http://www.ck12.org Chapter 12. Fluid Mechanics
FIGURE 12.14
Free-Body-Diagram.∑F=^0 →FN+Fb−mg=^0 →FN=mg−Fb.
A force equal to the normal forceFNapplied to the cube will keep the cube in equilibrium anywhere beneath the
water surface. The weight of the iron block is
mirong=ρironV g=(
7 , 860
kg
m^3)
( 0. 01563 m^3 )(
9. 81
m
s^2)
= 1 , 205. 2 → 1 , 205 N
→FN=mg−Fb= 1 , 205 − 153 = 1 , 052 N.The links below demonstrate Archimedes Principle.
http://demonstrations.wolfram.com/ForcesExertedOnAnImmersedObject/
http://www.youtube.com/watch?v=VDSYXmvjg6M
Floating Objects
Archimedes’ principle is no different for objects that are only partially submerged. The weight of the volume of
fluid displaced is still equal to the buoyant force. But since the object is in equilibrium while floating, the upward
force on the object is the buoyant force.
Illustrative Example 2
a. The density of yellow pine is 420mkg 3. What volume of water at a temperature of 4◦Cdoes a cylindrical log of pine
with dimensionsr= 15. 0 cmand lengthL= 7. 60 mdisplace in order to float? SeeFigure12.15.
Answer: The weight of the logmloggmust equal the buoyant force (the weight of displaced water)mwaterg.