PHYS1110H, 2011 Fall.
Shijie Zhong
Fluid Motion
Hydrostatic equilibrium
€
€
First, let’s define density and pressure.
Density is the mass per unit volume.
m
ρ= .
(1)
V
The unit is kg/m3. (1) is for 3-D body. Sometimes, it is convenient to define
surface density for 2-D object (e.g., a sheet of paper) as mass per unit area
(i.e., kg/m2), or linear density for 1-D object (e.g., a rope) as mass per unit
length (i.e., kg/m).
Pressure is the normal force per unit area exerted by a fluid.
F
P= .
(2)
A
Pressure has the unit of N/m2 or Pascal (Pa). Pressure is a scalar. At a given
point in fluid, pressure is exerted in all directions equally. That is, it does not
depend on direction.
Atmosphere has a pressure. At the sea-level, this pressure is 1 atmosphere or
atm and is 1.013x105 Pa or 1 bar.
Force from pressure: Force is a vector, while pressure is scalar. For a force
produced by pressure, we need to consider a surface with a certain area and
direction. The direction of the surface is that perpendicular to the surface,
but there are two such directions. Often we choose the one pointing outward
from a volume, and call it as normal vector of the surface, n (which is a unit
vector).


F = −PAn .
(3)
For example, for an object in the
water as shown in the left figure,
the force from pressure acting on
the surface with area dA and
normal
direction n, is


dF = −PndA
(4)
€
€
1
As expected, the force dF points toward the object.
The total force or the net force acting on the object is an integral for the
entire surface of the object.



Fnet = ∫ dF = − ∫ Pn dA ,
(5)
where both P and n can vary from a place to another place on the surface.
€
Hydrostatic equilibrium: For a fluid to remain at rest, the net force
everywhere in the fluid must be zero. This condition is hydrostatic
equilibrium.
€
€
Hydrostatic equilibrium with gravity
Consider a column of water with height h and basal area A. The weight of
the column of water is ρAhg. At the base of the column, there must be a
pressure P that produces a vertical force at the base to balance this weight.
According to (3) for the force at the base, we should have the force balance:
(6)
PA = ρAhg .
(7)
P = ρgh .
If the pressure at the surface of the water is P0 (e.g., atmospheric pressure),
then the pressure at the base of the water column is P = P0 + ρgh .
Notice that on the sidewall of the column, the force from pressure is along
horizontal direction, because the normal vector is horizontal, and the
horizontal forces cancel each other.
€
Take a derivative of (7) with respect to h leads to
dP
= ρg .
(8)
dh
€
2
This is the change in pressure with depth h.
Example 1: Density and pressure in the Earth’s atmosphere are related as
ρ=P/(h0g), where h0 is a constant and is ~8.2 km, and g is gravitational
acceleration. Find how P varies with the height y above the ground.
€
€
€
Answer: Set up a y coordinate pointing up with its origin at the ground.
Notice that in (8), h is the depth and increases downward. Using (8) for this
problem, we will have dy=-dh, and dP / dy = −ρg . Therefore, we have
dP / dy = −P /h0 , and the solution to this equation is
P = cexp(−y/h0 ) , where c is a constant that can be determined by
considering P=P0 at y=0 (i.e., the surface).
€
P = P0 exp(−y/h0 ) .
The atmospheric pressure decreases with increasing height.
Pascal’s law: when there is an increase in pressure at any point in a confined
fluid, there is an equal increase at every other point in the container.
The Pascal’s law applies in hydraulic lift, automobile brakes, and so on.
Archimedes’ principle and buoyancy
Consider a volume of water, V, with arbitrary geometry in a water tank (the
figure above to the left). The volume of water has a weight of mg=ρwVg.
Because of hydrostatic equilibrium (i.e., no motion), this weight must be
balanced by pressure force from the surrounding water to the surface of this
volume, i.e., from equation (5). The pressure force comes from all the
direction and is always perpendicular to the surface (or along -n of the
3
surface). Some of pressure force may cancel each other, but the net force
must be equal in magnitude and opposite in direction to the weight for
hydrostatic equilibrium. This net force from the pressure is called buoyancy
force.




Fnet = ∫ Pn dA = −mg = −ρ w Vg .
€
Now if we replace the volume of water, V, with another media, a solid or a
balloon (the above figure to the right). Then the same buoyancy force exerts
onto the solid or balloon, because for the integral in equation (9), pressure
from the surrounding water and n remain unchanged. This buoyancy force in
terms of magnitude is always equal to
Fb = ρ w Vg .
€
(9)
(10)
Note ρw is the density of the water. Then (6) says that the buoyancy is equal
to the weight of displaced water. This is Archimede’s principle.
Here is a simple proof of (10). Consider a cube with side length l that is
submerged in the water with density ρw. Suppose that the top of cube is a
depth of h, and then the depth of the bottom of the cube is l+h. The pressure
at any given depth b below the water surface is ρwgb, according to (7). The
net force from (5) is




Fnet = ∫ Pn dA = (−Pt A + Pb A)e y + (−Pr A + Pl A)e x ,
(11)
€
4
where A is the surface area for each side and is l2, Pt=ρwgh, and Pb=ρwg(h+l)
are the pressures at the top and bottom of the cube, and Pr= Pl=ρwg(h+l/2)
are averaged pressure on the right and left walls. Substituting them to (11):




Fnet = [−ρ w ghl 2 + ρ w g(h + l)l 2 ]e y = ρ w l 3 ge y = ρ w Vge y .
(12)
€
This proved (10). Notice that we had neglected the front and back walls, as
likely the left and right walls, the force from pressure on these two walls
cancels each other.
5