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The density of a cube.

  1. May 24, 2009 #1
    One end of a light thin string is attached to the bottom of a 3.7 g, 42 cm3 cork. The other end is attached to a 58 g cube. The arrangement is placed in water and it is found that with 82% of the cork submerged the cube is held in equilibrium. What is the density of the cube in g/cm3?

    I need help figuring out this problem.
     
  2. jcsd
  3. May 24, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi college boy19! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
     
  4. May 24, 2009 #3
    okay,
    the density of an object is its mass divided by its volume:

    D=m/v

    so for the cube all i have in the equation is :
    D= 58g/v

    So i know that in order to get the density i need the volume of the cube.

    this is where i am stuck.
     
  5. May 24, 2009 #4

    tiny-tim

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    Hi college boy19! :smile:
    That's not the only way to get the density …

    you can also work it out from the forces on the cork :wink:
     
  6. May 24, 2009 #5
    Well the forces acting on the cork are the buoyant force which is the the weight of the cork which is mg which is:

    BF= 3.7*g
    but gravity in SI units is 9.8kg/m^3 and I am using CGS units so gravity is 98g/cm^3
    So the weight of the cork is

    BF= 3.7(98)=362.6

    the buoyant force = to the weight of the displaced fluid
     
  7. May 24, 2009 #6

    tiny-tim

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    No, the forces acting on the cork are the buoyant force, the weight, and the tension in the string.
     
  8. May 24, 2009 #7
    the buoyant force:
    362.6
    the weight:
    mg=362.6
    how do i figure out the tension in the string?

    is it the buoyant force - the weight of the cube
     
  9. May 24, 2009 #8

    tiny-tim

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    How can the buoyant force be equal to the weight?

    How is buoyant force defined?
    You have two equations for it …

    one for the cork, and one for the mass …

    that should be enough to solve the whole problem. :smile:
     
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