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Static 3rdsecondary Chapter Three Vectors and Moments-55-
2
3 -1
-4
0 ,1
The unit vector: A vector whose norm is 1
2 22 2
2 22 2
it is denoted by " i " vector and " j " vector where :
i 1,0 i x y 1 0 1
j 0 ,1 j x y 0 1 1
j
i 1,0O
y'
y
xx'
1 1 2 2
2 2 1 1 2 1 2 1 2 1 2 1
Very important note: If A x ,y And B x ,y
AB B A x ,y x ,y x x , y y x x i y y j
Vectors and Moments
F ir st: Vectors
As we said before, Vectors is a physical quanti ty which has both magni tude and direction
Shapes of vectors
1stshape:We can write vector as a ,b 2ndshape:we can write vector as a ai b j
So for example: F 2 ,3 means F 2i 3 j Or H -4 ,-1 means H -4i j
This is called the position vector for any point
----------------------------------------------------------------------------------------------------------------------
The length of the vectors Norm of the vector
2 2
2 2
If A x ,y Then A x y 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 -4 5
----------------------------------------------------------------------------------------------------------------------
The unit vectors" i " and " j "
----------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Three Vectors and Moments-5-
o
Let A and B be two non Zero vectors , Let us represent them by OC and OD ,Then the
angle COD where 180 is called the smallest angleBetween these two vectors while the angle COD
of measu
ore more than 180 is called the biggest angle between them.
O fig 1
ab
D C
O
fig 2
ab
DC
Small angle
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 .
A B A B Cos owhere 0 180
Scalar Algebraic Product Vector Product
A
B
o30
I n this chapter, we wil l discuss the two kinds of vector multipl ication which are:
F ir st kind: Algebraic Product
It is denoted by
Some defini tions(1) The angle between the two vectors:
Very important note:
But we prefer to use the small angle
(2) The scalar product of two vectors :
----------------------------------------------------------------------------------------------------------------------
Example (1)
Answer
We can get the small angle between the two vectors A and B by
Representing them by drawing
Or the same point " see figures 1 and 2 "
If is the measur
the angle between two vectors :
Outwards I nwards
o
o
e of the angle the two vectors 0 180
the measure of the angle between them 360
small
So big
oFind A B where A is a vector of magnitude 5 and direction 30 north of west and B is a
vector of magnitude 8 towards south
o
o
A 5 And B 8 , and the measure of the small angle betweenA and B is 120 So A B A B Cos
A B 5 8 Cos120 -20
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Static 3rdsecondary Chapter Three Vectors and Moments-5-
B 40
A 65
Example (2)
Answer
----------------------------------------------------------------------------------------------------------------------
Example (3)
Answer
----------------------------------------------------------------------------------------------------------------------
Example (4)
Answer
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
oA B A B Cos -135 15 12 Cos Cos -0.75 138 36'
The vector A is of magnitude 65 towards west , B is a vector of magnitude 40 and direction
5east of south and with angle where Tan , Then find A B
12
oA B A B Cos 90 A B A B -Sin-5
A B 65 40 -100013
5
12
13
1 2
3
1 2
ABC is a right angled triangle at B where AB 6 cm , BC 8 cm , The vectors F , F andF of magnitude 150 , 200 , 250 gm.wt act along BA , BC and CA , Find :
1 F F
2 3 1 32 F F 3 F F
o o
1 2
1 2 1 2
2 3
1 ABC 90 : F F 150 200 Cos90 0
Note When F F F F 0 directly
2 The angle between F and F must be inward Or outward
So to get the measure of the smaller angle
2 3
2 3
o
2 3 2 3
1 3 1 3
Between F and F , we have to draw the ray BC
ACD is the angle between F and F
-8F F F F Cos 180 200 250 -Cos 50000 -40000
10
3 F F F F
6
Cos 150 250 2250010
CD B
A
E
3F 1F
2F
8 cm
10 cm 6 cm
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Static 3rdsecondary Chapter Three Vectors and Moments-5-
The Algebraic Projection of a vector in the dir ection of another vector
Definition
Conclusion: The algebraic Projection of a vector in the direction of another vector:
----------------------------------------------------------------------------------------------------------------------
Example (5)
Answer
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
B B
AA
DD
CCO
O
The projection of B in the
Direction of A is B Cos
B
AO A
B
D
A B
C
D
C
o
The projection of B in the direction
of A is B Cos 90 0 o
The projection of B in the direction
of A is B Cos 180 - B
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
o
A is a vector where A 30 and B is a vectors B 20 and the angle between A and
B is 75 , then find the algebraic projection of A in the direction of B and also the algebraic
Projec
tion of B in direction of A .
o
o
The algebraic projection of A in the direction of B A Cos 30 Cos75 7.768
The algebraic projection of B in the direction of A B Cos 20 Cos75 5.176
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Static 3rdsecondary Chapter Three Vectors and Moments-5-
B 12
A 18
o30
o60
Example (6)
Answer
----------------------------------------------------------------------------------------------------------------------
Example (7)
Answer
o
o
A is a vector of magnitude 18 and direction 60 north of east , B is another vector of
magnitude 12 and direction 30 south of west , Find the algebraic projection of the two
vectors A and B in the direction of the other .
o o o o
o
The smaller angle between the two vectors 30 90 30 150
The algebraic projection of A in the direction of B :
A Cos 18 Cos150 -9 3
The algebraic projection of B in the di
o
rection of A :
B Cos 12 Cos150 -6 3
A
BC
D
10
7.5
A
BC
D
10
7.512.5
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
a Don' t forget that the direction of the two vectors
must be either inwards Or outwards , so we must
extend AC and get the angle , so the algebraic
projection of CB in the direct
o
2 2
ion of AC :
CB Cos CB Cos 180 - CB Cos
10 4where AC 7.5 10 12.5 cm And Cos
12.5 5
-4So - CB Cos 10 -8
5
b The algebraic projection of BD in the
d
7.5irection of BA : BD Cos 12.5 7.5
12.5
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Static 3rdsecondary Chapter Three Vectors and Moments-6-
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 Then A B is ve
b If is an obtuse angle Then A B is ve
c If is a right angle Then A B is Zero and if :
A B
o
o o
0
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Static 3rdsecondary Chapter Three Vectors and Moments--
a a Cos i a Sin j a Sin
a Cos
a
o
o o
if the magnitude of the vector a is 3 and 110 , find the Component form.
3 Cos 110 3Sin 110 -1.026 2.82
Example:
a i j i j
F 4i 2 j in the direction of the vector
Rule (5)
----------------------------------------------------------------------------------------------------------------------
Rule (6)
----------------------------------------------------------------------------------------------------------------------
Rule (7)
Perpendicular resolution of vector Geometric Component
----------------------------------------------------------------------------------------------------------------------
Rule (8)
Example: Find the algebraic component of the vector AB where A -2,3 And B 1,-1
Answer
2 2
ABThe algebraic Component of F in the direction of AB is F
AB
And AB B A i j -2i 3 j 3i 4 j And AB 3 -4 5
3i 4 jAB 12 8F 4i 2 j5AB
0.85
i i j j 1 , i j j i 0
1 2 1 2 1 1 2 2 If a a i a j and b b i b j Then a b a b a b
aThe algebraic Component of F in the direction of a is equal to F
a
1 2 1 2 1 1 1 2 2 1 2 2
1 1 1 2 2 1 2 2 1 1 2 2
st st nd nd
1 2 1 2
Proof :
a b a i a j b i b j a i b i a i b j a j b i a j b j
a b i i a b i j a b j i a b j j a b a b
a b a i a j b i b j 1 1 2 2
Conclusion :
Don' t forget : i i j j 1 , i j j i 0
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Static 3rdsecondary Chapter Three Vectors and Moments--
If A 3i 4 j and B 5i 12 j , Then find A B and the measure of the angle
included between the two vectors .
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 .
If A 3i j and B 2i k j , Then find the value of k so that A B
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)
Answer
-------------------------------------------------------------------------------------------------------------------
Example (2)
Answer
-------------------------------------------------------------------------------------------------------------------
Example (3)
Answer o
o
A B 3i j 2i k j 3 2 1 k 6 k And A B 90
Cos90 0 A B A B Cos A B 0
6 k 0 k 6
2 2
2 2 2 2
-1 o
There is no angle given, then : A B 3i 4 j 5i 12 j 15i 48 j 15 48 63
A 3 4 5 And B 5 12 13
A B 63 63 63A B A B Cos Cos Cos 14 15'13 5 65 65A B
2 2 2 2
The algebraic projection of A in the direction of B : A Cos So we have to get Cos
A B 3 i j 0i 4 j 3 0 1 4 4
A 3 1 2 And B 0 4 4
A BA B A B Cos Cos
A B
-1 o4 1 1
Cos 602 4 2 2
1The algebraic projection of A in the direction of B : A Cos 2 1
21The algebraic projection of B in the direction of A : B Cos 4 2
2
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Static 3rdsecondary Chapter Three Vectors and Moments--
If A 2i 5 j and B i k j , Then find the value of k so that A // B
o
If A k i j and B 12i 5 j , Then find the value ofk that makes the measure of the
Angle between A and B equals 45
Example (4)
Answer
----------------------------------------------------------------------------------------------------------------------
Example (5)
Answer
-------------------------------------------------------------------------------------------------------------------
o
2 2 2 2 2
2 o 2
2 2 2 2
A // B 0 And A B A B Cos
Where A B 2i 5 j i k j 2 5k
And A 2 5 29 And B 1 k k 1
2 5k 29 k 1 Cos0 2 5k 29 k 1
So by squaring both: 2 5k 29 k 1 4 20k 25k 29k 2
22
9
4k 20k 25 0 2k 5 0 2k 5 k 2.5
o
2 2 2 22
22 2
22
A B A B Cos45 1
Where A B k i j 12i 5 j 12k 5
And A k -1 k 1 And B 12 -5 13
1 16912k 5 13 k 1 So by squaring both: 12k 5 k 1
22
169k 169144 120k 25 288 240k 50 169k
2
2
2
169 0
119k 240k 119 0 17k 7 7k 17 0
7 -17 k "agreed" And k "refused" 17 7
b b 4acNote : We can use formula to find k : k
2a
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Static 3rdsecondary Chapter Three Vectors and Moments--
ABCD is a trapezium in which AD // BC , m A m B , AD 16 cm , BC 21 cm2
And AB 12 cm , Find :
1 The algebraic projection of AD in the direction of CB
2 The algebraic projection of BD in the di
rection of CB
3 The algebraic projection of CD in the direction of AB
Example (6)
Answer
C
D
21 cm
16 cm
12 cm
A
B
C
D
16 cm
16 cm
12 cm
A
B
C
D16 cm
12 cm
A
B
5 cm
12 cm
5 cm
12 cm13 cm
16 cm
o1 The measure of the angle between AD and CB is equal 180
Thus the
of AD in the direction of CB is
AD Cos The angle between AD and CB
algebraic projection
o16Cos180 -16
2 Don' t forget that the direction of the two vectors
must be either inwards Or outwards , so we must
extend CB
o
2 2
and get the angle , so the algebraic
projection of BD in the direction of CB :
BD Cos BD Cos 180 - BD Cos
16 4where BD 16 12 20 cm And Cos
20 5
So -
-4BD Cos 20 -16
5
3 We must extend CD , so that angle EDF is
the angle between CD And AB
The algebraic projection of CD in the direction of AB :
CD Cos 180
o12
- CD Cos -13 -12 cm13
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Static 3rdsecondary Chapter Three Vectors and Moments--
ABCD is a rectangle which AB 6 cm , BC 8 cm , find :
5
a -2 AB AC c BC DB d AC BD2
2 2AC 8 6 10 cm
a -2 AB AC -2 AB AC Cos
6-2 6 10 -72
10
5 5b BC DB BC DB2 2
5BC BD C
2
o
2 2
os 180
5- BC BD Cos
2
5 -88 10 -160
2 10
c In AMB: AC 6 8 10 cm
1AM AC 5 cm Median from right angle
2
Also BM AM 5 cm Prop
2 2 2
erties of a rectangle
5 5 6 7So in AMB : Cos
2 5 5 25
7AC BD AC BD Cos 10 10 28
25
Example (8)
Answer
-------------------------------------------------------------------------------------------------------------------
C
DA
B
8 cm
C
DA
B
C
DA
B
10 cm
10 cm
8 cm
8 cm
6 cm
8 cm
6 cm
6 cm
M
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Static 3rdsecondary Chapter Three Vectors and Moments--
oABCD is a parallelogram in which m BAC 90 , BC 2AB 8cm , find:
1 1
a CA CB b AD DC c AB 3 AD d 7 BD AC 3 2
2 2
o
AC 8 4 4 3 cm
4 3a CA CB CA CB Cos 4 3 8 48
8
1 1 1b AD DC AD DC
3 2 6
1AD DC Cos 180
6
o
1- AD DC Cos6
1 4 8- 8 4 -6 8 3
c AB 3 AD 3 AB AD Cos 90
3 4 8 -Sin
4-32 3 -1
8
2 2
6 3
1d In AMB: AM AC 2 3 cm
2
MB 2 3 4 2 7 cm
BD 4 7 7 BD AC 7 BD AC Cos
2 37 4 7 4 3 336
2 7
Example (9)
Answer
-------------------------------------------------------------------------------------------------------------------
C
DA
B
8 cm
8 cm
4 cm 4 3
C
DA
B
8 cm
8 cm
4 cm 4 3
C
DA
B
8 cm
8 cm
4 cm 4 3
C
DA
B
8 cm
8 cm
4 cm2 3
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Static 3rdsecondary Chapter Three Vectors and Moments-6-
2 2
2 2
AB BC AB BC Cos where :
AB B A 7 ,5 4 ,1 3 ,4 And AB 3 4 5
And BC C B -5 ,-4 7 ,5 -12 ,-9 And BC -12 -9 15
3,4 -12 ,-9AB BC -36 36Cos5 15 75AB BC
-2425
The algebraic component of AB in the direction of BC is equal to :
-12 ,-9 3 -12 4 -9BCAB 3,4 -4.8
15 15BC
The algebraic component of BC in the direction of AB is equ
al to :
3 ,4 3 -12 4 -9ABBC -12 ,-9 -14.45 5AB
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
-------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Three Vectors and Moments--
Second kind: Vector Product
It is denoted by A B A B Sin k where : Rule
A A And B B and is the measure of the small angle between A and B and k is
the unit vector perpendicular to the plane containing the two vectors A and B .
The direction of the un
it vector k 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 k as shown in figures
the vector product A B is a And not and its direction is
Perpendicular to the plane containing the vectors A and B in the direction which is
determined according to
Note that : vector scalar
the right hand rule and the unit vector in the direction of A B
A B is the vector k which is equal to k
AB Sin
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Static 3rdsecondary Chapter Three Vectors and Moments--
Properties of the Vector Product between two vectors
Rule (1)
Note :
For the two vectors A B and B A have the same magnitude But have different direction
----------------------------------------------------------------------------------------------------------------------
Rule (2)
o o o
If A // B , Then A B 0
for , In this case is equal to Zero or 180 and hence Sin 0 Sin 180 0
If A B 0 , Then either A 0 Or B 0 Or
Note: A // B
----------------------------------------------------------------------------------------------------------------------
Rule (3)
----------------------------------------------------------------------------------------------------------------------Rule (4)
A B - B A A B Sin k
o
o
The vector product of any vector into itself is equal zero as Sin0 0
So for any vector : a a zero
Proof: a a a a Sin 0 k Zero
The Right hand system
o o
In the given figure:
OX and OY are two perpendicular directions
i and j are two unit vectors in these directions respectively:
So i j i j Sin90 1 1 Sin90 k k
Where k is the unit vector
perpendicular to plane OXY containing i and jAnd in the direction of the right hand rule.
From this: we can say : i j k k i j j k i
And when they are in th
e anticlock wise :
From this: we can say : j i -k i k - j k j -i
But i i j j k k 0
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Static 3rdsecondary Chapter Three Vectors and Moments--
Rule (5)
Example: 3 a 2b 6 a b a 6 b 6 a b ----------------------------------------------------------------------------------------------------------------------
Rule (6)
----------------------------------------------------------------------------------------------------------------------
Rule (7)
----------------------------------------------------------------------------------------------------------------------
Important Remark
----------------------------------------------------------------------------------------------------------------------
1 2 1 2 1 2 2 1
st nd nd st
If a a i a j and b b i b j Then a b a b a b k
Or a b 1 2 2 1 k
1 2 1 2 1 1 1 2 2 1 2 2
1 1 1 2 2 1 2 2
1 2 2 1 1 2 2 1
Proof : a b a i a j b i b j a i b i a i b j a j b i a j b j
a b i i a b i j a b j i a b j j
0 a b k a b -k 0 a b a b k
Don' t forget : i i j
j 1 , i j k & j i -k
Where i , j , k are the right system of unit vectors
For any two vectors a and b and for any scalar m:
m a b a m b m a b
For any three vectors a , b and c , The distributive law holds:
a b c a c b c
O
D N
C
A
B
We know that A B AB Sin
If OC represents the vector A And if OD represents the vectorB
And AB Sin OC OD Sin The surface area of the parallelogram OCND
Twice the surface area of OCD
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Static 3rdsecondary Chapter Three Vectors and Moments--
Comparison between Scalar and Vector products
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Static 3rdsecondary Chapter Three Vectors and Moments--
AB B A -1,5 2 ,1 -3 ,4 -3i 4 j BA - AB 3i 4 j
AC C A -4 ,-1 2 ,1 -6 ,-2 -6 i 2 j CA - AC 6 i 2 j
BC C B -4 ,-1 -1,5 -3 ,-6 -3i 6 j CB - BC 3i 6 j
AB AC
-3i 4 j -6 i 2 j 6 24 k 30k
BA CB 3i 4 j 3i 6 j 18 12 30k
BC CA -3i 6 j 6 i 2 j -6 36 30k
A 3i 2 j , B -i j , C i 5 j And D 6i a AB B A -4i 3 j And CD D C 5i 5 j AB CD 20 15 k 5k
b AC C A -2i 7 j & BA A B
4i 3 j & DC C D -5i 5 j
BA DC -i 2 j AC BA DC -2i 7 j -i 2 j -4 7 3k
c BC C B 2i 4 j And AB AC -4i 3 j -2i 7 j 8 21 29
BC AB
AC 29 2i 4 j 58i 116 j
BC AB AC BD 58i 116 j 7 i j -58 812 k -870 k
Example (4)Answer
-------------------------------------------------------------------------------------------------------------------
Example (5)Answer
-------------------------------------------------------------------------------------------------------------------
Example (6)
Answer
-------------------------------------------------------------------------------------------------------------------
Let C mi n j And A C 21 -2i 5 j mi n j 21
-2m 5n 21 1
And C B -8k mi n j i j k -8k -m n -8 2
From 1 and 2 and by simultanous : m 3 And n 5 C 3i 5 j
If A 2 ,3 and B 1,-1 ,Then find the vector C such that: A C 21 & C B -8k
If A 2 ,1 , B -1,5 and C -4 ,-1 Then prove that: AB AC BA CB BC CA
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
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Static 3rdsecondary Chapter Three Vectors and Moments--
A -2 j , B 6 i 4 j , C 7i 4 j And D -2i 4 j
AC C A 7 i 6 j And BD D B -8i 8 j
AC BD 7 i 6 j -8i 8 j 56 48 k 104 k
Area of ABCD Area of ABC Area of ACD
W
1here Area of ABC AB AC Sin
2
But we know that : AB AC Sin AB BC
1Area of ABC AB BC
2
AB B A 6 i 2 j And BC C B i 8 j
AB BC 6 i 2 j i 8 j 48 2 k 50 k
1 1
Area of ABC AB BC 50 25 12 2
1 1Also Area of ACD AC AD Sin AC AD
2 2
AC C A 7 i 6 j And AD D A -3i 6 j
AC AD 7 i 6 j -3i 6 j 42 18 k 60k
Area of
1 1
ACD AC AD 60 30 22 2
Area of ABCD Area of ABC Area of ACD 25 30 55
Example (10)
Answer
----------------------------------------------------------------------------------------------------------------------
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
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Static 3rdsecondary Chapter Three Vectors and Moments-5-
1
F 2i j
A 1,-2
O 0 ,0
r
1 2 3The forces F 2i j , F -i 2 j And F 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 .
2 F -i 2 j
3 F 2i 7 j
1 F i j
A 2 ,-3
B 1,1
r
1 2 3The forces F i j , F -5i 2 j And F 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 resu
ltant about B .
ii Prove that the line of action of the resultant of these forces passes through the origin O .
B 1 2 3
1 2 3
To prove that the algebraic sum of the moments The moment of their resultant
We must solve in details "Proof of the theorem"
M r F F F
And R F F F i j -5i 2 j 2i -
B
1 1
2 2
2i 3 j
And as the force act at A :
r BA A B 2 ,-3 1,1 1,-4 i 4 j
M r R i 4 j -2i 3 j 3 8 k -5k 1
Also M r F i 4 j i j 1 4 k 5k
M r F i 4 j
3 3
B 1 2 3
-5i 2 j 2 20 k -18k
M r F i 4 j 2i 0 8 k 8k
M M M M 5k 18k 8k -5k 2
From 1 and 2 : The sum of the moments about B is equal to the moment of the resulta
O O
nt about B
To prove that the line of action of the resultant of these forces passes through O
We must prove that : M Zeroo And M r R
Where r OA A O 2 ,-3 0 ,0 2i 3 j
O M 2i 3 j -2i 3 j 6 6 k 0 The line of action passes through O
2 F -5i 2 j
3F 2i
1 2 3
O
O
R F F F 2i j -i 2 j 2i 7 j 3i 4 j
Where r OA A O 1,-2 0 ,0 1,-2 i 2 j
M r R i 2 j 3i 4 j -4 6 k 2k
And M Rd where M M 2
2 2
And R R 3 -4 5
M 2The length of the perpendicular from O to the line of action of the resultant R : d 0.4
R 5
Example (7)
Answer
-------------------------------------------------------------------------------------------------------------------
Example (8)
Answer
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Static 3rdsecondary Chapter Three Vectors and Moments--
1 F -2i 3 j
C 4 ,7
A 1,1
1r
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 .
2r
B 2 ,4
B C
B 1
C 2
To prove that the line of action of F is parallel to BC
Then we must prove that M M
M r F BA F -i 3 j -2i 3 j 3 6 k -3k
M r F CA F -3i 6 j -2i
B C
3 j 9 12 k -3k
So M M the line of action of F is parallel to BC
The distance between the line of action of F and the straight line BC
Is the distance between any poi
2 2
B B
B
nt B or C to F
M F d where M -3k 3 And R R 2 3 13
M 3 3 13d
F 1313
1 F 4i 3 j
C -5 ,6
A -3,2
1r
The force F 4i 3 j acts at A -3 ,2 , Let B -1,-2 and C -5 ,6 , By using moments
Prove that the line of action of F Bisects BC .
2r
B -1,-2
B C
B 1
C 2
To prove that the line of action of F Bisects BC
Then we must prove that M -M
M r F BA F -2i 4 j 4i 3 j -6 16 k -22k
M r F CA F 2i 4 j 4i 3
B C
j 6 16 k 22k
So M -M the line of action of F Bisects BC
Example (11)
Answer
-------------------------------------------------------------------------------------------------------------------
Example (12)
Answer
-------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Three Vectors and Moments--
1F 3i 4 j
O 0 ,0
A 1,2
1r
1 2Two forces F 3i 4 j acts at A 1,2 and F 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 .
2r
B -1,3
O 0 ,0
2F 5i j
O 1 1 1
O
O 2 2 2
O
The Moment about O 0 ,0 with A
M r F OA F
M i 2 j 3i 4 j 4 6 k -2k
The Moment about O 0 ,0 with B
M r F OB F
M -i 3 j 5i j 1 15 k -14k
Then the resul
O
tant of the moment
: There is a big difference between :
" " and " "
M -2k 14k -16 k
To find the equation of the line of action about A
first :
Note
The moment of the resultant The resultant of moment
1 1
3 R 3i 4 j 5i j 8i 3 j Then Tan The slope m And A 1,2
8
3 3 3And y y m x x 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
Example (13)
Answer
-------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Three Vectors and Moments-6-
ToO OVector Moment M Algebraic Moment " Magnitude" M
Rotation is Anti Clockwise
"from arrow of the force to point"
OM F d
Rotation is Clockwise
"from arrow of the force to point"
OM -F d
O
d
F
O
d
F
OF
O
O passes through the
line of action of F
No rotation
M Zero
try to rotate a door
from its hinge
A 1d
2d BO
ve
-ve
2
1
From the opposite figure :
Moment of F about B F d
Moment of F about A F d
Moment of F about O Zero
2nd: Algebraic Moment of Coplanar forces
I n this section , we wil l change :
In order to do that , we must add the signof the moment of any force with respect to any point
i f :
Conclusion
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Static 3rdsecondary Chapter Three Vectors and Moments--
on 2 180n
x z
A
CBy
o30 o30
o30o30
o30o30
M2
1
oProperties
m A m B m C 60
AB BC AC
CX AB AX XBAY BC BY YC
BZ AC AZ ZC
M is the point of intersection of medians
CM BM AM 2CM BM AM and MX MZ MY
MX MZ MY 1
To get the sides or angles of t
he triangle :
Sin law Cos lawa b c
Sin A Sin B SinC
2 2 2
2 2 2
2 2 2
a b c 2bcCos A
b a c 2acCos B
c a b 2abCosC
o
o
Properties
m A m B m C m D 90
AB BC CD AD
BD AC "Diagonals"
MA MB MC MD
m MAB m MAD 45
1MX MJ MK ML AB
2
AC 2 AB
Equilateral traingle
Properties of Some Geometr ical Shapes
Rule : The angle of any regular polygon is :
-------------------------------------------------------------------------------------------------------------------
SquareD
o45
o45
C
BX
K
L
A
J
M
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Static 3rdsecondary Chapter Three Vectors and Moments--
Rectangle
Parallelogram
Rhombus
Regular hexagon
o
2 2 2
Properties
m A m B m C m D 90
AB CD And BC AD
AC AB BC And AC BD
MA MB MC MD
1 1ML MJ AB and MK MX AD
2 2
1MX MJ MK ML AB
2AB BC
DZ BOAC
Properties
Diagonals bisects each other And Not equal
m A m C And m B m DAB CD And BC AD MA MC And MB MD
Properties
Sides are equal
Diadonals bisect each other
Diagonals are perpendicular " AC BD "MB MA
MX MY MZ MTAB
o o o
Properties
n 2 4Each angle 180 180 120
2 6
-------------------------------------------------------------------------------------------------------------------
-------------------------------------------------------------------------------------------------------------------
---------------------------------------------------------------------------------------------------------------------
C
BX
K
JM
D
L
A
A
C
B
D
Z
O
A
C
B
D
M
B
A
C
D
x y
zo
o30o30
o30o30
o120
o120o30
o60o60
2L
L 3
L
L
L
L
L
L
o30B
A
F E
D
C
L 3
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Static 3rdsecondary Chapter Three Vectors and Moments--
ExamplesCalculate the moment about P of each of these forces :
(1)
----------------------------------------------------------------------------------------------------------------------
(2)
----------------------------------------------------------------------------------------------------------------------
(3)
----------------------------------------------------------------------------------------------------------------------
(4)
----------------------------------------------------------------------------------------------------------------------
(5)
2mP
3 N
2 m
P
6.5 N
8 N
P
2.5 cm
4 N
P
5 cm
o30
x
5 m
o40
o50
x
o140
3.5 N
PM F d -3 2 -6 Newton meter
PM F d 2 6.5 13 Newton meter
PM F d -8 0 0 Newton meter
PM F d 2.5 4 10 Newton meter
We must draw a line perpendicular from P to
the line of action of the force
So draw PX
PM F d -3.6 3.214 -11.57 Newton meter
o o
We must draw a line perpendicular from P to the line of action
of the force
dSo draw PX Sin 40 d 5 Sin 40 3.214 m
5
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Static 3rdsecondary Chapter Three Vectors and Moments--
x
z
A
CB
y
o30o30
M2
1
x
A
CB
4 3
y
82 3
4 3
ED
810
4zo
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 abo
ut :
i The vertex A ii The point of intersection of the three medians
2 2
A
i AX 18 9 9 3
M 6 9 3 54 3 N.cm
ii AX 9 3
M is a point of intersection of medians "Concurrent"
XM 1 1 1XM XA XM 9 3 3 3XA 3 3 3
XM YM ZM 3 3
MM -13 3 3 6 3 3 8 3 3 3 3 Newton . cm
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
E
1DE BC 8 cm Draw EX AD
2
EX is a median where XD 4 cm
EX 4 3 cm
And EY // XD EY XDEY 4 cm And EZ 4 3 cm
M -4 3 4 3 10 4 2 3 4 3 -64 Newton . cm
Example (4)
Answer
-------------------------------------------------------------------------------------------------------------------
Example (5)
Answer
-------------------------------------------------------------------------------------------------------------------
8
13
6
18 cm
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Static 3rdsecondary Chapter Three Vectors and Moments--
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 .
B C
B
C
B C
1M -M And AC 10 2 cm And BE AC 5 2 cm
2
M 9 10 F 5 2
And M 6 10 60 N.cm
And M -M 90 5F 2 -60
1505F 2 150 F 15 2 Newton
5 2
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
2 2
A
B
C
BC 25 20 15 cm
DH 15In AHD : Tan DH 4 3 cm
AD 20
AH 5 cm And HB 15 cm
OB 4In HOB : Sin OB 15 12 cm
HB 5
15 20In ABC : BX 12 cm
25
M -10 4 6 20 8 N.cm
M -7 12 10 12 36 N.cm
M 4 15 10
21 8 270 N.cm
Example (7)
Answer
-------------------------------------------------------------------------------------------------------------------
Example (8)
Answer
-------------------------------------------------------------------------------------------------------------------
CB
DA
8
F
69
E
10 cm
o45
3
4 cm
15 cm
21 cm
5 cm
15 cm
B
A
C
O
hx
D
12 cm
10
4 7
6
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Static 3rdsecondary Chapter Three Vectors and Moments--
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 .
O
O
In AHD : AD 2HD 20 cm
AH 10 3 cm
1OX AH 5 3 cm
2
OB OD 10 3 cm
OY 5 3 cm
M 3 5 3 7 10 3 4 10 3 2 5 3
M -25 3 N .cm
ABCDHO is a regular hexagon whose side length is 20 cm , the forces of magnitude 1 , 2 , 3 , 4 , 5
And 6 Newton act along AB , BC , CD , DH , HO and OA respectively , Find the alg
ebraic sum
of moments of these forces about A and M The center of the hexagon .
A
A
HB 2HO 40 cm
OB 20 3 cm
1AY AX OB 10 3 cm
2
AH AC 20 3 cm
M 2 10 3 3 20 3 4 20 3 5 10 3
M 210 3 N .cm
Example (9)
Answer
----------------------------------------------------------------------------------------------------------------------
Example (10)
Answer
----------------------------------------------------------------------------------------------------------------------
20 cm
o60
o30
BA
D
C
H
O
10
10 3
10 cm
5 3
5 3
10
x
y2
4
7
3
10 3
o60
o30
BA
D
C
H
O
20
10 3
20
10
x
y
4
3
2
1
10 cm
10 cm
10 3
20
40 cm
20 36
5
M
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Static 3rdsecondary Chapter Three Vectors and Moments-66-
oABCD is a parallelogram in which m ABC 60 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 al
gebraic sum of moments of these forces about each of A , B , C , D .
A
B
C
D
In ACD : Let AD x AC 3 AD 12 cm
12AD 4 3 cm DC 2AD 8 3 cm
3
M 18 12 9 6 162 N.cm
M 2 3 4 3 9 6 6 12 -102 N.cm
M -14 6 6 12 -156 N.cm
M -14 6 18 12 2 3 4 3 108 N .cm
o
ABCD is a trapezuim in which AD // BC , m A 90 , AD AD 10 cm , BC 20 cm , theforces of magnitude k , 5 , F , 5 and 10 2 Newton act along BA , CB , DC , DA and BD
Respectively , If th
e 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 .
A
A
B D B
B
M 0
In ABD : BD 10 2 cm
1
AH BD 5 2 cm2
AE HD 5 2 cm "Square"
M 5 10 F 5 2 10 2 5 2
50 5F 2 100 0 5 2F 50
50F 5 2 Newton
5 2
And M M 1 Where M F 10 2 5 10
M 5 2 10 2 5 10 50 N.cm And
DM 5 10 K 10
Then from 1 : 50 10K 50 10K 0 K 0
Example (11)
Answer
----------------------------------------------------------------------------------------------------------------------
Example (12)
Answer
CHD
A
9
18
B
o60
E
14y
x
6 2 3
12
o60
F
o30
4 3
CB
DA
k F
5
10 2
20 cm
10 2H
E
5 2
o45 5 2o
45
o45
10 cm
10
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Static 3rdsecondary Chapter Three Vectors and Moments-6-
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 th
e moment of F about C is equal to -30 Newton.cm , Then find F and its direction .
A B
C A
2 2
M M 30 N.cm
The line of action of F is parallel to AB
And M -M
The line of action of F Bisects AC
In ABC : BC 26 10 24 cm
D is the mid point of AC , DE // AB
E is the mid point of
B
BC
And Moment about B is ve Then F must be ve
30M F 12 30 12 F F 2.5 Newton
12
ABCD is a rectangle where AB 12 cm , BC 16 cm , A force F acts in the plane of the rectangleand the moment of F about B is equal to its moment about D is equal to -240 Newton.cm . If the
momen
t of F about A is equal to 240 Newton.cm , Then find F and its direction .
B D
B A
B
2 2
M M -240 N.cm
The line of action of F is parallel to BD
And M -M
The line of action of F Bisects ABMoment about B is ve
Then F must be ve M - F BH
In AEO : EO 8 6 10 cm
In
B
HB 8EHB : Sin HB Sin EB 6 4.8 cm
EB 10
240
M - F 4.8 -240 -4.8 F F 50 Newton4.8
Example (13)
Answer
----------------------------------------------------------------------------------------------------------------------
Example (14)
Answer
13 cm
12 cm
13 cm10 cm
B
A
C
D
F
24 cm
12 cm
CB
H
O DA
10
4F6 cm
6 cm
8 cm
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Static 3rdsecondary Chapter Three Vectors and Moments-6-
D M
D
M
o o
M M 1
Where M 6 4 3 5 4 3 6 4 3 -20 3 N.cm
And 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
F 12 Newton
To find the resultant at A :
x 6 12Cos60 6 Cos60 9 ve
o o
22
1 o
y 12 Sin60 6 Sin60 9 3 ve
R 9 9 3 18 Newton
y 9 3Tan 3 Tan 3 60
x 9
R has a direction along AD
oABCD is a rhombus with side length 8 cm and m B 60 , 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 algebra
ic
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 .
Very important Note : There is a difference between when we find the resul tant at a point and
When we find the moment of the resul tant at a point
Example (15) Important
Answer
----------------------------------------------------------------------------------------------------------------------
B
A C
D
30 30
30 30
o60o60
o60o60
3030
4 3 4 33030
o60 o604 42 3
2 3
4
8 4 3
4
8 cm
26
F
13
6
5
M
2
6
o12 Sin60
o6 Sin60 F 12
6
o12 Cos 60o6 Cos 60o60
o60 6
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o
1 2
ABCD is a trapezuim in which AD // BC , m B 90 , AB 12 cm , BC 8 cm , AD 9 cm .
Forces of magnitude F , 60 , F and 30 gm.wt act along AB , BC , CD and DA respectively ,find the values
1 2 A C of F ,F given that M M 0 .
2 2
A 2
2 2 2
C 1 1 1
Draw AH DC and Draw DX to join BC at x
In XCD : DC 12 1 145 cm
AH 12In AHD : Sin AH 9
AD 145
12M F 9 60 12 0145
108 -720 145F -720 F F -80 N
108145
M 30 12 F 8 8 F 360 0 F
-360
-45N8
C C
2 2
2 2
M 72 N.cm where M F 8 4 6
8F 24 72 F 6 Newton
For Moment about B
In ABC : CH 6 3 3 5 cm
6 3 6BX cm
3 5 5
Also AC 8 6 10 cm
6 8BO 4.8 cm
10
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
Respectivel
y , 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 .
Example (16)
Answer
----------------------------------------------------------------------------------------------------------------------
Example (17)
Answer
C
B
x
O
D
A
9
5 cm
6 cm
DA
B
C
xH
8 cm 1 cm
12 cm2F
30
60
1F12 cm
9 cm
H 3 cm
5
4
k
F4 5
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