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Finding an increase in depth, pressure and thrust

  1. Sep 14, 2013 #1
    1. The problem statement, all variables and given/known data
    A cylindrical tank of radius 4cm is partly filled with water.A solid metal sphere of radius 3cm is lowered into the water by means of a thin wire until it is totally immersed. Find the increase in
    (i) the depth
    (ii) the pressure at a point on the base
    (iii) the thrust on the base(in terms of π )


    2. Relevant equations
    Pressure = (depth)(g)(density)

    3. The attempt at a solution
    (i)
    h = depth of water before sphere added

    Volume of water before sphere added = 16(π )(h)

    volume after sphere added = 16(π)(h) + (4/3)(π)(64)

    pressure = (1)gh (since the density of water is 1/cm^3)

    I'm not sure what to do next I'd presume I have to get the depth(h) somehow?
     
  2. jcsd
  3. Sep 14, 2013 #2

    rude man

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    Check that second term on the right ...
     
  4. Sep 14, 2013 #3
    Oh sorry V = 16πh + (4/3)π(27) = 16πh + 36π

    I'm still not sure how to get the height
     
  5. Sep 14, 2013 #4
    Would this be correct?
    (i)
    V = volume after sphere is added

    V = 16πh +36π

    but V also = 16π(x+h) ........... where x is the additional depth the water has risen)

    16π(x+h) = 16πh + 36π

    16πx + 16πh = 16πh +36π ............ 16πh cancels

    16πx = 36π

    x = 36/16 =2.25 cm

    (ii) convert to kg/m^3 and meters so our answer will be in pascals

    1000g(0.0225+h) - 1000gh = 1000g(0.0225) +1000gh - 1000gh = 22.5g Pa

    (iii) F = 22.5g(16π) = 360πg N

    (these are the answers my book gives)
     
    Last edited: Sep 14, 2013
  6. Sep 14, 2013 #5

    rude man

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    OK but you should include g = 9.8 in your answer.

    I also advise using Rt and Rs for radius of the tank and sphere instead of the number, until the very end. That way you can check units term-by-term as you go along.
     
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