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Homework Help: Object Weight given only difference in weights in and out of water

  1. Jul 27, 2009 #1
    1. The problem statement, all variables and given/known data

    A chunk of carbon steel with density, ρ = 7.84 g/cm3 , is completely submerged in fresh water. The chuck of steel weighs 39 N more in air than in water. Please answer the following:
    (a) Find the buoyant force acting on the chuck of steel.
    (b) Find the the volume of the chuck of steel.
    (c) What is the mass of the chuck of steel?

    2. Relevant equations

    Fw(air)=Fw(water) + 39 N



    BUT I dont know how to find the volume.

    density of air is 1.16 kg/m3
    density of water is 1.0x10^3

    3. The attempt at a solution

    once I get the VOLUME I can find the relative weights by using rho*g*v
    and Fw(water)+ 39N = Fw(air)
  2. jcsd
  3. Jul 27, 2009 #2
    The buoyant force (upward direction) in the air is less than in the water, therefore
    Fw(air)=Fw(water) -39 N
    And you know that (mg-Fw(air))/g = m
  4. Jul 27, 2009 #3


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    Weight of the displaced water = 39 N = m*g = density of water*volume*g.
  5. Jul 27, 2009 #4
    ok, a little hazy. Fw(water) is 39N=dens water*v*g? than V = .00398

    if so, Fw(water)-39N=Fw (air) than Fw(air) is /38.96??
  6. Jul 27, 2009 #5
    -38.96 (which cant be right) *rather than /38.96
  7. Jul 27, 2009 #6


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    Fw(in air ) = density of steel*volume*g.
  8. Jul 27, 2009 #7

    so what you are saying is that I use previously acquired volume of .00398 in this eq.

    I am sorry if I am frustrating. I mean well!

    OH no... do I convert 7.84 g/cm^3 to kg/m^3 ??

    if so 7.84 g/cm^3 (100cm*100cm*100cm) * 1kg = 7840 kg/m^3
    (1m^3) 1000 G

    Fw(air) = 7.84 g/cm3


    Fw (air) = 7840 kg/m^3 * .00398 m^3 * 9.81 m/s

    is that right?

    then to get weight in water I subgract (39N/9.81) ?
  9. Jul 27, 2009 #8


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    then to get weight in water I subgract (39N/9.81) ?
    The weight in water = F(air) - 39 N.
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