8/8/2019 Teaching Note-Chapter 4xx

1/23

8/8/2019 Teaching Note-Chapter 4xx

2/23

7/25/2008

2

The direction ofMO is determined using the right-handrule

Sense of rotation

Sense of rotation

Resultant Moment of a system of coplanar forces

OR M Fd + =

8/8/2019 Teaching Note-Chapter 4xx

3/23

7/25/2008

3

8/8/2019 Teaching Note-Chapter 4xx

4/23

7/25/2008

4

8/8/2019 Teaching Note-Chapter 4xx

5/23

7/25/2008

5

Cross Product

The cross product of two vectors A and B yields thevector C, written

= C A B

The magnitudeofC is,

sinC AB =

( )0 180

o o

The directionofC is,

( )sin CAB = =C A B u

8/8/2019 Teaching Note-Chapter 4xx

6/23

7/25/2008

6

Laws of Operation

A X B B X A

Rather,

A X B = - B X A

B X A = -C

Multiplication by a Scalar

a( A X B ) = (aA) X B = A X (aB) = ( A X B )a

Distributive Law

A X ( B + D ) = ( A X B ) +( A X D )

8/8/2019 Teaching Note-Chapter 4xx

7/23

7/25/2008

7

i x j = k i x k = -j i x i = 0j x k = i j x i = -k j x j = 0

k x i = j k x j = -i k x k = 0

( )( )(sin90 ) (1)(1)(1) 1i j = =o

Laws of Operations

Consider cross product of vector Aand B

AX B = (Axi + A

yj + A

zk) X (B

xi + B

yj + B

zk)

= AxB

x(i X i) + A

xBy

(i X j) + AxB

z(i X k)

+ AyB

x(j X i) + A

yBy

(j X j) + AyB

z(j X k)

+ AzB

x(k X i) +A

zBy

(k X j) +AzB

z(k X k)

= (AyB

zA

zBy

)i (AxB

z- A

zB

x)j + (A

xBy

AyB

x)k

8/8/2019 Teaching Note-Chapter 4xx

8/23

7/25/2008

8

x y z

x y z

A A A

B B B

=

i j k

A B

Therefore,

Moment of a force-vector formulation

The moment of a force F about point O(the moment axispassing through Oand perpendicular to the planecontaining Oand F)

O= M r F

r is a vector drawn from Oto any point lying on the line of action of F

8/8/2019 Teaching Note-Chapter 4xx

9/23

7/25/2008

9

Magnitude MO= rF sin

: angle between r and F

sin ( sin )O

M rF F r Fd = = =

Direction The right-hand rule

The moment created by F about Ois MO= rAF

Consequently,

O B C= = M r F r F

This is called theprinciple of transmissibility

8/8/2019 Teaching Note-Chapter 4xx

10/23

7/25/2008

10

Cartesian vector formulation

O x y z

x y z

r r r

F F F

= =

i j k

M r F

Resultant moment of a system of forces

( )OR

= M r F

8/8/2019 Teaching Note-Chapter 4xx

11/23

7/25/2008

11

8/8/2019 Teaching Note-Chapter 4xx

12/23

7/25/2008

12

8/8/2019 Teaching Note-Chapter 4xx

13/23

7/25/2008

13

Principle of moments

Varignons Theorem

The moment of a force about a point is equal to the sum

of the moments of the forces components about the point

1 2 1 2( )

O= + = = M r F r F r F + F r F

8/8/2019 Teaching Note-Chapter 4xx

14/23

7/25/2008

14

8/8/2019 Teaching Note-Chapter 4xx

15/23

7/25/2008

15

Moment of a force about a specified axis

8/8/2019 Teaching Note-Chapter 4xx

16/23

7/25/2008

16

(20N)(0.5m) 10N mOM= =

1) Scalar analysis

3cos (10N m) 6N m

5y OM M = = =

or,directly 0.3(20N m) 6N myM = =

In general,

If the line of action of a force F is perpendicular to anyspecified axis aa, the magnitude of the moment ofFabout the axis is

a a M Fd =

is the perpendicular or the shortest distance from the force line ofaction to the axisad

8/8/2019 Teaching Note-Chapter 4xx

17/23

7/25/2008

17

max( )

z M Fd =

z M Fd' =

Dot product ??

cosAB =A B

1 =i i 1 =j 1 =k k

0 =i j 0 =i k 0 =k j

8/8/2019 Teaching Note-Chapter 4xx

18/23

7/25/2008

18

Applications of dot product ??

1) The angle between two vectors or intersecting lines

1

cos AB

=

A B0 180

o o

cosAB =A B

2) The components of a vector parallel and perpendicular toa line

cosA A = = A u

cos (1) cos Au A = =A u

Therefore,

cos ( )A = = A u A u u

8/8/2019 Teaching Note-Chapter 4xx

19/23

7/25/2008

19

2) Vector analysis

{ }(0.3 0.4 ) ( 20 ) 8 6 N m

O A

= = + = + M r F i j k i j

( 8 6 ) 6N m y O a

M = = + = M u i j j

O= M r F

cosa O O a

M M = = M u

( )a a

M = r F u

( )a aM = u r F

or

The triple scalar product

8/8/2019 Teaching Note-Chapter 4xx

20/23

7/25/2008

20

( ) x y za a a a x y z

x y z

M u u u r r r

F F F

= + +

i j k

i j k

( )

x y za a a

a a x y z

x y z

u u u

M r r r

F F F

= =u r F

( )a aM = u r F

Express Maas Cartesian vector,

[ ( )]a a a a a

M M= = u u r F u

For a series of forces,

[ ( )] ( )a a a aM = u r F u = u r F

8/8/2019 Teaching Note-Chapter 4xx

21/23

7/25/2008

21

( )a a

M = u r F

8/8/2019 Teaching Note-Chapter 4xx

22/23

7/25/2008

22

8/8/2019 Teaching Note-Chapter 4xx

23/23

7/25/2008