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Buoyant Forces and Archimedes' Principle

  1. Apr 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A 62.0-kg survivor of a cruise line disaster rests atop a block of Styrofoam insulation, using it as a raft. The Styrofoam has dimensions 2.00 m X 2.00 m X 0.0900 m. The bottom 0.024 m of the raft is submerged.

    a) Draw a free-body diagram of the system consisting of the survivor and raft.

    b) Write Newton's second law for the system in one dimension. (Use B for buoyancy, w for the weight of the survivor, and wr for the weight of the raft. Set a = 0.)

    c) Calculate the numeric value for B. (Seawater density = 1025 kg/m3)

    d) Calculate weight wr of the Styrofoam.

    e) What is the density of the Styrofoam?

    f) What is the maximum buoyant force, corresponding to the raft being submerged up to its top surface?

    g) What total mass of the survivors can the raft support?


    2. Relevant equations

    [tex]\rho[/tex] = [tex]\frac{M}{V}[/tex]

    P = [tex]\frac{F}{A}[/tex]

    V = lwh

    P = P0 + [tex]\rho[/tex]gh

    [tex]\frac{\rhoobj}{\rhofluid}[/tex] = [tex]\frac{Vfluid}{Vobj}[/tex]


    3. The attempt at a solution

    a) I have a free-body diagram including:
    • Force of Gravity (downwards)
    • Buoyancy (upwards)

    Must I include the Normal force as well?

    b) I have:
    B = (w + wr) = 0
    B = w + wr = mg + mrg

    B = mwaterg = [tex]\rho[/tex]Vg = [tex]\rho[/tex]lwhg
    Vraft = (0.024 m)(2.00 m)(2.00 m) = 0.096 m^3

    g(m + mr) = [tex]\rho[/tex]waterVraftg

    The g's cancel.

    c) 62 kg + mraft = (1025 kg/m^3)(0.096 m^3)
    mraft = 36.4 kg

    But based on my equations, 98.4 would also equal the buoyancy. I think there's an error in my attempt to write Newton's second law for the system in (b).

    Any guidance would be appreciated.
     
    Last edited: Apr 8, 2009
  2. jcsd
  3. Apr 8, 2009 #2

    LowlyPion

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    I think you want to recheck your math here.
     
  4. Apr 8, 2009 #3
    I'm sorry -- that was a typo. I have corrected the math portion of the equation, but I'm still not sure if I'm going about this problem correctly.
     
  5. Apr 8, 2009 #4

    LowlyPion

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    You know the mass of the raft then. And you know its volume from the total thickness, so you should be able to figure both the density and the total potential water displacement to determine the total buoyancy available.
     
  6. Apr 9, 2009 #5
    Actually, I figured out a-f. Now I just need help with g.

    a) Buoyant force up, gravity down
    b) B = wr + w
    c) B = [tex]\rho[/tex]Vg = 964.32 N
    d) B = wr + w
    964.32 N = wr + 607.6 N
    wr = 356.72 N

    e) [tex]\rho[/tex] = M/V = 95.9 kg/m^3

    f) B = [tex]\rho[/tex]Vg = 338.3 N

    g) The total mass of survivors the raft can support...
    Thoughts: Weight of each survivor = mg. This, plus the weight of the raft counters the maximum buoyant force, calculated in f, right? So, (weight of survivors) + (weight of raft) = Maximum buoyant force?
     
  7. Apr 9, 2009 #6

    LowlyPion

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    That's correct. Think of buoyancy as your credit limit. The maximum amount of water the total volume of the raft can displace. You necessarily start out with the mass of the raft. Then to that you can add people to the raft until you have reached the credit limit - the total amount of water displaced by the raft. When you exceed the limit then you are going to need a bailout.
     
  8. Apr 9, 2009 #7
    Thank you...

    by the way, love the analogy!
     
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