Moment of a force along an axis
Couples
“If you find yourself in a hole, stop digging.” –Will Rogers Objectives
¢  Understand
the vector formulation for
finding the component of a moment along
an axis
¢  Understand the idea of a couple and the
moment it produces
2
Moments Along an Axis, Couples
Monday, September 24, 2012
1
Tools
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Basic Trigonometry
Pythagorean Theorem
Algebra
Visualization
Position Vectors
Unit Vectors
Reviews
Cross Products
Dot Products
3
Moments Along an Axis, Couples
Monday, September 24, 2012
Review
¢  A
moment is the tendency of a force to
cause rotation about a point or an axis
4
Moments Along an Axis, Couples
Monday, September 24, 2012
2
Moment about an Axis
¢  There
are times that we are interested in
the moment of a force that produces
some component of rotation about (or
along) a specific axis
¢  We can use all the we have learned up to
this point to solve this type of problem
5
Moments Along an Axis, Couples
Monday, September 24, 2012
Moment about an Axis
¢  First
select any point on the axis of
interest and find the moment of the force
about that point
¢  Using the dot product and multiplication
of the scalar times the unit vector of the
axis, the component of the moment about
the axis can be calculated
6
Moments Along an Axis, Couples
Monday, September 24, 2012
3
Moment about an Axis
¢  If
we have an axis a-a we can find the
component of a moment along that axis
by
(
)
   
M a−a = ua−a ua−a  M

where M is the moment
about any point on a-a
7
Moments Along an Axis, Couples
Monday, September 24, 2012
Problem F4-17
8
Moments Along an Axis, Couples
Monday, September 24, 2012
4
Example
¢  Given:
9
System as Shown
Moments Along an Axis, Couples
Monday, September 24, 2012
Example
¢  Required:
Moment of force F along axis
a-a
10
Moments Along an Axis, Couples
Monday, September 24, 2012
5
Example
¢  Solution:
(
   
M a − a = ua − a ua − a  M
11
)
Moments Along an Axis, Couples
Monday, September 24, 2012
Example
¢  Solution:
(
)
   
M a−a = ua−a ua−a  M


need ua−a and M

where M is the moment of the force
about some point on a-a
12
Moments Along an Axis, Couples
Monday, September 24, 2012
6
Example
¢  Solution:

 r
ua−a = 0 A
r0 A

r0 A = A − O
{ } { }
{ A} = {−3,−2,6} ft
{O} = {0,0,0} ft
rOA

  
r0 A = −3i − 2 j + 6k ft
(
)
r0 A = 7 ft

3 2  6 
ua−a = − i − j + k
7 7
7
13
Moments Along an Axis, Couples
Monday, September 24, 2012
Example
¢  Solution:
Now we have to select some
point on a-a and find the moment of the
force F about that point
¢  Please don’t mistake rOA for the moment
arm
r
¢  rOA is just the position
vector that we used
to generate the unit
vector along a-a
OA
14
Moments Along an Axis, Couples
Monday, September 24, 2012
7
Example
¢  The
origin is on a-a (not always but in
this problem it is) so we can find the
moment of F about the origin
rOA
15
Moments Along an Axis, Couples
Monday, September 24, 2012
Example
¢  First
we define a moment arm from the
origin to the line of action of F
rOA
16
Moments Along an Axis, Couples
Monday, September 24, 2012
8
Example
¢  We
can use rOB as our moment arm

rOB = { B} − {O}
rOA
{ B} = {0,4,0} ft
{O} = {0,0,0} ft


rOB = 4 j ft
( )
17
B
rOB
C
Moments Along an Axis, Couples
Monday, September 24, 2012
Example
¢  Now
we have to describe F as a
Cartesian
vector


F = Fu
 F
uF = uCB

 r
uCB = CB
rCB

rCB = { B} − {C }
{ B} = {0,4,0} ft
{C } = {4,0,−2} ft
rOA
rOB

 

rCB = −4i + 4 j + 2k ft
(
)
rCB = 6 ft

4 4  2 
uF = − i + j + k
6 6
6




F = −400i + 400 j + 200k lb
18
(
)
Moments Along an Axis, Couples
B
C
Monday, September 24, 2012
9
Example
¢  Taking
the moment of F about the origin
(O)
  
M = rOB ⊗ F





M = 4 j ft ⊗ −400i + 400 j + 200k lb







M = 4 j ⊗ −400i + 4 j ⊗ 400 j + 4 j ⊗ 200k



M = 1600k + 800i ft ⋅ lb
( ) (
) (
((
(
)
)
) (
rOA
)) ft ⋅ lb
rOB
B
C
19
Moments Along an Axis, Couples
Monday, September 24, 2012
Example
¢  Finding
the magnitude of the projection of
M along a-a
 
ua−a  M

3 2  6 
ua−a = − i − j + k
7 7
7



M = 1600k + 800i ft ⋅ lb
(
rOA
)
  ⎛ 3 ⎞
⎛ 6⎞
ua−a  M = ⎜ − ⎟ (800 ft ⋅ lb) + ⎜ ⎟ 1600 ft ⋅ lb
⎝ 7⎠
⎝ 7⎠
 
ua−a  M = 1028.57 ft ⋅ lb
rOB
B
C
20
Moments Along an Axis, Couples
Monday, September 24, 2012
10
Example
¢  Converting
the magnitude to a Cartesian
vector
(
  
ua−a ua−a  M
)

3 2  6 
ua−a = − i − j + k
7 7
7
 
ua−a  M = 1028.57 ft ⋅ lb
  



ua−a ua−a  M = −440.82i − 293.88 j + 881.63k ft ⋅ lb
(
) (
rOA
)
rOB
B
C
21
Moments Along an Axis, Couples
Monday, September 24, 2012
Example
¢  To
find the magnitude of the projection,
we can also use what is known as the
mixed product
rOA
rOB
B
C
22
Moments Along an Axis, Couples
Monday, September 24, 2012
11
Example
¢  Again,
starting with the force F the
moment arm rOB and the unit vector
along the axis of concern uOA
rOA
rOB
B
C
23
Moments Along an Axis, Couples
Monday, September 24, 2012
Example
¢  Taking
advantage of the properties of the
dot product and the cross product, we
can set up the problem in a matrix form
as
ux
rx
Fx
24
uy
ry
Fy
uz
rz
Fz
Moments Along an Axis, Couples
rOA
rOB
B
C
Monday, September 24, 2012
12
Example
¢  Notice
that since there are no vectors in
this expression, the result of reducing the
matric will be a scalar.
ux
rx
Fx
25
uy
ry
Fy
uz
rz
Fz
rOA
rOB
B
C
Moments Along an Axis, Couples
Monday, September 24, 2012
Example
¢  We
can use the same method that we
used when we calculated the cross
product
¢  Here we use it to find the magnitude of
the moment along the axis
ux
rx
Fx
26
uy
ry
Fy
uz
rz
Fz
Moments Along an Axis, Couples
rOA
rOB
B
C
Monday, September 24, 2012
13
Example
¢  Substituting
the coefficients of the unit
vector
3
−
7
rx
2
−
7
ry
6
7
rz
Fx
Fy
Fz
rOA
rOB
B
C
27
Moments Along an Axis, Couples
Monday, September 24, 2012
Example
¢  Substituting
the coefficients of the
moment arm
3
2 6
−
−
7
7 7
0 ft 4 ft 0 ft
Fx
28
Fy
Fz
Moments Along an Axis, Couples
rOA
rOB
B
C
Monday, September 24, 2012
14
Example
¢  Substituting
3
7
0 ft
the coefficients of the force
2
7
4 ft
−
−
6
7
0 ft
rOA
−400lb 400lb 200lb
rOB
B
C
29
Moments Along an Axis, Couples
Monday, September 24, 2012
Example
¢  Using
3
7
0 ft
−
2
7
4 ft
−
our method for reducing the matrix
6
7
0 ft
−400lb 400lb 200lb
3
7
0 ft
−
−400lb
2
7
4 ft
−
rOA
400lb
rOB
B
C
30
Moments Along an Axis, Couples
Monday, September 24, 2012
15
Example
¢  Using
3
7
0 ft
−
2
7
4 ft
−
our method for reducing the matrix
6
7
0 ft
−400lb 400lb 200lb
3
7
0 ft
2
7
4 ft
−
−
−400lb
(
400lb
rOA
)
3
− ( 4 ft ) ( 200lb) − ( 400lb) ( 0 ft )
7
2
− ( 0 ft ) ( −400lb) − ( 200lb) ( 0 ft )
7
6
+ ( 0 ft ) ( 400lb) − ( −400lb) ( 4 ft )
7
(
)
(
)
31
rOB
B
C
Moments Along an Axis, Couples
Monday, September 24, 2012
Example
¢  Using
3
7
0 ft
−
2
7
4 ft
−
our method for reducing the matrix
6
7
0 ft
−400lb 400lb 200lb
−
3
7
0 ft
2
7
4 ft
−
−400lb
−
rOA
400lb
3
2
6
800 ft • lb) − ( 0 ) + (1600 ft • lb)
(
7
7
7
rOB
B
C
32
Moments Along an Axis, Couples
Monday, September 24, 2012
16
Example
¢  The
magnitude of the moment produced
by F along a-a is then
3
7
0 ft
2
7
4 ft
−
6
7
0 ft
−
3
7
0 ft
2
7
4 ft
−
−400lb 400lb 200lb
−
−400lb
3
2
6
800 ft • lb − 0 + 1600 ft • lb
7
7
7
7200 ft • lb
=
= 1028.57 ft • lb
7
−
(
)
()
33
rOA
400lb
(
)
rOB
B
C
Moments Along an Axis, Couples
Monday, September 24, 2012
Example
¢  Take
care in identifying the unit vector of
the force and the position vector that is
the moment arm
3
7
0 ft
−
2
7
4 ft
−
6
7
0 ft
3
7
0 ft
2
7
4 ft
−
−400lb 400lb 200lb
−400lb
−
400lb
3
2
6
800 ft • lb − 0 + 1600 ft • lb
7
7
7
7200 ft • lb
=
= 1028.57 ft • lb
7
−
34
(
)
()
(
rOA
)
Moments Along an Axis, Couples
rOB
B
C
Monday, September 24, 2012
17
Example
¢  Remember
that the unit vector will not
have units and that the position vector
will have units
3
7
0 ft
2
7
4 ft
−
6
7
0 ft
−
3
7
0 ft
2
7
4 ft
−
−400lb 400lb 200lb
−400lb
−
400lb
3
2
6
800 ft • lb − 0 + 1600 ft • lb
7
7
7
7200 ft • lb
=
= 1028.57 ft • lb
7
−
(
)
()
35
(
rOA
)
Moments Along an Axis, Couples
rOB
B
C
Monday, September 24, 2012
Couples
¢  A
couple is a system of two forces
¢  The forces must satisfy the following
conditions for the force system to be a
couple
l  The
magnitudes of the forces must be
the same
l  The unit vectors of the forces must be
opposite in direction


u1 = ( −1) u 2
F1 = F2
36
Moments Along an Axis, Couples
Monday, September 24, 2012
18
Couples
¢  A
couple produces pure rotation on a
system
¢  The sum of the two forces will be equal to
0


u1 = ( −1) u 2
F1 = F2
37
Moments Along an Axis, Couples
Monday, September 24, 2012
Couples
¢  Most
importantly, a couple produces the
same amount of rotation about any
moment center


u1 = ( −1) u 2
F1 = F2
38
Moments Along an Axis, Couples
Monday, September 24, 2012
19
Couples
¢  We
will start with two forces in space who
satisfy the conditions of a couple
F1 = F2 
u F1 = ( −1) u F2
F1
F2
39
Moments Along an Axis, Couples
Monday, September 24, 2012
Couples
¢  We
will then pick some point in space as
the moment center
F1 = F2 
u F1 = ( −1) u F2
a
F1
F2
40
Moments Along an Axis, Couples
Monday, September 24, 2012
20
Couples
¢  We
can pick a point on each force’s line
of action and construct moment arms
from point a to a point on the life of action
on
b
each of these forces
rab
a
F
1 = F2

u F1 = ( −1) u F2
41
F1
F2
rac
c
Moments Along an Axis, Couples
Monday, September 24, 2012
Couples
¢  The
total moment generated by the two
forces about a is equal to
 
 
 
 

M a = rab ⊗ F1 + rac ⊗ F2
a


u F1 = ( −1) u F2
42
F1
F2
F1 = F2
Moments Along an Axis, Couples
b
rab
rac
c
Monday, September 24, 2012
21
Couples
¢  Substituting
the definition of F1 and F2
 






M a = rab ⊗ u1 F1 + rac ⊗ u2 F2
( )
( )
a


u F1 = ( −1) u F2
43
b
rab
F1
F2
rac
c
F1 = F2
Moments Along an Axis, Couples
Monday, September 24, 2012
Couples
¢  Using
our definition of the magnitudes in
a couple
 






M a = rab ⊗ u1 F1 + rac ⊗ u2 F1
( )
( )
a


u F1 = ( −1) u F2
44
F1
F2
F1 = F2
Moments Along an Axis, Couples
b
rab
rac
c
Monday, September 24, 2012
22
Couples
¢  And
the relationship of the unit vectors
 





M a = rab ⊗ u1 F1 + rac ⊗ −u1 F1
( )
(
a


u F1 = ( −1) u F2
45
)
b
rab
F1
F2
rac
c
F1 = F2
Moments Along an Axis, Couples
Monday, September 24, 2012
Couples
¢  Manipulating
the cross product
 
 


M a = rab − rac ⊗ u1 F1
(
) ( )
a


u F1 = ( −1) u F2
46
F1
F2
F1 = F2
Moments Along an Axis, Couples
b
rab
rac
c
Monday, September 24, 2012
23
Couples
¢  Now
if we generate another position
vector from c to b and by vector addition
we can say

 
 

rab = rac + rcb
a
rcb
 
 


M a = rab − rac ⊗ u1 F1
(
)
(
)


u F1 = ( −1) u F2
47
b
rab
F2
F1
rac
c
F1 = F2
Moments Along an Axis, Couples
Monday, September 24, 2012
Couples
¢  Rearranging
terms

 
 

rab − rac = rcb
a
 
 


M a = rab − rac ⊗ u1 F1
(
)
(
)


u F1 = ( −1) u F2
48
F1 = F2
Moments Along an Axis, Couples
b
rab
rcb
F2
F1
rac
c
Monday, September 24, 2012
24
Couples
¢  And
substituting in the expression for the
moment at a
 


M a = rcb ⊗ u1 F1

 
 

rab − rac = rcb
( ) ( )


u F1 = ( −1) u F2
49
a
b
rab
rcb
F2
F1
rac
c
F1 = F2
Moments Along an Axis, Couples
Monday, September 24, 2012
Couples
¢  So
no matter where a is, the moment of
the couple will depend on how far apart
the forces are
 


M a = rcb ⊗ u1 F1
( ) ( )
b
rcb
F1
F2
c
50
Moments Along an Axis, Couples
Monday, September 24, 2012
25
Couples
¢  The
moment of a couple can be
calculated by taking a moment arm from
line of action of one of the forces to the
line of action of the other force and
forming the cross product of the moment
arm and the force that we went to
b
rcb
F1
F2
c
51
Moments Along an Axis, Couples
Monday, September 24, 2012
Couples
¢  In
a two-dimensional problem, if you can
find the perpendicular distance between
the forces, you can calculate the
magnitude of the moment by multiplying
the perpendicular distance times the
magnitude of either one of the forces
¢  The sign will be given by the sense of
rotation
b
rcb
F1
F2
52
Moments Along an Axis, Couples
c
Monday, September 24, 2012
26
Homework
¢  Problem
4-50
¢  Problem 4-51
¢  Problem 4-53
53
Moments Along an Axis, Couples
Monday, September 24, 2012
27