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Liquids and densities help please! due tommorow

  1. Feb 24, 2008 #1
    1. The problem statement, all variables and given/known data

    A 2.04 kg piece of wood floats on water of density 1006.2 kg/m3. If the density of wood is 486.0 kg/m3 and the density of lead is 11300 kg/m3, what minimum mass of lead, hung from it by a string, will cause it to sink?

    2. Relevant equations
    Bouyant force= (density of water)(Volume displaced)(gravity 9.81)
    Volume= (Pie)(r)squared(height)
    Density= mass/volume


    3. The attempt at a solution

    I found the weight of the log by doing (9.81)(2.04)=20.0124N. Then i found volume of the log (2.04)/486=0.004197531 metres cubed. Then i found bouyant force upwards (1006.2)(.004197531)(9.81)= 41.43308N. Then I subtracted them to get how much weight is required for it just to stay a float 21.42N. I dont Know what to do after that help please.
     
  2. jcsd
  3. Feb 24, 2008 #2
    Well, you have mass of lead. Find Fb which should be equal to weight of lead. Also, judging from the volume in 2, you didn't mention that is was a cylinder in 1.
     
  4. Feb 24, 2008 #3
    I don't have the mass of lead only the density. and how do you know its a cylinder? and if it is a cylinder i still cant find the volume that way because i dont have the radius or height.
     
  5. Feb 24, 2008 #4
    Isn't that volume of cylinder?
     
  6. Feb 24, 2008 #5
    I know thats a volume of a cylinder but i dont have the radius or height to use that equation so i just used the density=mass/volume
     
  7. Feb 24, 2008 #6
    Well, you are right up to the last part of 21.398N. You also need to account for the Fb of the lead underwater.
     
  8. Feb 24, 2008 #7
    To find the fb of lead would i just do density of lead times volume of the log? I dont know how to solve for the Fb of lea what numbers to use. (11300)(0.004197531)(9.81)=465.3N
     
  9. Feb 24, 2008 #8
  10. Feb 24, 2008 #9
    The buoyant force of the lead is the weight of the water that it displaces.
     
  11. Feb 24, 2008 #10

    Doc Al

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    Staff: Mentor

    You're on the right track, but it's a bit trickier than that.

    It's always good idea to draw a diagram and label all the forces involved. The diagram would show the submerged wood and a piece of lead (also submerged) hanging down from it by a string.

    What forces act on the wood? The buoyant force acting up; the weight and string tension acting down. You've calculated what the string tension must be.

    What forces act on the lead? String tension and buoyant force acting up; the weight acting down. Use this information to calculate the volume and thus the mass of lead needed.
     
  12. Feb 24, 2008 #11
    Forget the string and think of the lead and wood just below the surface. The total buoyant force on the lead and wood must be equal to the weight of the lead and wood. The buoyant force is the weight of the displaced water; the volume of the displaced water is the same as the volume of the lead and wood. Set this up and see what you have.

     
  13. Feb 25, 2008 #12

    Doc Al

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    Staff: Mentor

    That's a perfectly fine way to go as well. :wink:
     
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