12.4. Archimedes’ Law http://www.ck12.org
Fbottom−Ftop=PbottomA−PtopA
= (ρghbottom)A−(ρghtop)A
=ρg(hbottom−htop)A
=ρgVNote that the area of a face times the height of the cube is equal to the volume of the cubeV.
Check Your Understanding
Show that the buoyant force is equal to the weight of the fluid displaced.
Answer:
Recall that the density is given byρ=Vm.
SubstituteVmforρinto the equationFb=ρgV.
Fb=ρgV=
(m
V)
gV=mg→Fb=mg, wheremis the mass of the fluid displaced andmgis its weight.Illustrative Example 1
a. A cube of iron with side length 25.0 cm is submerged in a tank of water with temperature 4◦Csuch that the bottom
of the cube is 1.00 m below the surface of the waterFigure12.13. Calculate the buoyant force on the iron cube.
FIGURE 12.13
Answer: The volume of the cube isV= ( 0. 25 m)^3 = 0. 015625 → 0. 0156 m^3. This is the same volume as the
displaced water. Recall that water at a temperature of 4◦Chas a density of 1, 000 kgm 3. The buoyant force is
thereforeFb=mg=ρV g=
(
1 , (^000) mkg 3
)
( 0. 01563 m^3 )(
- (^81) sm 2
)
= 153. 3 → 153 N.
b. What is the minimum force required to hold the cube in equilibrium while it remains in water? The density of
iron is 7, 860 kgm 3.
Answer: We begin with the Free-Body-Diagram ofFigure12.14.
As the Free Body-Diagram inFigure12.14 shows, the normal force equals the difference of the buoyant force and
the weight of the cube.