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Density and Buoyancy Problem

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data
    1. A block is 7 cm wide, 8 cm long, and 2 cm tall and has a mass of 67 g.
    a) What is its volume?
    b) What is its density?
    c) Floating in water (with the 7x8 face down), how deep will the bottom be (what is x in
    figure 2)?
    d) How much extra weight can it support without sinking?
    2. A 2 cm by 2 cm by 2 cm metal block has a mass of 80 g.
    a) What volume of water does it displace when it is submerged?
    b) What is the weight of the water displaced?
    c) What is the apparent weight of the block when measured under water?


    2. Relevant equations
    A=Lw.
    V=Ah=Lwh.
    density = m/v
    Fbuoyancy = (Mass displaced water)*g
    mblock*g = Fbuoyancy =mdisplaced water*g
    ---->density of block* volume of block = density water* Volume displaced water
    ----> DENSITY BLOCK*H = DENSITY WATER X
    Fbouyancy = density water*Volume of block

    3. The attempt at a solution

    1a)V=Ah=Lwh.
    (.08m)(.07m)(.02m)
    V=0.000112m^3

    1b)density = m/v
    .067kg/0.000112m^3 = 598.214kg/m^3

    c) Density block* h = densitywater*x
    (598.214kg/m^3)(.02m) =(1000kg/m^3) x
    x=0.019964m

    d)m displaced water = Vblock* density water
    Mdw = (.000112m^3)(1000kg/m^3)
    Mdw = 0.112kg Mblock = .067kg
    mass block - massdw = .045kg

    2a)
    V=Lwh
    V=(.02)(.02)(.02) = .000008m^3
    Density = 10,000kg/m^3
    Pbmetal*Vmetal = density water*Vdisplaced water
    (10,000kg/m^3)(.000008m^3) = (1000kg/m^3)( Vdw)
    Vdw = 0.00008m^3

    b)F bouyancy = density water * vblock*g
    =(1000kg/m^3)(.000008m^3)(9.8m/s^2)
    =0.0784N
    dw =(1000kg/^3)(.000008m^3) =.008kg
    Fbouyancy = Mdw*g= .0784N
    c) I dont know...
    F bouyancy = density water * vblock*g
    =(1000kg/m^3)(.000008m^3)(9.8m/s^2)
    =0.0784N



    Can someone check my work?
    Question 2a-c, Im unsure of. I need some explanation or some ideas to solve 2b and 2c.
     
  2. jcsd
  3. Nov 14, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    All good.

    Careful here. Since the metal cube is totally submerged, the volume of water displaced is just equal to the volume of the cube. Do not try to compute the volume of displaced water by setting its weight equal to the weight of the cube--it's not floating!

    OK.
    Imagine you suspended this metal cube by a string. The tension in the string is its "apparent" weight. In air, that tension just equals the weight of the cube. But when the cube is submerged in water, the apparent weight is reduced by the upward buoyant force. (Something floating would have an apparent weight of zero.)
     
  4. Nov 14, 2009 #3

    Borek

    User Avatar

    Staff: Mentor

    In vaccum to be precise. It may seem as a nitpicking, but bouyancy in the air means you will be off by about 0.1% (order of magnitude, exact value depends on the density of the object). If you need high accuracy that's not a thing to forget :tongue:

    See http://www.titrations.info/volumetric-glass-calibration for example calculation in the real lab situation (scroll down to the text on the grey background).

    --
    methods
     
  5. Nov 14, 2009 #4

    Doc Al

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    Staff: Mentor

    Good point. Shame on me! :tongue2:
     
  6. Nov 14, 2009 #5

    Borek

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    Staff: Mentor

    LOL it is the first time that I pointed out at yor mistake, so far it was always the other way around :rofl:
     
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