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Lead sinker problem

  1. Jul 1, 2004 #1
    This one is different, and it is driving me crazy almost as bad as the other two.

    A lead sinker is attatched to a round platic sphere. If the sphere floats so that it is half submerged, and has a weight of 1oz. (6.25 * 10^-2lb) and a radius of 1in, (8.33 * 10-2ft) find the volume and weight of the sinker if the specific gravity of lead is 11.3.
  2. jcsd
  3. Jul 1, 2004 #2


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    Treat the sphere and sinker as a single body. If it floats, it's in static equilibrium; the forces down equal the forces up. The forces down are:

    [tex]F_{down} = g(m_{sphere} + m_{sinker}) = g(1oz. + \rho _{sinker}V_{sinker})[/tex]

    [itex]\rho _{sinker}[/itex] is the density of the sinker, which you can determine given the specific gravity. So really, you have one equation, two unknowns ([itex]F_{down}[/itex] and [itex]V_{sinker}[/itex]). Now, the forces up are:

    [tex]F_{up} = g\rho _{water}(V_{sinker} + V_{sphere}/2)[/tex]

    Of course, you know [itex]V_{sphere}[/itex] and [itex]\rho _{water}[/itex], so the only unknowns in this are [itex]V_{sinker}[/itex] again, and [itex]F_{up}[/itex] but [itex]F_{up} = F_{down}[/itex], so you have 2 equations and 2 unknowns (or 3 and 3 if you want to look at [itex]F_{up} = F_{down}[/itex] as it's own equation, and [itex]F_{up}[/itex] and [itex]F_{down}[/itex] as different unknowns). This should be pretty simple to solve.
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