1 2force systems force, moment, couple and resultants

1

2 Force Systems

Force, Moment, Couple and Resultants

3

Force Definition

• 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)

F

F

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 is a vector quantity (why?)

Force can result from the action of electric fields, magnetic fields, and various other phenomena.

• Force is the action of one body on another. [Statics]

4

FORCE SYSTEMSFORCE SYSTEMS

Force is a vector F

Line of action is a straight line colinear with the force

coplanar if the lines of action lie on the same plane

Force System:

concurrent if the lines of action intersect at a point

parallel if the lines of action are parallel

y

x

AF

BF

CF

DF

pararell coplaner ?

Hand Print

Scalar

Vector

Unit Vector

Magnitude of Vector

Writing Convention

F F F

F

ˆ i i

F

F FF

same symbol

Recommended Style

In this course, you have to write in this convention.

ˆ i i

7


2-D Force Systems2-D Force Systems 3-D Force Systems3-D Force Systems

Moment

Couple

Resultants

Moment

Couple

Resultants

Vector (2D&3D)

Basic Concept

8

2V

Free Vectors: associated with “Magnitude” and “Direction”

: Direction

or V| |V

Magnitude:

V

or V

Vector :

2V

1V

1 2 V V V

1V

2V

V

1 2

1 2

= ( 1)

V V V

V V

V

W aV

W

| | | |W a V

parallelogram

( ) ( )

( )

( )

a bA ab A

a b A aA bA

a A B aA aB

Representation 1 2( )V V V

( ) ( )

A B B A

A B C A B C

1V 2V

V

triangle( 0)a M

| | | |M b V

( 0)b

2V

1V

+ˆ ˆA A e Ae

ˆ

(unit vector of )

A

Ae

A

A

9

Operation Addition #5

R A B B ACommutative

Vector

10

Operation Addition #6Vector

( ) ( )R A B C A B C A B CAssociative

11

Operation Scalar Multiplication #2

( ) ( )

( )

( )

a bA ab A

a b A aA bA

a A B aA aB

ˆ ˆA A e Ae

ˆ unit vector of e A

associative

distributive wrt scalar addition

distributive wrt vector addition

wrt = with respect to

13

Component Resolution of a Vector

A vector may be resolved into two components.

R A B

Vector

| | ? | | ?A B

14

Basic relations of Triangle

(C/6, law of cosine, sine)

2 2 2 2 cosc a b ab

sin sin sin

a b c

a

bc

Law of cosine

Law of sine

1V

2V

V

1

2

qb

15

1 ___V

2 ___V

(Law of sine)

(Law of sine)

Given V, and , find 1 2,V V

2 2 2 2 cosc a b ab

sin sin sin

a b c

a

bc

Law of cosine

Law of sine

Hint

b

16

special case: projection vectors are orthogonal to each other

Vector Component and Projection

a

: vector components of (along axis a and b)

1 2F F R

21, FF

R

: projections of (onto axis a and b)

aF

R

b

bF

R

aF

1F

2F

(generally)a bF F R

a

b

bF

R

aF

ba FF

, : orthogonal projections & vector components

bF

=

17

Rectangular Components• Most commonly used

22yx FFF

xF

yF

1tan | / |y xF F

yx FFF

ˆ ˆ x yF F i F j

x

y

q

)directions and (in

of components vector ,

yx

FFF yx

)directions and (in

of componentsscalar ,

yx

FFF yx

F

xF

yF

ˆ ˆ i j vector component = vector projection

i

jcosF sinF

-p b

18

x

y

F

Fx=? Fy=?

x

y

F

y

F

x

b

cosxF F

sinyF F

cosxF F

sinyF F

b x

yF

sin( )xF F

cos( )yF F

sin( )xF F

cos( )yF F

-b q

= sinF

( cos )

cos

F

F

minus

(b>90)

19

ˆ ˆ7.7 4.6i j

EXAMPLE 2-1EXAMPLE 2-1

Given the magnitude of the tensionin the cable, T = 9 kN, express T interms of unit vector i and j

j)sin9(i)cos9(T

j662.11

69i

662.11

109

kN ANS

T

i

j

x

y

2 210 6 11.662 mAB

Correct?3 S.F. ˆ ˆ7.72 4.63i j

20

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

arm AB

sin 45 90sin 45 63.64 ( )parP P N

cos 45 90cos 45 63.64 ( )perP P N

ANS

60

45

arm BC

sin 30 9sin 30 4.5 ( )perP P N

cos30 9cos30 7.794 ( )parP P N

ANS

15

P = 90 N

15

30

30

15

15

P = 90 N

parper

par per

Defining direction

21

2/2 Combine the two forces P and T, which act on the fixed structure at B, into a single equivalent force R

P=800 N (8cm)

1 6sin 60tan 40.9

3 6cos60

oo

o

T=600 N (6cm)

R

525 N (5.25cm) R

Graphics

Geometric

P

T

R

2 2 2 2 cosR P T PT 524 NR

sin sin

T R

48.6o

Vector Component (Algebraic)

ˆ800P i

ˆ ˆ600(cos sin )T i j

ˆ ˆ346 393R P T i j

2 2346 ( 393) 524 NR

1 393tan 48.6

346o

49o

Point of application is B

Correct?

22

Example Hibbeler Ex 2-1 #1

2 21 2 1 2 2 cos

213 NRF F F FF

Determine the magnitude and direction of the resultant force.

Geometric

2 150 212.55 sin sin sin sin115

39.101 39.1

RF F

Two forces is not acting at the same point.

23

2 21 2 1 22 cos

213 NRF F F FF

Geometric

2 150 212.55

sin sin sin sin11539.101 39.1

RF F

1

02

ˆ ˆ100cos15 100sin15 N

ˆ ˆ150sin10 150cos10 No

F i j

F i j

Vector Component (Algebraic)

1 2

1 1

ˆ ˆ122.64 173.60 N

173.60tan tan 54.761

122.64

R

Ry

Rx

F F F F i j

F

F

Good? (get full score?)

- more explanation

- mark answer

- 5S.F. Then 3S.F. ˆ ˆdirection of , ?i j

24

Good Answer Sheet

2 21 2 1 2

2 2

2

Using the law of cosine:

2 cos

100 150 2(100)(150)cos115

212.55 213 N #

Applying the law of sine:

150 212.55

sin sin sin sin11539.101 39.1 #

R

R

R

F F F FF

F

F F

a

O

Geometric

25

Point of Application

26


1

2

ˆ ˆ(600cos30 600sin30 ) N

ˆ ˆ( 400sin45 400cos 45 ) N

F i j

F i j

Vector

27


1 2

2 2

1 1

ˆ ˆ ˆ ˆ(600cos30 600sin30 ) ( 400sin45 400cos45 )

ˆ ˆ236.77 582.84 N

236.77 582.84

629.10 629 N #

tan ( ) tan (582.84 236.77)

67.891 67.9 #

R

R

R

Ry Rx

R

F F F

i j i j

F i

F

F F

F

j

Vector

28

1F

2F

F1y

F1x

F2y

F2x

Ry

Rx

y

xo

• Reference axis (very very important)– Many problems do not come with ref. axis.– Assignment based on convenience/experience

)ˆˆ()ˆˆ( 221121 jFiFjFiFFFR yxyx

jFFiFFjRiR yyxxyxˆ)(ˆ)(ˆˆ

2121

xxxx FFFR 21

yyyy FFFR 21

The calculations do not reveal the point of application of the resultant force.

R1. Graphically

• Vector summation (addition)– Three ways to be mastered

2. Geometrically

3. Vector component (algebraically)

2 2 21 2 1 22 cosR F F F F

2

sin sin( )

R F

Originally pass through O

In case where forces do not apply at the same point of application, you have to find it too!

29

Recommended Problem

2/9, H2-17, 2/12, 2/26, H2-28

31

Three Dimensional Coordinate System

ik

y

x

z

j Real-life Coordinate System is 3D.

Introduce rule for defining the 3rd axis - “right-hand rule”: x-y-z - for consistency in math calculation (cross vector)

How does 2D differs from 3D?

y

x

z

2D

32

ˆFn

- If you known the magnitude and all directional cosines, you can write force in the form of

Rectangular Components (3D)

y

x

z

Fk

j

i

xy 222

zyx FFFF

)cos( xx FF )cos( yy FF

)cos( zz FF

ˆˆ ˆ i j k

ˆˆ ˆˆ cos cos cosdef

F x y zn i j k

kFjFiFF zyxˆˆˆ

ˆˆ ˆ (cos cos cos ) x y zF F i j k

- cos(x), cos(y), cos(z) : “directional cosines” of

F

iFxˆ

jFyˆ

kFzˆ

- cos2(x)+cos2(y)+cos2(z) = 1

F

Fn is a unit vector in the direction of

z

(maybe +/-)

directional cosine Method

ˆFn

projection & component

33

Example Hibbeler Ex 2-8

2 2 2 cos cos cos 1 x y z

ˆ ˆ ˆcos cos cosx y zF F i F j F k

Find Cartesian components of F

2 2 2cos cos 60 cos 45 1

1cos

2

1 1cos ( ) 60° or 120

2

1cos

2

ˆ ˆ ˆ(200cos60 200cos60 200cos45 ) N

ˆ ˆ ˆ(100 100 141.42 ) N

ˆ ˆ ˆ(100 100 141 ) N #

i j k

i j k

F i j k

x

y

z

2 1cos

4

34

Given the cable tension T = 2 kN. Write the vector expression of T

)kcosjcosi(cosTT zyx x

y

z

A

B

xA

B

1.2cos x length of AB

2 2 21.2 0.5 (0.4 0.3) 1.3length of AB cos 0.92x

x

1) directional cosine method

directionl cosine = -0.92

Real directional cosine

35

zA

B

08.03.1

1.0

ABoflength

1.0cos z

yA

B

38.03.1

5.0

ABoflength

5.0cos y

x

y

z

A

B

y

x

y

z

A

B

z

Thus, ˆˆ ˆ2 ( 0.92 0.38 0.08 ) kNT i j k

ANS

36

cos xx

A

A cos y

y

A

A cos z

z

A

A

Directional Cosines by Graphics

cos2(x)+cos2(y)+cos2(z) = 1

37

- 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).

x

y

z

A (x1, y1, z1)

B (x2, y2, z2)F

ˆ FF F n

ABAB rrr

1 1 1ˆˆ ˆ Ar x i y j z k

kzjyixrBˆˆˆ

222

kzzjyyixxrABˆ)(ˆ)(ˆ)( 121212

212

212

212

121212

)()()(

ˆ)(ˆ)(ˆ)(

zzyyxx

kzzjyyixxFF

2 1 2 1 2 1

2 2 22 1 2 1 2 1

ˆˆ ˆ( ) ( ) ( )ˆ

( ) ( ) ( )AB

x x i y y j z z kn

x x y y z z

Two-Point Method

Positionvector

38Ans

x

y

z

B

A

0.50.4

0.31.2

2 1 2 1 2 1

2 2 2

2 1 2 1 2 1

ˆ ˆ ˆˆ

x x y y z zF F F

x x y y z z

AB

FAB

i j krF n

r

1 1 1, , 1.2,0,0.3A x y z

2 2 2, , 0,0.5,0.4B x y z

2 2 2

ˆ ˆ ˆ0.0 1.2 0.5 0.0 0.4 0.32

0.0 1.2 0.5 0.0 0.4 0.3

i j kF ˆ ˆ ˆ2 0.92 0.38 0.08

F i j k kN

2) 2-point construction

ABr

ˆ FF F n

39

Write vector expression of . Also determine angle x, y, z, of Twith respect to positive x, y and z axes

T

nTT

where n = unit vector from B to A

222 5)5.7(4

k5j5.7i4

k51.0j76.0i41.0 Thus

kN)k51.0j76.0i41.0(10T

ANS

cos 0.41x

76.0cos y

51.0cos z

66x

139y

59z

Consider: T as force of tension acting on the bar

40


Determine the magnitude and the coordinate direction angles of the resultant force acting on the ring

1 2

2 2 2

ˆ ˆ ˆ ˆ ˆ(60 80 ) (50 100 100 )

ˆ ˆ ˆ(50 40 180 ) lb

50 ( 40) 180 lb 191.05 191 lb #

R

R

F F F F j k i j k

i j k

F

Vector

41


50 40 180ˆ ˆ ˆ191.05 191.05 191.05

ˆ ˆ ˆ0.26171 0.20937 0.94216

RF R Ru F F

i j k

i j k

cos 0.26171 74.8

cos 0.20937 102

cos 0.94216 19.6 #

R

R

R

F x

F y

F z

u

u

u

Vector

42


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.

1 1 1 1 1 1 1

2 2 2 2

ˆ ˆ ˆcos cos cos

ˆ ˆ ˆ(300cos 45 300cos60 300cos120 ) N

ˆ ˆ ˆ(212.13 150 150 ) N

ˆ ˆ ˆ

x y z

x y z

F F i F j F k

i j k

i j k

F F i F j F k

ˆ(800 ) NRF j

Vector

43


1 2

2 2 2

2x 2 2

ˆ ˆ ˆ ˆ ˆ ˆ ˆ800 212.13 150 150

ˆ ˆ ˆ ˆ800 (212.13 ) (150 ) ( 150 )

R

x y z

y z

F F F

j i j k F i F j F k

j F i F j F k

Vector

44


2 2

2 2

2 2

dir. 212.13 700cos 108

dir. 650 700cos 21.8

dir. 0 700cos 77.6 #

x

y

z

2

2

2

2

2

2

dir. 0 212.13

dir. 800 150

dir. 0 150

212 N

650 N

150 N #

x

y

z

x

y

z

x F

y F

z F

F

F

F

Vector

45


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.

Force

46


2 2

ˆ ˆ ˆ ˆ ˆ(4 0) (0 0) (0 4) m 4 4 m

4 ( 4) 5.6569 m

100 N ( )

4 4ˆ ˆ100( )5.6568 5.6568

ˆ ˆ(70.711 70.711 ) N

AB

AB

ABAB

AB

AB

r i j k i k

r

rF

r

i k

F i k

Force

47


2 2 2

ˆ ˆ ˆ ˆ ˆ ˆ(4 0) (2 0) (0 4) m 4 2 4 m

4 2 ( 4) 6 m

120 N ( )

4 2 4ˆ ˆ ˆ120( )6 6 6

ˆ ˆ ˆ(80 40 80 ) N

AC

AC

ACAC

AC

AC

r i j k i j k

r

rF

r

i j k

F i j k

Force

48


2 2 2

ˆ ˆ(70.711 70.711 ) N

ˆ ˆ ˆ(80 40 80 ) N

ˆ ˆ ˆ(150.711 40 150.711 ) N

(150.711) (40) (150.711) N

216.86 217 N #

R AB AC

R

F F F

i k

i j k

i j k

F

Force

49

(b) Two Angles orienting the line of action of force are given (, )

y

x

z

F

xF

yF

zFk

j

i

xyF

Resolve into two components at a time

Fx = Fxy cos() = F cos() cos()

Fy = Fxy sin() = F cos() sin()

Fz = F sin()

Fxy = F cos()

Othorgonal projection Method

50

yx

z

F

Fxy

FxFy

Fz

65o50o

cos50 3.21 kNoxyF F

cos65 1.36 kNox xyF F

sin50 3.83 kNozF F

sin 65 2.91 kNoy xyF F

ˆ ˆ ˆ

ˆ ˆ ˆ 1.36 2.91 3.83 kN

x y zF F F

F i j k

i j k Ans

yx

z

TAC

B

15o

51

NTT oAB 77315cos

NTT oABy 1155.81cos

NTT oz 20715sin

NTT oABx 7645.81sin

ˆ ˆ ˆ

ˆ ˆ ˆ764 115 207 N

x y zT T T

T i j k

i j k Ans

1 10tan 81.5

1.5o

CAB

TZ

TAB

52

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.

53

• 3D Rectangular Component:

2/99 2/100 2/107 2/110

Recommended Problems

55

Operation Products

1. Dot Products

2. Cross Products

3. Mixed Triple Products

A B

A B

( )A B C

Vector

56

scalar product cosP Q PQ

P

Q

i

jk

ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( )x y z x y zA B A i A j A k B i B j B k

x x y y z zA B A B A B A B

(unit vector) ( three orthogonal vector )

ˆ ˆ ˆ ˆ 0

ˆ ˆˆ ˆ 0

ˆ ˆˆ ˆ 0

i j j i

i k k i

j k k j

ˆ ˆˆ ˆ ˆ ˆ 1i i j j k k

ˆ ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ

x x x y x z

y x y y y z

z x z y z z

A i B i A i B j A i B k

A j B i A j B j A j B k

A k B i A k B j A k B k

?P Q

:

ˆ ˆ ˆ P 2 3 4

ˆ ˆ ˆ 4 2 5

Example

i j k

Q i j k

?PQ

57

Application of Dot Operation

• Angle between two vectors

1cos

| || |

P Q

P Q• Component’izing Vector

line

U

e/ /U U U

/ /ˆˆ ˆ ( ) U e U e

ˆˆ ˆ= ( ) U U e U e

e

ˆ ˆ( )T e e

ˆ ˆ( )T T e e

which direction?

T

/ /U U

cosP Q PQ

/ /, ?U U

:

ˆ ˆ ˆ U 2 3 4

1 ˆ ˆ ˆˆ 3

Example

i j k

e i j k

58

yx

z

F

Fxy

FxFy

Fz

65o50o

cos50 3.21 kNoxyF F

cos65 1.36 kNox xyF F

sin50 3.83 kNozF F

sin 65 2.91 kNoy xyF F

ˆ ˆ ˆ

ˆ ˆ ˆ 1.36 2.91 3.83 kN

x y zF F F

F i j k

i j k Ans

yx

z

TAC

B

15o

59

ˆ ˆ ˆ

ˆ ˆ ˆ764 115 207 N

x y zT T T

T i j k

i j k

2 2 791.55 Nxz x zT T T Ans

1 10tan 81.5

1.5o

CAB

e

ˆ ˆ( )T e e

ˆ ˆ( )T T e e

ˆˆ764 207xzT i k

which direction??

TZ

TAB

60

A

B

A B

e

= C A B

right-hand rule(A then B)

“Cross Product” of Vectors

line which are perpendicular with both vectors

:

sin

magnitude

AB

21 ( )A B

AB

def

= ˆ ( | | | |sin ) C

A B e

61

• Commutative Law is not valid

Operation Cross Product

A B B A

A B B A

( ) ( ) ( ) ( )a A B aA B A aB A B a

( ) ( ) ( )A B C A B A C

Laws of Operations

• Associative wrt scalar multiplication

• Distributive wrt vector addition

? A B B A

? A B B A C A B

A

B

C

A

B

B A

( ) ?

( ) ?

( ) ?

B A

B A

B A

62

ˆ ˆˆ ˆ ˆ ˆ 0i i j j k k

ik

y

x

z

ˆˆ ˆi j k i

jk+

x-y-z complies with right-hand rule

j

ˆˆ ˆj k i ˆ ˆ ˆk i j

63

ˆˆ ˆ U U i U j U kzx y

ˆ ˆˆ ˆ ˆ ˆ( ) ( )U V U i U j U k V i V j V kz zx y x y

ˆˆ ˆ ˆ ˆ ˆ( ) ( ) ( )

ˆˆ ˆ ˆ ˆ ˆ( ) ( ) ( )

ˆ ˆ ˆ ˆˆ ˆ( ) ( ) ( )

U V i i U V i j U V i kx x x y x z

U V j i U V j j U V j ky x y y y z

U V k i U V k j U V k kz x z y z z

ˆ ˆ( ) ( )

ˆ( )

U V U V U V i U V U V jz z z zy y x xU V U V kx y y x

This term can be written in a determinant form

How to calculate cross product

ˆˆ ˆ V V i V j V kzx y

64

ˆˆ ˆ ˆ ˆ

x y z x y

x y z x y

i j k i j

U U U U U

V V V V V

ˆˆ ˆ

x y z

x y z

i j k

U V U U U

V V V

Cross Product

+ +

- - -

+

ˆˆ ˆ( ) ( ) ( )U V U V U V i U V U V j U V U V kz z z zy y x x x y y x

65

Why cross product?

• Mathematical Representation of Moments, Torque

• Perpendicular Direction• Area Calculation

A

x

y

zA

B

B

C

O

ˆOABCnB A

B A

Area = ?A B

66

Mixed Triple Product

ˆ ˆ ˆ

ˆ ˆ ˆ( ) ( )

ˆ ˆ ˆ( )

ˆ ˆ ˆ( ) ( ) ( )

x y z x y z

x y z

x y z

y z z y x z z x x y y z

i j k

U V W U i U j U k V V V

W W W

U i U j U k

V W V W i V W V W j V W V W k

( ) ( ) ( ) ( )x y z z y y x z z x z x y y zU V W U V W V W U V W V W U V W V W

( )x y z

x y z

x y z

U U U

U V W V V V

W W W

( )V W U

x y z x y z

x y z x y z

x y z x y z

U U U W W W

W W W U U U

V V V V V V

( ) ( )U V W W U V

67

Why mixed triple product?

• Mathematical Representation of Moments along the axis.

• Volume Calculation

,O FM

O

r

Fn

, ,o FM

B

C

ˆOABCn

A

: ( )Volume C B A

( ) C B A

Volume must always +

68

Mixed Triple Product Scalar

Operation Product Summary

Cross Product Vector

Dot Product Scalar

Vector

72

Homepage URLs

Statics official HP http://www.lecturer.eng.chula.ac.th/fmekmn/ (User: Prince Password: Caspian)

Session 1 HPhttp://pioneer.netserv.chula.ac.th/~lsawat/course/statics/

http://blackboard.it.chula.ac.th/ (after the end of registration period)

73


2-D Force Systems2-D Force Systems 3-D Force Systems3-D Force Systems

Moment

Couple

Resultants

Moment

Couple

Resultants

VectorBasic Concept

74

Force Definition

• 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)

F

F

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 is a vector quantity (why?)

Force can result from the action of electric fields, magnetic fields, and various other phenomena.

• Force is the action of one body on another. [Statics]

75

Force Representation

• Vector quantity– Magnitude– Direction– Point of application

10 NF

Force

Use different colours in diagrams• Body outline blue• Load red• Miscellaneous black

(dimension, angle, etc.)

76

Type of ForcesType of Forces

External force

Internal force

Reactive force

Applied force

Force

Force

Strain

Stress

Contact force

Body force

Force

Concentrated

Distributed

F

F

77

Cables & Springs

T

T

Cable in tension

F

s

F ks

spring constantk

Force

78

2/2 Combine the two forces P and T, which act on the fixed structure at B, into a single equivalent force R

P=800 N (8cm)

1 6sin 60tan 40.9

3 6cos60

oo

o

T=600 N (6cm)

R

525 N (5.25cm) R

Graphical

Geometric

P

T

R

2 2 2 2 cosR P T PT 524 NR

sin sin

T R

48.6o

Algebraic

ˆ800P i

ˆ ˆ600(cos sin )T i j

ˆ ˆ346 393R P T i j 2 2346 ( 393) 524 NR

1 393tan 48.6

346o

49o

Point of application is BCorrect?

79

Not OK. !

How to add sliding vectors (forces)?

AA

2F

1F

1F

2F

A 2F

1F

R

R

21 FFR

is applied at point AR

A 1F

R

Point of Application is wrong

Point of application

Principle ofTransmissibility

Still OK.

FF

F2

F1

F2

F1

FF

F2

F1

R1R2

RR


R

This graphical method can be used to find Line of action

Special case: Addition of Parallel Sliding Force

F F

F2

F1

R1

R2

R1

R2

line of action

The better and efficient way will be discussed later, when we learn the concept of “moment”, “couple”, and “resultant force”

81

1 1 cos60 200 NoxT T

1 1 sin 60 346 NoyT T

2 2 693 Nx yR R R

T

VD1

Ty

Tx

x

y

60

1ˆ ˆ200 346 NT i j

2ˆ400 NT i

1 2ˆ ˆ600 346 NR T T i j

Point of application,But no physical meaning

Move all forces to that concurrent point

1 o346tan =29.97

600

Application Point Ans

82

How to add sliding vectors (forces)?

A

A

2F

1F

1F

2F

R

21 FFR

is applied at point AR


There is better way to find the point of application(or line of action), but you have to learn the concept ofmoment and couples first.

84

moment axis

Moment is a vector

MomentIn 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”

(magnitude)summation

Direction

M Fd i i

i

M Fd

85

Moment Definition

xFx

y

z

yd

O

( )O zM• Moment is a vector quantity.

– Magnitude– Direction– Axis of Rotation

• 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

86

Moment about point A :

Mathematical Definition (3D)

r F

A

r

d

aM F

,A FM r F

-Magnitude:

-Point of application: point A

-Direction: right-hand rule

+

(Unit: newton-meters, N-m)

=M Fd

F

- 2D, need sign convention and be consistent; e.g. + for counter- clockwise and – for clockwise

2D

X

from A to point of application of the force

d

| || | sinr F

Fd

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

87

can be used with

more than

2 components

+

O

F

x

y d1

d2

xF

yF

Mo = -Fxd2+Fyd1

** OM r F

*1 2OM r F r F

1 2( )F F

Same?

Varignon’s Theorem (Principle of Moment)

1 2 1 2!

Yes r F r F r F F

r F

sum of moment (of each force) = moment of sum (of all force)

Useful with rectangular components

88

Principle of Transmissibility & Moment

A

O

O

r

d

a

F

M

X

YZ

FrM

M = Fr sin a = FdSliding force has the same moment

convenient

AX AYr F r F

- direction: same

- magnitude:

Principle of Transmissibility is based on the fact that“moving force along the line of action causes no effect in changing moment”

position vector: from A to any point on line of action of the force.

89

d

Sample 2/5 Calculate the magnitude of the moment about the base point O of 600N force in five different ways.

600N

400

4m

2mA

O

Solution I: 2D Scalar Approach

600N

400

4m

2m A

O

ood 40sin240cos4

35.4600 dFM O

2610 N-m Ans

m35.4

Solution II: 3D Vector Approach

ˆ 2610 N-m k

OM r F

ˆ ˆ ˆ ˆ2 4 600 cos40 sin 40o o i j i j

CW

x

y

CW or CCW? CWCorrect?

90

600N400

4m

2mA

O

F2

F1

Solution III: Varignon’s theorem

NF o 46040cos6001

NF o 38640sin6002

460(4) 386(2)OM

2610 N-m (CW)

600N

400

4m

2mA

O

F2

B

d1

F1

F2

F1C

d2

Solution IV: Transmissibility

od 40tan241 m68.5

1 1 2610 N-m (CW)OM F d

Solution V: Transmissibility

od 40cot422 m77.6

2 2 2610 N-m (CW)OM F d

+

91

EXAMPLE 2.8EXAMPLE 2.8

In raising the flagpole, the tension T in the cable must supply a Moment about O of 72 kN-m. Determine T.

15 m

30sin 60

25.981m

72 (12)sin 43.898Td T

8.65T kN

1 25.981tan 43.898

12 15

ANS

60o12 m

30 m

d

92


Determine the moment of the force about point O.

ˆ ˆ(400sin30 400cos30 ) N

ˆ ˆ(200 346.41 ) N

F i j

i j

Moment

Correct? ˆ ˆ , ?i j

93


ˆ ˆ ˆ

0.4 0.2 0 N-m

200 346.41 0

0.4 0.2 ˆ N-m200 346.41

ˆ ˆ98.564 98.6 N m

O

i j k

M r F

k

k k Ans

r

Moment

ˆ ˆ(400sin30 400cos30 ) N

ˆ ˆ(200 346.41 ) N

F i j

i j

400sin 30(0.2) 400cos 30(0.4)

40 138.56 98.564 N m

ˆ98.6 N m #

O

O

M

M k

Q

ˆ ˆ(0.4 0.2 ) mr i j

3D Vector Approach

Scalar Approach (Varignon’s theorem)

94

Couple- Couple is a summed moment produced by two force of equal magnitude but opposite in direction.

O

F

F

ad

+

M = F(a+d) – Fa = Fd

magnitude does not depend on distance a (point O),i.e. any point on the body has the same magnitude.

Couple Fd

- tendency to rotate the “whole” object.

- no effect on moving object as translation.

2D representations: (Couples)

C

couple is a free vector

C C

Effect of Pure Rotation

96

A

F

B

AF

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.

Procedure may be reversed to combine a force with a couple

Force-couple system

d

F

F

B

A

B F

C

Principle of transmissibility

No changes in net external effect

The direction and magnitude of Force can not be changed, only line of action (i.e. only change to other pararell line)

BAC r F

?C

97

Principle of Transmissibility is based on the fact thatmoving force along the line of action causes no effect in changing moment

FA

B

FA

B

CA

B F

CA

B FC r F

from new location (B)to old location (A)

r

F

A

BF

A

B

No Moment:Principle of Transmissibility

98

In the viewpoint of Mechanics,Result of force to these systemsare equal

Why using equivalent system?

AF

Force-couple system

B

A

B F

C

Principle of transmissibility

M

real (physical) system

equivalent system equivalent system

All force systems are equal.

99

A

B

D

M

M

M

,

,

,

A F

B F

D F

M

M

M

Understanding Force-Couple system

AF

B

A

BF

C

D D

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)

BAr F

DAr F

0

ABr F

C

DBr F

C

C

0

Equivalent System

100

Vector Diagram

F

F12m

70m

P

P

7012 PF

Ans

P703600

51.42 kNP

CCWCW

Be careful of the direction of moment

101

2/11 Replace the force F by an equivalent force-couple system at point O.

kN50F

250 .mm

++ cos 20 (0.1cos 25 0.25cos10 )F

sin 20 (0.1sin 25 0.25sin10 )o o oF

17.3 N-m

Ans

20

M50 kN

25

10

x

y 20

50 kN

oM

0.1m

0.25 m

Couple occurred when moving F to O= Moment of F about O

CCWˆ ˆ17.1 46.9 kNF i j

Correct?

102

o 90 90 90 270R F kN

)21(90)12(90)21(90Mo

mkN1080 ANS

Got the meaning?

Moving all 3 forces to point O

Sum of couples

Engine number 3 fails. Determine the force-couple system on the body about point o.

couples occuring when moving forces.

(direction: left)

(CW)R

M

sum of moments?

x

y

+

yF

103


Replace the current system by an equivalent resultant force and couple moment acting A.

100 400cos 45

382.84 N

600 400sin45

882.84 N

Rx x

Rx

Rx

Ry y

Ry

Ry

F F

F

F

F F

F

F

Resultant

104


2 2

2 2

1 1

( 382.84) ( 882.84)

962.27 962 N #

882.84tan tan

382.84

66.556 66.6 #

R Rx Ry

R

R

Ry

Rx

F F F

F

F

F

F

100(0) 600(0.4) 400sin45 (0.8) 400cos45 (0.3)

551.13 551 N m #

A

A

A

R A

R

R

M M

M

M

Q

Resultant

105

Ans

6020cos300 ob

0.213 mb

b

300N

20o

60 N-m

D

exactly cancelled

b cos20

107

• 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

2/6 Simplest Resultant

2F1F

1R

3F

R

-Add two at a time get line of action of R

-Add many do not get line of action of R

FFFFR

...321 , xx FR yy FR

22 )()( yx FFR )/(tan 1xy RR

xF1

1F

2F

3F

R

y

x

yF1

yF2

yF3

yR

xF2 xF3

xR

q


108

Easier way to get a resultant + its location

any forces + couples system

1) Pick a point (easy to find moment arms)

2F

1F

3F

d1

d2d3

1F

2F

3F

O

F1d1

2) Replace each force with a force at point O + a couple

F2d2

F3d3

R

O

4) Replace force-couple system with a single force

d=Mo/R

3) Add forces and moments

FR

O

Mo=(Fidi)

Mo=Rd

O

arbitrary

(forces + couples : same procedures)

any forces + couples system single-force + special single-couple (wrench)

2D

single-force system (no-couple)

or single-couple system

3D

where

0R

resultant

109

2/87 Determine the resultant and its line of action of the following three loads.

R = ( 2.4cos20 -1.5sin20 -3.6cos20 ) i +( -2.4sin20 -1.5cos20 +3.6cos20 ) j kN

M = -2.4*0.2cos20 -1.5*0.12cos20 -3.6*0.3cos20 kN-m

M

(force-couple system)

O R

Move 3 forces to point O,Sums their force and couples

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

+

why?

110

M = -1.635 kN-m

At point O (0,0)

MO

R

ˆ ˆ1.64 0.99 kNR i j

At point X (x,y)

NM

OR

R P

ˆ 1.635 k kNM

0 M N

ˆ ˆ0 ( )M xi yj R

ˆ ˆ0 ( )M xi yj R

ˆˆ ˆ ˆ ˆ( ) ( 1.64 0.99 ) 1.635xi yj i j k

0.99 1.64 1.635x y ( line of action )

couples cancelled

+

P

O (0,0)

P (x,y)

ˆ ˆ(0 ) (0 )r x i y j

Two equivalent systems Moment at any pointmust be the same on both system

Sys 1

Sys 2 ˆ ˆ( )xi yj R M

Pick Point O

Correct?

111

M = -1.635 kN-m

At point O (0,0)

MO

R

ˆ ˆ1.64 0.99 kNR i j

How to locate Point P

O (0,0)

ˆ 1.635 k kNM

+

d

2 2

| | | 1.635 |0.853 m

| | ( 1.64) ( 0.99)

Md

R

d

1 1 0.99tan tan 31.118

1.64y o

x

R

R

( sin , cos )d d

(0.441, 0.73)

tanP

P

y y

x x

( 0.73) 0.99

0.441 1.64

y

x

d2 2 2x y d

1( )tan

y x

Manually Canceling Couples

How to find line of action ?

or

P

PP

O

112

Equivalent System Definition

Two force-couple systems are equivalent

Equivalent System

system I system II

system I system II

( ) ( )

( ) ( )O O

R R

R R

F F

M M

MO

RO

R P

Sys 1 Sys 2

113

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.

++

jFiFR yx

i30sin250200200400R

j30cos250350

j567i925R

x

y

G

R

M

350(1.65) 250sin 30 (0.825)GM

mN690

114

For line of action of resultant

ˆˆ ˆ ˆ ˆ( ) (925 567 ) 690x i y j i j k

ˆ : 567 925 690k x y

At y = 0; x = +1.218 m. ANS

x

y

G

j567i925R

mN690

x

y

G

j567i925R

b

( ) | |y GR b M

567(b)=690

b=1.218 m

ˆ ˆ0 ( )M xi yj R

ˆ ˆ( )xi yj R M

Couple Cancellation

If you want to find only b (not line of action itself)

+ or - , you have to find out manually

Sys I

Two equivalent systems Moment at point Gmust be the same on both system

, ,G SysII G SysIM M

Sys II

Two equivalent systems(2D)

115

Determine the resultant (vector) and the point on x and y axes which must pass.

R

R

++x

y

G

jFiFR yx

ˆ25 20sin 30

ˆ ( 20cos30 30)

i

j

j3.47i15

25(5) 30(9)

(20cos30 )(9) (20sin 30 )(5)

oM

mkN351

116

x

y

O

j3.47i15R

mkN351

For line of action of resultant

oMRr

k351)j3.47i15()jyix(

351y15x3.47

If y = 0; x = 7.42 m.

ANS

x

y

O

x = 0; y = -23.4 m.

1 2force systems force, moment, couple and resultants

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