A raft is made of 12 logs lashed together. Each is 42 cm in diameter and has a length of 6.5 m. How many people can the raft hold before they start getting their feet wet, assuming the average person has a mass of 65 kg? Do not neglect the weight of the logs. Assume the specific gravity of wood is 0.55.

Any object,submerged in a liquid, receives an upward force, which is called the buoyant force, and is equal -according to stated by Archimedes'principle- to the weight of the volume of liquid removed by the object.

When this force is exactly equal to the weight of the object, this one barely floats.

In our case, if we don't neglect the weight of the raft, the total weight can be expressed as follows:

Fg = ρw*Vraft*g + N*mp*g (1)

The buoyant force, when the raft is completely submerged (so the people on it is just starting to get their feet wet), can be expressed as follows:

Fb = ρh₂o*Vraft*g (2)

As we know that the specific gravity of wood is 0.55, we deduct that the density of wood is 550 kg/m³.

The volume of the raft, can be found taking the volume of each log, as the volume of a cylinder, as follows:

Vlog = π*r²*l

⇒ Vlog = π*(0.21m)²*6.5m = 0.9 m³

⇒ Vraft = Vlog*12 = 10.8 m³

Equating (1) and (2), we get:

Fg = Fb ⇒ ρwood*Vraft*g + N*mp*g = ρh₂o*Vraft*g

Replacing by the values that we have just written above,we have:

(550 kg/m³*10.8 ³m + N*65kg)*9.8m/s² = 1000 kg/m³*10.8 m³*9.8m/s²

Simplifying common terms, rearrranging and solving for N:

N = 450 kg/m³*10.8 m³ / 65 kg = 74.76 = 74 people as a maximum.