144 8 Integration of Fields
isρg, whereρis the mass density of the liquid, andgis the gravity acceleration,
pointing downward, towards the center of the earth. Thus the force balance reads
ρ
dvμ
dt
+∇νpνμ=ρgμ. (8.95)
In a stationary situation, the time derivative of the velocity vanishes. Then one has
∇νpνμ=ρgμ. On the other hand, the force on the solid body, which is assumed to
be totally surrounded by the liquid, is given by (8.91). Replacing the surface integral
by the pertaining volume integral, with the help of the Gauss theorem, and making
use of the momentum balance, one obtains for the buoyancy force
F
buoy
μ =−
∫
V
ρgμd^3 r, (8.96)
where the integral is to be extended over the volume of the solid. Whenρandgare
constant, the integral yields the volumeVof the solid, irrespective of its shape, and
ρV=Mflis the mass of the fluid, contained in such a volume. Thus one has
Fμbuoy=−Mflgμ, (8.97)
which is just the Archimedes principle. Due to the minus sign in (8.97), this force acts
against gravity. The weight “felt” by the body with the massMs, immersed inside
the liquid is(Ms−Mfl)g.
How about the proof for the Archimedes principle for a floating body, that is only
partially immersed in the liquid, say in water? Imagine the floating body is cut at the
water level and a mass equal to that of the part cut off is placed at the center of gravity
of the part remaining under water. The force balance is still the same. Assuming that
a very thin layer of water is above the body, the considerations given above now
apply.
8.4.4 Torque on a Rotating Solid Body
The forceFν, per surface element, exerted by a fluid on the surface of a solid body is
proportional to−nκpκν,cf.(8.94). The torqueTμexerted by a fluid on a stiff solid
body is
Tμ=−εμλν
∮
∂V
rλnκpκνd^2 s. (8.98)
As before,∂Vindicates the closed surface of the solid body,nis its outer normal. It
is understood, thatr=0 coincides with the center of gravity of the body.
Examples for the computation of forces and torques are presented in Sects.10.5.2
and10.3