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Buoancy / Archimedes Principle

  1. Jan 19, 2005 #1
    I have run into a problem and am completly stumped after. Help me before i pull my hair out.
    What is the smallest number of whole logs (p (density) 725 kgm^-3) r= .0800m l=3.00m that can be used to build a raft that will carry four people each with a mass of 80.0 kg.

    so using (pi*r^2*l)*725 i found the mass of each log to be 43.7 kg
    BUt beyond this i run into huge headaches by trying to derive something that can include X # of 43.7 kg logs in relation to bouyant force and Weight

    please help
  2. jcsd
  3. Jan 20, 2005 #2


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    The buoyant force on the logs is the weight of the water they displace and the condition of minimum buoyancy will be met if the logs are just completely sumberged with the passengers just at the surface of the water. All you have to do is balance all the forces (weight of the logs + passengers against the buoyant force).
  4. Jan 21, 2005 #3
    Find the force generated by the 4 people on the raft, as well as the net force per log. That should set you on the right course.
  5. Jan 22, 2005 #4
    I have some questions about this problem too. Okay, I've got that:
    Force of buoyancy = Weight (of logs) + Weight (of load/people)

    Since each person's mass was 80.0 kg and there are 4 people, I went
    Weight (load) = (80.0kg) x (4) x (9.81m/s^2) = 3139.2 N (Force of the load)

    and the force of each log is
    (density of log) x (volume of log) x g = (725kg/m^3) x (pi) x (0.08m)^2 x (3m) x (9.81m/s^2)
    = 429 N

    But I'm not really sure what to do next....
  6. Jan 23, 2005 #5
    Once you have the total force applied by the people on the raft as well as the net force created by each log (the difference between mg and the bouyant force), you need to find the number of logs that will make the force applied by the people + the net force of all the logs = 0

    F(p) = force created by passengers
    F(l) = net force created by each log
    F(p) + n*F(l) = 0

    Solve for n.
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