How Many People Can a Raft Made of 9 Logs Hold Before Getting Wet?

  • Thread starter Thread starter superjen
  • Start date Start date
AI Thread Summary
A raft constructed from 9 logs, each with a diameter of 29.1 cm and a length of 5.7 m, is being analyzed for its capacity before occupants get wet. The calculations involve determining the volume of the logs, buoyant force, and the mass of the raft. Initially, the calculation suggested the raft could hold 46 people, which was incorrect due to an error in applying gravitational acceleration in the buoyant force equation. After reevaluating the numbers, it was clarified that the buoyant force should not be multiplied by 9.81, leading to a more accurate assessment of the raft's capacity. The discussion highlights the importance of precise calculations in determining buoyancy limits.
superjen
Messages
26
Reaction score
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!?
 
Physics news on Phys.org
superjen said:
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!?

From your numbers you have 3411 - 2047 = 1364 and divide by 69.3 gives a different number.
 
ahh! ok, i see now, !
makes sense, i didn't need to multiply it by 9.81!
Thanks :)
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top