# A floating metal tube

1. Mar 4, 2005

### elemnt55

A hollow steel tube (diameter = 3.84 cm) is sealed at one end and loaded with lead shot to give a total mass of 0.161 kg. When the tube is floated in pure water, what is the depth, z, of its bottom end? :rofl:

2. Mar 4, 2005

### xanthym

The hollow tube will experience a Buoyant Force due to displaced liquid given by Archimedes' Principle:
{Buoyant Force} = ρgV
where ρ is the liquid density, g the gravitational acceleration, and V the displaced liquid volume.

The tube will sink into the water until its sealed end reaches an equilibrium position where the buoyant force described above exactly equals the tube's weight "mg" on land. If the tube's sealed end sinks to a depth "z", the displaced liquid volume will be:
V = πr2z
so the equilibrium condition is satisfied by:
mg = {Buoyant Force} = ρgV = ρgπr2z
⇒ z = m/(πr2ρ)
For this problem, {m = 0.161 kg}, {r = (3.84 cm)/2 = 0.0192 m}, and {ρ = 1.0 g/cm^3 = 1000 kg/m^3}, so that:
z = (0.161)/{π(0.0192)2(1000)}
z = (0.139 m)

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