FORCE SYSTEMS
2-D Force Systems
3-D Force Systems
Force
Force
Moment,Couple
Moment, Couple
Resultants
Resultants
1
3D-Force Systems
Rectangular Components, Moment,
Couple, Resultants
2
Moment (3D)
moment
axis
X
Moment about
point P :
F
Y
d
-Magnitude: | r || F | sin   Fd
F2
P
O
M P,F  r  F
-Direction: right-hand rule
-Point of application: point O
(Unit: newton-meters, N-m)
In 3D, forces (generally) are not in the same plane.
In many cases on 3D, d (the perpendicular distance) is hard to
find. It is usually easier to find the moment by using the vector
approach with cross product multiplication.
4
Cross Product
M o,F  r  F
-
-
rz
iˆ
 rx
ˆj
ry
kˆ
rz
iˆ
rx
ˆj
ry
Fz
Fx
Fy
Fz
Fx
Fy
+
+
iˆ
ˆj
kˆ
 rx
ry
Fx
Fy
-
+
M o  (rxiˆ  ry ˆj  rz kˆ)  ( Fxiˆ  Fy ˆj  Fzkˆ)
 rx Fx (iˆ  iˆ)  rx Fy (iˆ  jˆ)  rx Fz (iˆ  kˆ ) 
 ry Fx ( jˆ  iˆ)  ry Fy ( jˆ  jˆ)  ry Fz ( jˆ  kˆ) 
Beware: xyz axis
must complies with
right-hand rule
 rz Fx (kˆ  iˆ)  rz Fy (kˆ  jˆ)  rz Fz (kˆ  kˆ)
M o  (ry Fz  rz Fy )iˆ  (rz Fx  rx Fz ) ˆj  (rx Fy  ry Fx )kˆ
5
Moment (Cross Product)
M o  (ry Fz  rz Fy ) iˆ  (rz Fx  rx Fz ) ˆj  (rx Fy  ry Fx ) kˆ
Physical Meaning

F
z
+
Fz

r
x
Fy
A
+
Fx
rz
y
My = +Fxrz - Fzrx
y
rx
O
Mx = - Fyrz + Fzry
z
+
ry
Mz= -Fxry +Fyrx
x
6
Moment
Moment About a Point #4
Resultant Moment of Forces
z
F3
F1
MO   (ri  Fi )
r1
r3
F2
r2
O
i
MO
y
x
7
Varignon’s Theorem (Principal of Moment)

F1

F3
- their moments about a point may be
found in two ways
r
O
- Two or more concurrent forces
A

F2
- for nonconcurrent forces see
Resultants sections (2D - 2/6, 3D- 2/9)

     
   
Mo  r  F1  r  F2  r  F3  ...  r  (F1  F2  F3  ...)


 r  ( F )

   
M o   (r  F )  r  R
- Sum of the moments of a system of concurrent forces about a
given point equals the moment of their sum about the same point
8
Determine
the vector expression for the

moment Mo of the 600-N force about point O.
The design specification for the bolt at O



would require this result. M
 r F
O
OA

rOA  (0.05  0.13sin60 ) ˆi
 (0.14  0.13cos60 ) ˆj  0.15kˆ

rOA  0.1626ˆi  0.205ˆj  0.15kˆ
F ?
Fz  600sin 45o
Fxy  600cos45o
Fx  Fxy sin600  600cos45o sin60o
Fy  Fxy cos600  600cos45o cos60o
F  600(0.612 iˆ  0.354 ˆj  0.707 kˆ )
M O ,F  rOA  F
iˆ
ˆj
kˆ
 0.1626 0.205 0.15
367
212
424
 55.2 iˆ  13.86 ˆj  40.8 kˆ
 367ˆi  212ˆj  424kˆ
N
Ans
9
Using: rOP (3D Vector)
MO
z
6m
x
O
y
rOP
400N
1.2m
0.8m
P
rOP   6iˆ  0.8 ˆj  1.2kˆ
F  400 ˆj
MO  rOP  F
i
ˆj
kˆ
 -6 0.8 1.2  480iˆ  2400kˆ
0 -400 0
N-m
Ans
10
Using: rOQ (3D vector)
z
x
rOQ
O
y
VD2
6m
MO
0.8m
400N
1.2m
Q
rOQ  6iˆ  0 ˆj  1.2kˆ
F  400 ˆj
MO  rOQ  F
i
ˆj
kˆ
 -6
0 1.2  480iˆ  2400kˆ
0 -400 0
N-m
Ans 11
Using: rOQ (Scalar  3Plane)
MO 
iˆ 
ˆj 
kˆ
M O , x   rz Fy  1.2  400  480
plus
MO, y  0
rx
M O , z   rx Fy  6  400  2400
rz
ˆ
ˆ
MO =480i+2400k
N-m
Ans
Not-Recommended Method
12
Moment
Example Hibbeler Ex 4-4 #1
Determine the moment
about the support at A.
rB  (1iˆ  3 jˆ  2kˆ ) m
r  (3iˆ  4 jˆ) m
C
rCB  rB  rC  ( 2iˆ  1jˆ  2kˆ ) m
13
Moment
Example Hibbeler Ex 4-4 #2
rCB  ( 2iˆ  1jˆ  2kˆ ) m
rCB
2iˆ  1ˆj  2kˆ
1
uF 

 ( 2iˆ  1jˆ  2kˆ )
rCB
( 2)2  ( 1)2  22 3
F  (60 N)uF  ( 40iˆ  20 jˆ  40kˆ ) N
14
Moment
Example Hibbeler Ex 4-4 #3
iˆ
M A  rC  F  3
ˆj
4
kˆ
0  (160iˆ  120 jˆ  100kˆ ) N  m
40 20 40
M A  (160)2  ( 120)2  (100)2  223.61  224 N  m
#
15
Moment
Example Hibbeler Ex 4-4 #4
M A  rC  F  rB  F
ˆj
iˆ
kˆ
M A  rB  F  1
3
2 Nm
40 20 40
M A  (160iˆ  120 jˆ  100kˆ ) N  m
16
z
x
Moment about Point
M A, F  r  F
y
 AB
Moment about line
( projection effect )
M A,F ,AB  ( M A,F  eˆAB ) eˆAB
18
Finding moment of force about (arbitary) axis 
M
M
iˆ
ˆj
kˆ
r  F  rx
Fx
ry
Fy
rz
Fz
nˆ
M O ,F


F
M o,F ,

r
O
M
M
Depend on line  only,
Not depend on point O
M o,F ,  ((r  F )  nˆ )nˆ



rx
ry
rz
( r  F  nˆ )  rx
ry
rz
 Fx
Fy
Fz
Fx
Fy
Fz



, ,  are the directional cosines of the unit vector nˆ
(i.e. nˆ  iˆ   ˆj   kˆ) 19

C  A B
C _|_ A and C _|_ B
M B ,F
B
F
 (rA  rB )  F _|_ nˆ 
A
M B ,F
rA  rB
(rA  rB )  F _|_ (rA  rB )
M A, F
M A, F
nˆ
B
F
A
Moment of F about point {A,B}
in the direction of 
(generally)
M A,F ,  (rA  F )  nˆ
M A,F ,  M B ,F , where A, B on line 
Moment of F in the direction of 
Moment of F projected to line 
Moment of F about line 
M A,F ,  M B,F ,  ...

M F ,
where A, B are any points on the line 
 ({(rA  rB )  rB }  F )  nˆ
 ((rA  rB )  F  rB  F )  nˆ
 ((rA  rB )  F )  nˆ  (rB  F )  nˆ
0
 M B ,F ,
Moment about axis is sliding vector.
20
Finding moment of force about (arbitary) axis 
M
M
iˆ
ˆj
kˆ
r  F  rx
Fx
ry
Fy
rz
Fz
nˆ
M O ,F


F
M o,F ,

r
O
M
M
Depend on line  only,
Not depend on point O
M o,F ,  ((r  F )  nˆ )nˆ



rx
ry
rz
( r  F  nˆ )  rx
ry
rz
 Fx
Fy
Fz
Fx
Fy
Fz



, ,  are the directional cosines of the unit vector nˆ
(i.e. nˆ  iˆ   ˆj   kˆ) 22

Moment
 rQX
 about Point P
  F   nˆ
moment
 axis
rQX  ( F   F )   nˆ
  rQX  F   rQX  F X  nˆ
F
  rQX  F   nˆ   rQX  F   nˆ
d
Moment about line  (Definition)
Line 
(moment axis)
Direction:
right-hand
rule
How to find “Moment about line ” ?
M  ,F  | F  | d
Hard to find
X  Plane
X
M  ,F
F
Point Q :
Q  Line 
Q  Plane
O
M P , F  rPX  F
Plane :
Plane _|_ nˆ
nˆ
M  , F : | F  | d
P
F 
F
d
Q
A
A : Any point
on line 
 M A, F  nˆ
M  , F | rQX  F  |   rQX  F   nˆ   rQX  F   nˆ  M Q , F  nˆ
Hard to find
Moment about line 
We will prove that
Line 
(moment axis)
F
(M A, F  nˆ ) M A, F ,
nˆ
X
M  ,F
A : Any point
on line 
is equal to
M Q , F ,
( M
Q,F
 nˆ

Moment of F about point {A,Q}
projected to line 
d
Q
A
M  , F  M Q , F  nˆ
 M A, F  nˆ
A : Any point on line 
M A, F ,   rAX  F   nˆ
( A  B) _|_ A
(rAQ  F ) _|_ rAQ
rAQ // nˆ
 rAQ  F _|_ nˆ
 (rAQ  rQX )  F   nˆ
  rAQ  F  rQX  F   nˆ
 (rAQ  F )  nˆ  (rQX  F )  nˆ
must prove
to be
0
M Q , F ,
Moment about line 
Line 
(moment axis)
(M A, F  nˆ ) M A, F ,
F
is equal to
nˆ
M Q , F ,
( M
Q,F
 nˆ

X
M  ,F
Point A is any point in the line 
d
Moment about axis is sliding vector.
Q
A
M A, F ,  M B , F ,  ...
where A, B are any points on the line 
Moment of F
about point A
in the direction of 
M  ,F
Moment of F
about line 
Moment of F in the direction of 
Moment of F projected to line 
27


Find M z of T (the moment of  about z-axis
T
y
passing through the base O )
rAB  15 ˆj  12iˆ  9kˆ
A
15 m
T  10
T = 10 kN
12iˆ  15 ˆj  9kˆ
12  15  9
2
2
2
A(0,15,0) B(12,0,9)
kN
Mo  r  T  15 ˆj 10(0.566iˆ  0.707 ˆj  0.424kˆ)
 150(0.566kˆ  0.424iˆ) kN-m
M z  (Mo  kˆ)  84.9 kN-m
O
x
z
kˆ
9m
12 m
B
Figure must be shown
M z  (Mo  kˆ) kˆ  84.9kˆ kN-m
M z , F  { (r  T )  kˆ } kˆ
M O , F  (r  T )
OK
Ans
OK
not OK
OK
28
2/133 A 5N vertical force is applied to the knob of the window-opener mechanism
when the crank BC is horizontal. Determine the moment of force about point A and
about line AB.
y
25cos30 mm
C’
nˆ AB
nˆ AB
D’
r
D’
r 75 mm
A
B’
50cos30
x
mm
rAD  (75cos30o )iˆ  75 ˆj
MA  r F
 (75cos 30iˆ  75 ˆj )  ( 5kˆ)
nˆ AB,1  cos30iˆ  sin 30kˆ
M AB  M A  nˆAB   162.26 N-mm
 375iˆ  325 ˆj
N-mm
Ans
nˆAB   (cos30iˆ  sin30kˆ)
  M A  nˆAB  + 162.26
M AB
M AB  ( M A  nˆ AB )nˆ AB
 126.26(cos 30 iˆ  sin 30o kˆ) N-mm
Ans
o
M AB  (M A  nˆ AB )nˆ AB
 126.26 (cos30o iˆ  sin 30o kˆ)  N-mm
29
Moment
Example Hibbeler Ex 4-8 #1
Determine the moments of this force about the x and a axes.
rA  ( 3iˆ  4 jˆ  6kˆ ) m
uˆ  iˆ
x
M x  uˆ x  (rA  F )
1
0
0
 3
4
6 Nm
40 20 10
 80 N  m
M x  80iˆ N  m
#
30
Moment
Example Hibbeler Ex 4-8 #2
3
4
uˆa  (  iˆ  jˆ) m
5
5
Ma  uˆa  (rA  F )
3 5 4 5
 3
4
40
20
0
6 Nm
10
3 4 6 4 3 6
 ( )

Nm
5 20 10 5 40 10
 120 N  m
Ma  Mauˆa  (72iˆ  96 jˆ) N  m
#
31
Moment
Example Hibbeler Ex 4-9 #1
Determine the moment MAB produced by F = (–600i + 200j – 300k) N,
which tends to rotate the rod about the AB axis.
M AB  uˆB  (r  F )
rB
0.4iˆ  0.2 ˆj
uˆB  
rB
0.42  0.22
uˆ  0.89443iˆ  0.44721 jˆ
B
Vector r is directed from any
point on the AB axis to any
point on the line of action of
the force.
r  rD  0.2 jˆ m
32
Moment
Example Hibbeler Ex 4-9 #2
M AB  uˆB  (rD  F )
0.89443 0.44721
0

0
0.2
0
600
200
300
0.89443
0
 0.2
600
300
 53.666 N  m
M AB  M AB uˆB
 ( 48.0iˆ  24.0 jˆ) N  m #
33
Moment
Example Hibbeler Ex 4-9 #3
Vector r is directed from any point on the AB axis to any point on
the line of action of the force.
try r  rBC  0.2iˆ  0.2 jˆ  0.3kˆ m
M AB  uˆB  (rBC  F )
0.89443 0.44721
0

0.2
0.2
0.3
600
200
300
 53.665 N  m
rBC
34

F
M A, F
M F ,A  r  F
position vector:
from A to point of application
of the force
r
X
A
Y
d
Z

p
p  line :
F
O
M p , F , = M o , F , 
nˆ

F

r
r
A
d
Y
Z
X
position vector:
from A to any point on
line of action of the force.
M F ,  { ( r  F )  nˆ }nˆ
position vector:
from any point on line  to
any point on tline of action of
the force.
36
parallel
with line 
nˆ
F2
 M  ?
F1
O
{
3
{(r  F )  nˆ }
i
}nˆ
i=1
P
F3
F1 intersects
with that axis.
F2 is parallel
with that axis.
 { (r  F3 )  nˆ }nˆ
(
Why?

M  , F1 = roo  F1  nˆ  0
M P, F2 _|_ F2 and F2 / / nˆ

M P, F2 _|_ nˆ
Forces which interest or parallel with axis,
do not cause the moment about that axis
37
Couple

M
Couple is a summed moment
produced by two force of equal
magnitude but opposite in direction.

F
B

rB

r

rA
A
d

F
O
MO  rA  F  rB  (F )  (rA  rB )  F
  
M  r F
magnitude and direction
Do not depend on O
from any point on line of the action
to any point on the other line of action
Moment of a couple is the same about all point
 Couple may be represented as a free vector.
39
The followings
are equivalent
couples


M
M
F
F


F

M

M
F

F
d/2
2F
2F
F
Every point has the equivalent
moment.
2D representations: (Couples)
couple is
a free vector
M
M
M
M
40
- Couple tends to produce a “pure” rotation of the body about an
axis normal to the plane of the forces (which constitute the
couple); i.e. the axis of the couple.

M1

M2

 F2

 F1

F2

M

F1

M2

M

M1

F 
F
- Couples obey all the usual rules that govern vector quantities.
- Again, couples are free vector. After you add them (vectorially), the
point of application are not needed!!!
- Compare to adding forces (i.e. finding resultant), after you add the forces
vectorially (i.e. obtaining the magnitude and direction of the resultant), you
still need to find the line of action of the resultant (2D - 2/6, 3D - 2/9).
41
30 N
60
30 N
60
x
y
0.05 m
25 N
1) Replace the two couples
with

a single couple M that still
produces the same external
effect on the block.


2) Find two forces F and  F on
two faces of the block that
parallel to the y-z plane that
will replace these four forces.
25 N
(forces act parallel to y-z plane)
z
(25)(0.1)= 2.5 N-m
60
1.8
2.23

sin  sin 60o
M

y
z
M  1.82  2.52  2(1.8)(2.5) cos 60o  2.23 N-m
60
(30)(0.06)= 1.8 N-m
   44.3o
M
F
M 2.23

 22.3 N
d
0.10
42
Moment
Example Hibbeler Ex 4-13 #1
Replace the two couples acting on the pipe column by
a resultant couple moment.
M1  d  F  (150 N)(0.4 m)  60 N  m
M  (60iˆ) N  m
1
M2  rDC  FC
3 
4
 (0.3iˆ)  125  ˆj  kˆ  N  m
5 
5
 30(iˆ  jˆ)  22.5( iˆ  kˆ ) N  m
 (22.5 ˆj  30kˆ ) N  m
43
Moment
Example Hibbeler Ex 4-13 #2
MR  M1  M2  (60iˆ  22.5 jˆ  30kˆ ) N  m #
44
y
MO,240N-m
O
MO,F
r
z
200mm
250mm
30O
1200N
240N-m
x
Vector Diagram
F  1200(cos30o ˆj  sin30o kˆ)
r  0.2iˆ  0.25kˆ
M O ,F  r  F
iˆ
ˆj
kˆ
 0.2
0
0.25  -260iˆ  120 ˆj  208kˆ
0 1039 600
(
M O , 240N-m  240 cos30o ˆj - sin 30o kˆ
M O  M O ,F  M O ,240 N m  - 260iˆ  328 ˆj  88kˆ

N-m Ans
45
Concepts #1
Review
• Vectors can be manipulated by scalar multiplication,
addition, subtraction, dot product, cross product and
mixed triple product. Vectors representing can be
classified into free, sliding and fixed vectors.
• Position vectors describe the position of a point relative
to a reference point or the origin.
• Statically, force is the action of one body on another.
In dynamics, force is an action that tends to cause
acceleration of an object. To define a force on rigid
bodies, the magnitude, direction and line of action
are required. Thus, the principle of transmissibility is
applicable to forces on rigid bodies.
46
Concepts #2
Review
• To define a moment about a point, the magnitude,
direction and the point are required. To define a
moment about an axis, the magnitude, direction and
the axes are required. To define a couple, the
magnitude and direction are required.
47
Chapter Objectives Descriptions #1
• Use mathematical formulae to manipulate physical
quantities
– Specify idealized vector quantities in real worlds
and vice versa
– Obtain magnitude, direction and position of a vector
– Manipulate vectors by scalar multiplication,
addition, subtraction, dot product, cross product and
mixed triple product
– Describe the physical meanings of vector
manipulations
• Obtain position vectors with appropriate
representation.
48
Chapter Objectives Descriptions #2
• Use and manipulate force vectors
– Identify and categorize force vectors
– Describe the differences between force
representation in rigid and deformable bodies
– Identify and represent forces in real worlds with
sufficient data and vice versa
– Manipulate force vectors
49
Chapter Objectives Descriptions #3
• Use and manipulate moment vectors
– Identify and categorize moment vectors
– Describe the differences between moments about
points, moments about axes and couple
– Identify and represent moments in real worlds with
sufficient data and vice versa
– Manipulate moment vectors
50
Review Quiz #1
Review
• Use mathematical formulae to manipulate physical
quantities
– Give 4 examples of vector quantities in real world.
– In how many ways can we specify a 2D/3D
vector? Describe each of them.
– How can we prove that two vectors are parallel?
– What are the differences between the vector
additions by the parallelogram and triangular
constructions?
– Even though we can manipulate vectors
analytically, why do we still learn the graphical
methods?
51
Review Quiz #2
Review
• Use mathematical formulae to manipulate physical
quantities
– What are the mathematical definitions of dot, cross
and mixed triple products?
– What are the physical meanings of addition,
subtraction, dot product, cross product and mixed
triple product?
– What are the meanings of associative, distributive
and commutative properties of products?
– What are the differences between 2D and 3D
vector manipulation?
52
Review Quiz #3
Review
• Obtain position vectors with appropriate
representation.
– Given points A and B, what information do you
need to obtain the position vector and what name
will you give to the position vectors and distance
vector between the two points?
53
Review Quiz #4
Review
• Use and manipulate force vectors
– For the following forces – tension in cables, forces
in springs, weight, magnetic force, thrust of rocket
engine, what are their classification in the
following force types – external/internal,
body/surface and concentrated/distributed forces?
– If a surface is said to be smooth, what does that
mean?
– What are the differences between force
representation in rigid and deformable bodies?
– What are the additional cautions in force vector
manipulation that are not required in general
vector manipulation?
54
Review Quiz #5
Review
• Use and manipulate moment vectors
– Give 5 examples of moments in real world and
approximate them into mathematical models.
– What information do you need to specify a
moment?
– What is the meaning of moment direction?
– If a force passes through a point P, what do you
know about the moment of the force about P?
– What are the differences between physical
meanings of moments about points, moments
about axis and couples?
55
Review Quiz #6
Review
• Use and manipulate moment vectors
– As couples are created from forces, why do we
write down the couple vectors instead of forces in
diagrams?
– Given a couple of a point P, what do you know of
the couple about a different point Q?
– If we know moments about different points or
axes, why can’t we add components of moments
as in vector summation?
– Why can we simply add couple components
together?
56
Resultant
Resultant Definition
• The “force-couple systems” or “force systems” can be
reduced to a single force and a single couple (together
called resultant) that exert the same effects of
– Net force
 Tendency to translate
– Net moment  Tendency to rotate
• Two force-couple systems are equivalent if their
resultants are the same.
57
Force – Couple Systems

F
B

F

F
A
B

r
A
C  r F

F
B
A

F
No changes in the net
external effects
C  r F

: Couple of F about point B
from new location (point B)
calculated the same way as
“Moment of Point B by the
force F at the old position”

to any point on the line of action of F
(which applied at the old point)
59
R
MO
Vector diagram
Move 3 forces to point O
(
F1200  1200 -sin10o iˆ  cos10o ˆj
F800  800iˆ

F200  200iˆ
R   F  F1200  F800  F200  792iˆ  1182 ˆj N
Ans
MO   M  MO,1200  MO,800  MO,200
r1200  0.075iˆ - 0.220kˆ
r800  0.1 ˆj  0.55kˆ
r200  -0.1 ˆj - 0.55kˆ
 r1200  F1200  r800  F800  r200  F200
 260iˆ  45.8 ˆj  88.6kˆ  440 ˆj  80kˆ  110 ˆj  20kˆ
(
 (
 (

 260iˆ  504 ˆj  28.6kˆ
60
N-m
Ans
Resultant
Example Hibbeler Ex 4-15 #1
Replace the current system by
an equivalent resultant force
and couple moment acting at
its base, point O.
F1  800kˆ N
F2  300uCB  300(rCB rCB )
 0.15iˆ  0.1jˆ 
 300 

2
2
 ( 0.15)  (0.1) 
 ( 249.62iˆ  166.41 ˆj ) N
4ˆ 3 ˆ
M  500(  j  k )
5
5
 ( 400 ˆj  300kˆ ) N  m
61
Resultant
Example Hibbeler Ex 4-15 #2
FR   F 


FR  F1  F2
FR  ( 249.62iˆ  166.41ˆj  800kˆ ) N
F  ( 250iˆ  166 jˆ  800kˆ ) N
R
#
62
Resultant
Example Hibbeler Ex 4-15 #3
MR   M 
 O

MRO   MC   MO
MRO  M  (rC  F1 )  (rB  F2 )
MRO  ( 400 ˆj  300kˆ )  (1kˆ )  ( 800kˆ ) 
MRO  ( 0.15iˆ  0.1 ˆj  kˆ )  ( 249.62iˆ  166.41 jˆ)
MRO  ( 400 ˆj  300kˆ )  (0) 
MRO  ( 166.41iˆ  249.62 ˆj  0.0005 kˆ )
MRO  ( 166iˆ  650jˆ  300kˆ ) N  m
#
63
Recommended Problems
• 3D Moment and Couples:
2/124 2/125 2/129 2/132
64
Wrench
Resultant


F1

F2
M2 
M3

M1
O
O

F3
1) Pick a point (easy
to find moment arms)


F1

F2
- r1 start from ___________
to_____________

R
O

F3
2) Replace each force with a
force at point O + a couple
  
M1  r1  F1

M
3) Add the forces
vectorially to get the
resultant force (since the
forces are concurrent now)
and add the couple
vectorially to get the
resultant couple
   

R  F1  F2  F3  ...   F




 
M  M1  M 2  M 3  ...   (r  F )
66

F1
   

R  F1  F2  F3  ...   F




 
M  M1  M 2  M 3  ...   (r  F )

F2
O

F3
R x   Fx
R y   Fy
R z   Fz
R  ( Fx ) 2  ( Fy ) 2 ( Fz ) 2
Vector
Scalar
(2D * 3Plane)
iˆ
ˆj
M  r  F  rx
Fx
ry
Fy
 
M x  (r  F ) x
kˆ
( ry Fz  rz Fy )iˆ  ( rz Fx  rx Fz ) ˆj
rz 
( rx Fy  ry Fx )kˆ
Fz
 
M y  (r  F ) y
 
M z  (r  F ) z
M  M x2  M y2  M y2
67

R

M
- The choice of point O is arbitrary;
the resultant couple will not be the same for
each point O selected (in general), but the
resultant force will be the same.
O
Ex)

R
- The resultant couple cannot be cancelled
by moving the resultant force (in general).
r

M
X
M which // R, cannot be cancelled
N  r R
M which |_ R, can be cancelled.
Wrench Resultant (not very useful)
- All force systems can be represented with a wrench resultant as
shown in the figures


R

M
Positive wrench
R

M
Positive if
right-hand rule
Negative wrench
68
for N   M 1
How to find Wrench Resultant

M

M1

R
M-R
plane

M
d

R

M1 N

M2
O
O
O

R
d
(move to right)

R

M2
Vector approach see ex. 2/16
with M-R plane

M
R
O

Plane Containing R
and
M1
nˆR
How to find M1, M2
( knowing M , R )
M 2  { M  nˆR } nˆR
M1  M  M 2
69
The simplest form of force-couple system
3D
any forces + couples system
single-force + single couple

R

M
(which // with each other)
wrench resultant
2D
any forces + couples system
single-force system (no-couple)
OR single-couple system

F1

F2
O

F3
Why 2D is different from 3D?
70

F1
Special cases: Wrench Resultant

F2
O
1) Coplanar: 2D (Article 2/6)

F3

F1

F2
O

F3
2) Concurrent force:
the resultant will pass through the point of concurrency.
No resultant moment at concurrent point. Pick the point
of concurrency!
z
3) Parallel forces (not in same plane):
single-force system
(no-couple)
x
O
y
OR single-couple system
71
Sample problem 2/13
Find the resultant
z
50 N
Move all force to point O
70 N-m
100 N-m
80 N
50 N
80 N
x
R  (80  80)iˆ  (100  100) ˆj
 (50  50)kˆ  0 N
96 N-m
1.2
O
1.6
M  80(1.2) ˆj  50(1.6)iˆ  100(1)kˆ
100 N
 70iˆ  96 ˆj  100kˆ = 10iˆ N-m
1
100 N
y
pass thru O: no need to calculate couple
Ans
R0
M  80(1.2) ˆj  100(1)kˆ  50(1.6)iˆ  70iˆ  96 ˆj  100kˆ
 10iˆ N-m
72
50 N
z
Find the resultant
Move all force to point O
x
R  (50  500  200  300) ˆj  350 ˆj N
O
500 N
M O  50(0.35)iˆ  50(0.5)kˆ
- 200(0.35)kˆ
.35
y
300 N
200 N
.35
R |_ M
- 300(0.35)iˆ
Moving R
 - 87.5iˆ -125kˆ N-m can erase M
0.5
0.5
completely
z
New point: (x,y,z)
x
R
Couple cancelled: MO  (- xiˆ - yjˆ - zkˆ)  R  0
Equivalent System (at O): ( xiˆ  yjˆ  zkˆ)  R  M
O
O
y
M
Which quadrant?
(350x)kˆ  (350z)iˆ  87.5iˆ  125kˆ
x
-125
-87.5
 -0.357 z 
 0.250
350
-350
y : any value
73
74
Find the wrench resultant, give coordinates on x-y plane that
the wrench resultant acts.
Solution 0 (Wrong)
Move all force to point O
z
R  20iˆ  40 ˆj  40kˆ N
R
y
MO  0.06kˆ  40 ˆj  (0.1iˆ  0.08 ˆj)  40kˆ
 2.4iˆ  4 ˆj  3.2iˆ  0.8iˆ  4 ˆj
O
MO
x
z
Move R to point P (x,y,z), to cancel the couple
Couple cancelled: MO  (- xiˆ - yjˆ - zkˆ)  R  0
Equivalent System (at O): M  ( xiˆ  yjˆ  zkˆ)  R
O
y
R
x
O
P
z
y
x
40 y  40 z  0.8
20 z  40 x  4
y  z  0.02
z  2 x  0.2
40 x  20 y  0
2x  y  0
2 x  z  0.02
z  2 x  0.2
unable to solve!!
Generally in 3D, we can not change force-couple
system to single-force system.
75
Find the wrench resultant, give coordinates on x-y plane that
the wrench resultant acts.
Solution 1: Direct Method
z
Move all force to point O
R
y
R  20iˆ  40 ˆj  40kˆ N
MO  0.06kˆ  40 ˆj  (0.1iˆ  0.08 ˆj)  40kˆ
 2.4iˆ  4 ˆj  3.2iˆ  0.8iˆ  4 ˆj
O
MO
x

M

M1

R

M2
O
nˆR 
1
202  402  402
( 20iˆ  40 ˆj  40kˆ 
M O ,//  ( M O  nˆR ) nˆR
144
(
)
60
negative wrench
1
(20iˆ  40 ˆj  40kˆ)
60
 (0.8iˆ  1.6 ˆj  1.6kˆ)
M O ,|_  M O  ( M O  nˆR )nˆR  1.6iˆ  2.4 ˆj  1.6kˆ
76
Find the wrench resultant, give coordinates on x-y plane that
the wrench resultant acts.

M

M1
R  20iˆ  40 ˆj  40kˆ N

R
M O ,//  ( M O  nˆR )nˆR  (0.8iˆ  1.6 ˆj  1.6kˆ) N-m

M2
M O ,|_  M O  ( M O  nˆR )nˆR  1.6iˆ  2.4 ˆj  1.6kˆ
O
new point P: (x,y,z)

M1 N
O

M2
MO  0.8iˆ  4 ˆj

R
M O,|_  N  0
old point O: (0,0,0)
( xiˆ  yjˆ  zkˆ)  R  M o,|_  0
( xiˆ  yjˆ  zkˆ)  (20iˆ  40 ˆj  40kˆ)
 1.6iˆ  2.4 ˆj  1.6kˆ
line of action
40 y  40 z  1.6
20 z  40 x  2.4
40 x  20 y  1.6
y  z  0.4
2 x  z  0.12
y  z  0.4  2 x  0.8
z  0 : x  0.6 y  0.4
Ans
77
Find the wrench resultant, give coordinates on x-y plane that
the wrench resultant acts.
Solution 2: Equivalent System
Assume (x,y,0) is the point where wrench passes.
M O,Sys1  0.06kˆ  40 ˆj  (0.1iˆ  0.08 ˆj )  40kˆ
 2.4iˆ  4 ˆj  3.2iˆ  0.8iˆ  4 ˆj
MO,Sys1  MO,Sys 2
RSys1  RSys 2
R  20iˆ  40 ˆj  40kˆ N
z
M O ,Sys 2  ( xiˆ  yjˆ)  R  M P
 (40 y)iˆ  (40x) ˆj  (40x  20 y)kˆ  M P
y
R
x
O
Parallel
Condition
P
y
MP
x
M O ,Sys 2  (40 y 
M P  M nˆR  M
1
(20iˆ  40 ˆj  40kˆ)
60
M (+ or – is ok)
M ˆ
2M
2M ˆ
)i  ( 40 x 
) j  (40 x  20 y 
)k
3
3
3 78
Find the wrench resultant, give coordinates on x-y plane that
the wrench resultant acts.
M O ,Sys1  2.4iˆ  4 ˆj  3.2iˆ  0.8iˆ  4 ˆj
M O ,Sys 2  (40 y 
M ˆ
2M ˆ
2M ˆ
)i  ( 40 x 
) j  (40 x  20 y 
)k
3
3
3
MO,Sys1  MO,Sys 2
M
 0.8
3
2M
40 x 
 4
3
2M
40 x  20 y 
0
3
40 y 
x  0.06
y  0.04
The coordinate in x-y
plane, where wrench
resultant passes
M  2.4
Magnitude: 2.4 N-m
R  20iˆ  40 ˆj  40kˆ N (negative wrench)
1
M  2.4 (20iˆ  40 ˆj  40kˆ) N-m
60
P : (0.06,0.04) m
Ans
Direction: opposite with R
(negative wrench)
79
Find the wrench resultant, give coordinates on x-y plane that
the wrench resultant acts.
Solution 3: wrench condition
Move forces to P (x,y,0)
R  20iˆ  40 ˆj  40kˆ N
| R | 202  402  402  60
M P  (  xiˆ  yjˆ)  20iˆ
+(  xiˆ  yjˆ  0.06kˆ)  40 ˆj
+((0.1  x )iˆ  (0.08  y ) ˆj )  40kˆ
z
 (0.8  40 y)iˆ  (40x  4) ˆj  (20 y  40x)kˆ N-m
wrench
condition
y
MP
x
O
R
R
MP
P
y
P
nˆM P 
x
nˆR  nˆMP
nˆ R 
1
(20iˆ  40 ˆj  40kˆ)
60


1
(0.8  40 y )iˆ  (40 x  4) ˆj  (20 y  40 x)kˆ
| MP |
(0.8  40 y ) 2  (40 x  4) 2  (20 y  40 x ) 2
Take it as
the other unknown 80
Find the wrench resultant, give coordinates on x-y plane that
the wrench resultant acts.
1
nˆ R 
nˆR  nˆMP
nˆM P 
60
(20iˆ  40 ˆj  40kˆ)
1
(0.8  40 y )iˆ  (40 x  4) ˆj  (20 y  40 x)kˆ
| MP |
20 ˆ 40 ˆ 40 ˆ
1
i
j k 
(0.8  40 y )iˆ  (40 x  4) ˆj  (20 y  40 x)kˆ
60
60
60
| MP |
M
 0.8  40 y
3
2M
 40 x  4
3
2M
 20 y  40 x
3
M (+ or – is ok)
x  0.06
y  0.04
M  2.4
R  20iˆ  40 ˆj  40kˆ N (negative wrench)
1
M  2.4 (20iˆ  40 ˆj  40kˆ) N-m
60
P : (0.06,0.04) m
Ans
The coordinate in x-y
plane, where wrench
resultant passes
Magnitude: 2.4 N-m
Direction: opposite with R
(negative wrench)
81
Hibbeler Ex 4-136
M
The three forces acting on the
block each have a magnitude of
10 N. Replace this system by a
wrench and specify the point
where the wrench intersects the
z axis, measured from point O.
d
d
R  F3
y
(10  2)(cos 45o )
d 

 2
10
|R |
| M|_ |
M
x
M //
M|_
erasable
(dir:  )
 d  2 2 m
R  10 jˆ N
Positive wrench
M//  (10  2cos 45)( jˆ)  10 2 jˆ N-m
Ans
82
FR  10 jˆ N
M  M jˆ N  m
Hibbeler Ex 4-136
||
M||
D
FR
||
A
M
d
B
O,Sys1
  M O,Sys2
(rOB  F1)  (rOA  F2 )  (rOB  F3 )  (rOD  FR )  M|| jˆ
(rAB  F1)  (rOA  F2 )  (rOD  FR )  M|| jˆ
( 2kˆ )  10(cos 45iˆ  sin45 jˆ)  (6 jˆ  2kˆ)  ( 10 jˆ)
 (dkˆ )  ( 10 jˆ)  M jˆ
10 2iˆ  10 2 jˆ  20iˆ  10diˆ  M|| jˆ
||
y -direction:
10 2  M||
 M||  10 2  14.142 N  m
x -direction:
10 2  20  10d  d  0.58579 m
83
Equivalent System
Example Hibbeler Ex 4-136 #1
The three forces
acting on the
block each have a
magnitude of 10
N. Replace this
system by a
wrench and
specify the point
where the wrench
intersects the z
axis, measured
from point O.
84
Equivalent System
Example Hibbeler Ex 4-136 #2
 F


F

system I
system II 

F1  F2  F3  FR
(10 N)(cos 45iˆ  sin 45 jˆ)  (10 jˆ N)  (10 N)(  cos 45 iˆ  sin 45 jˆ )  FR
F  10 jˆ N
R
85
Equivalent System
Example Hibbeler Ex 4-136
FR  10 jˆ N
#3
M  M jˆ N  m
||
||
 M


M

O,system I
O,system II 

(rOB  F1 )  (rOA  F2 )  (rOA  F3 )  (rOD  FR )  M|| jˆ
(rAB  F1 )  (rOA  F2 )  (rOD  FR )  M|| jˆ
( 2kˆ m)  (10 N)(cos 45iˆ  sin 45 jˆ)  (6 jˆ  2kˆ ) m  ( 10 jˆ N) 
(dkˆ )  ( 10 ˆj N)  (M jˆ N  m)
||
86
Equivalent System
Example Hibbeler Ex 4-136 #4
( 10 2iˆ  10 2 jˆ  20iˆ) N  m  (10 N)diˆ  M|| jˆ
y -direction:
( 10 2) N  m  M||
x -direction:
M||  10 2 N  m  14.142 N  m Wrench:
ˆj N
F


10
R
( 10 2  20) N  m  (10 N)d
M  14.1jˆ N  m
d  0.58579 m
||
d  0.586 m
Ans
87
• สรุปแนวทางการสอนทีจ่ ะเน้ น
• หน่วยให้ใส่ ตลอด โดยในบรรทัดแรกให้ใส่ ตามตัวแปรที่ยกมาตั้ง ส่ วน
บรรทัดต่อๆ ไป รวมเป็ นหน่วยเดียวไว้ทา้ ยบรรทัดได้
• - สัญลักษณ์ สาหรับ moment/couple ให้ใช้ลูกศรที่มีวง
บอกการหมุน ไม่ใช้หวั ลูกศรคู่ และให้กาหนดแกน xy พร้อมทิศทวน
เข็มเป็ นบวกในการวาดรู ปแกน
• - การนิยาม ให้ระวังให้มีการนิยามตัวแปรโดยการเขียนหรื อวาดรู ป
88
Equivalent System
Reduction Summary
Single force +
single couple
General force systems
2D force systems
Single force or
single couple
simplest
systems
3D force systems
Wrench
89
A flagpole is guyed by 3 cables. If the
tensions in the cables have the same
magnitude P (N), replace the forces
exerted on the pole with an eqivalent
wrench and determine the resultant
force R and the point where the axis of
the wrench intersects the x-z plane
nˆF  nˆM
Assume (x,0,z) is the point where wrench passes.
y
F1
P
(9iˆ  20 ˆj  12kˆ)
25
a 92  202  122
a(15iˆ  20 ˆj ) P
P
F2  P
 (15iˆ  20 ˆj )  (3iˆ  4 ˆj )
5
a 152  202 25
F1  P
F3
F2
z
F2  P
x
P: (x,0,z)
a(9iˆ  20 ˆj  12kˆ)
a(24 ˆj  18kˆ)
a 242  182
RSys  F1  F2  F3 


P
P
(24 ˆj  18kˆ)  (4 ˆj  3kˆ)
30
5
3P ˆ
(2i  20 ˆj  kˆ)
25
90
P
M P  (  xiˆ  20ajˆ  zkˆ)  ( 9iˆ  20 ˆj  12kˆ)
25
P
 (  xiˆ  20ajˆ  zkˆ)  (3iˆ  4 ˆj )
5
P
 (  xiˆ  24ajˆ  zkˆ)  ( 4 ˆj  3kˆ)
5
P
{(240a  20 z )iˆ  (9 z  12 x ) ˆj  (20 x  180a )kˆ}
25
P
 {( 4 z )iˆ  ( 3z ) ˆj  (4 x  60a )kˆ}
5
P
 {( 72a  4 z )iˆ  ( 3x ) ˆj  (4 x )kˆ}
5
MP 
y
F1
F3
z
x
P: (x,0,z)
P
3 x  6 z ˆ
{( 24a  12 z )iˆ  (
) j  ( 24a  12 x )kˆ}
5
5
3P
 x  2z ˆ

{( 8a  4 z )iˆ  (
) j  ( 8a  4 x )kˆ}
5
5
MP 
F2
91
RSys  F1  F2  F3 
MP 
3P
 x  2z ˆ
{( 8a  4 z )iˆ  (
) j  ( 8a  4 x )kˆ}
5
5
nˆF  nˆM
1
9 5
(2iˆ  20 ˆj  kˆ)  
wrench condition
1
3P
 x  2z ˆ
{( 8a  4 z )iˆ  (
) j  ( 8a  4 x)kˆ}
| MP | 5
5
M (+ or – is ok)
y
F1
3P ˆ
(2i  20 ˆj  kˆ)
25
F3
F2
z
x
P: (x,0,z)
1
M
1
M
1
M
3P
2
( 8a  4 z ) 
5
9 5
3P (  x  2 z )
20

5
5
9 5
3P
1
( 8a  4 x )  
5
9 5
812
a
405
806
z
a
405
3P
8
8 5
M 9 5
(
a)  
Pa
5
405
75
x
92
y R
MO
z O
2.4m
R  F800  F1600  2400 ˆj N
(
MO  r  F

800
(
 r F

1600N
0.9m
800N x
VD1
1600
(

 2.4iˆ  800 ˆj  2.4iˆ  0.9kˆ  1600 ˆj
 1440iˆ  5760kˆ N-m
R  MO  0  R
|
MO

M has no component in the
direction of R.
We can move R to new position
to eliminate this couple completely
93
y R
R  2400 ˆj N
MO
z O
MO  1440iˆ  5760kˆ N-m
1600N
0.9m
2.4m
800N x
VD1
y R
MO
 MO  r  R
MO  (-r )  R  0
(
 (
1440iˆ  5760kˆ  rxiˆ  ry kˆ  2400 ˆj

1440iˆ  5760kˆ  2400rx kˆ  2400ryiˆ
rx  2.4
rx  0.6
z O
R
r
x
VD2
Ans
94
z
MO
a
O
R
F
R  Fiˆ  Fkˆ
x
(

M O  aiˆ  (  F  kˆ  bjˆ  ckˆ  Fiˆ
b
F
c
y
VD1
 F (a  c) ˆj  Fbkˆ
We move R to the new location (x,y,z)
where there is no couple.
MO  (-r )  R  0
 MO  r  R
F (a  c) ˆj  Fbkˆ  ( xiˆ  yˆj  zkˆ)  ( Fiˆ  Fkˆ)
F (a  c) ˆj  Fbkˆ   yFiˆ  F ( x  z) ˆj  Fykˆ
iˆ : 0   yF  y  0
kˆ : Fb  Fy  y  b
Generally b != 0, how come?
Generally in 3D, we can not
change force-couple system
to single-force system.
95
MO
a
O
R
F
x
b
R  Fiˆ  Fkˆ
F
c
MO  F (a  c) ˆj  Fbkˆ
y
nˆR 
VD1
(
R
1 ˆ ˆ

(i  k )
|R|
2

1 ˆ ˆ 1 ˆ ˆ
M  M O  nˆR nˆR  {( F ( a  c  ˆj  Fbkˆ) 
(i  k )}
(i  k )
2
2
Fb ˆ ˆ

(i  k )
2
z
x
R
 Fb  ˆ
ˆj - Fb kˆ
M per  M O - M  - 
i

F
a

c
(


2
 2 
MO
O r
VD2
M per  r  R
M
(
 (
 ( Fb 2  iˆ  F ( a  c  ˆj  ( Fb 2  kˆ  xriˆ  yr ˆj  Fiˆ  Fkˆ
y
xr  a  c
yr  b 2

Ans
Note: we can calculate wrench just in 1 step see sample 2/16.
96
Recommended Problems
• 3D Resultants:
2/140 2/142 2/149 2/150
98
Equivalent System

F1

F2
N1
CP
R
CP
R
N3
A
A

F3
N2
?
B
B
P
P
C
C
force-couple System B
force-couple System A
In Statics Mechanics, we treat these two systems are equivalent if and only if
RSys1  RSys 2
R  R
(Pure Tendency to translate)
M Sys1  M Sys 2 (for all point)
CP  CP
(Pure Tendency to rotate)
(in fact, just any one point is ok)
( i.e. M P  M P )
(just one point, and can be any point)
Equivalent System
useful for
reducing any force-couple system
=> simplest resultant
- General (3D) Force System
- Concurrent Force System
- Parallel Force System
- Coplanar Force System (2D System)
101
General-3D Force Systems
simplest system
z

y
R
x
O
P
y
MP
RSys1  RSys 2
x
M Sys1  M Sys 2
(for any one point)
102
Concurrent Force Systems

F2
F1
O

F3
RSys1  RSys 2
FR   F
(and no couple)
simplest system

O
R
M Sys1  M Sys 2
(for any one point)
No benefit to use,
because it is satisfied by default
(moment at O)
103
Coplanner System

F1
simplest system
x
x

F2
M1
y
O

y
O

F3
RSys1  RSys 2
R
for most case (99.9%)
M Sys1  M Sys 2
(for any one point)
FR   Fi
MRO   Mi   (ri  Fi )
(Moment at point O)
104
105
Equivalent System
Example Hibbeler Ex 4-16 #1
Determine the
magnitude,
direction and
location on the
beam of a
resultant force
which is equivalent
to the system of
forces measured
from E.
106
Equivalent System
Example Hibbeler Ex 4-16 #2
  Fx,sys II   Fx,sys I  FR cos  (500 N)cos60  (100 N)  350 N
  Fy ,sys II   Fy ,sys I  FR sin  (500 N)sin60  (200 N)  233.01 N
FR  420.47 N  420 N
107
Equivalent System
Example Hibbeler Ex 4-16 #3
tan 
FR sin 233.01 N

FR cos
350 N
  33.653  33.7  33.7 CW
Ans
  ME ,sys II   ME ,sys I 
FR sin (d )  (500 N)sin60(4 m)  (100 N)(0.5 m)  (200 N)(2.5 m)
(233.01 N)(d )  1182.25 N
 d  5.0730 m  5.07 m
Ans
108
Equivalent System
Example Hibbeler Ex 4-19 #1
Determine the magnitude and direction of a resultant
equivalent to the given force system and locate its point of
application P on the cover plate.
System I
System II
109
Equivalent System
Example Hibbeler Ex 4-19 #2
FA  300kˆ lb, rA  8iˆ ft,
FC  150kˆ lb,
FB  200kˆ lb, rB  8 jˆ ft
r  ( 8 sin 45iˆ  8 cos 45 jˆ) ft
C
  Fsys II   Fsys I 


FR  FA  FB  FC
 FR  650kˆ lb
Ans
  MO,sys II   MO,sys I 


MRO  rA  FA  rB  FB  rC  FC
MRO  (8iˆ ft)  ( 300kˆ lb)  ( 8 jˆ ft)  ( 200kˆ lb) 
MRO  ( 8 sin 45iˆ  8 cos 45 ˆj ) ft  ( 150kˆ lb)
MRO  2400 ˆj  1600iˆ  848.53iˆ  848.53 ˆj lb  ft
MRO  (751.47iˆ  1551.5 jˆ) lb  ft
(1)
110
Equivalent System
Example Hibbeler Ex 4-19 #3
MRO  r  FR  ( xiˆ  yjˆ) ft  ( 650kˆ lb)
MRO  ( 650 yiˆ  650 xjˆ) lb  ft
(2)
From (1)  (2)
( 650 yiˆ  650 xjˆ) lb  ft  (751.47iˆ  1551.5 ˆj ) lb  ft
Equating iˆ and ˆj components
1551.5 lb  ft
x
 2.3869 ft  2.39 ft
650 lb
751.47 lb  ft
y
 1.1561 ft  1.16 ft
650 lb
Ans
Ans
111
Equivalent System
Reduction 3D System to a Wrench #1
FR about O & MRO  FR about O & (M||  M )
FR about O & (M||  M )  FR about P & M||
112
Equivalent System
Reduction 3D System to a Wrench #2
FR about P & M|| 
FR about P & M|| about P (free vector)
113