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Buoyant Force with two attached blocks

  1. Dec 3, 2008 #1
    1. The problem statement, all variables and given/known data
    As shown in the figure below, a wooden cube measuring 20.0 cm on each side floats in water with 90.0% of its volume submerged. Suspended by a string below the wooden cube is a metal cube. The metal cube measures 10.0 cm on each side and has a specific gravity of 6.00.
    a) Taking the density of water to be 1000 kg/m^3, what is the density of the wooden cube?
    b) What is the tension in the string between the cubes? Assume the string itself has negligible mass and volume. Use g = 9.8 m/s^2.
    c) The pair of blocks is now placed in a different liquid. When the blocks are at equilibrium in this new liquid, the buoyant force acting on the wooden cube is exactly the same as the buoyant force acting on the metal cube. What is the density of this new liquid?

    2. Relevant equations
    1) Fb=p x V x g
    2) P(object)/P(water)=V(displaced)/V(object)

    3. The attempt at a solution
    I've only been trying to do part A (I've spent 1.5 hours on it already) by using equation 2, but I've only gotten this: (8.00m^3 x 90%)p(obj)= (1000kg/m^3)(8.00m^3), which means p(obj)=1111.1, but that doesn't take into account the second block, so I tried adding the [p(metal obj) x V (disp)] to the right side of the equation, but then I get 1944, which is wrong as well.
  2. jcsd
  3. Dec 3, 2008 #2


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    First draw a force diagram (in physics always draw a diagram!)

    a, The first step is to calcualte the apparent weight of the metal block. This is the weight (due to gravity) minus the flotation. Then you need a force diagram for the wooden block; you have it's own weight + the apparent weight of the metal block pulling down and the flotation pushing up.

    Remember the flotation force is the weight of water displaced (good old Archimedes)
  4. Dec 4, 2008 #3
    I got the W(app) to be 58.8N, which makes means that the V(water) displaced is 7.2kg [m(wood)-58.8N=(90%)(V(wood)) x (p(water)] (This is how I tried to go about it using the free body diagram). From this, I determined that the W(wood)=70.56, meaning M(wood)=11.76N [70.6N-58.8N]. Using this to find the density, I got it to be 1633.3N/M^3 [M(wood)/V(water displaced)], but this appears incorrect and brings me back to square one seeing as I have only one attempt at an answer left. I think the problem might be in my apparent weight calculation [mg-Fb(on metal block)], because then the apparent weight should equal the tension force between the two blocks (?).
  5. Dec 4, 2008 #4


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    I think you are almost there;

    Volume of metal cube = 10x10x10cm = 1000cc
    density = 6g/cc so mass = 6000g, it displaces 1000g of water, so it's apparent weight is 5000g

    The wooden cube is 20x20x20 = 8000cc and it displaces 90% of this = 7200g water
    The force down on the cube is it's own weight + apparent weight of the metal
    The force up is the weight of water displaced.
    Since it isn't moving these must be equal.

    So 8000(cc)*density(g/cc) + 5000(g) = 7200(g) check units balance
    density = (7200-5000)g/8000cc = 0.27 g/cc
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