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2001, W. E. Haisler
Chapter 3: Conservation of Linear Momentum
Conservation of Linear Momentum
Chapter 3
The linear momentum of a mass (m) with a velocity
v  vxi  v y j  vzk is defined to be mv . Important: linear
momentum is a vector.
 Accumulationof

linear momentum
  Linear momentum   Linear momentum 

 
 


entering
system

leaving
system

 
 

within
system

 during time period t   during time period t 
 

during time period t  
P   (mv)   (mv)   F
out
ext
in
t
1
2001, W. E. Haisler
Chapter 3: Conservation of Linear Momentum
or, for a time period t:
Psyst
t t
P
syst t
  (mv)in t   (mv)out t   Fext t
For a differential volume element xyz with density ,
the linear momentum
P is given by ( xyz )v .
2
2001, W. E. Haisler
3
Chapter 3: Conservation of Linear Momentum
Consider mass flow in the x direction with density  and
velocity vx flowing into the volume xyz through the
surface yz during a time increment of t. Total mass
entering and leaving through the “x face” in time t is:
y
y
(  vx ) yzt
x
(x,y,z)
z
x
z
x
(  vx )
yzt
x+ x
2001, W. E. Haisler
Chapter 3: Conservation of Linear Momentum
4
For the “left x-face”, the total mass input is:
( vx ) yzt
x
and its momentum is:
( vxv ) yzt .
x
For the “right x-face”, the total mass output is:
( vx )
yzt
xx
and its momentum is:
( vxv )
yzt
xx
vxv  momentum flux in x direction
= momentum of mass flow in x direction (per unit area
and per unit time)
2001, W. E. Haisler
5
Chapter 3: Conservation of Linear Momentum
The momentum of the mass flow will tend to try to move the control
volume (body) and therefore external surface tractions (forces/unit
area) may be required to keep the body in equilibrium. For example,
consider mass flowing through a pipe:
pipe
flow out
flow in
Think of the fluid within the pipe as the "system." For such a system,
the pipe boundary must exert forces on the fluid boundary. These
normal boundary forces on the fluid are what make the fluid turn as it
flows through the pipe. The freebody of the fluid in the pipe is shown
by:
2001, W. E. Haisler
6
Chapter 3: Conservation of Linear Momentum
fluid boundary
g
fluid flow
g
fluid boundary
forces on fluid
boundary
Note that the boundary forces are distributed along the pipe in some
fashion that has yet to be determined. Because of viscosity effects, we
will find that the pipe will also exert shear (or drag type) forces on the
fluid which tend to slow the fluid near the pipe boundary.
Traction forces will exist in both solid and fluid systems. Consider a
frame problem from ENGR 211:
2001, W. E. Haisler
8 kN
7
Chapter 3: Conservation of Linear Momentum
y
8 kN
M
x
V
2m
2m
10 kN
F
10 kN
5m
Reactions at
support
F = 10 kN
V = 8 kN
M = -20 kNm
5m
Freebody of structure
t(n)
n
M
F
=
=
V
Actual force
distribution
Idealized force and
moment resultants
At some point x
Actual force
distribution
Traction Vector
(force per unit area)
At some point x in the beam, there is a distribution of normal (axial
and shear forces) that can be represented by a "traction" vector.
2001, W. E. Haisler
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Chapter 3: Conservation of Linear Momentum
Now consider a general shaped body (system) with forces applied to
its external surface as shown below:
F3
F3
F1
F4
F4
F1
y
F
F1
-n
x
z
A
F1
A
F1
F1
F2
n
F1
Freebody 1
F2
F
Freebody 2
With a cutting plane "A" which has an outward unit normal n on
freebody 2, make two freebody diagrams. At the cut section we must
place internal reactions (forces) as shown. In general, these forces will
be forces distributed over the entire area of the cut section.
For freebody 2, consider the force F applied to an area A that has
unit normal n . Now define the traction t(n ) on this surface as
2001, W. E. Haisler
Chapter 3: Conservation of Linear Momentum
t(n )
9
F
 lim
A0 A
The notation t(n ) means the traction vector (vector force per unit area)
acting on the surface with unit normal n . This vector is not
necessarily normal to the surface. t(n ) is shown below:
t ( n )
A
n
A
n
t (n )
t  n   lim
F
F
  lim
  t n 
A
A
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Chapter 3: Conservation of Linear Momentum
10
For freebody 1, we note that the force at a given point must be equal
and opposite to the force on freebody 2 and hence the tractions must
also be equal and opposite. The normal on FB 1also points in the
opposite direction to FB 2. The traction on the surface with unit
normal  n is denoted as t(-n ) and we have that t(-n )   t(n )
Now lets consider a differential volume system in the fluid (or solid).
Just as for the pipe example, the differential system (xyz) can
also have tractions on its boundary. Define the tractions on the x-face
as shown below:
2001, W. E. Haisler
11
Chapter 3: Conservation of Linear Momentum
y
t( i )
t( i )
y
x
t( i )
y
x x
t( i )
x
(x,y,z)
t( i )
z
x
x
z
components of
t( i )
t(i )  t(i ) x i  t(i ) y j  t(i ) z k
z
The traction vector on the left x face is denoted by: t
(i) x
right face by t
(i) xx
.
and on the
Note: Traction is defined as positive when
it acts in the positive coordinate direction (just like fluid velocity v ).
2001, W. E. Haisler
12
Chapter 3: Conservation of Linear Momentum
The traction vector on the +i face may be written in terms
of its components as: t(i)  t(i) i  t(i) j  t(i) k . The
x
y
z
subscripts on the vector components mean the following:
1st subscript: face the force acts on (i means x)
2nd subscript: direction of the force.
For example, t(i) is the traction on the i face acting in the y
y
direction.
For the tractions acting on the -i and +i faces, the sum of the
external forces (on x-faces only) is given by
[ t(i)  t(i)
]yz .
x
xx
2001, W. E. Haisler
Chapter 3: Conservation of Linear Momentum
13
We should also consider body forces that act on the volume.
Assume that the body force per unit mass (usually due to
gravity) is given g  g xi  g y j  g zk . The total body force
(vector) on the volume during a time increment t is
( gxyz)t
2001, W. E. Haisler
Chapter 3: Conservation of Linear Momentum
14
Now, add the momentum terms in the y and z directions:
In the above, we have looked at momentum associated with
mass flow in the x direction only. We can write similar
momentum terms for mass flow in the y and z directions.
Similarly, we can define the forces on the volume due to
tractions on the y and z faces.
Now, the change in momentum with respect to time:
( v
  v )xyz
t t
t
The conservation of linear momentum (including linear
momentum terms, tractions and boundary forces for each of
the 3 coordinate directions) becomes
2001, W. E. Haisler
( v
Chapter 3: Conservation of Linear Momentum
15
  v ) xyz
t t
t
=[( vxv ) (vxv )
] yzt
x
xx
+ [(v yv ) (v yv )
] xzt
y
yy
+ [(vz v) (vz v)
] xyt
z
zz
+ [(t(i) t(i)
)xz]t
)yz] t + [(t( j) t( j)
y
yy
x
xx
+ [(t(k )  t(k )
) xy] t
z
zz
+ (  g xyz) t
Divide by xyzt and take the limit of each term;
x, y, z, t  0, to obtain
2001, W. E. Haisler
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Chapter 3: Conservation of Linear Momentum

(

v
v
)

(

v
v
)

(

v
v
)
y
 ( v ) 
x 
z

x
y
z
t




















t tj t
 i  k g
x y z
Use the calculus product rule on the 1st and 2nd term in above
equation. The second term (double underlined) becomes


(

v
v
)
 ( vxv )
 ( vz v ) 
y




x
y
 z 










(

v
)

 ( vx )
 ( vz )  
y


v

v

v
   v
v




v


v
y y
z  z 
x
y
 z   x  x








2001, W. E. Haisler
17
Chapter 3: Conservation of Linear Momentum
and the first term (single underline) becomes



(

v
)
 

 v    v 
t
 t 
t

Group these two expansions as follows
 ( vx )  ( v y )  ( vz )   
 v
v


x
y
 z    t 











v

v

v
  vx   v y   vz    v
x
y
 z   t





 ( vx )  ( v y )  ( vz ) 

Conservation of mass is:




x
z
 z 
t







so that the two triple-underlined terms (multiplied by v ) are
conservation of mass and sum to zero.


2001, W. E. Haisler
18
Chapter 3: Conservation of Linear Momentum
Thus, after incorporating conservation of mass (continuity)
into the linear momentum equation, and rearranging the
result, we obtain:



t

t

t

v

v

v
(
i
)
(
j
)

v
   v x v y v z    g  
 (k )
x
y
z 
x y z
t

Note that the above is a vector equation. Note that since
conservation of mass was incorporated into the above,
this linear momentum equation automatically satisfies
conservation of mass. Also, the term in brackets can be
written in vector notation:
[...](v )v
so that
 t(i)  t( j)  t(k )

v
   (v )v   g  

x y z
t