# Solve for Equations of Motion Conservation of linear momentum is a little
bit different then conservation of mass. The main thing that you will recognize this is
newtons law, it is the sum of the forces equals the mass times the acceleration. So what forces
are we talking about. Well we have the normal forces, and we use sigma to indicate them,
and we have the tangential forces, and we use tau to indicate that. So lets take a look
on one plane of a differential Q, and our coordinate system is going to be such that
this is z, this is x, and this is y. The first thing that we are going to look at is this
normal force that comes out of the plane, normal to the plane. We call this force sigma
sub xx, and if you look at the two subscripts the way they are defined. Is the first subscript
is the direction of the normal to the plane. So if you look at it. It is in the x-direction.
Now lets look at the same plane, but lets look at the tangential forces. So our tangential
forces would go this way along the plane, and this way along the plane. So lets name
them. Lets call this one number 1, and this one number 2. If you recall the firsts subscript
is the direction normal to the plane. So that is going to be x. The second subscript is
the direction of the force, or stress. So for normal forces those two subscripts are
going to be the same, but if we look at number 1. The direction id in Z. If we look at number
2 again it is in the normal is in the x plane, but the stress is in the y plane. You might
look at these and say why are these stresses positive, when they are in the negative direction…
The reason for that is that we consider the normal to the plane as the positive direction.
So what that means in this particular case is we shift our coordinate system such that
the positive coordinate system looks like this. This is x, this is y, and this is z.
So we basically rotated the axis. Now we can look at a differential element using the different
forces, and using a Taylor series expansion, and what we get in differential form is the
surface forces. So this is our normal. This is one of our tangential forces. This is the
second tangential force and this is all multiplied by the volume of our differential element.
We can do the same thing with the y forces. Here however it will be sigma yy, tau xy,
tau zy. We will also would have the surface forces in the z direction. In addition the
other kind of force we have are body forces. These body forces we generally assume are
gravity. So now when we sum up our differential forces we have both our surface forces and
our body forces. Notice those arrows, that indicates that these are vectors, which means
we have them in the x-direction, the y-direction, and the z-direction. Now using newtons second
law, and I am just going to look at it in the x direction. The sum of these forces in
the x-direction have to be the mass times the acceleration in the x-direction. What
this leads us to is the equations of motion, and the equations of motions are fundamental
equations that we use in fluids. We just simplify them for our particular needs. So this is
in the x-direction, rho,gx, plus our normal force, plus our tangential forces. Remember
there are 2 of them. Tau zx/dz, and this has to equal rho, du/dt, plus u, du/dx, plus v,
du/dy, plus w, du/dz. So this part right here are the sum of the forces in the x direction,
and this part right here is the acceleration in the x-direction. Where this is the local
acceleration, and these are the convective term. Since it is a vector we would have to
do it in all 3 directions. So for example in the y direction plus our other forces is
going to equal rho. Now it is dv/dt, plus u*dv/dx, plus v*dv/dy, plus w*dv/dz), and
when you put this all together all three of these in the x-direction, y-direction, and
z-direction we get the equation of motion.

### One thought on “Solve for Equations of Motion”

• November 25, 2012 at 9:49 pm