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Need help with a buoyancy problem

  1. Jun 4, 2014 #1
    1. The problem statement, all variables and given/known data
    A cube of side length 3 cm floats in water (density = 1g/cm^3) with 1 cm floating above the water. What is the density of this cube?


    2. Relevant equations
    Fbouyant = ρfluid*g*Vsubmerged
    Fg= mg

    3. The attempt at a solution

    I am having problems figuring out what the submerged volume of the cube is to plug into the formula. I have a formula that states that volume submerged for an object = Az, where A = the area of the object and z = the depth that the object is submerged so:

    V=Az = 6(3cm^2) * 2cm (that is submerged) = 108cm^3 It seems that this is wrong though since the volume should be 18cm^3 to give me the right answer.

    I then use the formula

    Fb = Fg for an object that is floating
    ρfluid*g*Vsubmerged = ρobject*g* Vtotal of object

    cancelling out the g from both sides gives a ratio that allows me to calculate for the density of the object. The answer is supposed to be 2/3 g/cm^3 which I am not getting.

    Any guidance would be appreciated :)
     
  2. jcsd
  3. Jun 4, 2014 #2

    Nathanael

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    Homework Helper

    One side of the cube is [itex]9cm^2[/itex] and the part that is submerged is [itex]2cm[/itex] in height.

    So what's the volume of a rectangular-cube (probably not the right name for it) with a base of [itex]9cm^2[/itex] and a height of [itex]2cm[/itex]?
     
  4. Jun 4, 2014 #3

    TSny

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    Gold Member

    Hi AlphaLima. Welcome to PF!

    In the formula V = Az, A is not the total surface area of the cube. See this link .
     
  5. Jun 4, 2014 #4
    L * W * H = 3*3*2 = 18cm^3. Thank you guys! This makes a lot more sense now :)
     
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