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Buoyancy force

  1. Dec 7, 2004 #1
    The bottom of a steel "boat" is a 6.00 m x 9.00 m x 5.00 cm piece of steel(density of steel = 7900 kg/m^3) . The sides are made of 0.460 cm-thick steel.

    what minimum height must the sides have for this boat to float in perfectly calm water? in cm

    i have that F_B (buoyancy force) is equal to W_boat (weight of boat) is equal to W_B + 2*W_s1 + 2*W_s2. where w is the weights and it equal rho*g*V. F_B = density of water*g*total volume and total volume is equal to 6*9*(h+ .05)

    any suggestions would really help
  2. jcsd
  3. Dec 7, 2004 #2


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    Give this a try:

    [tex]Weight_{bottom} + 2\rho_{steel} g \ h (L + W) = \rho_{water} g (L-2\times0.046)(W-2\times0.046)(h-0.05)[/tex]
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