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Vacuum buoyancy problem

  1. Oct 9, 2005 #1
    A very powerful vacuum cleaner has a hose 2.75 cm in diameter. With no nozzle on the hose, what is the weight of the heaviest brick that the cleaner can lift?

    So I need to find the pressure and then multiply by the area of the hose... but if P= density*g*h... would I use the density for air? What would the height be?

    Am I approaching this the right way?
  2. jcsd
  3. Oct 9, 2005 #2

    Andrew Mason

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    The 'very powerful' means that it produces practically a perfect vacuum in the hose. So you have to find the total force at the hose opening, assuming that the pressure inside the hose is 0,

    One atm = 101325 Pa (N/m^2)

    Use F = PA where A = area of hose opening. You don't need to know the density of the air.

  4. Oct 9, 2005 #3
    So I assume the pressure is equal to 1 atm? Or 0? I'm confused... If it was zero then the Force is 0...
  5. Oct 9, 2005 #4
    oh wait, nevermind... i got it! thanks!
  6. Oct 10, 2005 #5

    Andrew Mason

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    I get:

    [tex]F_{net} = P_{atm}A_{hose} - P_{hose}A_{hose}[/tex]

    [tex]F_{net} = 1.01325e5 * 5.94(e-4) N - 0 = 59.6 N[/tex]

    or the weight of a 6 kg. brick. This is a force that would accelerate a 6 kg brick at 9.8 m/sec^2. Imagine a tornado 'hose' the size of a house and you can see why a tornado can send cars sailing through the air.

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