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1 cube two fluids

  1. Mar 14, 2016 #1
    1. The problem statement, all variables and given/known data
    A layer of oil that has a density of 930kg/m^3 is floating on the surface of water in a container if a wooden cube with a length of 4 cm becomes submerged where it's lower half is in water and it's upper half is in oil
    the cube's density is 960kg/m^3 find
    A) the buoyancy force affecting the cube
    B) the length of the cube that is submerged in water

    2. Relevant equations
    density=m/v
    buoyancy=density*volume*gravity
    cubic volume=L^3

    3. The attempt at a solution
    so I know the cube's density is less than water so it floats
    so I use FB=Fg. 960=m/0.04^3. m=0.06144
    FB = 0.06144*9.81=0.6027N is that correct?
    B)FB=Fg
    1000*Vsubmegred*9.81=0.6027
    V=0.000061m^3
    0.04*0.04*L=V
    L=0.038 now this is my problem the answer sheet says that the submerged part's length is 1.71*10^-2
    I can't seem to find the significance of oil and it's density what am I doing wrong?
     
  2. jcsd
  3. Mar 14, 2016 #2

    SteamKing

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    Remember, the submerged part of the cube is said to be half in the water and half in the oil.

    Treat the total submergence as a variable T and work out what T must be so that the cube is floating in equilibrium with the oil and water.
     
  4. Mar 14, 2016 #3
    well I have seen the problem again and am sorry because I translated it wrong it doesnt say half it just says the lower surface is in water and the upper surface is in oil
     
  5. Mar 14, 2016 #4

    SteamKing

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    OK, so T may not be evenly split between the water and the oil. You still should be able to find a relationship between the two parts given the densities of each fluid and the fact that the cube is floating in equilibrium.
     
  6. Mar 14, 2016 #5
    The closest thing on my mins is
    Fnet=FBwater+FBoil-Fgcube=0 but that still doesnt get me the answer
     
  7. Mar 14, 2016 #6

    SteamKing

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    Have you adjusted your calculations from what you posted in the OP?
     
  8. Mar 14, 2016 #7
    nope
     
  9. Mar 14, 2016 #8

    SteamKing

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    Why not?
     
  10. Mar 14, 2016 #9
    What am I required to change?I think that FBwater is correct and the mass of the cube is correct too I can also obtain the volume of of the part of the cube in oil but I dont know what to do with it
     
  11. Mar 14, 2016 #10
    Your original post does not include the buoyant force exerted by the oil. Add that according to your equation in #5.
     
  12. Mar 14, 2016 #11
    well after hours of crying and thinking why did I go into the advanced highschool program I think I've found the answer and it goes like this
    FBwater+FBoil-Fg=0
    1000*0.04*0.04*L*9.81+930*(0.04-h)*0.04*0.04*9.81-0.602=0
    L=0.0171 m.
    I feel good now btw
     
  13. Mar 14, 2016 #12
    Always feels nice to help. We can feel a little bit better @SteamKing #peaceout
     
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