Question

A raft is made of 9 logs lashed together. Each is 41 cm in diameter and has a length of 6.9 m. How many people can the raft hold before they start getting their feet wet, assuming the average person has a mass of 63 kg? Do not neglect the weight of the logs. Assume the specific gravity of wood is 0.61

Answer 1

Volume of cylindrical log= *r^2*h, where r is radius and h is length.

Here, r = diameter/2 = 41/2 =20.5 cm = 0.205 m.

Also, l = 6.9 m

So, volume= *0.205*0.205*6.9 = 0.911 m3

Mass=density*volume= 610*0.911= 555.71 kg.

Now, forces on the 9 logs include:

i) Weight of the 9 logs= 9*mg downward, where m is mass of the log and g is gravitational acceleration

=9*555.71*9.8 = 49013.622

ii)Weight of people= N*Mg downward, where N is number of persons, M is mass of each person and g is gravitational acceleration.

So, weight of people= N*63*9.8 = 617.4N

iii)Buoyant force = 9Vdg upward, where V is volume of each log, d is density of water and g is gravitational acceleration.

So, buoyant force= 9*0.911*1000*9.8 = 80350.2 newtons.

For equilibrium, these forces must be balanced.

So, 80350.2= 617.4*N + 49013.622

=>617.4N=80350.2-49013.622=31336.58

=>N=31336.58/617.4=50.7

So, required number of person = 50