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Fluid mechanics buoyancy force

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data

    I've attached the problem

    I'm kind of lost as to how to even start this problem.


    For part )

    I think my first approach is to sum up forces in the vertical direction.
    Which would yield F-W1-W2=ma=0

    Where:
    F=buoyancy force
    W1=weight of submerged volume of oak
    W2=weight of the steel thing.

    W1=(Specific weight of oak)*(L-h)*A
    (L-h)*A is essentially the volume of the submerged oak or the volume of the displaced fluid.

    Can someone verify if this is the right approach?

    Also, how do I find the specific weight of oak & steel given the specific gravity?
     

    Attached Files:

  2. jcsd
  3. Sep 16, 2012 #2

    rude man

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    Why just the submerged part of the oak? Doesn't gravity work on the top part too?

    For the rest you just need to review what the definition of specific gravity is and how it relates to weight.
     
  4. Sep 16, 2012 #3
    Oops I got that confused. It's the whole oak buoy, notjust the submerged part.

    F is the weight of the displaced volume.
     
  5. Sep 16, 2012 #4

    rude man

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    Correct, now use the given specific gravities to balance the forces and solve for h.
     
  6. Sep 16, 2012 #5
    I'm trying to find the density from the SG. But what I'm confused about is in our book there's only specific gravity equations for liquids and gases. Oak and steel are solid materials. But in my textbook they used the same equation for liquid to solve for the density of oak and steel which is just (1000kg/m^3)*SG of oak or steel. Why is it that they assume oak and steel are liquids?
     
  7. Sep 16, 2012 #6

    rude man

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    You are given the specific gravities for both the oak and the steel. That is all you need to get the weights of the buoy and the counterweight. Did you look up the definition of s.g.?
     
  8. Sep 16, 2012 #7
    Yes, but I need to convert it to density or specific gravity before it can be much use.

    Weight of oak is specific weight of oak multiplied by the volume of the oak. To get the specific weight of oak I need to multiply 0.71*1000*9.81
    Where 9.81 m/s^2 is gravity
    1000kg/m^3 is density
    0.71 is the SG of oak

    But I don't understand why I have to multiply by 1000.
     
  9. Sep 17, 2012 #8

    rude man

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    S.G. is a ratio. It is the ratio of the weight of a given substance in relation to the weight of the same volume of fresh water at 4 deg. C.

    So 1 cc of wood weighs 0.71 times as much as 1 cc of water.

    You need to multiply 0.71 by 1000 because 1 cubic meter of water has a mass of 1000 kg. You need to use the units your prof or whoever tells you to use, in this case the widely-used mks system (meter, kilogram, second). So you now know that 1 cubic meter of oak has a mass of 0.71 x 1000 = 710 kg.

    Next, you need to convert mass to weight, and that's what the 9.81 number does. It converts 710 kg to the right number of Newtons, the unit of force or weight in the mks system.

    You now have what you need to finish the exercise. Remember - everything must be in mks, which means buoyancy must also be expressed in Newtons.
     
  10. Sep 17, 2012 #9
    Thanks a bunch.

    If you can check my work, that'd be awesome.

    ƩF=Fb-Wo-Ws=ma=0, Ws=Fb-Wo
    Fb=buoyancy force
    Wo=weight of oak
    Ws=weight of steel



    Volume of immersed oak=.1*.1*(4-.6)=.034m^3
    Vs=volume of steel=Ws/γs
    γs=specific weight of steel

    γ=specific volume of seawater=1.02*1000*9.81=10006.2 N/m^3
    Fb=γ*(Volume of seawater)=10006.2*(.034+Ws/γs) << The parenthesis is the volume of the displaced fluid which is the volume of seawater taken up by the steel&oak
    10006.2*(.034+Ws/(7.85*9.81*1000))=340.21+.13Ws

    Plugging the above back into Ws=Fb-Wo
    gives Ws=340.21+.13Ws-Wo=340.21+.13Ws-γo*Vo
    where γo=specific weight of oak and Vo=volume of oak
    Vo=.1*.1*4
    γo=.71*1000*9.81

    Ws=340.21+.13Ws-(.71*1000&9.81)(.1*.1*4)
    Ws=70.81N
     
  11. Sep 18, 2012 #10

    rude man

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    Specific volume does not have units of N/m^3 so I'm not sure where you're going with this.

    Frankly, I get dizzy trying to tie in s.g. in this problem. Most of us are used to dealing in density and I suggest you do the same. Density ρ = mass/volume.
    The relation is ρ = 1000*(s.g.) kg/m3. So for example, ρ of sea water = 1000*1.02 = 1020 kg/m3. S.g. is not in absolute MKS units but is related to water at 4 deg C. That makes it an awkward parameter. Switch to density.

    Now rewrite your first equation using density. Buoyancy Fb now becomes A(L-h)ρg. If you insist on using s.g., Fb = 1020A(L-h)g(s.g.).
     
  12. Sep 18, 2012 #11
    Hey sorry specific volume was typo. It's supposed to be specific weight

    A general equation I used for specific weight=ρ*g = SG*1000*g which has the units N/m^3. It's just the way we do it in class, but it's essential the same as using density except multiplying it by the gravity.

    The "Vs=volume of steel=Ws/γs" is the volume of seawater displaced by the steel. Don't I need to consider that since the force of buoyancy is the volume of the displaced seawater (=volume of steel&oak submerged in the seawater) multiplied by seawater's specific weight?
     
  13. Sep 18, 2012 #12
    i attached another image I saw in my book. It's a similar problem and they seemed to take into consideration the volume of the steel.
     

    Attached Files:

  14. Sep 18, 2012 #13

    rude man

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    I have to eat crow! You're quite right, you have to include the volume of the steel.

    And when I doI get the same answer you did! Congrats and sorry again for my blunder.

    Now go ahead with part b?
     
  15. Sep 18, 2012 #14
    Thanks for all the help by the way.

    I actually had a question on part b that I was going to ask my professor.

    Part b should be pretty simple, but when they refer to the spar buoy, does that include the steel piece as well or just the oak? It's my first time seeing the term spar buoy, but from google images it seems a spar buoy includes the metal piece as well.
     
  16. Sep 18, 2012 #15

    rude man

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    The idea is simply that the salt water is replaced by fresh water. That will cause h to change. The buoy and the steel are the same as in part a.
     
  17. Sep 18, 2012 #16
    Okay I just did it. The weight of the oak and steel remain unchanged.

    Weight of oak = (.1*.1*4)(.71*1000*9.81)=278.6N
    Total weight of oak and steel = 278.6+70.7=349.3N

    Volume of steel = Ws/γs
    = 70.7 / (7.87*9.81*1000)=9.162 * 10^-4 m^3

    Volume of submerged oak = .1*.1*(4-h)

    Fb=Volume of submerged liquid * (1.00*1000*9.81)=9810*volume of submerged liquid
    = 9810 (.1*.1*(4-h) + 9.162 * 10^-4 ) = 349.3N
    h=0.53 m or 53 cm

    I think from this problem, I can assume that the same object placed in any kind of liquid would have the same buoyancy force? Since the weight remains unchanged.
     
  18. Sep 18, 2012 #17

    rude man

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    You got the right answer (h = 0.53 m) but your statement is only correct as long as the object doesn't sink.

    If the object sinks, Fb is not = W since there is another force in play, to wit, the bottom of the lake or whatever pushing up on the object.

    In all cases Euclid's statement is correct of course. Smart boy for a guy living 2500 years ago!
     
  19. Sep 18, 2012 #18
    Oh yeah, I completely forgot the other 2 cases. Thanks for the help.
     
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