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How much ice to float a 80 Kg person

  1. Apr 18, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the smallest block of ice which when floating on salt water will carry a person of mass 80 Kg

    Densities:

    Salt Water = 1025
    Ice = 917

    2. Relevant equations

    F_b=P_f * V_f * G

    3. The attempt at a solution

    I figured F_b would have to be equal to the force of the person down which is equal to 784 N. Then plugging that into the equation and solving for V_f I get .078. I used the Salt Water for P_f. Is there somewhere that I need to plug in Ice?
     
  2. jcsd
  3. Apr 18, 2007 #2

    Hootenanny

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    Don't forget the weight of the ice...:wink:
     
  4. Apr 18, 2007 #3
    ok so I tried to add the weight of ice in here is what I get.

    like before I have the equation.

    W_p * G = 784

    This is the force of the pearson. Next I have

    784 = Volume of the ice * Density of water * Gravity

    Then to add in the weight of ice I do this.

    Volume of the ice * Density of Ice * Gravity + 784 = Volume of ice * density of water * gravity

    This makes an equation has everything cancel out.
    What am I doing wrong?
     
  5. Apr 18, 2007 #4
    (vol ice*(density ice)+m)g=volume ice(density SALT water)*g Now the question is ice salty? I think not. On the left hand side we have the total weight of the ice plus man, on the right the weight of water displaced.
     
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