### Force System

```2 Force Systems
Force, Moment, Couple and
Resultants
1
Force Definition
F
F
• Force is a vector quantity (why?)
• Force is the action of one body on another. [Statics]
• Force is an action that tends to cause acceleration of an object.
[Dynamics]
• The SI unit of force magnitude is the newton (N).
• One Newton is equivalent to one kilogram-meter per second
squared (kg·m/s2 or kg·m · s – 2)
Examples of mechanical force include the thrust of a rocket
engine, the impetus that causes a car to speed up when you step on
the accelerator, and the pull of gravity on your body.
Force can result from the action of electric fields, magnetic
fields, and various other phenomena.
3
FORCE SYSTEMS

F
Force is a vector
Line of action is a straight
line colinear with the force
Force System:
concurrent if the lines of
action intersect at a point

FA

FB
y

FC

FD
x
parallel if the lines of action
are parallel
coplanar if the lines of action
lie on the same plane
pararell  coplaner ?
4
Writing Convention
Hand
Print
Scalar
F
Vector
F F F
Unit Vector
iˆ
Magnitude of
Vector
F
In this course, you have to
write in this convention.
iˆ i
same
symbol
i
F
F
F
Recommended Style
FORCE SYSTEMS
Vector (2D&3D)
Basic Concept
2-D Force Systems
3-D Force Systems
Moment
Moment
Couple
Couple
Resultants
Resultants
7
Free Vectors: associated with “Magnitude” and “Direction”
parallelogram
Magnitude:
Vector :
| V | or V
V or V

V1
V
W
V1
( a  0)
M
| M | b | V |
V2
| W | a | V |
V
=V1  ( 1)V2
V1
V

aV
V  V1  V2
V2
V2
 : Direction
W
(V  V1  V2 )
V  V1  V2
Representation
(b  0)
AB  B A
+
V2
V2
V1
A  (B  C )  ( A  B)  C
a(bA)  (ab )A
(a  b )A  aA  bA
a( A  B )  aA  aB
triangle
A  A eˆ  Aeˆ
eˆ A 
A
A
(unit vector of A)
8
Vector
Operation
Commutative
R  AB  B A
9
Vector
Operation
Associative
R  A  B  C  ( A  B )  C  A  (B  C )
10
wrt = with respect to
Operation
Scalar Multiplication #2
a(bA)  (ab )A
associative
(a  b )A  aA  bA
a( A  B )  aA  aB
A  A eˆ  Aeˆ
eˆ  unit vector of A
11
Vector
Component
Resolution of a Vector
| A | ?
| B | ?
A vector may be resolved
into two components.
R  AB
13
Basic relations of Triangle
(C/6, law of cosine, sine)
Law of cosine
a

c  a  b  2ab cos 
2

b

c
2
2
Law of sine
a
b
c


sin  sin  sin 
14
2
Hint
V
V2



a


1
V1
b

c
Given V,  and , find V1 ,V2
Law of cosine
V1  ___
(Law of sine)
V2  ___
(Law of sine)
c2  a 2  b2  2ab cos 
Law of sine
a
b
c


sin  sin  sin 
15
Vector Component and Projection
b

R

Fb

R
 
F1 , F2 : vector components of
(along axis a and b)
F1  F2  R

F2

F1
a

Fa
Fa
: projections of
Fb

R
(onto axis a and b)
Fa  Fb  R (generally)
special case: =
projection vectors are
orthogonal to each other
 
Fa , Fb

R
b

Fb
: orthogonal projections & vector components

Fa
a
16
Rectangular Componentsiˆ
• Most commonly used
y
vector component = vector projection
Fx  F cos 
ˆj

Fy
ˆj 
Fy  F sin 

F


Fx
  
F  Fx  Fy
F  Fx iˆ  Fy ˆj
F  Fx2  Fy2
iˆ
x
  tan | Fy / Fx |
1
 

Fx , Fy  vectorcomponentsof F
(in x and y directions)

Fx , Fy  scalar componentsof F
(in x and y directions)
17
Fx=?
Fy=?

F
y
F
  F ( cos  )
 F cos 
p
x

Fy  F cos(p   )
x

Fx  F sin(p   )
= F sin 
Fx   F cos
y
minus
(>90)

F
Fy  F sin 

x

F
y

y


Fx  F cos
Fy  F cos(    )
Fy  F sin 
Fx  F sin(    )
x
18
EXAMPLE 2-1
Given the magnitude of the tension
in the cable, T = 9 kN, express T in
terms of unit vector i and j
AB  102  62  11.662 m

T  (9 cos) ˆi  (9 sin) ˆj
  10   ˆ
  6  ˆ
  9 
  i   9 
  j
  11.662 
  11.662 
y
T
ˆj
3 S.F.

ˆi
x
7.72 iˆ  4.63 ˆj
 7.7 iˆ  4.6 ˆj
Correct?
kN
ANS
19
We are using robot arm to put the cylindrical part into a hole.
Determine the components of the force which the cylindrical part exerts on the
robot along axes
(a) parallel and perpendicular to arm AB
(b) parallel and perpendicular to arm BC
15
15
par
per
30
par
60
Defining
direction
45
per
30
15
15
P = 90 N
arm AB
P = 90 N
Pper  P cos45  90cos45  63.64N (
)
Ppar  P sin45  90sin45  63.64 N (
arm BC
)
Ppar  P cos30  9cos30  7.794 N (
)
Pper  P sin30  9sin30  4.5N (
)
ANS
ANS
20
2/2 Combine the two forces P and T, which act on the fixed structure
at B, into a single equivalent force R
Graphics
P=800 N (8cm)


R
T=600 N (6cm)
R  525 N (5.25cm)
Geometric
6sin 60o
  tan
 40.9o
o
3  6cos60

R
Vector Component (Algebraic)
T  600(cosiˆ  sin  ˆj )
R  P  T  346iˆ  393 ˆj
R2  P2  T 2  2PT cos 
R  524 N
P
1
P  800iˆ
  49o
T
R

sin  sin 

  48.6o
T
R  3462  ( 393) 2  524 N
  tan 1
393
 48.6o
346
Correct?
Point of application is B
21
Example Hibbeler Ex 2-1 #1
Determine the magnitude and direction of the resultant force.
Two forces is not acting
at the same point.
FR  F12  F22  2F1F2 cos 
Geometric
 213 N
F2
F
150
212.55
 R 

sin sin
sin sin115
22
  39.101  39.1
Vector Component (Algebraic)
Geometric
Good?
(get full score?)
- more explanation
direction of iˆ , jˆ ?
- 5S.F. Then 3S.F.
FR  F12  F22  2F1F2 cos 
 213 N
F1  100cos15iˆ  100sin15 jˆ N
F  150sin100 iˆ  150cos10o jˆ N
2
F2
F
150
212.55
 R 

sin sin
sin sin115
  39.101  39.1
FR   F  F1  F2  122.64iˆ  173.60 jˆ N
  tan
1
FRy
FRx
 tan1
173.60
 54.761
122.64
23
Geometric
Using the law of cosine:
O
FR  F  F  2F1F2 cos 
2
1
2
2
 1002  1502  2(100)(150)cos115
FR  212.55  213 N
#
Applying the law of sine:
F2
F
150
212.55
 R


sin sin
sin sin115
  39.101  39.1
#

24
Point of Application
25
Vector
Example Hibbeler Ex 2-6 #1
F1  (600 cos30iˆ  600 sin30 jˆ) N
F  ( 400 sin 45iˆ  400 cos 45 jˆ) N
2
26
Vector
Example Hibbeler Ex 2-6 #2
FR  F1  F2
FR  (600cos30iˆ  600 sin30 jˆ)  ( 400 sin 45iˆ  400cos 45 jˆ)
F  236.77iˆ  582.84ˆj N
R
FR  236.772  582.842
 629.10  629 N
#
  tan1(FRy FRx )  tan1(582.84 236.77)
 67.891  67.9
#
27
• Reference axis (very very important)
R2  F12  F22  2F1F2 cos
– Many problems do not come with ref. axis.
– Assignment based on convenience/experience
– Three ways to be mastered
1. Graphically
2. Geometrically

F2
sin(  )
Originally pass
through O
y

F2

F1
F2y
F1y

R
Ry
3. Vector component
(algebraically)
  
R  F1  F2  ( F1xiˆ  F1 y ˆj )  ( F2 xiˆ  F2 y ˆj )
Rxiˆ  Ry ˆj  ( F1x  F2 x )iˆ  ( F1 y  F2 y ) ˆj
Rx  F1x  F2 x   Fx
Ry  F1y  F2 y   Fy
R
sin
o
x
F1x
F2x
Rx
The calculations do not reveal the
point of application of the resultant
force.
In case where forces do not apply at the same
point of application, you have to find it too!
28
Recommended Problem
2/9, H2-17, 2/12, 2/26, H2-28
29
Three Dimensional Coordinate System
y
ˆj
z
Real-life Coordinate System is 3D.
kˆ
iˆ
Introduce rule for defining the 3rd axis
- “right-hand rule”: x-y-z
- for consistency in math calculation
(cross vector)
x
How does 2D differs from 3D?
y
2D
z
x
31
Rectangular Components (3D)
ˆj
z
kˆ
Fz kˆ
z
x
Fx iˆ

F
Fy ˆj
projection & component
y
Fx  F cos( x )
Fy  F cos( y )
nˆF
Fz  F cos( z )
y
F  Fx2  Fy2  Fz2

F  Fxiˆ  Fy ˆj  Fz kˆ
nˆF
F  F (cos xiˆ  cos y ˆj  cos z kˆ)
x

(maybe +/-)
- cos(x), cos(y), cos(z) : “directional cosines” of F
-
iˆ
cos2(x)+cos2(y)+cos2(z)
=1
def
nˆF  cos  xiˆ  cos  y ˆj  cos  z kˆ

nˆ F is a unit vector in the direction of F
- If you known the magnitude and all directional cosines, you can write
force in the form of iˆ ˆj kˆ
directional cosine Method
32
Example Hibbeler Ex 2-8
Find Cartesian components of F
cos2  x  cos2  y  cos2 z  1
cos2   cos2 60  cos2 45  1
cos2  
1
4
1
cos  
2
1
cos   
2
1
2
  cos1(  )  60° or 120
z
x
y
F  F cos x iˆ  F cos y jˆ  F coszkˆ
 (200 cos 60iˆ  200 cos 60 jˆ  200 cos 45kˆ ) N
 (100iˆ  100 jˆ  141.42kˆ ) N
F  (100iˆ  100 jˆ  141kˆ ) N
#
33

Given the cable tension T = 2 kN. Write the vector expression of T
1) directional cosine method
z
B
y
Real
directional
cosine
x
A

T  T (cosx ˆi  cosy ˆj  cosz kˆ )
cos  x 
length of AB 
1.2
length of AB
1.22  0.52  (0.4  0.3)2  1.3
x
directionl
cosine = -0.92
B
x
A
cos x  0.92
34
z
B
B
y
y
A
x
z
B
y
A
 cos  y 
0. 5
0 .5

 0.38
lengthof AB
1.3
y
B
z
z
A
A
 cos  z
x
Thus,
0 .1
0 .1


 0.08
lengthof AB
1. 3
T  2 (0.92 iˆ  0.38 ˆj  0.08 kˆ ) kN
ANS
35
Directional Cosines by Graphics
A
cos  x  x
A
cos  y 
Ay
A
cos2(x)+cos2(y)+cos2(z) = 1
cos  z 
Az
A
36
- Usually, the direction of force is not given using the directional
cosines. Need some calculation.
- Two examples
(a) Two points on the line of action of force is given (F also given).

 
rAB  rB  rA
z
Two-Point
Method

F
B (x2, y2, z2)
A (x1, y1, z1)
B
y
x
F  F nˆF
rA  x1iˆ  y1 ˆj  z1kˆ

r  x iˆ  y ˆj  z kˆ
2
2
Position
vector
2

rAB  ( x2  x1 )iˆ  ( y2  y1 ) ˆj  ( z2  z1 )kˆ
nˆ AB 
( x2  x1 )iˆ  ( y2  y1 ) ˆj  ( z2  z1 )kˆ
( x2  x1 )2  ( y2  y1 ) 2  ( z2  z1 ) 2

( x2  x1 )iˆ  ( y2  y1 ) ˆj  ( z 2  z1 )kˆ
FF
( x2  x1 ) 2  ( y2  y1 ) 2  ( z2  z1 ) 2
37
z
2) 2-point construction
0.5
B
y
0.4
rAB
F  F nˆF
A
1.2
0.3
x
A  ( x1, y1, z1   (1.2,0,0.3
B  ( x2 , y2 , z2   (0,0.5,0.4
x2  x1  iˆ  ( y2  y1  ˆj  ( z2  z1  kˆ
(
rAB
F  FnˆF  F
F
2
2
2
rAB
( x2  x1   ( y2  y1   ( z2  z1 
0.0  1.2  iˆ  ( 0.5  0.0  ˆj  ( 0.4  0.3 kˆ
(
F 2
2
2
2
( 0.0  1.2   ( 0.5  0.0   ( 0.4  0.3
(
F  2 0.92iˆ  0.38 ˆj  0.08kˆ

Ans
kN
38

T
Write vector expression of . Also determine angle x, y, z, of T
with respect to positive x, y and z axes
Consider: T as force of tension acting on the bar

 T  T nˆ
where nˆ = unit vector from B to A
4ˆi  7.5ˆj  5kˆ

4 2  (7.5) 2  5 2
 0.41ˆi  0.76ˆj  0.51kˆ
Thus
cos x  0.41

T  10 ( 0.41ˆi  0.76ˆj  0.51kˆ ) kN
 x  66
cos  y   0.76
 y  139
cos  z  0.51
z  59
ANS
39
Vector
Example Hibbeler Ex 2-9 #1
Determine the
magnitude and the
coordinate direction
angles of the resultant
force acting on the
ring
FR   F  F1  F2  (60 jˆ  80kˆ )  (50iˆ  100 jˆ  100kˆ )
 (50iˆ  40 ˆj  180kˆ ) lb
FR  502  ( 40)2  1802 lb  191.05  191 lb
#
40
Vector
Example Hibbeler Ex 2-9 #2
uFR  FR FR
50 ˆ
40 ˆ
180 ˆ

i
j
k
191.05
191.05
191.05
 0.26171iˆ  0.20937 jˆ  0.94216kˆ
cos   uFR x  0.26171
cos   uFR y  0.20937
cos   uFR z  0.94216
  74.8

  102
  19.6
#
41
Vector
Example Hibbeler Ex 2-11 #1
Specify the coordinate
direction angles of F2 so
that the resultant FR
acts along the positive y
axis and has a
magnitude of 800 N.
FR  (800 ˆj ) N
F1  F1 cos  x1iˆ  F1 cos  y 1 jˆ  F1 cos  z1kˆ
 (300 cos 45iˆ  300 cos 60 jˆ  300 cos120kˆ ) N
 (212.13iˆ  150 jˆ  150kˆ ) N
F2  F2 x iˆ  F2 y jˆ  F2 z kˆ
42
Vector
Example Hibbeler Ex 2-11 #2
FR  F1  F2
800 ˆj  212.13iˆ  150 jˆ  150kˆ  F2 x iˆ  F2 y jˆ  F2 z kˆ
800 ˆj  (212.13  F2x )iˆ  (150  F2 y ) jˆ  ( 150  F2 z )kˆ
43
Vector
Example Hibbeler Ex 2-11 #3
x dir.
0  212.13  F2 x
y dir. 800  150  F2 y
z dir.
0  150  F2 z
F2 x  212 N
F2 y  650 N
F2 z  150 N
x dir.
212.13  700 cos  2
y dir.
650  700 cos  2
z dir.
0  700 cos  2

#
 2  108
2  21.8
 2  77.6 #
44
Force
Example Hibbeler Ex 2-15 #1
The roof is supported by cables as shown. If the cables exert
forces FAB = 100 N and FAC = 120 N on the wall hook at A as
shown, determine the magnitude of the resultant force acting at A.
45
Force
Example Hibbeler Ex 2-15 #2
rAB  (4  0)iˆ  (0  0) jˆ  (0  4)kˆ  m  4iˆ  4kˆ m
rAB  42  ( 4)2  5.6569 m
FAB
rAB
 100 N (
)
rAB
4
4
iˆ 
kˆ )
5.6568
5.6568
 (70.711iˆ  70.711kˆ ) N
 100(
FAB
46
Force
Example Hibbeler Ex 2-15 #3
rAC  (4  0)iˆ  (2  0) jˆ  (0  4)kˆ  m  4iˆ  2 jˆ  4kˆ m
rAC  42  22  ( 4)2  6 m
FAC
rAC
 120 N (
)
rAC
4ˆ 2ˆ 4 ˆ
i  j  k)
6
6
6
 (80iˆ  40 jˆ  80kˆ ) N
 120(
FAC
47
Force
Example Hibbeler Ex 2-15 #4
FR  FAB  FAC
 (70.711iˆ  70.711kˆ ) N 
 (80iˆ  40 ˆj  80kˆ ) N
 (150.711iˆ  40 ˆj  150.711kˆ ) N
FR  (150.711)2  (40)2  (150.711)2 N
 216.86  217 N
#
48
(b) Two Angles orienting the line of action of force are given (, )
Othorgonal
projection Method
Resolve into two components at a time
kˆ

Fz
ˆj
z

F

Fy

Fx
y

Fz = F sin()
Fxy = F cos()

Fxy

iˆ
Fx = Fxy cos() = F cos() cos()
x
Fy = Fxy sin() = F cos() sin()
49
z
Fz
F
Fx 65o
x
50o F
y
y
Fxy
Fz  F sin50o  3.83 kN
Fxy  F cos50o  3.21 kN
Fx  Fxy cos65o  1.36 kN
Fy  Fxy sin65o  2.91 kN
F  Fx iˆ  Fy ˆj  Fz kˆ
 1.36iˆ  2.91 ˆj  3.83kˆ kN
Ans
50
 CAB  tan 1
10
 81.5o
1.5
z
y
TAB
C A
15o
TZ
x
B
T
Tz  T sin 15o  207N
TAB  T cos15o  773N
Ty  TAB cos81.5o  115N
Tx  TAB sin 81.5o  764N
T  Tx iˆ  Ty ˆj  Tz kˆ
 764iˆ  115 ˆj  207 kˆ N
Ans
51
2/110 A force F is applied to the surface of the sphere as shown.
The 2 angles (zeta, phi) locate Point P, and point M is the
midpoint of ON. Express F in vector form, using the given
x-,y- z-coordinates.
52
Recommended Problems
• 3D Rectangular Component:
2/99 2/100 2/107 2/110
53
Vector
Operation Products
1. Dot Products
AB
2. Cross Products
AB
3. Mixed Triple Products
A  (B  C )
55
scalar product
P  Q  PQ cos

P

kˆ

Q
ˆj
iˆ
iˆ  iˆ  ˆj  ˆj  kˆ  kˆ  1
(unit vector)
iˆ  ˆj  ˆj  iˆ  0
iˆ  kˆ  kˆ  iˆ  0
ˆj  kˆ  kˆ  ˆj  0
( three orthogonal vector )
A  B  ( Ax iˆ  Ay jˆ  Az kˆ)  (Bx iˆ  By jˆ  Bz kˆ)
 Ax iˆ  Bx iˆ  Ax iˆ  By jˆ  Ax iˆ  Bz kˆ 
 Ay jˆ  Bx iˆ  Ay jˆ  By jˆ  Ay jˆ  Bz kˆ 
 Az kˆ  Bx iˆ  Az kˆ  By jˆ  Az kˆ  Bz kˆ
A  B  Ax Bx  Ay By  Az Bz
Example :
P  2iˆ  3 jˆ  4kˆ
Q  4iˆ  2 jˆ  5kˆ
P Q  ?
PQ  ?
56
Application of Dot Operation
• Angle between two vectors
• Component’izing Vector
U
U//
eˆ
P  Q  PQ cos
 P Q 
  cos 

 | P || Q | 
1
U  U //  U

U
Example :
U  2iˆ  3 jˆ  4kˆ
(
1 ˆ ˆ ˆ
i  j k
3
U //  (eˆ Uˆ ) eˆ
eˆ 
U = U  (eˆ Uˆ ) eˆ
U// ,U  ?

which direction?
eˆ
T  (T  eˆ)eˆ
T
(T  eˆ)eˆ
57
z
Fz
F
Fx 65o
x
50o F
y
y
Fxy
Fz  F sin50o  3.83 kN
Fxy  F cos50o  3.21 kN
Fx  Fxy cos65o  1.36 kN
Fy  Fxy sin65o  2.91 kN
F  Fx iˆ  Fy ˆj  Fz kˆ
 1.36iˆ  2.91 ˆj  3.83kˆ kN
Ans
58
 CAB  tan 1
z
y
10
 81.5o
1.5
TAB
C A
TZ
15o
x
B
T
T  Tx iˆ  Ty ˆj  Tz kˆ
 764iˆ  115 ˆj  207kˆ N
which direction??
eˆ
T  (T  eˆ)eˆ
(T  eˆ)eˆ
Txz  764iˆ  207kˆ
Ans
Txz  Tx2  Tz2  791.55 N
59
“Cross Product” of Vectors
A B
magnitude :
AB sin 
right-hand rule
(A then B)
B

eˆ
A
C=A  B
def
=
1 (
AB 2
)
AB
( | A| | B |sin  ) eˆ
C
line which are
perpendicular
with both vectors
60
Operation Cross Product
A  B ? B  A
A B 
?
Laws of Operations
C  AB
B A
• Commutative Law is not valid
AB  B A
(
AB   B A
(B  A) ?

( B )  A ?
B
A
B
A
B  (  A) ?
• Associative wrt scalar multiplication
B  A  C
a( A  B )  (aA)  B  A  (aB )  ( A  B )a
A  (B  C )  ( A  B )  ( A  C )
61
x-y-z complies with right-hand rule
y
ˆj
z
kˆ
iˆ
x
iˆ  ˆj  kˆ
iˆ
ˆj  kˆ  iˆ
kˆ  iˆ  ˆj
+
kˆ
ˆj
iˆ  iˆ  ˆj  ˆj  kˆ  kˆ  0
62
How to calculate cross product
U  U x iˆ U y ˆj U z kˆ
V  V x iˆV y ˆj V z kˆ
U V  (U x iˆU y ˆj U z kˆ)  (V x iˆV y ˆj V z kˆ)
 U V (iˆ  iˆ)  U V (iˆ  ˆj )  U V (iˆ  kˆ)
x x
x y
x z
 U V ( ˆj  iˆ)  U V ( ˆj  ˆj )  U V ( ˆj  kˆ)
y x
y y
y z
 U V (kˆ  iˆ)  U V (kˆ  ˆj )  U V (kˆ  kˆ)
z x
z y
z z
U V  (U y V zU z V y )iˆ  (U x V zU z V x) ˆj
(U x V yU y V x)kˆ
This term can be written in a determinant form
63
Cross Product
iˆ
ˆj
kˆ
U V  U x U y U z
Vx Vy Vz
-
-
Vz
Vx
-
ˆj
ˆj
iˆ
kˆ
iˆ
 Ux U y Uz Ux U y
Vx
Vy
+
Vy
+
+
U V  (U y VzUz V y)iˆ  (U z V xU x V z) ˆj (U x V yU y V x)kˆ
64
Why cross product?
• Mathematical Representation
of Moments, Torque
• Perpendicular Direction
• Area Calculation
nˆOABC

B A
B A
z
C
A
y
A
B
Area = A? B
B
O
x
65
Mixed Triple Product
iˆ
ˆj
kˆ
U  (V  W )  (U x iˆ  U y jˆ  U z kˆ )  Vx
Vy
Vz
W x Wy
Wz
 (U x iˆ  U y jˆ  U z kˆ ) 
(VyWz  VzWy )iˆ  (VxWz  VzWx ) jˆ  (VxWy  VyWz )kˆ 


U  (V  W )  Ux (VyWz  VzWy )  Uy (VxWz  VzWx )  Uz (VxWy  VyWz )
Ux
Uy
Uz
Ux
U  (V  W )  Vx
Vy
Vz
  Wx Wy
W x Wy
Wz
Vx
Uy
Vy
Uz
W x Wy
Wz
Wz  U x
Uy
Uz
Vz
Vy
Vz
Vx
U  (V  W )  W  (U  V )  V  (W  U )
66
nˆ
Why mixed triple product?
M O ,F
• Mathematical Representation
of Moments along the axis.

M o,F ,

F

r
O
• Volume Calculation
Volume : C  ( B  A)
A
C
nˆOABC
B
C  ( B  A)
Volume must always +
67
Vector
Operation Product Summary
Dot Product
Scalar
Cross Product
Vector
Mixed Triple Product
Scalar
68
69
Homepage URLs
 Statics official HP
http://www.lecturer.eng.chula.ac.th/fmekmn/
 Session 1 HP
http://pioneer.netserv.chula.ac.th/~lsawat/course/statics/
http://blackboard.it.chula.ac.th/
(after the end of registration period)
72
FORCE SYSTEMS
Vector
Basic Concept
2-D Force Systems
3-D Force Systems
Moment
Moment
Couple
Couple
Resultants
Resultants
73
Force Definition
F
F
• Force is a vector quantity (why?)
• Force is the action of one body on another. [Statics]
• Force is an action that tends to cause acceleration of an object.
[Dynamics]
• The SI unit of force magnitude is the newton (N).
One newton is equivalent to one kilogram-meter per second squared
(kg·m/s2 or kg·m · s – 2)
Examples of mechanical force include the thrust of a rocket engine, the impetus
that causes a car to speed up when you step on the accelerator, and the pull of
Force can result from the action of electric fields, magnetic
fields, and various other phenomena.
74
Force
Force Representation
• Vector quantity
– Magnitude
– Direction
– Point of application
Use different colours in diagrams
• Body outline
 blue
 red
• Miscellaneous
 black
(dimension, angle, etc.)
F
10 N
75
Type of Forces
F
External force
F
Applied force
Reactive force
Force
Internal force
Stress
Strain
Concentrated
Contact force
Force
Force
Body force
Distributed
76
Force
Cables & Springs

k  spring constant
T
T
s
F
Cable in tension
F  ks
77
2/2 Combine the two forces P and T, which act on the fixed structure
at B, into a single equivalent force R
Graphical
P=800 N (8cm)


R
T=600 N (6cm)
R  525 N (5.25cm)
Geometric
6sin 60o
  tan
 40.9o
o
3  6cos60

R
T
R

sin  sin 

R  524 N
  48.6o
T
R  P  T  346iˆ  393 ˆj
P  800iˆ
T  600(cosiˆ  sin  ˆj )
R2  P2  T 2  2PT cos 
P
1
Algebraic
  49o
R  3462  ( 393) 2  524 N
  tan 1
Correct?
393
 48.6o
346
Point of application is B
78

F2
How to add sliding vectors (forces)?

F2

R

R is applied at
Principle of
Transmissibility

F1
A
A
  
R  F1  F2

F1
point A
Point of application
Not OK. !
Still OK.

R
Point of Application
is wrong
A

F1

R
A

F1

F2
79
Special case: Addition of Parallel Sliding Force
F
F
F1
F2
F1
F
F2
F
F
F2
F1
F1
F
R2
F2
R1
line of
action
Point of application
R2
R
R1
R2
R
R1
R
This graphical method can be used to find Line of action
The better and efficient way will be discussed later, when we
learn the concept of “moment”, “couple”, and “resultant force”
y
T
Ty
60 Tx
Move all forces to
that concurrent point
T1  200iˆ  346 ˆj
VD1
N
x
T1x  T1 cos60o  200 N
T1 y  T1 sin60o  346 N
T2  400iˆ
N
R  T1  T2  600iˆ  346 ˆj
N
R  Rx2  Ry2  693 N
Point of application,
But no physical meaning
  tan 1
346
=29.97o
600
Application Point
Ans
81

F2
A
How to add sliding vectors (forces)?

F2

F1
A

R

F1
  
R  F1  F2

R is applied at
point A
Point of application
There is better way to find the point of application
(or line of action), but you have to learn the concept of
moment and couples first.
82
Moment
In addition to the tendency to move a body, force
may also tend to rotate a body about an axis
From experience (experiment)
magnitude depends only on “F” and “d”
M  Fd
(magnitude)
summation
M   Fd
i i
i
Direction
moment
axis
Moment is a vector
84
z
Moment Definition
• Moment is a vector quantity.
– Magnitude
– Direction
– Axis of Rotation
•
•
•
•
•
•
(MO )z
O
y
x
dy
Fx
The unit of moment is N·m
The moment-arm d (perpendicular distance)
The right-hand rule
determined by vector cross product
Sign convention: 2D +k or CCW is positive.
Moment of a force or torque
85
Mathematical Definition (3D)
from A to point of application of the force
M A,F  r  F

F

M

-Direction: right-hand rule
-Point of application: point A
X
r
A
d
-Magnitude: | r || F | sin   Fd
(Unit: newton-meters, N-m)
r F

F
2D
- 2D, need sign convention and be
consistent; e.g. + for counterclockwise and – for clockwise
+
MFd
d
86
can be used with
Varignon’s Theorem (Principle of Moment)
more than
2 components
M  r  F1  r  F2
*
O
Same?
Yes !
F1  F2 )
M  r  (F
**
O
(
r  F1  r  F2  r  F1  F2

 r F
The moment of a force about any point is equal to the sum of
the moments of the components of the force about that point
sum of moment
(of each force) =
moment of sum
Fy
Useful with rectangular components
(of all force)

F
Fx
d2
Mo = -Fxd2+Fyd1
y
+
O
d1
x
87
Principle of Transmissibility & Moment
Principle of Transmissibility is based on
the fact that
“moving force along the line of action
causes no effect in changing moment”
O

F

M

r
   convenient
M  r F
X
A
d
position vector:
from A to any point on
line of action of the force.
Y
Z
rAX  F  rAY  F
- direction: same
- magnitude:
O
Sliding force
has the same moment
M = Fr sin  = Fd
88
Sample 2/5 Calculate the magnitude of the moment about
the base point O of 600N force in five different ways.
2m
2m
4m
A
400
Solution I: 2D Scalar Approach
d  4 cos40o  2 sin 40o
 4.35 m
 2610
O
x
Solution II: 3D Vector Approach
MO  r  F
(

CW
(
  2610kˆ N-m
N-m Ans
CW or CCW?
Correct?
600N
 2iˆ  4 ˆj  600 cos 40o iˆ  sin 40o ˆj
M O  F  d  600 4.35
O
d
400
y
600N
4m
A
CW
89

2m
A
F2
4m
F1
400
600N
O
B
F1
F2
d1
2m A
4m
400
600N
O
Solution III: Varignon’s theorem
Solution IV: Transmissibility
+
F1  600cos40o  460N
F2  600sin 40o  386N
MO  460(4)  386(2)
 2610 N-m (CW)
d1  4  2 tan40o  5.68 m
MO  F1d1  2610 N-m (CW)
C F1
d2
F2
Solution V: Transmissibility
d2  2  4 cot 40o  6.77 m
MO  F2d2  2610 N-m (CW)
90
EXAMPLE 2.8
In raising the flagpole, the tension T in the cable must supply a
Moment about O of 72 kN-m. Determine T.
tan 1
25.981
 43.898
12  15
30sin 60
30 m

12 m
 25.981m
d
o 60

15 m
72  Td  T (12)sin 43.898
 T  8.65 kN ANS
91
Moment
Example Hibbeler Ex 4-7 #1
Determine the moment of the force
F  (400 sin30iˆ  400 cos30 jˆ) N
 (200iˆ  346.41jˆ) N
Correct?
iˆ , ˆj ?
92
Moment
Example Hibbeler Ex 4-7 #2
3D Vector Approach
iˆ
r
ˆj
kˆ
MO  r  F  0.4
0.2
0
200 346.41 0

0.4
0.2
200 346.41
kˆ
N-m
N-m
 98.564kˆ  98.6kˆ N  m Ans
F  (400 sin30iˆ  400 cos30 jˆ) N
 (200iˆ  346.41jˆ) N
r  (0.4iˆ  0.2 jˆ) m
Scalar Approach (Varignon’s theorem)
 MO  400 sin 30(0.2)  400 cos 30(0.4)
 40  138.56  98.564 N  m
MO  98.6kˆ N  m
#
93
Couple
- Couple is a summed moment produced by
two force of equal magnitude but opposite in
direction.
M = F(a+d) – Fa = Fd
d
a

F
+
magnitude does not depend on distance a (point O),
i.e. any point on the body has the same magnitude.

F
Couple  Fd
O
Effect of Pure Rotation
- tendency to rotate the “whole” object.
- no effect on moving object as translation.
2D representations: (Couples)
C
C
couple is a free vector
C
94
Force-couple systems
- Line of action of a force on a body may be changed if a couple is
added to compensated for the change in the tendency to rotate of
that body.
C ?
C  rBA  F
No changes in net external effect
B
A

F
Principle of transmissibility


F
 B
F
d
A

F
B


F
C
A
Force-couple system
The direction and magnitude of Force can not be changed, only line of action
(i.e. only change to other pararell line)
Procedure may be reversed to combine a force with a couple
96
C  r F
from new location (B)
to old location (A)
B
B
r
A
F
A
No Moment:
Principle of Transmissibility
C
F
B
B
A
F
F
A
C
F
A
Principle of Transmissibility
B
is based on the fact that
moving force along
the line of action causes no effect in changing moment
A
B
F
97
Why using equivalent system?
B
B
A


F
C
A
Principle of transmissibility
Force-couple system
All force systems are equal.
real (physical)
system


F
In the viewpoint of Mechanics,
Result of force to these systems
are equal
M
equivalent system

equivalent system
98
Understanding Force-Couple system
Moment about point B of force F
= tendency of force F to rotate the object at point B
 couple occurs when moving Force F from A to B
( couple occurs when moving Force F parallel to
its line of action to the point B)
0
Equivalent System
M A, F  0
M B ,F  rBA  F
M D ,F  rDA  F
B
B

F
D
A


F
C
D
A
M A  rAB  F 
MB 
C
C
M D  rDB  F 
C
99
Be careful of the
direction of moment
P
70m
P
Vector Diagram
F
12m
F
CW  CCW
F 12  P  70
3600  70 P
P  51.42 kN Ans
100
2/11 Replace the force F by an equivalent force-couple system at point O.
F  50 kN
50 kN
y
20
0.25 m
10
250 mm.
0.1m
25
x
50 kN
M
20
Couple occurred when moving F to O
= Moment of F about O
+ M 
o
F cos 20 (0.1cos 25  0.25cos10 )
 F sin 20o (0.1sin 25o  0.25sin10o )
 17.3 N-m
F  17.1iˆ  46.9 ˆj kN
CCW
Correct?
Ans
101
Engine number 3 fails. Determine the force-couple system on the
Moving all 3 forces to point O
F
y
o
(direction: left)
R   F  90  90  90  270 kN
couples occuring when moving forces.
y
sum of moments?
M o  90(21)  90(12)  90(21)
x
R
  1080 kN  m
(CW)
ANS
Sum of couples
M
Got the meaning?
102
Resultant
Example Hibbeler Ex 4-14 #1
Replace the current system by
an equivalent resultant force
and couple moment acting A.
FRx   Fx   
FRx  100  400 cos 45
FRx  382.84 N
FRy   Fy   
FRy  600  400 sin 45
FRy  882.84 N
103
Resultant
Example Hibbeler Ex 4-14 #2
F  F 2  F 2 
Rx
Ry
 R

FR  ( 382.84)2  ( 882.84)2
FR  962.27  962 N
  tan
1
FRy
FRx
#
882.84
 tan
382.84
1
  66.556  66.6
#
 MR   M A  
 A

MRA  100(0)  600(0.4)  400 sin 45(0.8)  400cos 45(0.3)
MRA  551.13  551 N  m #
104
b cos20
b
exactly cancelled
D
20o
60 N-m
300N
300 b  cos 20o  60
b  0.213 m
Ans
105
2/6 Simplest Resultant
• Resultant of many forces-couple is the simplest force-couple combination
which can replace the original forces/couples without changing the external
effects on the body they act on

F1

F2

R

F2
y
F3 y

F3
F2 y

R1
Ry

F3
Point of application
  get
line of action of R
-Add many  do not get line
of action of R

F1
F1 y

R

F1x
   

R  F1  F2  F3  ...   F
R  ( Fx ) 2  ( Fy ) 2
x
F2 x
Rx
F3 x
Rx   Fx , Ry   Fy
  tan1 (Ry / Rx )
107
Easier way to get a resultant + its location

F1
arbitrary
d1 O
d3

F2

F1
F1d1
O
d2

F3
1) Pick a point (easy to find
moment arms)

R
d=Mo/R
F2d2

F2

F3


R  F
O
Mo=(Fidi)
F3d3
2) Replace each force with a force
at point O + a couple
(forces + couples : same procedures)
2D
O
any forces + couples system resultant
 single-force system (no-couple)
or single-couple system  where 


Mo=Rd
R 0 
4) Replace force-couple system
with a single force
3D
any forces + couples system
 single-force + special single-couple (wrench)
108
2/87 Determine the resultant and its line of action of the
why?
Move 3 forces to point O,
Sums their force and couples
M
O
R
(force-couple system)
Note: R is the
same regardless
with the location
point we move
the force to
Note: M depends
on the location
where we move
the force to
M = -2.4*0.2cos20 -1.5*0.12cos20
-3.6*0.3cos20 kN-m
R = ( 2.4cos20 -1.5sin20 -3.6cos20 ) i
+( -2.4sin20 -1.5cos20 +3.6cos20 ) j kN
109
At point O (0,0)
R  1.64 iˆ  0.99 ˆj kN
M
Sys 1
O
R
M = -1.635 kN-m
P
M   1.635 kˆ kN
At point X (x,y)
0M N
couples
cancelled
Correct?
O (0,0)
0  M  ( xiˆ  yjˆ)  R
P (x,y)
0  M  ( xiˆ  yjˆ)  R
Sys 2
O
R
R
M
N
P
( xiˆ  yjˆ)  R  M
r  (0  x)iˆ  (0  y) ˆj
Two equivalent systems
Pick Point O
Moment at any point
must be the same on both system
( xiˆ  yjˆ)  (1.64iˆ  0.99 ˆj)  1.635kˆ
0.99 x  1.64 y  1.635
( line of action )
110
At point O (0,0)
M
O
R
R  1.64 iˆ  0.99 ˆj kN
P
d
M = -1.635 kN-m
M   1.635 kˆ kN
Manually Canceling Couples
d
|M |

|R|
| 1.635 |
( 1.64)  ( 0.99)
2
2
How to find line of action ?
How to locate Point P
O (0,0)
tan 1 
Ry
Rx
 tan 1
0.99
 31.118o
1.64

d 
 0.853 m
P
(d sin  , d cos  )
 (0.441, 0.73)
y  yP
 tan 
x  xP
O
or
d
y  (0.73) 0.99

x  0.441
1.64
y(

P
1
)x
tan 
x2  y 2  d 2
111
Equivalent System
Equivalent System Definition
Sys 2
Sys 1
O
R
M

O
R
P
Two force-couple systems are equivalent
(FR )system I  (FR )system II
(MRO )system I  (MRO )system II
112
A car stuck in the snow.
Three students attempt to
free the car by exert forces
on the car at point A, B and
C while the driver’s actions
result in a forward thrust of
200 N as shown in picture.
Determine
1) the equivalent force-couple system at the car center of mass G
2) locate the point on x-axis where the resultant passes.
( F  ˆi  ( F  ˆj

R  ( 400 200 200 250sin30  ˆi
 ( 350 250cos30  ˆj

R 
x
R
y



R  925 ˆi  567 ˆj
y
M
G
x
+
MG  350(1.65)  250sin30 (0.825)
 690 N  m
113
For line of action of resultant
y
y

R  925 ˆi  567 ˆj

R  925 ˆi  567 ˆj
b
690 N  m
G
G
x
x
Sys II
Sys I
( xiˆ  y ˆj )  (925 iˆ  567 ˆj )  690 kˆ
Couple Cancellation
kˆ : 567 x  925 y  690
At y = 0; x = +1.218 m.
ANS
If you want to find only b (not line of action itself)
Two equivalent systems
(2D)
+ or - , you have to
find out manually
Ry (b) | MG |
567(b)=690
b=1.218 m
0  M  ( xiˆ  yjˆ)  R
Two equivalent systems
Moment at point G
must be the same on both system
M G ,SysII  M G ,SysI
( xiˆ  yjˆ)  R  M
114


Determine the resultant R (vector) and the point on x and y axes which R
must pass.
y
+
G
( F  ( F 
 ( 25  20sin 30  iˆ

R 
x
ˆi 
y
ˆj
 ( 20cos 30  30) ˆj
x
M o  25(5)  30(9)
 (20cos30 )(9)  (20sin 30 )(5)
  351 kN  m
  15 ˆi  47.3 ˆj
115
y
y

R  15 ˆi  47.3 ˆj
 351 kN  m
O
x
O
x
For line of action of resultant

 
r  R  Mo
(x ˆi  y ˆj)  (15ˆi  47.3 ˆj)   351kˆ
47.3 x  15 y   351
If
y = 0;
x = 7.42 m.
x = 0;
y = -23.4 m.
ANS
116
117
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