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Homework Help: Archimedes principle and floating slab

  1. Apr 22, 2005 #1
    A styrofoam slab has a thickness of 12.6 cm
    and a density of 470 kg/m 3 . What is the area
    of the slab if it floats just awash (top of slab
    is even with the water surface) in fresh water
    when a 65.7 kg swimmer is aboard? Answer
    in units of m 2 .

    im trying to use pressure=density of water times thickness of slab times 9.8 and mass=density of water times thickness of slab times area

    i think im on the correct path here but im sure as how to account for the force of the person on the slab in these equations

    A tall water cooler tank is standing on the
    floor. Some fool punched two small holes
    in the tank's wall, one hole at a height of
    34 cm above the floor and the other hole
    60 cm directly above the first hole and 94 cm
    above the floor. Each hole produces a jet of
    water that emerges in a horizontal direction
    but eventually hits the floor at some distance
    from the tank.
    If the two water jets (emerging from each
    hole) hit the floor at exactly the same spot,
    how high H is the water level in the tank
    (relative to the room's floor)? Answer in
    units of cm.

    i cant even find the correct equations to use for this problem, can anyone offer some help
  2. jcsd
  3. Apr 22, 2005 #2


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    Yours is not the easiest approach. Use Archimiedes principle
  4. Apr 22, 2005 #3


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    Do you have Bernouilli's Equation relating the velocity change of a fluid to the pressure difference?
  5. Apr 22, 2005 #4

    ah, ive figured it out, area = Mass / Thickness * (density of water - density of slab)

    im still pretty lost on the second problem.
  6. Apr 22, 2005 #5

    P1 + qgh1 + ½qv1 = P2 + qgh2 + 1/2qv2

    im just confused on how to use it relating to this problem. p1 and p2 are the initial pressures on the holes, i think, so p1 = 1000 times 9.8 times 34 and p2= 1000 times 9.8 times 60

    but then you plug these in and solve for h, correct, where h would be the height of the water level in the tank

    edit; ah you solve for v instead in the above equation
  7. Apr 22, 2005 #6


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    You should be able to use this equation to get the initial velocity of each water jet leaving the tank. For each jet individually you have no height difference, a pressure difference between the inside and outside of the tank, and essentially zero velocity in the tank. When you get your two initial velocities for the two holes you have two projectiles hitting the ground in the same place.

    While inside the tank, you have essentially no velocity and the pressure difference is related to the height difference
    Last edited: Apr 22, 2005
  8. Apr 22, 2005 #7
    ugh, nothing im doing is coming out right for this. for p1 and p2 im putting in h times g times 1000, filling in ½qv1 with 1/2 times 1000 times v1, and qgh with 1000 times 9,8 times height on tank.

    this is just not coming out right at all
  9. Apr 23, 2005 #8


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    It sounds to me like you are working in the wrong direction. The first thing you need to do is figure out how long it takes a drop of water to reach the floor from the height of each hole. From that you can calculate the relative horizontal velocity of the two jets of water. From there I think you can find the pressure ratio inside the tank at the two holes, and I know you can find the pressure difference. Knowing both of those will get you to the answer.
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