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Archimedes problem

  1. Oct 5, 2014 #1
    1. The problem statement, all variables and given/known data
    A rod 6 meters in length has specific gravity 25/36. One end of the rod is tied to a 5 meter rope, which in turn is attached to the floor of a pool 10 meters deep. Find the length of the part of rod, which is out of water.

    The answer is 1 meter.

    2. Relevant equations
    mg=V(imm)p(l)g=V(rod)p(rod)g

    F(of buoyancy)=Tension + mg

    3. The attempt at a solution
    I don't really know what to do, but here's what I did anyway.

    I tried using V(imm)*density of fluid=V(rod)*density of rod. But then I thought that there should be a tension component in the force equation as well. I further tried ignoring tension, and using the above mentioned equation to figure out how much of the rod would be outside. I got 2meters. I figured I could use trig to figure out the angle of inclination with the water level, but I couldn't really use it.
     
  2. jcsd
  3. Oct 5, 2014 #2
    Sorry forgot the figure
     

    Attached Files:

  4. Oct 5, 2014 #3

    SteamKing

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    Hint: Analyze the rod as a free body. Since you know the density of the rod (or specific gravity, in this case), you should be able to determine if the rod floats. If the rod is capable of floating, you should be able to write equations of statics using the end tied to the rope as your reference. Knowing other facts, like the depth of the pool, you should be able to calculate how much of the rod sticks out of the pool when it is in equilibrium.
     
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