- #1
superjen
- 26
- 0
A raft is made of 9 logs lashed together. Each is 29.1 cm in diameter and has a length of 5.7 m. How many people can the raft hold before they start getting their feet wet. Assume the average person has a mass of 69.3 kg and that the density of water and wood is 1000 kg/m3 and 600 kg/m3, respectively
[okay, what i did was i found the volume of one log.
V = (pi)(r^2)H
r is used 0.1455 (29.1 converted to m then divided by 2 to get radius)
h i used 5.7m , then times it by 9.
(i got an answer of 3.411875)
then i found FB. FB = pwaterVg (1000)(3.41188)(9.81)
since i knew the density of wood, and the volume, i found the mass
m = pwoodV
got an answer of 2047.125
then i used this equation FB = [ x*mass of one person + mass of 9 logs] g
when I am done, i get an answer of 46 people. its wrong. can someone show me where I am going wrong!?
[okay, what i did was i found the volume of one log.
V = (pi)(r^2)H
r is used 0.1455 (29.1 converted to m then divided by 2 to get radius)
h i used 5.7m , then times it by 9.
(i got an answer of 3.411875)
then i found FB. FB = pwaterVg (1000)(3.41188)(9.81)
since i knew the density of wood, and the volume, i found the mass
m = pwoodV
got an answer of 2047.125
then i used this equation FB = [ x*mass of one person + mass of 9 logs] g
when I am done, i get an answer of 46 people. its wrong. can someone show me where I am going wrong!?