Chap. 2 Force Vectors
Chapter Outline
„
„
„
„
„
„
„
„
„
Scalars and Vectors
Vector Operations
Vector Addition of Forces
Addition of a System of Coplanar Forces
Cartesian Vectors
Addition and Subtraction of Cartesian Vectors
Position Vectors
Force Vector Directed along a Line
Dot Product
2
Scalar and Vectors
3
Vector Operations
„
Multiplication and Division by a Scalar
„
Vector Addition
4
Vector Operations
„
Vector Subtraction
* draw the vector from the head of B to the head of A
5
Vector Operations
„
Resolution of a Vector
1. 已知R和A求B
2. 已知R和A、B的作用線
6
Vector Operations
„
Vector Addition of Forces
7
p.29, Problem 2-17
Resolve the force F2 into components acting along the u and v axes
and determine the magnitudes of the components.
9
p.31, Problem 2-29
The beam is to be hoisted using two chains. If the resultant
force is to be 600 N directed along the positive y axis,
determine the magnitudes of forces FA and FB acting on each
chain and the angle θ of FB so that the magnitude of FB is a
minimum. FA acts at 30° from the y axis as shown.
10
11
Addition of a System of Coplanar
Forces
„
„
Cartesian Vector
F = Fx i + Fy j
Resultants
FR = F1 + F2 + F3
= (F1x + F2x + F3x ) i
+ ( F1y + F2y + F3y ) j
= (FRx ) i + (FRy ) j
12
Addition of a System of Coplanar
Forces
FR x = ΣFx
FRy = ΣFy
2
FR = FRx + FRy 2
θ = tan
−1
FRy
FRx
13
p.41, Problem 2-50
The three forces are applied to the bracket. Determine the range of
values for the magnitude of force P so that the resultant of the
three forces does not exceed 2400 N.
14
15
Cartesian Vectors
„
„
Right-Handed Coordinate System
Rectangular Components of
a Vector
A=Ax+Ay+Az
„
„
A
A
uA=
=
A
A
Unit Vector
Cartesian Unit Vector
i,j,k
16
Cartesian Vectors
„
Cartesian Vector Representation
A = Ax i + Ay j + Az k
„
Magnitude
|A| =
2
2
Ax + Ay + Az
2
=
2
A' + Az
2
A在xy平面上的投影
17
Cartesian Vectors
„
Direction
α, β, γ 及方向餘弦(direction cosine)
Ay
A
A
cos γ = z
cosα = x cosβ =
A
A
A
„
單位向量 (unit vector)
Ay
A Ax
A
i+
j+ z k
uA = =
A A
A
A
= cosα i + cosβ j + cosγ k
又 cos 2α + cos 2β + cos 2γ = 1
A = Au A
= Acosα i + Acosβ j + Acosγ k
= Ax i + A y j + Az k
18
Cartesian Vectors
„
Addition and Subtraction of Cartesian
Vectors
R = A+B = (Ax+Bx)i + (Ay+By)j
+ (Az+Bz)k
R’= A-B = (Ax - Bx)i + (Ay - By)j
+ (Az - Bz)k
„
Concurrent Force Systems
FR = Σ F = ΣFx i + ΣFy j + ΣFz k
19
p.55, Problem 2-84
Determine the coordinate direction angles of F1 and FR.
20
21
Position Vectors
„
x, y, z Coordinates 座標 (位置向量)
(0,2,0)
(6,-1,4)
(4,2,-6)
22
Position Vectors
„
Position Vector 位置向量
相對於固定點O的向量
r = xi + yj +zk
„
Relative Position Vector
23
Force Vector Directed along a Line
AB = rB - rA
AB
F = Fu F = F ( )
AB
24
p.62, Ex.2-15,
Determine the resultant force acting at A.
A(0,0,4 )
B(4,0,0 )
C (4,2,0 )
AB = (4,0,−4) L AB = 5.657 m
AC = (4,2,−4) L AC = 6.0m
AB
F AB = FAB‧
AB
100
=
(4,0,-4)
5.657
= ( 70.71 , 0 , - 70.71 ) N
AC
F AC = FAC‧
AC
120
=
(4,2,-4)
6
= ( 80 , 40 , - 80 ) N
∴ R = FAB + F AC = (150.71 , 40 , - 150.71) N
R = 150.712 + 40 2 + (-150.71) 2 = 217 N
25
p.65, Problem 2-93
The chandelier is supported by three chains which are concurrent
at point O. If the force in each chain has a magnitude of 300 N,
express each force as a Cartesian vector and determine the
magnitude and coordinate direction angles of the resultant force.
26
27
Dot Product
A ⋅ B = AB cosθ , 0 o ≤ θ ≤ 180 o
ex: Work = F ⋅ r
„
Cartesian Vector Formulation
i ⋅i = 1
j ⋅i = 0
i⋅k = 0
j⋅k = 0
i⋅ j = 0
k ⋅i = 0
k ⋅k =1
k⋅ j =0
j ⋅ j =1
A⋅ B = ( Ax i + Ay j + AZ k ) ⋅ (Bx i + By j + BZ k )
= Ax B x + Ay B y + Az B z
28
Dot Product
„
Applications
1. angle
θ = cos-1(
A⋅ B
)
AB
when θ = 90o , cosθ = 0 , A ⋅ B = 0
⇒ A⊥B
2. components of a vector parallel and perpendicular to a line
A|| = A cos θ = A ⋅ u
∴ A|| = A cos θ ⋅ u = ( A ⋅ u ) ⋅ u
又 A⊥ = A − A||
A⊥ = A sin θ
29
p.72, Ex. 2-17,
Determine the parallel and perpendicular components of
the force to member AB.
A( 0 , 0 , 0 ) B( 2 , 6 , 3 )
AB
(2,6,3)
1
=
= (2,6,3)
AB
2 2 + 6 2 + 32 7
= ( F ⋅ u AB )u AB
u AB =
F AB
6 1
= 300 ⋅ ⋅ (2,6,3)
7 7
= ( 73.5 , 220 , 110.2 ) N
F ⊥ = F - F AB
= ( - 73.5 , 80 , - 110.2 ) N
FAB = 73.52 + 220 2 + 110.2 2 = 257 N
F⊥ = (−73.5) 2 + 80 2 + (−110.2) 2 = 154.7 N
30
p.76, Problem 2-116
Two forces act on the hook. Determine the angle between
them.Also, what are the projections of F1 and F2 along the y axis?
31
32
p.78, Problem 2-133
Two cables exert forces on the pipe. Determine the magnitude of the
projected component of F1 along the line of action of F2.
33
34