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Buoyancy problem

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data
    With the nose above the water, about 95% of the body is submerged. Calculate the power expended by a 50-kg woman treading water in this position. Assume that the average density of the human body is about the same as water (p = pw = 1 g/cm3 ) and that the area A of the limbs acting on the water is about 600 cm2.


    2. Relevant equations Bear with me...
    FB = gVpw <-- force up, due to buoyancy
    Fg = gVp <-- force down due to gravity
    FD = Fg - FB = gV(p - pw ) <-- net downward force (assuming the object is more dense than water)

    Mass of water accelerated per unit time (m) = Avpw
    Momentum given to the water per second = mv
    ^^ these two equations make absolutely no sense to me, but when they are put together:
    FR = pwAv2 <-- force produced that should be equal to FD to counter the downward force so you can float, and since the units (density x area x velocity2) turn out to be essentially mass x acceleration (ie a force), $#!t makes sense...

    FD = FR = gV(p - pw ) = pwAv2

    Rearranged... v = *sqrt* of: gV(p - pw) / Apw

    KE/sec = Power generated by limbs, P = 1/2 mv2

    Substituting eq'ns for m & v...

    P = 1/2 *sqrt* of: [W (1 - pw/p)]3 / Apw

    Here W is the weight of the object (W = gVp).


    3. The attempt at a solution
    First, the question says to assume that density of the body is the same as water... so why is it necessary to do work to remain afloat? In the last equation, if you divide the density of the body and water, it equals 1, and 1-1 = 0, therefore P = zero... :uhh:
     
  2. jcsd
  3. Nov 14, 2009 #2

    Delphi51

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    A force must be exerted on the body to keep 5% of it out of the water.
    To make the water push up on the body, the hands must push down on the water with the same force. It is rather like a rocket hovering. We know the thrust (force) but want the power = F*v. Do we estimate the average speed of the hand? Why does the area of the hands matter? It seems to me we need another property of water that indicates how fast we must push on the water with that hand area to get the force we want. Something to do with viscosity or just the mass of the water that the hands move?
     
  4. Nov 15, 2009 #3
    Power = F*v? I think it's better to go with:

    P = 1/2 *sqrt* of: [W (1 - pw/p)]3 / Apw

    We have weight, density of water & body, area of limbs, so technically we should be able to solve for P... but as I said the density of water & the woman's body (as mentioned in the question) are the same... I must be misunderstanding something...

    Nothing in my textbook mentions viscosity, so it could be the mass of the water that the hands move that matters...

    Btw, the answer is P = 7.8 W.
     
  5. Nov 15, 2009 #4

    Delphi51

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    I have no idea what your equation is saying - looks like weight divided by force, which would give units of mass, not power.

    Interesting to work backwards from the answer . . .
    The force needed to keep 5% of the body out of the water is .05mg = 24.5 N.
    P = Fv, v = P/F = 7.8/24.5 = 0.318 m/s.
    That would be some kind of average because about half of the time the hands are not pushing but are retracting for the next push, so during the push perhaps a speed of 0.6 m/s. Quite reasonable, but how do we figure out how fast you have to push to get an average force of 24.5 N? Could it be just the F = ma required to accelerate the water up to speed? How much water? It would be 600 cm² at the hand and how far away from the hand? It still seems to me that the viscosity determines how hard you have to push at any given speed. The person would have to flap at super speed to remain afloat in air.
     
  6. Nov 16, 2009 #5
    I explained the equation in my first post - it's just P = 1/2 mv2. And it's from the text book, so I'm pretty sure I should use it.

    Here's my textbook explaining the equation (7.5 - pg. 87-88):

    http://books.google.com/books?id=e9...ed to remain afloat Whether an animal&f=false

    Tell me if the link works out... After eqation (7.10), I don't understand what the text book is saying.
     
  7. Nov 16, 2009 #6

    Delphi51

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    Thanks very much for the link, Sodr! A peek into a fascinating textbook.
    I simplified what it says to this case where the density of the animal and water is the same. So eqn 7.10 just says F = .05mg = 24.5 N as we had already figured.
    Equation 7.11 says the mass of water moved per second is M = ρAv, which makes sense if you think of the velocity as d/t so your cross sectional area A times d is the volume and multiplying by ρ gives the mass still divided by the time. The assumption is that if your hand moves 10 cm, it makes 10 cm of water move at the speed of the hand. Wish I'd thought of that!
    The next equation just says that Mv is the change in momentum (remember M is the mass per second) so F = Mv = ρAv². We know everything in that except the v, so we can find the velocity of the hand necessary to produce the 24.5 N force.

    Finally, it says the power needed to move the water is ½Mv², which seems very odd until you remember that M is the mass per second, so this is the kinetic energy per second - the right units anyway! It does bother me a bit but I can't pin down why. If you slap the M and v formulas in, it works out to
    Power = ½Fv. Ah, that's what's bothering me - that ½ doesn't seem right! But you can now easily work out your answer.
     
  8. Nov 18, 2009 #7
    I like what I'm hearing... I'll tackle this question shortly.
     
  9. Nov 18, 2009 #8
    I need help with the units...

    We have 24.5 N... so should we convert 600cm^2 and 1g/cm^3 into meters? How do I do this?

    600cm^2 * 1g/cm^3 = 600 g/cm
    600g/cm * 100cm/m * 1kg/100g = 60 kg/m

    Nevermind.
     
    Last edited: Nov 18, 2009
  10. Nov 18, 2009 #9
    Thank You! Here's a present:
     
    Last edited by a moderator: Sep 25, 2014
  11. Nov 18, 2009 #10

    Delphi51

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    Thanks!
     
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