Vectors and Moments

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Vectors and Moments First: Vectors As we said before, Vectors is a physical quantity which has both magnitude and direction

Shapes of vectors 1st shape: We can write vector as  a ,b 

2nd shape: we can write vector as a  aiˆ  b ˆj So for example: F   2 ,3  means F  2iˆ  3 ˆj Or H   -4 ,-1 means H  -4iˆ  ˆj -4 3

-1

2

This is called the position vector for any point ----------------------------------------------------------------------------------------------------------------------

The length of the vectors “ Norm of the vector” If A   x , y  Then

A  x 2  y 2 is the length of vector A and it is denoted by Norm A

Example : If A   3 ,-4  Then the norm  length  of

A 

 3

2

  -4   5 2

----------------------------------------------------------------------------------------------------------------------

The unit vectors " i " and " j " y The unit vector: A vector whose norm is 1

0 ,1

it is denoted by " i " vector and " j " vector where : i   1,0   

i  x y 

j   0 ,1  

2

2

j  x y  2

2

 1

2

0 

 0   1 2

2

  1  1 2

x'

j

O

x i  1,0 

y' ---------------------------------------------------------------------------------------------------------------------Very important note: If A   x1 , y1  And B   x2 , y2  AB  B  A   x2 , y2    x1 , y1    x2  x1 , y2  y1    x2  x1  i   y2  y1  j

Static – 3rd secondary

- 55 -

Chapter Three – Vectors and Moments

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In this chapter, we will discuss the two kinds of vector multiplication which are:

Scalar  Algebraic  Product

Vector Product 

First kind: Algebraic Product It is denoted by Some definitions (1) The angle between the two vectors: Let A and B be two non  Zero vectors , Let us represent them by OC and OD ,Then the C angle  COD where   180 o is called the smallest angle D Between these two vectors while the angle  COD of measure more than 180 o is called the biggest angle between them. b a  O fig  1 Very important note: We can get the small angle between the two vectors A and B by C D Representing them by drawing the angle between two vectors : Outwards Or Inwards the same point " see figures  1 and  2 " Small angle b a  o If  is the measure of the small angle the two vectors  0    180 O fig  2  o So the measure of the big angle between them  360  

But we prefer to use the small angle (2) The scalar product of two vectors : The scalar product of two vectors is a scalar quantity which is equal to the norm of the first vector multiplied by Cosine the angle between them .

B  A

A

B Cos 

where 0    180 o

---------------------------------------------------------------------------------------------------------------------Example (1) Find A B where A is a vector of magnitude 5 and direction 30o north of west and B is a vector of magnitude 8 towards south Answer A A  5 And B  8 , and the measure of the small angle between o

A and B is 120 So A Static – 3rd secondary

A

B  A

B Cos 

B  5  8 Cos120 o  -20 - 55 -

30 o

B Chapter Three – Vectors and Moments

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Example (2) If A and B are two vectors such that A  15 , B  12 and A

B  -135 , Then find

The measure of the angle between A and B Answer A B  A B Cos   -135  15  12 Cos  Cos   -0.75     138 o36' ---------------------------------------------------------------------------------------------------------------------Example (3) The vector A is of magnitude 65 towards west , B is a vector of magnitude 40 and direction 5 east of south and with angle  where Tan   , Then find A B 12 Answer o A B  A B Cos  90     A B  A B  - Sin  -5 A B  65  40   -1000  A  65 13 13 5 B  40



12 ---------------------------------------------------------------------------------------------------------------------Example (4) ABC is a right angled triangle at B where AB  6 cm , BC  8 cm , The vectors F1 , F2 and

F3 of magnitude 150 , 200 , 250 gm.wt act along BA , BC and CA , Find : 1 F1 F2  2  F2 F3  3  F1 F3 Answer 1  ABC  90 o :  F1 F2  150  200  Cos 90 o  0 Note When F1  F2  F1

 2  The angle between F2 and

 3  F1

F3  F1

Static – 3rd secondary

D

C

8 cm

F1 B

F2

  ACD is the angle between F2 and F3 F3  F2

F3



Between F2 and F3 , we have to draw the ray BC

 6 cm

10 cm

F3 must be inward Or outward

So to get the measure of the smaller angle

 F2

E A

F2  0 directly

 -8  F3 Cos  180 o     200  250   -Cos    50000    -40000  10  6 F3 Cos   150  250   22500 10

- 55 -

Chapter Three – Vectors and Moments

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The Algebraic Projection of a vector in the direction of another vector Definition The algebraic projection of vector B in the direction of the vector A is defined to be the scalar Quantity B Cos  , where B  B and  is the measure of the smaller angle between A and B D

D

B

B

 O

A

O

C

 C

A

The projection of B in the Direction of A is  B Cos  D

B

A

B

B

O

A

D

C

The projection of B in the direction of A is  B

C A

The projection of B in the direction

Cos 90 o  0

of A is  B

Cos 180 o  - B

Conclusion: The algebraic Projection of a vector in the direction of another vector:

The magnitude of the first vector × Cos the angle between the two vectors So

A

B  ABCos 





A B  A  BCos    A The algebraic projection of B in the direction of A ---------------------------------------------------------------------------------------------------------------------Example (5) A is a vector where A  30 and B is a vectors B  20 and the angle between A and

B is 75o , then find the algebraic projection of A in the direction of B and also the algebraic Projection of B in direction of A . Answer The algebraic projection of A in the direction of B  A Cos  30 Cos75o  7.768 The algebraic projection of B in the direction of A  B Cos  20 Cos75o  5.176 Static – 3rd secondary

- 55 -

Chapter Three – Vectors and Moments

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Example (6) A is a vector of magnitude 18 and direction 60 o north of east , B is another vector of magnitude 12 and direction 30o south of west , Find the algebraic projection of the two vectors A and B in the direction of the other . Answer A  18 The smaller angle between the two vectors   30 o  90 o  30 o  150 o  The algebraic projection of A in the direction of B : A Cos   18 Cos150 o  -9 3

30

60 o

o

B  12

The algebraic projection of B in the direction of A :

B Cos   12 Cos150 o  -6 3 ---------------------------------------------------------------------------------------------------------------------Example (7) ABCD is a rectangle in which AB  7.5 cm , BC  10 cm , Find :  a  The algebraic projection of the vector CB in the direction of AC

 b  The algebraic projection of the vector BD

in the direction of BA Answer A  a  Don' t forget that the direction of the two vectors must be either inwards Or outwards , so we must

D

7.5

extend AC and get the angle , so the algebraic



B

projection of CB in the direction of AC :

10

CB Cos   CB Cos 180 o     - CB Cos  where AC 

7.5 

2

  10   12.5 cm And Cos   2

 -4  So - CB Cos   10     -8  5 

 b The algebraic projection of BD

direction of BA : BD Cos   12.5 

Static – 3rd secondary

10 4  12.5 5 A

in the

D

12.5

7.5 7.5  7.5 12.5

- 55 -

C

 B

C 10

Chapter Three – Vectors and Moments

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Properties of the Scalar Product between two vectors Rule (1) A

B  A

B Cos   B

A Cos   B

Note (1): The units for measuring the scalar quantity A

A

B is equal to the unit of :

A multiplied by the unit of B

For example: If A is a unit vector , whose norm is the Newton and B is a displacement whose norm is meter Then the unit of the calr Product A

B is Newton meter

Note (2): The scalar product of two Non  Zero vectors is  ve Or -ve quantity Or zero according to the measure of the  between the two vectors

 a  If 

is an acute angle 0 < θ < 90 o  Then A

 b  If 

is an obtuse angle 90 o < θ < 180 o  Then A

 c  If 

is a right angle θ =

π  Then A 2

B is  ve B is  ve

B is Zero and if :

A B  0 , Then either A  0 Or B  0 Or A  B ----------------------------------------------------------------------------------------------------------------------

Rule (2) The scalar product of any vector into itself is equal to the square of its magnitude a  a 2 where a  a

So for any vector a

---------------------------------------------------------------------------------------------------------------------For any two vectors a and b and for any scalar m: Rule (3)

m a  Example: 3 a

b  a

2b  6 a

m b   m  a

b  a

6 b 6

a

b

 b



---------------------------------------------------------------------------------------------------------------------Rule (4) For any three vectors a , b and c , The distributive law holds:

a  b



c  a

 

c  b

c



----------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

- 56 -

Chapter Three – Vectors and Moments

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i i  j j 1 , i j  j i 0 ---------------------------------------------------------------------------------------------------------------------If a  a1 ˆi  a2 ˆj and b  b1ˆi  b2 ˆj Then a b  a1 b1  a2 b2 Rule (6)

Rule (5)

Proof : b   a1 ˆi  a2 ˆj 

b ˆi  b ˆj   a ˆi b ˆi  a ˆi b ˆj  a ˆj b ˆi  a ˆj  a b ˆi ˆi   a b ˆi ˆj   a b  ˆj ˆi   a b  ˆj ˆj   a b  a b Conclusion : a b   a ˆi  a ˆj   b ˆi  b ˆj   1  1  2  2 a

1 1

1

1

2

1

2

1

2

1

1

2

2

st

1

2

1

2

1

2

2

1 1

st

nd

2

b2 ˆj

2

nd

2

 Don' t forget : i i  j j  1 , i j  j i  0  ----------------------------------------------------------------------------------------------------------------------

Rule (7)

Perpendicular resolution of vector “ Geometric Component” a 



 

a Cos i 



a Sin j

a

a Sin

 a Cos

Example : if the magnitude of the vector a is 3 and   110 o , find the Component form. a  3 Cos 110 o i  3Sin 110 o j  -1.026 i  2.82 j ----------------------------------------------------------------------------------------------------------------------

Rule (8) a

The algebraic Component of F in the direction of a is equal to F

a

Example: Find the algebraic component of the vector F  4i  2 j in the direction of the vector AB where A   -2 ,3  And B  1,-1 Answer The algebraic Component of F in the direction of AB is F

AB AB



 



And

AB  B  A  i  j  - 2i  3 j  3i  4 j And

 F

AB AB



 4i  2 j

Static – 3rd secondary



AB 

3

2

  -4   5 2

 3i  4 j   12  8  0.8 5

5

- 56 -

Chapter Three – Vectors and Moments

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Mixed examples on the whole lesson The analysis being referred to two perpendicular directions and i and j are the unit vectors in these two directions  Example (1) A  3i  4 j and B  5i  12 j , Then find A B and the measure of the angle

If

included between the two vectors . Answer B   3iˆ  4 ˆj 

There is no angle given, then : A A 

3

2

 4   5 2

And

5

B 

2

 5iˆ  12 ˆj   15i

2

 48 j 2  15  48  63

  12   13

63 63 63     Cos -1  14 o15' 13  5 65 65 A B ------------------------------------------------------------------------------------------------------------------Example (2)

 A

B  A

B Cos   Cos  

A

2

B



If A  3 i  j and B  4 j , Then find the algebraic projection of each of the two vectors In the direction of each other .

Answer The algebraic projection of A in the direction of B : A

B 

A 

 3

 A

2



3 ˆi  ˆj

 0iˆ  4 ˆj  

  1  2

B  A

2

And

A Cos   So we have to get Cos

3 0   1  4   4

B 

B Cos   Cos  

0 

2

 4  4

A

B

A

B

2



4 1 1     Cos -1  60 o 24 2 2

1 1 2 1  The algebraic projection of B in the direction of A : B Cos   4   2 2 ------------------------------------------------------------------------------------------------------------------Example (3)  The algebraic projection of A in the direction of B :

A Cos   2 

If A  3i  j and B  2i  k j , Then find the value of k so that A  B Answer A B   3iˆ  ˆj   2iˆ  k ˆj   3  2   1 k   6  k And A  B    90 o  Cos 90 o  0   A

B  A

B Cos    A

B 0

 6 k 0   k 6 Static – 3rd secondary

- 56 -

Chapter Three – Vectors and Moments

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Example (4) A  2i  5 j and B  i  k j , Then find the value of k so that A // B Answer

If

A // B Where And

A A 

   0 o  And B   2i  5 j 

2

2

B  A

A

B Cos 

 i  k j   2  5k

  5   29 2

B 

And

 1

 k   k 2  1

2

2

 2  5k  29  k 2  1 Cos0 o   2  5k  29  k 2  1 So by squaring both:

 2  5k 

2

 29  k 2  1  4  20k  25k 2  29k 2  29

 4k 2  20k  25  0    2k  5   0  2k  5  k  2.5 ---------------------------------------------------------------------------------------------------------------------Example (5) 2

If A  k i  j and B  12i  5 j , Then find the value of k that makes the measure of the Angle between A and B equals 45o A Where And

B  A A A 

B Cos 45 o     1

B  k i  j 

k 

Answer

2

12i  5 j   12 k  5

  -1  k 2  1 2

And

B 

12 

2

  -5   13 2

1 169 2 2  So by squaring both:  12 k  5   k  1 2 2 169k 2  169 144  120k  25   288  240k  50  169k 2  169  0 2 2  119k  240k  119  0    17k  7 7k  17   0  12 k  5  13 k 2  1 

k

7 "agreed" And 17

k



-17 "refused" 7

b  b 2  4ac 2a -------------------------------------------------------------------------------------------------------------------

Note : We can use formula to find k :

Static – 3rd secondary

k

- 56 -

Chapter Three – Vectors and Moments

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Example (6)

ABCD is a trapezium in which AD // BC , m   A   m   B  

 2

, AD  16 cm , BC  21 cm

And AB  12 cm , Find :

1The algebraic projection of

AD in the direction of CB

 2 The algebraic projection of

BD in the direction of CB

 3 The algebraic projection of

CD in the direction of AB

Answer 1 The measure of the angle between AD and CB is equal 180 o Thus the algebraic projection of AD in the direction of CB is



AD Cos  The angle between AD and CB

A



16 cm

D

12 cm

 16Cos180 o  -16

B

 2  Don' t forget that the direction of the two vectors

A

must be either inwards Or outwards , so we must extend CB and get the angle  , so the algebraic

C

21 cm 16 cm

12 cm

D

12 cm

projection of BD in the direction of CB : BD Cos   BD Cos 180 o     - BD Cos  where BD 

16 

2

  12   20 cm And Cos   2



B

16 4  20 5

 -4  So - BD Cos   20     -16  5 

 3 We must extend CD

A

, so that angle  EDF is

5 cm

16 cm

16 cm

D  12 cm

12 cm

C

13 cm

the angle between CD And AB  The algebraic projection of CD in the direction of AB :

B

 12  CD Cos  180 o     - CD Cos   -13     -12 cm  13 

Static – 3rd secondary

- 56 -

16 cm

5 cm

C

Chapter Three – Vectors and Moments

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Example (7) ABCD is a square , the length of whose side is 8 cm , find :

 2 BC 



1   c  BC -2 DB  AD  4  Answer CD  AB CD Cos   8  8 Cos180 o  64   -1  -64

a

AB

a

AB

b

 2 BC 

CD

b

1 1   AD  2      BC 4 4  

AD 



d 

A

BC

 -2 DB   -2 BC

8 cm

D

8 cm

C D

1 BC AD Cos  2 1   8  8 Cos0 o  32 2 B A

c

 AC   BD 

8 2

DB Cos   -2  8  8 2 Cos135 o  128 B

C

 A

d 

 AC   BD  

AC

8 cm

D

BD Cos   8  8  Cos 90 o  0

C B -------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

- 55 -

Chapter Three – Vectors and Moments

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Example (8) ABCD is a rectangle which AB  6 cm , BC  8 cm , find :

a

 -2 AB 

AC 

a

b

8

 -2 AB  BC

2

c

AC

BC

5   DB  2  Answer

  6   10 cm 2

AC

 -2 AB

AC 

6 

2

AC Cos 

So in  AMB :

 AC   BD  

10 cm

6 cm

B A

8 cm



B

2

8 cm



6 cm



M

2

B

C

D

A

2

  5   6  7  2  5  5  25

C D

10 cm

6 cm

  8   10 cm

1 AC  5 cm  Median from right angle  2 Also BM  AM  5 cm  Properties of a rectangle  2

D



 AM 

5 Cos  

8 cm

A

6  -2  6  10   -72 10 5  5 DB  DB     BC 2  2 5  BC BD Cos  180 o    2 5  - BC BD Cos  2 5 -8   8  10   -160 2 10

 c  In  AMB:

 AC   BD 

d 

8 cm

C

7  28 25 -------------------------------------------------------------------------------------------------------------------



Static – 3rd secondary

AC

BD Cos   10  10 

- 55 -

Chapter Three – Vectors and Moments

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Example (9) ABCD is a parallelogram in which m   BAC   90o , BC  2AB  8 cm , find:

a

CA

8  4

AC 

 a  CA b

c

b 

CB 2

CB





1   c  AB  DC  2  Answer



3 AD  3 AB

 -32 3 

 MB 



 7 BD

AM 

2 3 

2



B



4 3

B

C

8 cm

A



8 cm

D

4 3

B

C

8 cm

A

B

D

8 cm

4 cm

4 cm

C

8 cm

A

4  -16 3 8

 BD  4 7 BD

4 3

4 cm

2

D



AD Cos  90 o   

  4   2 7 cm

 7 BD   AC  8 cm

4 cm

1 AC  2 3 cm 2

  AC   7

d  A

 3  4  8   - Sin 

 d  In  AMB:

3 AD

 4 3 cm

4 3  CA CB Cos   4 3  8   48 8 1  1 DC  DC     AD 2  6  1    AD DC Cos  180 o    6  1  - AD DC Cos  6 1 4 8  - 84  6 8 3

1   AD  3 

AB

2

1   AD  3 

8 cm



D

2 3

8 cm

C

AC Cos 

2 3  336 2 7 ------------------------------------------------------------------------------------------------------------------7 4 7 4 3 

Static – 3rd secondary

- 55 -

Chapter Three – Vectors and Moments

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Example (10) If ABCD is a square , then prove that : AB  BC





BD  Zero

Answer

 AB  BC   AB

BD  AB

BD Cos135o  BC

BD  BC

8 cm

A

BD

D

BD Cos 45 o

 -1   1   88 2  88 2   Zero  2  2

C

B

------------------------------------------------------------------------------------------------------------------Example (11) Prove that for any two vectors a and b : a  b a  b  a 2  b 2 where a  a



 



and b  b

a  b  a  b   a

a a

Answer b  b a b

b  a 2  b2

------------------------------------------------------------------------------------------------------------------Example (12) Let a  3i  2 j , b  -i  4 j And c  -2 j

  



1 Then find with respect to i and j the vector: - c   2 b -3a  c   5 Answer Let a  3i  2 j , b  -i  4 j And c  -2 j 1 Then find with respect to i and j the vector: - c   2 b -3a  c  5

  

 2 b   -3a   -6 b a   -6 3  -1  2  4   -30 1 1  - c   2 b   -3a   c  - c   -30  c  - c  6 c  5 5



 -7 c  -7  -2j   14 j

-------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

- 55 -

Chapter Three – Vectors and Moments

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Example (13) ABC is a triangle in which A   3,2  , B   -4 ,3  and C   -3,-6  , Then find m   BAC  Answer AB

AC  AB

 Cos   BAC  

AC Cos   BAC 

AB

AC

AB

AC

B

AB  B  A   -4 ,3    3 ,2    -7 ,1  -7 i  j  AB 

 -7 

2

A

  1  50 2

AC  C  A   -3 ,-6    3 ,2    -6 ,-8   -6 i  8 j  AC  AB

C

 -6    -8   10 2

2

AC   -7 i  j 

 -6 i  8 j   42  8  34

34      61o16' 10 50 ------------------------------------------------------------------------------------------------------------------Example (14) If A   1,1 , B   2 ,3  and C   5 ,-1 , Then prove that AB  AC , Then find the algebraic   Cos   BAC  

Component of F  7i  4 j Newton in the two directions AB and AC .

Answer AB  B  A   2 ,3    1,1  1,2   i  2 j And

AB 

And AC  C  A   5 ,-1   1,1   4 ,-2   4i  2 j And AB

AC   i  2 j 

 4i  2 j   4  4  0

 1

2

 2  5

AC 

2

4 

2

 2  2 5 2

    90 o   AB  AC

The algebraic component of F in the direction of AB is equal to :  F

AB AB



 7i  4 j



 i  2 j  3 5

5 Newton

And The algebraic component of F in the direction of AC is equal to :  F

AC



 7i  4 j



 4i  2 j   2

5 Newton 2 5 AC -------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

- 55 -

Chapter Three – Vectors and Moments

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Example (15) If A   4 ,1 , B  7 ,5  and C   -5 ,-4  , Find Cos  where  is the angle between the two Vectors AB and BC , Then determine the algebraic Component of each of the two vectors in the direction of the other .

Answer AB

BC  AB

BC Cos  where :

AB  B  A  7 ,5    4 ,1   3 ,4 

AB 

And

And BC  C  B   -5 ,-4   7 ,5    -12 ,-9   Cos  

AB

BC

AB

BC



And

3

2

 4   5

BC 

2

 -12 

2

  -9   15 2

 3 ,4   -12 ,-9   -36  36  -24 5  15

75

25

The algebraic component of AB in the direction of BC is equal to :  AB

BC BC

  3 ,4 

 -12 ,-9   3  -12   4  -9   -4.8 15

15

The algebraic component of BC in the direction of AB is equal to :  BC

AB

  -12 ,-9 

 3 ,4   3  -12   4  -9   -14.4

5 5 AB -------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

- 56 -

Chapter Three – Vectors and Moments

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Second kind: Vector Product It is denoted by  Rule  A A

And B  B



A B  A

B



Sin kˆ where :

and  is the measure of the small angle between A and B and kˆ is

the unit vector perpendicular to the plane containing the two vectors A and B . The direction of the unit vector kˆ is determined according to the right hand rule which states that: If the curved fingers of the right hand indicates the rotation of the vector A towards vector B Throught the smaller angle , then the thumb will indicate the direction of kˆ as shown in figures

Note that : the vector product A  B is a vector  And not scalar

 and its direction is

Perpendicular to the plane containing the vectors A and B in the direction which is determined according to the right hand rule and the unit vector in the direction of A  B A B is the vector kˆ which is equal to kˆ  AB Sin

Static – 3rd secondary

- 56 -

Chapter Three – Vectors and Moments

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Properties of the Vector Product between two vectors Rule (1)



 

A  B - B  A  A

Note : For the two vectors A  B





B Sin  kˆ

 and  B  A  have the same magnitude But have different direction

----------------------------------------------------------------------------------------------------------------------

Rule (2) If A // B , Then A  B  0 for  , In this case is equal to Zero or 180 o and hence Sin 0o  Sin 180o  0 Note : If A  B  0 , Then either A  0 Or B  0 Or A // B ----------------------------------------------------------------------------------------------------------------------

Rule (3)

The vector product of any vector into itself is equal zero as Sin0 o  0 So for any vector : a  a  zero Proof:

a  a 



a



a Sin 0 o kˆ  Zero

---------------------------------------------------------------------------------------------------------------------Rule (4) The Right hand system

In the given figure: OX and OY are two perpendicular directions ˆi and ˆj are two unit vectors in these directions respectively: So ˆi  ˆj  ˆi

ˆj Sin90 o  1  1  Sin90 o  kˆ  kˆ

Where kˆ is the unit vector perpendicular to plane OXY containing ˆi and ˆj And in the direction of the right hand rule. From this: we can say : ˆi  ˆj  kˆ

kˆ  ˆi  ˆj

ˆj  kˆ  ˆi

And when they are in the anticlock wise : From this: we can say : ˆj  ˆi  -kˆ

ˆi  kˆ  - ˆj

kˆ  ˆj  -iˆ

But ˆi  ˆi  ˆj  ˆj  kˆ  kˆ  0 ‫اى ح اج ة ش ب ه ب ع ض‬

Static – 3rd secondary

- 56 -

Chapter Three – Vectors and Moments

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Rule (5)

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For any two vectors a and b and for any scalar m:

Example:

m a   b  a  m b   m  a  b  3 a  2b  6  a  b   a  6 b  6  a  b 

---------------------------------------------------------------------------------------------------------------------For any three vectors a , b and c , The distributive law holds: Rule (6)

 a  b 



 

c  a  c  b  c



---------------------------------------------------------------------------------------------------------------------Rule (7) If a  a ˆi  a ˆj and b  b ˆi  b ˆj Then a  b   a b  a b  kˆ 1

2

1

2

1

2

2

1

Or a  b   1st  2 nd  2 nd  1st  kˆ

Proof : a  b   a1 ˆi  a2 ˆj    b1ˆi  b2 ˆj   a1 ˆi  b1ˆi  a1 ˆi  b2 ˆj  a2 ˆj  b1ˆi  a2 ˆj  b2 ˆj  a1 b1 ˆi  ˆi   a1 b2 ˆi  ˆj   a2 b1  ˆj  ˆi   a2 b2  ˆj  ˆj 

 

 0  a1 b2 kˆ  a2 b1 - kˆ  0   a1 b2  a2 b1  kˆ  Don' t forget : i  i  j  j  1 , i  j  kˆ & j  i  - kˆ    Where ˆi , ˆj , kˆ are the right system of unit vectors





----------------------------------------------------------------------------------------------------------------------

Important Remark

N

D

B We know that A  B  AB Sin 

O



If OC represents the vector A And if OD represents the vector B And

AB Sin  OC

OD

A

C

Sin  The surface area of the parallelogram OCND

 Twice the surface area of  OCD ----------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

- 56 -

Chapter Three – Vectors and Moments

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Comparison between Scalar and Vector products

Static – 3rd secondary

- 56 -

Chapter Three – Vectors and Moments

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Let A  6 ˆi  15 ˆj ,

B  5iˆ  3 ˆj

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,

Example (1) C  -iˆ  ˆj , find :

b   A  B  C  A  B  C c B   A  C   d  C  B   A Answer Let ˆi , ˆj , kˆ  be a right system of unit vectors  a   A  B   6 ˆi  15 ˆj    5iˆ  3 ˆj   11iˆ  12 ˆj   A  B   C   11iˆ  12 ˆj    -iˆ  ˆj   1  2  2  1  kˆ   -11  12  kˆ  kˆ  b   A  B   6 ˆi  15 ˆj    5iˆ  3 ˆj    -18  75  kˆ  -93kˆ  A  B  C  -93kˆ  -iˆ  ˆj    -93kˆ -iˆ    -93kˆ ˆj  a

st

 93 kˆ

nd

nd

st

ˆi Cos 90 o  93 kˆ

ˆj Cos 90 o  0  0  0

 A  C   6 ˆi  15 ˆj    -iˆ  ˆj    -6  15  kˆ  9kˆ  B   A  C    5iˆ  3 ˆj   9kˆ  45  ˆi  kˆ   27  ˆj  kˆ   45  - ˆj   27  ˆi   -27iˆ  45 ˆj  d  C  B   -iˆ  ˆj    5iˆ  3 ˆj   6 ˆi  15 ˆj    3  5  kˆ  6 ˆi  15 ˆj   48  kˆ ˆi   120  kˆ  ˆj  c

 48 ˆj  120iˆ ------------------------------------------------------------------------------------------------------------------Example (2) ˆ ˆ ˆ ˆ If A  i  3 j , B  mi  2 j , A  B  -14kˆ , Then find the value of m .

Answer ˆ A  B  -14 k   ˆi  3 ˆj    miˆ  2 ˆj   -14 kˆ    -2  3m  kˆ  -14 kˆ  -2  3m  -14  m  4 ------------------------------------------------------------------------------------------------------------------Example (3) If A  -2iˆ  5 ˆj and B  3iˆ  4 ˆj , find the vector C in the plane A and B such that: A  C  -14kˆ And C  B  6kˆ Answer Let C  miˆ  n ˆj And A  C  -19 kˆ   -2iˆ  5 ˆj    miˆ  n ˆj   kˆ  -14 kˆ   -2n  5m  kˆ  -14 kˆ   -2n  5m  -14    1 And

C  B  6 kˆ   miˆ  n ˆj    3iˆ  4 ˆj   kˆ  6 kˆ   4m  3n  6     2 

From  1 and  2  and by simultanous : m  3 Static – 3rd secondary

And n  2 - 55 -

  C  3iˆ  2 ˆj Chapter Three – Vectors and Moments

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Example (4) If A   2 ,3  and B  1,-1 , Then find the vector C such that: A C  21 & C  B  -8 kˆ Answer Let C  miˆ  n ˆj And A C  21    -2iˆ  5 ˆj   miˆ  n ˆj   21  -2m  5n  21     1 And

C  B  -8 kˆ   miˆ  n ˆj    ˆi  ˆj   kˆ  -8 kˆ   - m  n  -8     2 

From  1 and  2  and by simultanous : m  3 And n  5   C  3iˆ  5 ˆj ------------------------------------------------------------------------------------------------------------------Example (5) If A   2 ,1 , B   -1,5  and C   -4 ,-1 Then prove that: AB  AC  BA  CB  BC  CA Answer AB  B  A   -1,5    2 ,1   -3 ,4   -3iˆ  4 ˆj  BA  - AB  3iˆ  4 ˆj

AC  C  A   -4 ,-1   2 ,1   -6 ,-2   -6 ˆi  2 ˆj  CA  - AC  6 ˆi  2 ˆj BC  C  B   -4 ,-1   -1,5    -3 ,-6   -3iˆ  6 ˆj  CB  - BC  3iˆ  6 ˆj AB  AC   -3iˆ  4 ˆj    -6 ˆi  2 ˆj   6  24  kˆ  30kˆ BA  CB   3iˆ  4 ˆj  3iˆ  6 ˆj   18  12   30kˆ BC  CA   -3iˆ  6 ˆj  6 ˆi  2 ˆj    -6  36   30kˆ ------------------------------------------------------------------------------------------------------------------Example (6) If A   3 ,-2  , B   -1,1 , C  1,5  and D  6 ,0  , Then find:

 a  AB

 CD

 b  AC



 BA  DC



 c  BC



 AB



AC BD

Answer A  3iˆ  2 ˆj , B  -iˆ  ˆj , C  ˆi  5 ˆj And D  6iˆ  a  AB  B  A  -4iˆ  3 ˆj And CD  D  C  5iˆ  5 ˆj  AB  CD   20  15  kˆ  5kˆ

 b  AC

 C  A  -2iˆ  7 ˆj & BA  A  B  4iˆ  3 ˆj

BA  DC  -iˆ  2 ˆj 

 c  BC

 C  B  2iˆ  4 ˆj

 BC   AB BC  AB



& DC  C  D  -5iˆ  5 ˆj



AC  BA  DC   -2iˆ  7 ˆj    -iˆ  2 ˆj    -4  7   3kˆ And

AB

AC   -4iˆ  3 ˆj  -2iˆ  7 ˆj   8  21  29

  58iˆ  116 ˆj  AC  BD   58iˆ  116 ˆj 7 ˆi  ˆj    -58  812  kˆ  -870 kˆ

AC  29  2iˆ  4 ˆj

------------------------------------------------------------------------------------------------------------------Static – 3rd secondary

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Let A  i  j

a A

C



,

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B  -i  2 j and

B

b  A



 B C

Example (7) C  4 j , Then find:



 a  a  c   i  j    4 j    4  0  kˆ  4kˆ



 a c

b



b  4kˆ

 -i  2 j   -4  kˆ

b  c   -i  2 j    4 j   -i  2 j

c a

b  i  j 

d  b

c   -i  2 j 



c  B

 A



B C

d  B





C A  A



B C

Answer

  j   -4 0   8 0   0   a   b  c    i  j    -iˆ  2 j    - 2  1 kˆ  -3kˆ ˆi  8 kˆ

 -i  2 j   -1  2  -3  b   -3c   -3  b  c   -3  -i  2 j    4 j   -3  -4  0  kˆ  12kˆ



 

4 j   0  8  8



   



a



b  -3



 b c a  a b c  8 a  -3 c  -24 a  c  -24  4kˆ  -96kˆ ------------------------------------------------------------------------------------------------------------------Example (8) If A  3i  4 j , B  4i  5 j , Then find the vector C where C // A , C  B  62kˆ Answer Let C  ai  b j , So C // A   C  A 0   ai  b j    3i  4 j    -4 a  3b  kˆ   - kˆ  4 a  3b   0   4 a  3b  0    1 C  B  62kˆ   ai  b j    4i  5 j   62kˆ 

 5a  4b  kˆ  62kˆ

 5a  4b  62     2  , Then from  1 and  2  : b  -8 & c  6  C  6 i  8 j ------------------------------------------------------------------------------------------------------------------Example (9) If A   2 ,3  And B   m ,2  , find the value of m in each of the following cases:  a  A  B  -5 kˆ b A  B

 c  The measure of the angle between A a b

and B equals 45o Answer  2i  3 j    mi  2 j   -5kˆ    4  3m  kˆ  -5kˆ   4  3m  -5   m  3

 2i  3 j   mi  2 j   0  2m  6  0   m  -3 A B  2i  3 j   mi  2 j    Cos 45o  1  2m  6   c  Cos  2 2 2 2 2 A B 13 m 2  4  2   3  m   2  2 By squaring both sides :  13  m2  4   2  2m  6   13m2  52  2  4m 2 24m  36  A B



 5m 2  48 m  20  0 Static – 3rd secondary

  m  10  5m  2   0   m  10 - 55 -

Or m 

-2 5

Chapter Three – Vectors and Moments

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Example (10) If A  0 ,-2  , B  6 ,-4  , C  7 ,4  and D   -2 ,4  are vertices of a quadrilateral , then find AC  BD and calculate the area of ABCD Answer A  -2 ˆj , B  6 ˆi  4 ˆj , C  7iˆ  4 ˆj And D  -2iˆ  4 ˆj AC  C  A  7 ˆi  6 ˆj And BD  D  B  -8iˆ  8 ˆj

 AC  BD  7 ˆi  6 ˆj    -8iˆ  8 ˆj    56  48  kˆ  104 kˆ Area of ABCD  Area of  ABC  Area of  ACD Where Area of  ABC 

1  AB  AC  Sin 2

But we know that : AB  AC  Sin  AB  BC 1 AB  BC 2 AB  B  A  6 ˆi  2 ˆj And BC  C  B  ˆi  8 ˆj  AB  BC   6 ˆi  2 ˆj   ˆi  8 ˆj    48  2  kˆ  50 kˆ  Area of  ABC 

1 1 AB  BC   50  25     1 2 2 1 1 Also Area of  ACD   AC  AD  Sin  AC  AD 2 2 AC  C  A  7 ˆi  6 ˆj And AD  D  A  -3iˆ  6 ˆj  AC  AD  7 ˆi  6 ˆj    -3iˆ  6 ˆj    42  18  kˆ  60 kˆ  Area of  ABC 

1 1 AC  AD   60  30     2  2 2 Area of ABCD  Area of  ABC  Area of  ACD  25  30  55 --------------------------------------------------------------------------------------------------------------------- Area of  ACD 

Static – 3rd secondary

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Chapter Three – Vectors and Moments

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Second :

Moments (Torque)

Vector Moment

Algebraic Moment

Introduction : One person is sitting at the end of a see - saw Can balance the weight of the two people sitting closer in The turning effect produced by a force depends on its distance from the pivot point. It is easier to close the door if you push at the edge , this is because you are applying force further away from the pivot point The hinge  you would need a greater force closer to the hinge to produce the same turning effect. The moment of a force measures turning effect of the force on the body on which it is acting the moment of a force depends on the magnitude of the force and its distance from the axis of rotation . -------------------------------------------------------------------------------------------------------------------

1st : Vector Moment of a force about a point The moment of a force F about O  denoted by M o is defined to be a vector quantity r  F 

F

Rule: M o  r  F  OA  F

O

Where r is the position vector for any point  A  Say  on the

A

r

line of action of the force F with respect to O . Notes :

1 The above definition makes M o



is equal to r  F and not F  r because : r  F  - r  F

 2  OA  A  O  3  The magnitude of a moment :



F

d

Rule: M o  F d

O

A

r Where d is the perpendicular distance from the point to the line of action of F Static – 3rd secondary

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Chapter Three – Vectors and Moments

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Example (1)

Let ˆi and ˆj are two unit vectors in the two perpendicular directions OX and OY respectively A force F  4i  3 j acts at the point A   2 ,3  , Calculate the moment of this force about : The point B   -1,2  , Also find the magnitude of this moment .

Answer ˆ Let k be a unit vector perpendicular to both ˆi and ˆj such that ˆi , ˆj ,kˆ form a right hand





system of unit vectors . M B  r  F  BA  F Where BA  A  B   2 ,3    -1,2    3 ,1  3iˆ  ˆj  M  r  F   3iˆ  ˆj    4iˆ  3 ˆj    -9  4  kˆ  -13kˆ B

Then the magnitude of

M B  13

F  4iˆ  3 ˆj

d B  -1,2 

r

A  2 ,3 

Note : To find the magnitude of any moment : Remove kˆ and the -ve Sign ------------------------------------------------------------------------------------------------------------------Example (2) A force F  3i  4 j acts at the point C   1,2  ,Then find :

1The Moment of this force about point A   5 ,4   2 The length of the perpendicular from A on the line of action of this force . Answer M A  r  F  AC  F Where AC  C  A   1,2    5 ,4    -4 ,-2   -4iˆ  2 ˆj  M  r  F   -4iˆ  2 ˆj    3iˆ  4 ˆj    -16  6  kˆ  -10 kˆ

F  3iˆ  4 ˆj

A

Then the magnitude of

M A  10

And the magnitude of F 

 3

d 2

 4  5 2

A  5 ,4 

r

C 1,2 

10 2 5 -------------------------------------------------------------------------------------------------------------------

So

M  F d   10  5d   d 

Static – 3rd secondary

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Chapter Three – Vectors and Moments

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Example (3)

A force F  -2i  j acts at the point B  7 ,2  ,Then find : The length of the perpendicular from A  3 ,-1 on the line of action of this force .

Answer M A  r  F  AB  F Where AB  B  A   7 ,2    3 ,-1   4 ,3   4iˆ  3 ˆj  M  r  F   4iˆ  3 ˆj    -2iˆ  ˆj    4  6  kˆ  10 kˆ

F  -2iˆ  ˆj

A

Then the magnitude of

M A  10

And the magnitude of F 

 -2 

d 2

  1  5 2

A  3,-1

B 7 ,2 

r

10 2 5 5 ------------------------------------------------------------------------------------------------------------------Example (4) The two forces F1  2i  j and F2  mi  2 j acts at A   3 ,-1 and B   -1,3  respectively M  F d   10  5d   d 

So

Find m such that the sum of the moments of these two forces vanishes at O  0 ,0 

Answer M OA  r1  F1  OA  F1 Where OA  A  O   3 ,-1   0 ,0    3 ,-1  3iˆ  ˆj  M  r  F   3iˆ  ˆj    2iˆ  ˆj    3  2  kˆ  5 kˆ OA

And

1

1

M OB  r2  F2  OB  F2

F2  miˆ  2 ˆj

B  -1,3 

r2

F1  2iˆ  ˆj

r1

A  3,-1

O 0 ,0 

Where OB  B  O   -1,3   0 ,0    -1,3   -iˆ  3 ˆj

 M OB  r2  F2   -iˆ  3 ˆj    miˆ  2 ˆj    2  3m  kˆ

The sum of the moments vanishes :  M OA  M OB  0



 5 kˆ   2  3m  kˆ  0  5kˆ  2kˆ  3m kˆ  0  7kˆ  3m kˆ  0  by kˆ



7 3 ------------------------------------------------------------------------------------------------------------------ 7  3m  0   m 

Static – 3rd secondary

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Chapter Three – Vectors and Moments

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Example (5)

The two forces F1  mi  3 j and F2  Li  5 j acts at A   2 ,5  and B   1,-3  respectively Determine the value of m and L such that the sum of the moments of these two forces with respect to the origin and with respect to D   5 ,-2  vanishes .

First : With respect to the Origin  0 ,0 

Answer

M A  r1  F1  OA  F1 Where OA  A  O   2 ,5    0 ,0    2 ,5   2iˆ  5 ˆj  M  r  F   2iˆ  5 ˆj    miˆ  3 ˆj   6  5m  kˆ A

1

B 1,-3 

r1

r2

A  2 ,5 

O 0 ,0 

1

M B  r2  F2  OB  F2

And

F1  miˆ  3 ˆj

F2  Liˆ  5 ˆj

Where OB  B  O   1,-3   0 ,0   1,-3   ˆi  3 ˆj  M  r  F   ˆi  3 ˆj    Liˆ  5 ˆj    -5  3L  kˆ B

2

2

The sum of the moments vanishes :  M A  M B  0   6  5m  kˆ   -5  3L  kˆ  0  6kˆ  5 m kˆ  5kˆ  3L kˆ  0

  by kˆ 

 3L  5m  -1     1 Second : With respect to the D  5 ,-2  M A  r1  F1  DA  F1 Where DA  A  D   2 ,5    5 ,-2    -3 ,7   -3iˆ  7 ˆj  M  r  F   -3iˆ  7 ˆj    miˆ  3 ˆj    -9  7m  kˆ A

1

B 1,-3 

r2

r1

A  2 ,5 

D  5 ,-2 

1

M B  r2  F2  OD  F2

And

F1  miˆ  3 ˆj

F2  Liˆ  5 ˆj

Where DB  B  D   1,-3    5 ,-2    -4 ,-1  -4iˆ  ˆj  M  r  F   -4iˆ  ˆj    Liˆ  5 ˆj    20  L  kˆ B

2

2

The sum of the moments vanishes :  M A  M B  0   -9  7m  kˆ   20  L  kˆ  0  by kˆ  -9  7m  20  L  0





  by kˆ 

 L  7m  -11     2  From  1 and  2  : By simultanous

 -1 L  7m  -11  3 

3L  3m  -1

 -3L  5m  1 

3L  21m  -33 -16 m  -32

 m  2 And by substitution in any equation: L  3 Static – 3rd secondary

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Chapter Three – Vectors and Moments

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Example (6)

Find the force F which acts at A   3 ,-2  such that its moment at the origin vanishes and the Algebraic measure of its moment about B  -1,2  equals -8 kˆ .

Answer Let the force F  xiˆ  y ˆj

F  xiˆ  y ˆj

First : with respect to the Origin  0 ,0  M O  r1  F  OA  F Where OA  A  O   3 ,-2    0 ,0    3 ,-2   3iˆ  2 ˆj  M  r  F   3iˆ  2 ˆj    xiˆ  y ˆj    3 y  2x  kˆ O

B  -1,3 

1

M O  0   3 y  2x  kˆ  0

And

A  3,-1

r2

 3 y  2x  0     1

Second : with respect to the B  -1,2 

r1 O 0 ,0 

M B  r2  F  BA  F Where BA  A  B   3 ,-2    -1,2    4 ,-4   4iˆ  4 ˆj  M  r  F   4iˆ  4 ˆj    xiˆ  y ˆj    4 y  4x  kˆ B

And

1

M B  -8 kˆ   4 y  4x  kˆ  -8 kˆ

 4 y  4x  -8   2 

 2 y  2x  -4     2  Then by subtracting  2   1 :  - y  -4  y  4 And x  -6



Then F  -6 ˆi  4 ˆj

-------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

- 56 -

Chapter Three – Vectors and Moments

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Theorem of Coplanar forces The algebraic sum of the moments of a system of forces acting at a point about any point in the space is equal to the moment of the resultant of these forces about the same point . Proof Let F1 , F2 , F3 , ............ Fn be a finite number of forces acting at A and Let O be any point in the space .  M o  r  F1  r  F2  r  F3  ..........  r  Fn



 M o  r  F1  F2  F3  ..........  Fn  Mo  r  R

Where R

A

r



is the resultant of the forces

F3



F2

O

F1

R

 The sum of the moments of forces about O is equal to the moment of the resultant of these forces about O . The length of the perpendicular from any point to the line of action of the Resultant of forces M Where M= M And R  F1  F2  F3  ..........  Fn R ------------------------------------------------------------------------------------------------------------------M  Rd 

 d

Very important remarks If the moment of F about A  M A   The moment of F about B  M B  Then AB // The line of action of F



If the moment of F about A  M A   - The moment of F about B  M B  Then the line of action of F passes through the mid - point of AB

MA

MB

A

B



Steps to solve different kinds of moment's problem :

1 To prove that the line of action of the resultant forces  R  bisect AB , Then : Find the sum of moments of the given forces at A Find the sum of moments of the given forces at B So if the Sum of M A  - the sum of M B Then the line of action of the resultant forces  R  bisect AB

 2  To prove that  R  // AB  We must prove that : M A  M B Then  M A  Zerooo  3  If the line of action of the forces  R  passes through A  Then Also If the line of action of the forces  R  passes through B   M B  Zerooo Static – 3rd secondary

- 56 -

Chapter Three – Vectors and Moments

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Example (7)

The forces F1  2i  j , F2  -i  2 j And F3  2i  7 j act at A   1,-2  , Find the moment of each of these forces about the origin and hence find the length of the perpendicular from the origin to the line of action of the resultant .

Answer R  F1  F2  F3   2iˆ  ˆj    -iˆ  2 ˆj    2iˆ  7 ˆj   3iˆ  4 ˆj Where r  OA  A  O   1,-2    0 ,0   1,-2   ˆi  2 ˆj M  r  R   ˆi  2 ˆj    3iˆ  4 ˆj    -4  6  kˆ  2 kˆ O

And

A 1,-2 

r

M  R d where M  M O  2 And R  R 

 3

2

F2  -iˆ  2 ˆj

O 0 ,0 

  -4   5 2

F3  2iˆ  7 ˆj

F1  2iˆ  ˆj M 2  The length of the perpendicular from O to the line of action of the resultant R : d    0.4 R 5 ------------------------------------------------------------------------------------------------------------------Example (8) The forces F1  i  j , F2  -5i  2 j And F3  2i act at A   2 ,-3   i  Prove that : The algebraic sum of the moments of these forces about B   1,1 is equal to the moment of their resultant about B .  ii  Prove that the line of action of the resultant of these forces passes through the origin O .

Answer To prove that the algebraic sum of the moments  The moment of their resultant We must solve in details "Proof of the theorem" A  2 ,-3  M B  r  F1  F2  F3



And

F3  2iˆ



R  F1  F2  F3  ˆi  ˆj    -5iˆ  2 ˆj   2iˆ  -2iˆ  3 ˆj

r

And as the force act at A :  r  BA  A  B   2 ,-3   1,1  1,-4   ˆi  4 ˆj  M B  r  R   ˆi  4 ˆj    -2iˆ  3 ˆj    3  8  kˆ  -5 kˆ    1 B 1,1

F2  -5iˆ  2 ˆj F1  ˆi  ˆj

Also M 1  r  F1   ˆi  4 ˆj    ˆi  ˆj   1  4  kˆ  5kˆ

M 2  r  F2   ˆi  4 ˆj    -5iˆ  2 ˆj    2  20  kˆ  -18kˆ M 3  r  F3   ˆi  4 ˆj    2iˆ    0  8  kˆ  8kˆ

 M B  M 1  M 2  M 3  5kˆ  18kˆ  8kˆ  -5kˆ     2 

From  1 and  2  : The sum of the moments about B is equal to the moment of the resultant about B To prove that the line of action of the resultant of these forces passes through O

We must prove that : M O  Zeroo

 And MO  r  R Where r  OA  A  O   2 ,-3   0 ,0   2iˆ  3 ˆj  M O   2iˆ  3 ˆj    -2iˆ  3 ˆj   6  6  kˆ  0   The line of action passes through O Static – 3rd secondary

- 55 -

Chapter Three – Vectors and Moments

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Example (9) The forces F1  4i  j , F2  -2i  3 j , F3  i  5 j and F4  3i  7 j act at the origin

Point  O  , Prove that the line of resultant of these forces passes through A -3 ,4  , find also

the sum of the moments of these forces about C   2 ,-5  , then determine the length of the perpendicular drawn from C to the line of action of the resultant of these two forces . Answer To prove that the line of resultant of these forces passes through A  -3 ,4  F1  4iˆ  ˆj

We must prove that : M A  Zeroo  And M A  r1  R Where r1  AO  O  A   0 ,0    -3 ,4   3iˆ  4 ˆj And

O 0 ,0 

R  F1  F2  F3  F4

r1

 R   4iˆ  ˆj    -2iˆ  3 ˆj   ˆi  5 ˆj    3iˆ  7 ˆj   6 ˆi  8 ˆj  M A   3iˆ  4 ˆj    6 ˆi  8 ˆj    -24  24  kˆ  0

A  -3,4 

 The line of action passes through A

r2 C  2 ,-5 

F2  -2iˆ  3 ˆj F3  ˆi  5 ˆj

F4  3iˆ  7 ˆj

And as the force act at C :  M C  r2  R  CO  R Where CO  O  C   0 ,0    2 ,-5   -2iˆ  5 ˆj

 M C  r  R   -2iˆ  5 ˆj   6 ˆi  8 ˆj   16  30  kˆ  -14 kˆ M C  R d where M C  M C  14 And R  R 

6    8  2

2

 10

M A 14   1.4 R 10 ------------------------------------------------------------------------------------------------------------------d

Static – 3rd secondary

- 55 -

Chapter Three – Vectors and Moments

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Example (10) If F1  14i  4 j , F2  Li  6 j are two forces and the sum of their moment about A  0 ,0  is equal to -40 kˆ and the sum of their moments about B   0 ,9  is equal to 176 kˆ , find the value of L and the length of the perpendicular drawn from A to the line of action of the resultant of these two forces . Answer M  -40 kˆ And M  176 kˆ A

B

Let the two forces F1 and F2 act at point  x , y 

M A  r1  R

C  x ,y

M A  r1  R Where r1  AC  C  A   x , y    0 ,0   xiˆ  y ˆj

r2

And R  F1  F2   14iˆ  4 ˆj    Liˆ  6 ˆj   14  L  ˆi  10 ˆj

 M A  r1  R   xiˆ  y ˆj     14  L  ˆi  10 ˆj  -40 kˆ   10 x  14 y  y L  kˆ 

r1

B 0 ,9 

F1  14iˆ  4 ˆj F2  Liˆ  6 ˆj

A 0 ,0 

 10 x  14 y  y L  -40    1

M B  r2  R Where r2  BC  C  B   x , y   0 ,9   xiˆ   y  9  ˆj

 M B  r2  R   xiˆ   y  9  ˆj    14  L  ˆi  10 ˆj  176 kˆ   10 x   14 y  y L  126  9L   kˆ 

 10 x  14 y  y L  126  9L  176     2 

By substituting  1 in  2  : -40  126  9L  176  9L  90   L  10  F1  14i  4 j And  F2  10i  6 j M A  R d where M A  M A  40 And R  R 

 24   10  2

2

 26

M A 40 20   R 26 13 ------------------------------------------------------------------------------------------------------------------d

Static – 3rd secondary

- 55 -

Chapter Three – Vectors and Moments

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Example (11) The force F  -2i  3 j acts at A   1,1 , Let B   2 ,4  and C   4 ,7  , By using moments Prove that the line of action of F is parallel to BC and find the distance between them . Answer

To prove that the line of action of F is parallel to BC Then we must prove that M B  M C

A 1,1

M B  r1  F  BA  F   -iˆ  3 ˆj    - 2iˆ  3 ˆj    3  6  kˆ  -3kˆ

r2

M C  r2  F  CA  F   -3iˆ  6 ˆj    - 2iˆ  3 ˆj    9  12  kˆ  -3kˆ So M B  M C

F1  -2iˆ  3 ˆj

r1

  the line of action of F is parallel to BC

B  2 ,4 

 The distance between the line of action of F and the straight line BC

C  4 ,7 

Is the distance between any point  B or C  to F  M B  F d where M B  -3kˆ  3 And R  R 

 2  3 2

2

 13

MB 3 3 13   F 13 13 ------------------------------------------------------------------------------------------------------------------Example (12) The force F  4i  3 j acts at A   -3 ,2  , Let B   -1,-2  and C   -5 ,6  , By using moments d

Prove that the line of action of F Bisects BC . Answer

To prove that the line of action of F Bisects BC

A  -3,2 

Then we must prove that M B  -M C

M B  r1  F  BA  F   -2iˆ  4 ˆj    4iˆ  3 ˆj    -6  16  kˆ  -22 kˆ M C  r2  F  CA  F   2iˆ  4 ˆj    4iˆ  3 ˆj    6  16  kˆ  22 kˆ

So

M B  - MC

  the line of action of F Bisects BC

F1  4iˆ  3 ˆj

r2 r1 B  -1,-2 

C  -5 ,6 

-------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

- 55 -

Chapter Three – Vectors and Moments

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Example (13) Two forces F1  3i  4 j acts at A   1,2  and F2  5i  j acts at B   -1,3  , Then find The resultant of the moment about the origin and the equation of the line of action of this Resultant .

Answer The Moment about O  0 ,0  with A

A 1,2 

M O  r1  F1  OA  F1

 M O   ˆi  2 ˆj    3iˆ  4 ˆj    4  6  kˆ  -2 kˆ

F1  3i  4 j

r1

The Moment about O  0 ,0  with B

O 0 ,0 

M O  r2  F2  OB  F2

B  -1,3 

 M O   -iˆ  3 ˆj    5iˆ  ˆj   1  15  kˆ  -14 kˆ

F2  5i  j

Then the resultant of the moment Note : There is a big difference between : "The moment of the resultant" and "The resultant of moment"  M  -2 kˆ  14 kˆ  -16 kˆ

r2 O 0 ,0 

O

To find the equation of the line of action about A 3 first : R   3iˆ  4 ˆj    5iˆ  ˆj    8iˆ  3 ˆj  Then Tan    The slope  m And A   1,2  8 3 3 3 And y  y1  m  x  x1    y  2   x  1   y  x   2  8  8 8 8 Then the equation of action of R about A is : 8 y  3x  13 -------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

- 55 -

Chapter Three – Vectors and Moments

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2nd : Algebraic Moment of Coplanar forces In this section , we will change :



Vector Moment M O



To   Algebraic Moment " Magnitude"  M O 

In order to do that , we must add the sign of the moment of any force with respect to any point if : O O

F

d

O

d

O passes through the

F F Rotation is Anti  Clockwise

line of action of F  No rotation

Rotation is Clockwise

"from arrow of the force to point"

"from arrow of the force to point"

M O  Zero

MO  F d

M O  -F d

try to rotate a door from its hinge

Conclusion From the opposite figure : Moment of F about B  F d 2

O

Moment of F about A  F d1 Moment of F about O  Zero

d2

B

 ve

A

d1 -ve

Static – 3rd secondary

- 56 -

Chapter Three – Vectors and Moments

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Properties of Some Geometrical Shapes n2

o Rule : The angle of any regular polygon is :    180  n  Equilateral traingle

A

Properties m   A   m   B   m   C   60 o

30 o 30 o

AB  BC  AC

x

CX  AB  AX  XB AY  BC  BY  YC

1 2

M

o

BZ  AC  AZ  ZC M is the point of intersection of medians

30 30 o

B

CM BM AM 2      CM  BM  AM MX MZ MY 1 To get the sides or angles of the triangle :

z

30 o 30 o

C

y

and MX  MZ  MY

Sin law a b c   Sin A Sin B SinC

Cos law

a 2  b 2  c 2  2bcCos A b 2  a 2  c 2  2a cCos B

c 2  a 2  b 2  2abCosC -------------------------------------------------------------------------------------------------------------------

Square K

D

C

Properties m   A   m   B   m   C   m   D   90 o AB  BC  CD  AD BD  AC "Diagonals"

L

J M

MA  MB  MC  MD m   MAB   m   MAD   45 MX  MJ  MK  ML 

o

A

1 AB 2

45o 45o

B X

AC  2 AB

Static – 3rd secondary

- 56 -

Chapter Three – Vectors and Moments

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Rectangle

K

D

C

Properties m   A   m   B   m   C   m   D   90 o

L

AB  CD And BC  AD

 AC 

2

  AB    BC  2

2

And AC  BD

J

M

A D

B

X

MA  MB  MC  MD C 1 1 O ML  MJ  AB and MK  MX  AD 2 2 1 MX  MJ  MK  ML  AB 2 Z B AB  BC A DZ  BO  AC ------------------------------------------------------------------------------------------------------------------D C Parallelogram Properties

M

Diagonals bisects each other And Not equal

A

m   A   m   C  And m   B   m   D 

B

AB  CD And BC  AD MA  MC And MB  MD ------------------------------------------------------------------------------------------------------------------C

Rhombus

z

o

Properties Sides are equal Diadonals bisect each other

B

D

Diagonals are perpendicular " AC  BD " y x MB  MA MX  MY  MZ  MT  AB A --------------------------------------------------------------------------------------------------------------------L C B Regular hexagon o o 30

120

Properties

L

L

 n  2   180 o  4  180 o  120 o Each angle  2

30 o o 30 30 o 30 o

A

6

L 3

Static – 3 secondary

- 56 -

D

L 3

L rd

60 o 60 o

2L

L F

120

o

30

L

o

E

Chapter Three – Vectors and Moments

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Examples

Email : [email protected]

Calculate the moment about P of each of these forces : (1) 3N P

M P  F  d  -3  2  -6 Newton meter

2m

----------------------------------------------------------------------------------------------------------------------

(2) P

2m

M P  F  d  2  6.5  13 Newton meter 6.5 N

----------------------------------------------------------------------------------------------------------------------

(3) M P  F  d  -8  0  0 Newton meter

8N P

---------------------------------------------------------------------------------------------------------------------We must draw a line perpendicular from P to (4) the line of action of the force x 4N So draw PX o 30 2.5 cm M P  F  d  2.5  4  10 Newton meter 5 cm P ---------------------------------------------------------------------------------------------------------------------We must draw a line perpendicular from P to the line of action (5) 3.5 N of the force d 140 o o So draw PX  Sin 40   d  5 Sin 40 o  3.214 m o 40 5 5m x M P  F  d  -3.6  3.214  -11.57 Newton meter 50 o

Static – 3rd secondary

- 56 -

Chapter Three – Vectors and Moments

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Example (1) ABCD is a square of side length 2 m , The forces of magnitude 8 , 12 , 16 , 10 , 6 2 , 4 2 Newton Act along AB , BC , DC , AD , CA and BD respectively , Find the sum of their moment about :

 i  The vertex A  ii  About the Mid - point of BC  iii  The intersection point of the two diagonals . Answer

i 

10

A

2m

D

M A  -10  0  16  2  12  2  8  0  6 2  0  4 2  2

2m M A  -32  24  8  Zero

8

4 2

B

 ii  To get Y F :

2m

C

12 10 2 m

A 1 1 1   YF   2 2 :y f:1m 1 1 M y  -16  1  12  0  8  1  10  2  6 2  4 2 2 2

D

6 2

16

4 2

8

F o

B

 iii 

16

2m

45o 45o

1: 1 : 2

M y  -16  8  20  6  4  -26 Newton meter

6 2

M O  -10  1  16  1  12  1  8  1

45 45o

12 10

A

y

45o 45o 1m

2m

C

D

M O  -10  16  12  8  -6 Newton meter 16

O 8

B Static – 3rd secondary

- 56 -

6 2

4 2

45o 45o

C 12

Chapter Three – Vectors and Moments

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Example (2) ABCD is a rectangle in which AB  12 cm , BC  16 cm , The forces of magnitude F , 2 F , 3 F And 4 F kg.wt. act along AB , CB , CD and AD respectively , Find N  BC such that the sum of The moment about N vanishes .

Answer Sum of M N  0 "Vanishes" Everyline from N must be  to each force M N  -4 F  12  3F   16  x   F  x

12 cm M N  0   -48 F  48 F  3F x  F x  0

4F

M

A

D

F

3F

 -2F x  0   x  0

N

2F

B C  The distance between B and N is Zero x 16  x  So N lies on B ---------------------------------------------------------------------------------------------------------------------Example (3) A lamina in the form of a circle whose diameter AC is of length 10 cm , AB and AD are two

Chords in the lamina whose lengths are 6 and 8 cm respectively and in two different sides of AC The two forces 5 and 9 Newton act along AB and AD , what is the direction of rotation of the lamina about an axis perpendicular to the lamina and passes through :

 i  The center  M 

 ii  The point C . Answer

In  ABC : m   B   90 o " Semi - circle"

B

Draw ME  AB , MF  AD 1  ME // BC  ME  BC  4 cm 2 1 Also MF  CD  3 cm 2 M M  -5  4  9  3  7 Newton . cm

8

E 5

A F

M C  -5  8  9  6  14 Newton . cm

M

C

5

5

6

9

D

 The lamina rotates about M in the direction ADCB And rotates about C in direction ADCB also ----------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

- 55 -

Chapter Three – Vectors and Moments

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Example (4) ABC is an equilateral triangle , The length of whose side is 18 cm , Forces of magnitude 8 , 6 , 13 Newton act along AB , BC , AC , Find the algebraic sum of the moments of these forces about :

 i  The vertex A

 ii  The point of intersection of the three medians Answer

i 

AX 

18 

2

 9   9 3 2

A

 M A  6  9 3  54 3 N.cm

 ii 

AX  9 3

13

M is a point of intersection of medians "Concurrent" XM 1 1 1   XM  XA  XM   9 3  3 3 XA 3 3 3  XM  YM  ZM  3 3

y

z

1

8

2

M

B

x

6

M M  -13 3 3  6  3 3  8  3 3  3 3 Newton . cm

30 o 30 o

C

18 cm ------------------------------------------------------------------------------------------------------------------Example (5) ABC is an equilateral triangle , The length of whose side is 16 cm , D is the mid - point of AB ,

Forces of magnitude 2 3 , 4 3 , 10 Newton act along AB , CB , CD , Find the algebraic sum of the moments of these forces about the mid - point of AC Answer 1 DE  BC  8 cm  Draw EX  AD 2  EX is a median where XD  4 cm

A

x

 EX  4 3 cm And

4 3

D

EY // XD  EY  XD

 EY  4 cm And EZ  4 3 cm

8

2 3

10

M E  -4 3  4 3  10  4  2 3  4 3  -64 Newton . cm B

E

8

o

y

z

4

C

4 3 -------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

- 55 -

Chapter Three – Vectors and Moments

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Example (6) ABCD is a rectangle in which AB  5 cm and BC  12 cm , The forces of magnitude 8 , 15 , 17 10 , 26 Newton act along BA , BC , CD , DA and CA respectively , Find :

 i  A point H  AB such that the algebraic sum of the moments about H equal 317 Newton.cm in ABCD direction .

 ii  O  BC such that the algebraic sum of the moments about O equal 116 Newton.cm in DCBA direction .

Answer

 i  AC   5 

2

  12   13 cm 2



HM In  AHM : Sin   AH  HM  AH Sin  12  HM   5  x   13

5x

 x  3 cm EO  ii  In  EOC : Sin   OC

26

8

x

5

15 B

C 12 cm

12 13

  AH  5  3  2 cm in the direction ABCD 19

A

5  EO  OC Sin    EO   12  x   13 5 M O  10  5  17  12  x   26  12  x    8x 13  -116  254  17 x  2  12  x   2  8x  35 x  490 

M

17

317  50  10x  204  15x  2  5  x   12

 19 x  57 

D

H

M H  10   5  x   17  12   15  x   26  5  x   

19

A



D

 26 E

8 B

O x

15

12  x

17



5

C

 x  14 cm

 CO  12  14  - 2 cm  O  BC , O  BC Where CO  2 cm ----------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

- 55 -

Chapter Three – Vectors and Moments

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Example (7) ABCD is a square with side length 10 cm , The forces of magnitude 6 , 8 , 9 and F Newton act along AB , CB , CD and AC respectively , Find F if the sum of the algebraic moments of these forces about B and C are equal in magnitude and differ in signs . Answer 1 M B  -M C And AC  10 2 cm And BE  AC  5 2 cm 2 A M B  9  10  F  5 2

And And

10 cm

M C  6  10  60 N.cm

F

M B  -M C  90  5F 2  -60

 5F 2  150   F 

D

9

E

6

150  15 2 Newton 5 2 45o

B

C

8 ------------------------------------------------------------------------------------------------------------------Example (8)

ABC is a right angled triangle at B were AB  20 cm , AC  25 cm , D  AC where AD  4cm , let DH  AC where H  AB , The forces of magnitude 4 , 6 , 7 and 10 Newton act along AB , BC AC and DH respectively , Find the sum of the algebraic moments of these forces about A , B and C Answer

BC 

 25 

2

  20   15 cm 2

A

DH 15 In  AHD : Tan     DH  4   3 cm AD 20  AH  5 cm And HB  15 cm OB 4 In  HOB : Sin     OB  15   12 cm HB 5 15  20 In  ABC : BX   12 cm 25 M A  -10  4  6  20  8 N .cm

4 cm D



5 cm 4

h



3



O

15 cm

7

x 21 cm

10 12 cm

6

B

15 cm

C

M B  -7  12  10  12  36 N .cm M C  4  15  10  21  8  270 N .cm -------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

- 55 -

Chapter Three – Vectors and Moments

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Example (9) ABCDHO is a regular hexagon whose side length is 10 cm , the forces of magnitude 3 , 7 , 4 and 2 Newton act along AB , CB , CD and HD respectively , Find the algebraic sum of moments of these forces about O .

Answer In  AHD : AD  2HD  20 cm  AH  10 3 cm 1  OX  AH  5 3 cm 2  OB  OD  10 3 cm

2

H

y

D 60

10 cm

o

4

10

5 3

10 3

20 cm

O

C

 OY  5 3 cm M O  3  5 3  7  10 3  4  10 3  2  5 3

5 3

 M O  -25 3 N .cm

10

7

30 o

x

B 3 ---------------------------------------------------------------------------------------------------------------------Example (10) ABCDHO is a regular hexagon whose side length is 20 cm , the forces of magnitude 1 , 2 , 3 , 4 , 5 A

And 6 Newton act along AB , BC , CD , DH , HO and OA respectively , Find the algebraic sum of moments of these forces about A and M The center of the hexagon  . Answer HB  2HO  40 cm 4 H  OB  20 3 cm 1 5 20  AY  AX  OB  10 3 cm 20 2 30 o  AH  AC  20 3 cm M O 40 cm M A  2  10 3  3  20 3  4  20 3  5  10 3 6 20 3 y  M A  210 3 N .cm 10 10 3

D 60 o

10 cm

3 10 cm

10 3

C

2

20

A

1

B

10 3

x ----------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

- 55 -

Chapter Three – Vectors and Moments

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Example (11) ABCD is a parallelogram in which m   ABC   60o and AC  BC , AC  12 cm , the forces of magnitude 14 , 18 , 9 , 6 and 2 3 Newton act along AB , CB , CD , DA and AC respectively . Find the algebraic sum of moments of these forces about each of A , B , C , D . Answer F In  ACD : Let AD  x  AC  3 AD  12 cm 12  AD   4 3 cm   DC  2 AD  8 3 cm 3 14 A y B M A  18  12  9  6  162 N .cm 60 o M B  2 3  4 3  9  6  6  12  -102 N .cm M C  -14  6  6  12  -156 N .cm

2 3

6

4 3

12 60 o

M D  -14  6  18  12  2 3  4 3  108 N .cm

18

30 o

D

H

x

C

9

E ---------------------------------------------------------------------------------------------------------------------Example (12) ABCD is a trapezuim in which AD // BC , m   A  90 o , AD  AD  10 cm , BC  20 cm , the

forces of magnitude k , 5 , F , 5 and 10 2 Newton act along BA , CB , DC , DA and BD Respectively , If the sum of the forces moment about A vanishes and if the sum of the forces moment about B equals that about D , find the values of F and K . Answer MA 0 20 cm In  ABD : BD  10 2 cm

5

1  AH  BD  5 2 cm 2  AE  HD  5 2 cm "Square"

B 45

o

45o 10 cm

M B  M D     1 Where

M B  F  10 2  5  10

M B  5 2  10 2  5  10  50 N .cm And Static – 3rd secondary

10

C

F

10 2

A

50  5 2 Newton 5 2

Then from  1 : 50  10K  50

10 2

5 2

50  5F 2  100  0  5 2F  50

And

H

k

M A  5  10  F  5 2  10 2  5 2

 F

45o

5 2

D

E

M D  5  10  K  10

  10K  0   K  0 - 666 -

Chapter Three – Vectors and Moments

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Example (13) ABC is a right angled triangle at B and AB  10 cm , AC  26 cm , A force F acts in the plane of The triangle and the moment of F about A is equal to its moment about B is equal to 30 Newton.cm If the moment of F about C is equal to -30 Newton.cm , Then find F and its direction . Answer M A  M B  30 N.cm A  The line of action of F is parallel to AB 13 cm And M C  -M A

 The line of action of F Bisects AC In  ABC : BC 

 26 

2

D

10 cm

  10   24 cm 2

13 cm F

 D is the mid  point of AC , DE // AB  E is the mid  point of BC And

B

Moment about B is  ve  Then F must be  ve

12 cm

12 cm

C

24 cm 30 M B  F  12   30  12 F   F   2.5 Newton 12 ---------------------------------------------------------------------------------------------------------------------Example (14) ABCD is a rectangle where AB  12 cm , BC  16 cm , A force F acts in the plane of the rectangle

and the moment of F about B is equal to its moment about D is equal to -240 Newton.cm . If the moment of F about A is equal to 240 Newton.cm , Then find F and its direction . Answer M B  M D  -240 N.cm 8 cm A O  The line of action of F is parallel to BD And

6 cm

M B  -M A

 The line of action of F Bisects AB

In  EHB :

Sin 

2

C

B

  6   10 cm 2

HB 8   HB  Sin   EB   6  4.8 cm EB 10

 M B  - F  4.8   -240  -4.8 F   F 

Static – 3rd secondary

4F

6 cm

Then F must be  ve  M B  - F  BH

8

10



H

 Moment about B is  ve

In  AEO : EO 



D

240  50 Newton 4.8

- 666 -

Chapter Three – Vectors and Moments

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Example (15) Important ABCD is a rhombus with side length 8 cm and m   B   60 o , the forces of magnitude 6 , 5 , 13 , F and 6 Newton act along BA , CB , CD , AD and AC respectively , If the sum of the algebraic mesures of moments of these forces about D equals the sum of the algebraic measures of moments of these forces about the point of intersection of the rhombus diagonals , Find the magnitude of F . Also find the magnitude and the direction of the resultant of the forces which are acting at A only . Answer

M D  M M     1 Where And

D

M D  6  4 3  5  4 3  6  4 3  -20 3 N.cm M M  13  2 3  5  2 3  6  2 3  F  2 3

So from  1 : -20 3  4 3  2 3 F  2 3F  24 3

4 3 30 30 30 30

F 4 3 2 3 2 o o 60 o 4 4 60 60 o 4 6 60 30 A 60 o 30 M 4 60 o C 2 3

6

To find the resultant at A :

9

2



 9 3



2

8 cm

6

   x  6  12Cos60 o  6 Cos60 o  9  ve     y  12 Sin60o  6 Sin60 o  9 3  ve   R

13

8

2

 F  12 Newton

4 3

30 30

5

B 12 Sin60

 18 Newton

o

F  12

6 Sin60 o

y 9 3   3    Tan 1 3  60 o x 9 R has a direction along AD  Tan 

6

60 o

6 Cos 60 o

60 o

12 Cos 60 o 6

Very important Note : There is a difference between when we find the resultant at a point and When we find the moment of the resultant at a point ----------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

- 666 -

Chapter Three – Vectors and Moments

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Example (16) ABCD is a trapezuim in which AD // BC , m   B   90 o , AB  12 cm , BC  8 cm , AD  9 cm . Forces of magnitude F1 , 60 , F2 and 30 gm.wt act along AB , BC , CD and DA respectively , find the values of F1 ,F2 given that M A  M C  0 . Answer Draw AH  DC and Draw DX to join BC at x

In  XCD : DC 

12 

2

  1  145 cm 2

AH 12 In  AHD : Sin     AH  9  AD 145 12  M A  F2  9   60  12  0 145 

9 cm

A



30 F2

12 cm F1 60

B



C

8 cm

D

12 cm

H

x

1 cm

108 -720  145 F2  -720  F2    F2  -80 N 108 145

-360  -45 N 8 ---------------------------------------------------------------------------------------------------------------------Example (17) M C  30  12  F1  8   8 F1  360  0   F1 

ABCD is a rectangle in which AB  8 cm and BC  6 cm , H  AB such that BH  3 cm , forces Of magnitude F , 4 , 9 , K , 5 and 4 5 Newton act along DA , AB , BC , DC , CA and HC Respectively , Find : F and K given that the algebraic measure of the moments of these forces About C  72 N.cm in the direction ABC and vanishes about B . Answer M C  72 N.cm where M C  F  8  4  6  8F  24  72   F  6 Newton

k

D

C

For Moment about B In  ABC : CH 

O

6 

2

  3   3 5 cm 2

6 3 6  BX   cm 3 5 5 Also AC   BO 

8

2

A

  6   10 cm 2

9

5

F

x

5 cm

4

H

4 5

6 cm

B

3 cm

6 8  4.8 cm 10

 M B  -K  6  5  4.8  4 5 

6  F 8 5

 -6 K  24  24  8 F  0  -6 K  8  6   0   K  8 Newton Static – 3rd secondary

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Chapter Three – Vectors and Moments