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Buoyancy, depth a block will sink in two different liquids

  1. Apr 6, 2014 #1
    1. The problem statement, all variables and given/known data

    4. The bottom half of a tank is filled with water (ρ = 1.0 x 103 kg/m3), and the top half is filled with oil (ρ = 0.85 x 103 kg/m3). Suppose that a rectangular block of wood of mass 5.5 kg, 30 cm long, 20 cm wide and 10 cm high is placed in this tank. How deep will the bottom of the block be submerged in the water? (Buoyancy)

    2. Relevant equations

    [tex]F_{net} = B - F_g[/tex]

    [tex]B = ma[/tex]

    3. The attempt at a solution

    I thought I would solve for the rate at which the block of wood accelerates in oil. Using that, I would solve for the volume of fluid displaced using the equation [tex]B = ma[/tex] Here is my full work: http://imgur.com/7qcnbbY

    Is my reasoning correct?
     
    Last edited: Apr 6, 2014
  2. jcsd
  3. Apr 6, 2014 #2
    I don't think you did it right.

    The problem wants to know the final equilibrium condition. So the block will have part of it's volume submerged in oil and part of it's volume submerged in water. Can you write a single equation for this?
     
  4. Apr 6, 2014 #3

    rude man

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    My post deleted ...

    ... and resuscitated:

    The fun part of this problem is whether there exists more than one stable floating position.
    Rather than pursue this further for the time being I suggest assuming the safest orientation which is
    .3m = horizontal
    .2m = horizontal
    .1m = vertical

    This permits summing of vertical forces = 0 easily.

    I found a good paper which goes into the orientation stability question in some - er - depth:

    www.cns.gatech.edu/~predrag/courses/PHYS-4421.../buoyancy.pdf [Broken]
     
    Last edited by a moderator: May 6, 2017
  5. Apr 6, 2014 #4

    SteamKing

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    Unfortunately, this link is broken.

    You do raise an interesting point, but the OP lacks sufficient detail to allow a coherent analysis. We are not provided with the dimensions of the tank and consequently cannot determine the depth of water or oil in which the block can float.
     
    Last edited by a moderator: May 6, 2017
  6. Apr 6, 2014 #5

    rude man

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    Sorry, here's the pdf file attached.

    It doesn't matter what the dimensions of the tank are so long as there is enough oil depth to cover the top of the block, and enough water depth to let the block float rather than sink to the bottom. Of course, the tank also has to be large enough to house the block horizontally.

    I agree, the problem is insufficiently defined.
     

    Attached Files:

  7. Apr 7, 2014 #6
    Thanks for the replies everyone. I talked to my teacher and he said that you could assume that the depth of oil is 10+ cm and the depth of water is 10+ cm. He also said that the net force is zero when the block is submerged in the two liquids, and that I need to find the fraction of each that it's in and that I can determine that by their relative densities.

    My attempt with this information was to find the buoyant force knowing it's equal to mg of the block as Fnet = B - Fg = 0. Then I solved for the total density of the fluid displaced. Then I used a system of 2 equations where the first was (density of water * volume in water) + (density of oil * volume in oil) = (volume of object * total fluid density) and the second equation was (volume in oil + volume in water = 6mm^3). That gave me the volume of the block in each liquid. Does that work or have I made a mistake?
     
  8. Apr 7, 2014 #7
    Your approach looks better. The only mistake I think I see is in your 2nd equation. What is your reasoning for using this equation?
     
  9. Apr 7, 2014 #8
    My reasoning is that the total volume of the block is the portion submerged in water plus the portion submerged in oil.
     
  10. Apr 7, 2014 #9
    Sorry, I guess I was thinking you gave us 3 equations:

    1) ∑F = B - mg = 0
    2) ?
    3) V1 + V2 = Vtotal

    1) and 3) are correct but 2) looks like an attempt to set up 1) but you didnt do it correctly.
     
    Last edited: Apr 7, 2014
  11. Apr 7, 2014 #10
    Ah, I see. I thought you were talking about the system of equations. I can't say I've really reasoned out the second equation much, but I think I can defend it by saying that the mass of water displaced plus the mass of the oil displaced is equal to the total mass of fluid displaced by the block.
     
  12. Apr 7, 2014 #11

    rl.bhat

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    When the block in equilibrium, the net pressure on the block is zero.
    Since the density of the block is more than the density of liquid, it will submerge completely in the liquid and partly in the water.
    Now net upward pressure on the block = pressure due to liquid + pressure due to water.
     
  13. Apr 7, 2014 #12

    rude man

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    I'm not sure of your approach eitther. I worry about your " Then I solved for the total density of the fluid displaced. " What do you mean by that? Some kind of average density of the fluids displaced? Not sure how that would pan out.

    I used only one equation myself.

    Perhaps you can post your result, but if you do, to avoid math errors, use symbols:
    h = total height of block in solution (i.e. vertical dimension of block; I would pick h = 10cm.),
    m = mass of block
    ρo = oil density
    ρw = water density
    A = area of top (or bottom) of block

    ... or whatever symbols you like. But avoid number-crunching.
    Remember, we're looking for the distance between the oil-water level and the bottom of the block.
     
  14. Apr 7, 2014 #13
    That equality is trivial but that's not what you wrote in post #6. Don't you think you could just use 1) to get your equation?
     
    Last edited: Apr 7, 2014
  15. Apr 7, 2014 #14

    rude man

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    BTW the block volume is .1m x .2m x .3m = 0.006 m^3 = 6,000,000 mm^3 = 6e6 mm^3.
     
  16. Apr 7, 2014 #15
    Thanks guys. I just finished my third attempt, could anyone tell me if the third time is indeed the charm?

    http://imgur.com/C5nbBtL
     
    Last edited: Apr 7, 2014
  17. Apr 8, 2014 #16
    Looks like you got it although a small typo referring to h3 at the end there.

    Another comment I would have is you could have taken a small short cut and just gave the buoyant force as the volume of each liquid displaced. You actually derived the same thing from first principles which is fine, but rather than derive this every time you do these types of problem just jump ahead and write it out directly. Also you can get more complicated kinds of problems where you will want to avoid proving the same result because it will be time consuming.
     
  18. Apr 8, 2014 #17

    rude man

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    Looks good, though I got h2 =0.044m.
    Congrats for hanging in!
     
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