without sinking from a 40-cm-diameter Styrofoam sphere in water? Assume the volume of the mass is negligible compared to that of the sphere.

Homework Equations



FB=poVog
Volume of sphere: 4/3 π r3
p=m/V

The Attempt at a Solution



I tried this:
mo/3.35 m3=300 kg/m3
Therefore,
mo=1005 kg

So,
To find the exact amount of mass added to the Styrofoam sphere to sink in water:
1005k + m2/3.35 m3=1000 kg/m3

Therefore:
m2=2345 kg

This is obviously an obscenely large mass to make Styrofoam sink in water. Any help is appreciated.


Quixote
 
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