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Buoyancy and Archimedes' Principle

  1. Jan 18, 2005 #1
    If an object floats in water, its density can be determined by tying a sinker on it so that both the object and sinker are submerged. Show that the specific gravity, (Density_substance)/(density_water at 4 degrees), is given by (w)/(w_1 - w_2), where w is the object's weight in air alone, w_1 is the apparent weight when a sinker is tied to it , and the sinker is submerged, and w_2 is the apparent weight when both the object and sinker are submerged.

    Could anyone give me some pointers on what to do?

  2. jcsd
  3. Jan 18, 2005 #2


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    Archimedes' principle : buoyant force on an object is equal to the weight of the water displaced. The weight of the water displaced is of course equal to the volume of the object being immersed multiplied by the density of water.

    For this problem, you're expected to disregard the buoyancy due to air, that is, treat the weights measured in air as "true" weights.

    Define : [itex]w_s, v_o, v_s, \rho_w, \rho_o, m_o, g[/itex] as true weight of sinker, volume of object, volume of sinker, density of water, density of object, mass of object and gravitational accleration respectively. The other definitions are as given in the problem statement.

    Set up 3 equations like so :

    [tex]w = (m_o)(g) = (v_o)(\rho_o)(g)[/tex] ---(1)

    [tex]w_1 = (w + w_s) - (v_s)(\rho_w)(g)[/tex] ---(2)

    [tex]w_2 = (w + w_s) - (v_o + v_s)(\rho_w)(g)[/tex] ---(3)

    Take (2) - (3) :

    [tex]w_1 - w_2 = (v_o)(\rho_w)(g)[/tex] --- (4)

    Take (1)/(4) :

    [tex]\frac{w}{w_1 - w_2} = \frac{\rho_o}{\rho_w}[/tex]

    And you're done. If you need further explanation please post.
    Last edited: Jan 18, 2005
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