Question

A raft is made of 10 logs lashed together. Each is 32.0 cm in diameter and has a length of 6.50 m. How many people (whole number) can the raft hold before they start getting their feet wet, assuming the average person has a mass of 65.0 kg? Do not neglect the weight of the logs. Assume the density of wood is 600 kg/m³

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Density of water = = 1000 kg/m³

Density of wood = _w = 600 kg/m³

Mass of average person = m = 65 kg

Number of people the raft can hold = N

Diameter of the log = D = 32 cm = 0.32 m

Length of the log = L = 6.5 m

Number of logs = n = 10

Total volume of the logs = V

V = nD²L/4

V = (10)(0.32)²(6.5)/4

V = 5.227 m³

We want to find the number of people the raft can hold before they start getting their feet wet that is the logs of the raft are completely submerged in the water.

By balancing the buoyancy force on the raft to the weight of the raft and the people,

Vg = _wVg + Nmg

V = _wV + Nm

V( - _w) = Nm

(5.227)(1000 - 600) = N(65)

N = 32.17

N = 32 people

The raft can hold 32 people before they start getting their feet wet.