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Wize University Physics Textbook (Master) > Fluid Mechanics

Buoyancy

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Buoyancy


Buoyant force is an upward force exerted on an object either partially submerged or completely submerged in a fluid due to pressure differences between top and bottom of the object.

Archimedes's Principle

Archimedes' principle says that the amount of buoyant force equals to the weight of the liquid that the object displaces while floating/sinking.

Fb=Mdisplacedg=ρfVdisplacedg\boxed{F_b=M_{displaced}g=\rho_{f}V_{displaced}g}


Hence, buoyant force has the following equation:

Fb=ρfVsubg\boxed{F_b=\rho_fV_{sub}g}

  • ρf\rho_f is the density of fluid
  • VsubV_{sub} is the submerged volume of the object
  • gg is the gravitational acceleration
Wize Tip
When an object is only partially submerged, it is only displacing part of its own volume. So the buoyant force only depends on the part of the object that is in water. Note that VdisplacedV_{displaced} is the same as VsubV_{sub}.

Watch Out!
Buoyant force is also shown by BB or FBF_B.


PAGE BREAK

Usually there is a competition between buoyant force and gravity. At static equilibrium condition when the object doesn't move, the net force on the object is zero. So:

ρfVsubg=mg=ρoVog\rho_fV_{sub}g=mg=\rho_oV_og

  • Whereρo\rho_ois the density of the object and VoV_ois the volume of the object


Wize Concept
If the object is fully immersed in water and then it is released:
  • If ρo>ρf\rho_o>\rho_f, then mg>Fbmg>F_b and the object sinks
  • If ρo<ρf\rho_o<\rho_f, then mg<Fbmg<F_b and the object floats on the surface of the fluid

Exam Tip
For a floating object, the fraction of its volume submerged in the fluid is given as:

Fraction Submerged=ρoρf\text{Fraction Submerged}=\dfrac{\rho_{o}}{\rho_f}


Watch Out!
The apparent weight is the force of gravity minus the buoyant force.





Example: Hanging Boxes in Water


Boxes AA, BB and CC have the same size. AA and CC are located at the same level but BB is deeper in water. If mA>mB>mCm_A>m_B>m_C, compare:

a) Buoyant force
b) Force exerted by the water on top of each box
c) Tension in the rope
d) Apparent weight



Part a)

The buoyant force is always:

B=ρfluid Vsub gB=\rho_{fluid} \ V_{sub} \ g
All blocks have the same size and therefore the same VsubV_{sub}, which means that the buoyant forces are all equal:

BA=BB=BCB_A=B_B=B_C


Part b)

The force exerted by the water on top of each box is due to the pressure of the column of water:

F=ρwaterghF=\rho_{water}gh

Therefore we have:
FB>FA=FCF_B>F_A=F_C because hB>hA=hCh_B>h_A=h_C


Part c)

To find tension, we need to look at the free body diagram of each box.

For each box we have:

∑F=ma\sum F=ma

B+T−mg=0B+T-mg=0

T=mg−BT=mg-B

We know that BB is the same for all boxes (from part a).

Therefore we have:

TA>TB>TC T_A>T_B>T_C because mA>mB>mCm_A>m_B>m_C


Part d)

The apparent weight is the real weight minus the Buoyant force:

W=mg−BW=mg-B

Again, we know that BB is the same for all boxes (from part a).

Therefore we have:

WA>WB>WCW_A>W_B>W_C because mA>mB>mCm_A>m_B>m_C

Example: Wooden Cylinder Floating in Oil and Water


A cylindrical block of wood 3.503.50 cm high is submerged partly in water and partly in oil. It is submerged so that the flat circular face is facing vertically upwards, and completely covered by the oil. Oil, with a density of 930930 kg/m3 floats on water. The density of wood is 960960 kg/m3. At what position from the bottom of the block is the water-oil interface?


Let's draw the free body diagram for the cylinder:


At equilibrium, the net force on the cylinder should be zero:

∑F=ma\sum F=ma

FB water+FB oil−mg=0F_{B \ water}+F_{B \ oil}-mg=0

ρwaterVin waterg+ρoilVin oilg=mwoodg\rho_{water}V_{in \ water} \cancel g+\rho_{oil}V_{in \ oil} \cancel g=m_{wood} \cancel g


Let's define the fraction of the volume in water as:

fw=fraction in water=Vin waterVtot   →   Vin water=fwVtotf_w=\text{fraction in water}=\dfrac{V_{in \ water}}{V_{tot}} \ \ \ \to \ \ \ V_{in \ water}=f_wV_{tot}

and the fraction of the volume in oil as:

fo=fraction in oil=Vin oilVtot   →   Vin oil=foVtotf_o=\text{fraction in oil}=\dfrac{V_{in \ oil}}{V_{tot}} \ \ \ \to \ \ \ V_{in \ oil}=f_oV_{tot}


Substituting these in, and using mwood=ρwoodVtotm_{wood}=\rho_{wood}V_{tot} , the force equation becomes:

ρwaterVin water+ρoilVin oil=mwood\rho_{water}V_{in \ water} +\rho_{oil}V_{in \ oil} =m_{wood}

ρwaterfwVtot+ρoilfoVtot=ρwoodVtot\rho_{water}f_w\cancel {V_{tot}} +\rho_{oil}f_o\cancel{V_{tot}} =\rho_{wood}\cancel{V_{tot}}

ρwaterfw+ρoilfo=ρwood\rho_{water}f_w +\rho_{oil}f_o =\rho_{wood}


Now, the fractions submerged in oil and water should add up to one, that is fo+fw=1f_o+f_w=1.

This means that fo=1−fwf_o=1-f_w and our equation becomes:

ρwaterfw+ρoil(1−fw)=ρwood\rho_{water}f_w +\rho_{oil}(1-f_w) =\rho_{wood}


Let's solve for the fraction in water:

ρwaterfw+ρoil−ρoilfw=ρwood\rho_{water}f_w +\rho_{oil}-\rho_{oil}f_w =\rho_{wood}

fw(ρwater−ρoil)=ρwood−ρoilf_w(\rho_{water} -\rho_{oil}) =\rho_{wood}-\rho_{oil}

fw=ρwood−ρoilρwater−ρoilf_w=\dfrac{\rho_{wood}-\rho_{oil}}{\rho_{water} -\rho_{oil}}

Put the numbers in:

=960−9301000−930=0.43=43%=\dfrac{960-930}{1000-930}=0.43=43\%


Therefore the height in water is 43%43\% of the total height:

h=0.43×3.5=1.5h=0.43\times3.5=1.5 (cm)

Practice: Floating Helium Balloon


A large Helium balloon has a volume of 20.020.0 m3. The density of air is 1.291.29 kg/m3 and the density of Helium is 0.1790.179 kg/m3. The mass of the balloon material is 5.05.0 kg.
a) What is the buoyant force acting on the balloon?