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Homework Help: Having trouble with Hydrostatics problem

  1. Oct 9, 2004 #1
    Hi everyone!

    The problem statement reads: A 3.15kg piece of wood (Spec. Grav. = 0.5) floats on water. What is the minimum mass of lead, hung from it by a string will cause it to sink?

    Heres my reasoning. Since the log is floating, I know that it is in equilibrium, and therefore the Bouyant force Fb = force due to gravity. Thus Fb = 3.15kg *9.8m/s^2.

    But I also know that an object will float on water if [tex]\rho_w[/tex] > [tex]\rho_o[/tex]

    I know the specific gravity as 0.5 so I know that the density of the wood is one half that of the water!! So ive concluded that I must add enough lead to make the density of wood + density of added lead = density of water to get a minimum to sink. My only problem is, if density is m/V, how would i know what V to use?

    Ok so let me think this--- I know mass of wood, and I know the denisty of it is 1/2 of water. So set 1/2 water density = m/V where m is the mass of the wood. Then solve for V and ill know the woods volume. But i Have no idea where to go from here! Sure i can look up density of lead in the book, but i dont know how much MASS to add! Would anyone happen to know how to approach this? THANKS!!!!!!
  2. jcsd
  3. Oct 9, 2004 #2
    Ok heres a little more thought into it--- Lead is a defined material that will always have the same density no matter how small a chunk or large a chunk you have. So BY itself even the tinniest piece of lead will sink, But interestingly when you add that piece to the wood there will be an amt that the wood will still float.
  4. Oct 9, 2004 #3

    Doc Al

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    Staff: Mentor

    This is a perfectly good way to approach this problem: Set the average density equal to that of water:
    [tex]\frac{m_{wood} + m_{lead}}{V_{wood} + V_{lead}} = \rho_{water}[/tex]
    Since you know that [itex]\rho = m/V[/itex], you can rewrite this equation in terms of masses and densities only and solve for the only unknown, the mass of the lead.
  5. Oct 12, 2004 #4
    THank you doc al, using that information heres what I did:

    Since we know the mass as 3.15kg, and we know SG = 0.5, then we know m/V = 500 kg/m^3 (half density of water) and thus can solve for Volume of wood = 0.0063 m^3.

    Since the woods floating, we have a statics problem with Fnet = 0. Thus we need to add a mass of lead that increases the net density of wood + lead = to that of water to get a MIN for the wood/lead combo to sink.

    Therefore I set the avg density of water ill denote with p as:

    p(water) = [m(wood) + m(lead)] / [(V(wood) + V(lead)]

    Since we know p = m/V, I attempted rewriting this equation in terms of the masses and densities only and solve for the only unknown, the mass of the lead:

    So thus 1000 = (3.15kg + mlead) / (0.0063 m^3 + Vlead)

    Although this eqn has 2 unknowns, we still know the common ratio of mlead/Vlead = density of lead = 11.3E3 kg/m^3

    Thus... Multiplying both sides by (0.0063 +Vlead) we have

    3.15kg + mlead = 6.3 +1000Vlead

    then: 3.15kg + 1000Vlead = mlead

    Divide both sides by V and get

    3.15/Vlead + 1000 = 11.3E3

    Solve for V lead then sub back into mlead/Vlead = 11.3E3 and get that mlead = 3.4558 kg

    But then....

    I looked at it a little differently instead and thought to increase the total mass of the (wood + lead) to equal the mass of the displaced water,
    thus the buoyant force will be cancelled at this point (and a little more mass will make it sink...)

    since we know the volume of the wood, we know when it is totally submerged the displaced water volume will be the same, and we can then determine the buoyant force of the water = mg = (volume displaced water * water density)g = (mass of wood + mass of lead)*g
    (volume displaced water * water density) = (mass of wood + mass of lead)

    and so I got Mlead = 2.00 kg (using sea water density 1.025 x 10^3...)

    Would this method be wrong? Is this first answer of 3.45kg the more accurate one then? thanks
  6. Oct 13, 2004 #5
    Well this is my last bump for the thread, I would reaaaaly appreciate anyone checking the validity of my method. Thank you :smile:
  7. Oct 13, 2004 #6
    the buoyant force also acts upon the lead. so you must take it into account. it will lead back to what Doc Al has given you.
    The total weight of lead and wood = The buoyant force acts on them
    Last edited: Oct 13, 2004
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