1. . . Basic Hydrostatic Equationz
(x+x, y+y, z+z)
The sum of the forces on an arbitrary static
fluid element is:
SF = 0 (*)
or Body forces + surface forces = 0 . z y
(x,y,z) y x
An arbitrary fluid element(i.e. any fluid element) can be bounded
or set apart by the surface of a cube. x
Engineers employ mathematical descriptions to model and learn about the real world.
Let’s look at a small amount of fluid inside a large body of fluid in order to develop an equation
for hydrostatic systems. We can identify this fluid matter or the fluid element by placing an imaginary
box around the region of interest. The “box” must be much larger than molecular size and yet small
enough to discern differences in properties like pressure and density.
We are particularly interested in the forces within a fluid system, so let’s look at the forces on this
typical fluid element. The forces will either act on the surface(i.e. one of the 6 faces) or on the matter
throughout the entire volume. All the surface and body forces must sum to zero since the system is
at equilibrium. This is true because we are talking about a static system.
In our analysis, we will look at terms for the surface and body forces and substitute into equation (*).

2. . . . . Basic Hydrostatic Equationz
Pressure is the surface force acting in this problem. We
must consider the pressure acting on each of the 6 faces.
Look at the left face, which is an x-plane or yz-plane
located at x and outlined in yellow.
(x+x, y+y, z+z) . Fx z y
Remember the force due to pressure acts normal to
the surface. The force on this face has been indicated
by an arrow. It tends to cause the fluid to move in the
plus x-direction. The size of this force will be the
pressure at x times the area. The area is DyDz, so the
force is (Px) DyDzi.
(x,y,z) y x x
x+Dx .
(x+x, y+y, z+z)
Now consider the force on the right face or x-plane,
located at x+x. The border of this surface is now
outlined in yellow. Surface force due to pressure
on this faces tries to push the element in the negative
x-direction. The pressure at x+x is (Px+x) and the
area is DyD z. Force equals (Px+x) DyDz(-i). .
(x,y,z)
Surface forces in x-direction: (Px - Px+x) DyDzi

3. . . . . Basic Hydrostatic Equationz
(x+x, y+ y, z+ z)
We have looked at the surface forces on 2 of the 6
faces. The front face bordered in green is called
the y-plane at y. Pressure causes a force in the
plus y-direction(into the sheet). Its magnitude
will be the pressure at y, (Py), times the area of the
front which is xz. Hence, the force is equal
to (Py) xz j. . z y
(x,y,z) Fy y x x
x+Dx .
On the back face, the force is: (Py+y) xz(- j). . z
Surface forces in y-direction: (Py - Py+y) xz j y
If we follow the same procedure for the remaining two faces(top and bottom), we will have
all the surface forces acting on the element.
Surface forces in z-direction: (Pz - Pz+z) xy k

4. Gradient P Basic Hydrostatic Equation 
The only body force to be considered is that due
to gravity. This is equal to the force due to gravity per unit volume g times the volume of the cubic
fluid element, xDyDz. Recall that the sum of the body forces and the surfaces forces acting on
the fluid will equal zero. Substituting,
(g) xDyD z + (Px - Px+x) DyD zi + (Py - Py+y) xz j + (Pz - Pz+z) xy k = 0
Dividing through by xDyDz gives:
g + Px - Px+x_ i + Py - Py+y_ j + Pz - Pz+z_ k = 0
x y z
Now we take the limit as x, Dy, and Dz go to zero.
Lim [g + Px - Px+x_ i + Py - Py+y_ j + Pz - Pz+z_ k ] = 0
x x y z
y0
z0
This becomes: g - P i - P j - P k = 0
x y z
Notice that we have used the definition of the partial
derivative to get the second, third and fourth terms. You
might want to see the MAP on partial derivatives.
Or g = P i + P j + P k
x y z 
Gradient P
g = P
The gradient is a vector that indicates the direction of greatest change(increase) with displacement.
So, in a hydrostatic system, the direction of greatest increase in pressure is the same direction as the
acceleration due to gravity. Notice that the rate of change in pressure with distance varies directly with .

5. Forces on Submerged Surfaces
The pressure acts over the entire surface of
the submerged object. It is a distributed
force. Locally, each small element of area dA
on the surface is flat. The orientation of the
area element is represented by a unit normal
vector n, perpendicular to dA.
dA = n dA  = n dA n is a unit normal vector 
Liquid dA
For a very small area on the surface, the infinitesimal force associated with pressure is:
dF = - p dA = - p dA n
The magnitude of the force is simply the pressure at that position times the area. Its direction
is opposite to the normal vector.
We integrate dF to find the force associated with pressure. To see an application now, click on the hyperlink
for Example 1. Other example may be found on the web page for Chapter 2.

6. . o dF The resultant force on the surface is the “sum” dF
The resultant force on the surface is the “sum”
of all the distributed forces.
Note that p is generally not
constant.
FR =  dF = -   p dA
  A . o
The point of application of the resultant force is
chosen so that the torque on the object is the
same as the distributed force. 
dM = r x dF
M =  r x dF = -  r p dA (r x n)
  A FR