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Wooden Cube submerged length and the carry limit

  1. Mar 13, 2016 #1
    1. The problem statement, all variables and given/known data
    A wooden cube with the length of 10 cm is and a density of 700 kg/m^3 is floating on water
    A)Find the submerged parts length of the cube
    B) find the maximum added mass to the block before it becomes totally submerged

    2. Relevant equations
    FB=density*volume*gravity
    FB=Fg
    density=m/v




    3. The attempt at a solution
    A)

    mass of the cube = 700*0.1^3=0.7
    so Fg = 0.7*9.81=6.867N
    FB=Fg=6.867
    1000*V*9.81=6.867
    V=0.0007
    L^3=0.0007
    L=0.008. The answer sheet says that L submerged must be 7 cm
    B) m2=Fb-m1
    m2=1000*0.1^3*9.81-0.7=98.1-0.7=97.4Kg. The answer sheet says that the added mass is 0.3 kg
    am lost guys welp
     
  2. jcsd
  3. Mar 13, 2016 #2
    Sorry guys I messesd up on the second question solved it now but still I need A)
     
  4. Mar 13, 2016 #3

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    When calculating the mass of the displaced water, you don't need to multiply by g, since later on you're going to divide by g.
    You are assuming the cube shrinks on all sides to produce the volume of displacement. This is not so. At least two of the cubes dimensions are going to be the same. The only dimension which varies is the draft. Draw a picture of this cube if you get confused.
    You need to get in the habit today of showing units in your calculations.

    What happens when you multiply mass by g? Do you get kilograms as a result?

    Always get in the habit of checking your arithmetic for mistakes.
     
  5. Mar 13, 2016 #4
    I don'
    I don't get what you mean I was trying to get the volume of the submerged part by doing that is the volume correct?
     
  6. Mar 13, 2016 #5
    oh wait I totally get you now so 0.0007=0.1*0.1*L
    and L = 0.07 m thank you!
     
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