Question

A raft is made of 13 logs lashed together. Each is 45 cm in diameter and has a length of 6.5 m .

1) 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 0.60.

Please Recheck work when finished! and be clear please/

Answer 1

from the given data

volume of each log = pi*r^2*L

= pi*0.225^2*6.5

= 1.0338 m^3

volume of 13 logs, V = 13*1.0338

= 13.45 m^3

density of logs = 0.6*1000

= 600 kg/m^3

mass of logs, m_Log = density*volume

= 600*13.45

= 8070 kg

let N is the number of persons can stand on the raft with they start getting their feet wet

in the equilibrium,

net force acting on raft = 0

Fnety = 0

B - m_Log*g - N*mp*g = 0 (here B is buoyant force)

==> N*mp*g = B - m_Log*g

N = (B - m_Log*g)/(mp*g)

= (rho_water*V*g - m_Log*g)/(mp*g)

= (rho_water*V - m_Log)/(mp)

= (1000*13.45 - 8070)/65

= 82.8

so, 82 is no of people can stand on the raft