Mechanics (1)
BAS 006
Lecture 3: Force Systems (3D)
Prof. Dr. Mohamed Safwat Saad El-Din
Basic Science Engineering Department
2013
1
Lecture 3
1
Mechanics (1)
Delta University
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
Chapter Objectives
Parallelogram Law
Cartesian vector form
Dot product and angle between 2 vectors
2
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
1
Delta University
Faculty of Engineering
Chapter Outline
1.
2.
3.
4.
5.
6.
7.
8.
9.
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
3
Lecture 3
Mechanics (1)
Delta University
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
2.5 Cartesian Vectors
Right-Handed Coordinate System:
A rectangular or Cartesian
coordinate system is said to be
right-handed provided:
• Thumb of right hand points in
the direction of the positive z
axis
• z-axis for the 2D problem would
be perpendicular, directed out
of the page.
4
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
2
Delta University
Faculty of Engineering
2.5 Cartesian Vectors
Rectangular Components of a Vector:
• A vector A may have one, two or
three rectangular components along
the x, y and z axes, depending on
orientation
• By two successive application of the
parallelogram law
•
A = A’ + Az
•
A’ = Ax + Ay
• Combing the equations,
A can be expressed as
•
A = Ax + Ay + Az
5
Lecture 3
Mechanics (1)
Delta University
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
2.5 Cartesian Vectors
Unit Vector
• Direction of A can be specified
using a unit vector
• Unit vector has a magnitude of 1
• If A is a vector having a
magnitude of A ≠ 0, unit vector
having the same direction as A
is expressed by uA = A / A. So
that
A = A uA
6
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
3
Delta University
Faculty of Engineering
2.5 Cartesian Vectors
Cartesian Vector Representations
• 3 components of A act in the positive i,
j and k directions
A = Axi + Ayj + AZk
*Note the magnitude and direction
of each components are separated,
easing vector algebraic operations.
7
Lecture 3
Mechanics (1)
Delta University
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
2.5 Cartesian Vectors
Magnitude of a Cartesian Vector
• From the colored triangle,
A = A'2 + Az2
• From the shaded triangle, A
A' =
2
x
+ Ay2
• Combining the equations
gives magnitude of A
A = Ax2 + Ay2 + Az2
8
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
4
Delta University
Faculty of Engineering
2.5 Cartesian Vectors
Direction of a Cartesian Vector
• Orientation of A is defined as the
coordinate direction angles α, β
and γ measured between the tail
of A and the positive x, y and z
axes
0° ≤ α, β and γ ≤ 180 °
• The direction cosines of A is
cos α =
Lecture 3
A
Ax
cos γ = z
A
A
Ay
cos β =
A
9
Mechanics (1)
Delta University
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
2.5 Cartesian Vectors
Direction of a Cartesian Vector
• Angles α, β and γ can be determined by the inverse
cosines
Given
A = Axi + Ayj + AZk
then,
uA = A /A = (Ax/A)i + (Ay/A)j + (AZ/A)k
where
A = Ax2 + Ay2 + Az2
Lecture 3
Mechanics (1)
10
Prof.
M. Safwat
Eng.
Malek
Abuwarda
5
Delta University
Faculty of Engineering
2.5 Cartesian Vectors
Direction of a Cartesian Vector
• uA can also be expressed as
uA = cosαi + cosβj + cosγk
•
Since A = Ax2 + Ay2 + Az2
have
and uA = 1, we
cos 2 α + cos 2 β + cos 2 γ = 1
• A as expressed in Cartesian vector form is
A = AuA
= Acosαi + Acosβj + Acosγk
= A x i + A y j + A Zk
Lecture 3
Mechanics (1)
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11
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
Example:
Determine the coordinate angle γ for F2 and then express
each force acting on the bracket as a Cartesian vector.
12
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
6
Delta University
Faculty of Engineering
Solution
For Force F2
Since : cos2 α + cos2 β + cos2 γ= 1
Then:
However, it is required that γ2 > 90o
Thus: γ2 = cos-1 (-0.5) = 120 o
13
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Delta University
Faculty of Engineering
Solution
z
x
Fig. a
Fig. b
By resolving F1 and F2 into their x, y and z components as
shown in Figs. A and b respectively.
14
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
7
Delta University
Faculty of Engineering
Solution
F1 and F2 can be expressed in Cartesian vector form as:
F1 = 450 cos 45o sin 30o (-i) + 450 cos 45o cos 30o (+j)
+ 450 sin 45o (+k)
F1 = { -159 i +276 j + 318 k} N
And
F2 = 600 cos 45o i + 600 cos 60o j +600 cos 120o k
F2 = { -424 i +300 j - 300 k} N
15
Lecture 3
Mechanics (1)
Delta University
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
2.6 Addition and Subtraction of Cartesian Vectors
Concurrent Force Systems
• Force resultant is the vector sum
of all the forces in the system
FR = ∑F = ∑Fxi + ∑Fyj + ∑Fzk
16
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
8
Example:
Determine the magnitude and coordinate direction angles
of the resultant force acting on the bracket.
17
Eng. Malek Abuwarda
Solution
Resultant force:
By adding F1 and F2 vectorally, the Resultant force FR can
be obtained:
FR = F1 + F2
FR = { -159.01 i +275.57 j + 318.2 k} +
{ -424.26 i +300 j - 300 k}
= { -265.16 i +575.57 j – 18.20 k} N
The magnitude of FR is:
FR =
FR =
(FR )2x + (FR )2y + (FR )2z
(265.16)2 + (575.57 )2 + (18.20)2
= 633.97N
18
Eng. Malek Abuwarda
9
Solution
The coordinate direction angles of FR are:
 (FR )x 
−1  265.16 
o
 = 65.3
 = cos 
F
633
.
97


 R 
α = cos−1 
 (FR )y 
−1  575.57 
o
 = 24.8
 = cos 
 633.97 
 FR 
β = cos −1 
 (FR )z 
−1  18.20 
o
 = 88.4
 = cos 
F
633
.
97


 R 
γ = cos −1 
19
Eng. Malek Abuwarda
Delta University
Faculty of Engineering
Example 2.8
Express the force F as Cartesian vector.
20
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
10
Delta University
Faculty of Engineering
Solution
Since two angles are specified, the third angle is found by
cos 2 α + cos 2 β + cos 2 γ = 1
2
2
2
cos α + cos 60 o + cos 45 o = 1
2
2
cos α = 1 − (0.5 ) − (0 .707 ) = ±0 .5
Two possibilities exit, namely
α = cos −1 (0.5)= 60 o
α = cos −1 (− 0.5) = 120o
21
Lecture 3
Mechanics (1)
Delta University
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
Solution
By inspection, α = 60º since Fx is in the +x direction
Given F = 200N
F = Fcosαi + Fcosβj + Fcosγk
= (200cos60ºN)i + (200cos60ºN)j
+ (200cos45ºN)k
= {100.0i + 100.0j + 141.4k}N
Checking:
F = Fx2 + Fy2 + Fz2
=
(100.0)2 + (100.0)2 + (141.4)2
= 200N
22
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
11
Delta University
Faculty of Engineering
2.7 Position Vectors
x,y,z Coordinates
• Right-handed coordinate
system
• Positive z axis points
upwards, measuring the
height of an object or the
altitude of a point
• Points are measured relative
to the origin, O.
23
Lecture 3
Mechanics (1)
Delta University
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
2.7 Position Vectors
Position Vector
• Position vector r is defined as a fixed vector which
locates a point in space relative to another point.
• E.g.
r = xi + yj + zk
24
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
12
Delta University
Faculty of Engineering
2.7 Position Vectors
Position Vector
• Vector addition gives rA + r = rB
• Solving
•
r = rB – rA = (xB – xA)i + (yB – yA)j + (zB –zA)k
or r = (xB – xA)i + (yB – yA)j + (zB –zA)k
25
Lecture 3
Mechanics (1)
Delta University
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
2.7 Position Vectors
•
•
•
•
•
Length and direction of cable AB can be found by
measuring A and B using the x, y, z axes
Position vector r can be established
Magnitude r represent the length of cable
Angles, α, β and γ represent the direction of the
cable
Unit vector, u = r/r
26
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
13
Delta University
Faculty of Engineering
Example 2.12
An elastic rubber band is attached to points A and B.
Determine its length and its direction measured from A
towards B.
27
Lecture 3
Mechanics (1)
Delta University
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
Solution
Position vector
r = [-2m – 1m]i + [2m – 0]j + [3m – (-3m)]k
= {-3i + 2j + 6k}m
Magnitude = length of the rubber band
r=
(− 3)2 + (2)2 + (6)2
= 7m
Unit vector in the director of r
u = r /r
= -3/7i + 2/7j + 6/7k
28
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
14
Delta University
Faculty of Engineering
Solution
α = cos-1(-3/7) = 115°
β = cos-1(2/7) = 73.4°
γ = cos-1(6/7) = 31.0°
29
Lecture 3
Mechanics (1)
Delta University
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
2.8 Force Vector Directed along a Line
•
•
•
In 3D problems, direction of F is
specified by 2 points, through
which its line of action lies
F can be formulated as a
Cartesian vector
F = F u = F (r/r)
Note that F has units of forces (N)
unlike r, with units of length (m)
30
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
15
Delta University
Faculty of Engineering
2.8 Force Vector Directed along a Line
•
•
•
Force F acting along the chain can be presented as a
Cartesian vector by
- Establish x, y, z axes
- Form a position vector r along length of chain
Unit vector, u = r/r that defines the direction of both the
chain and the force
We get F = Fu
31
Lecture 3
Mechanics (1)
Delta University
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
Example 2.13
The man pulls on the cord with
a force of 350N. Represent this
force acting on the support A,
as a Cartesian vector and
determine its direction.
32
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
16
Delta University
Faculty of Engineering
Solution
End points of the cord are:
A (0m, 0m, 7.5m)
and
B (3m, -2m, 1.5m)
r = (3m – 0m)i + (-2m – 0m)j + (1.5m – 7.5m)k
= {3i – 2j – 6k}m
Magnitude = length of cord AB
r=
(3m )2 + (− 2m )2 + (− 6m )2
= 7m
Unit vector,
u = r /r
= 3/7i - 2/7j - 6/7k
33
Lecture 3
Mechanics (1)
Delta University
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
Solution
Force F has a magnitude of 350N, direction specified by u.
F = Fu
= 350N(3/7i - 2/7j - 6/7k)
= {150i - 100j - 300k} N
α = cos-1(3/7) = 64.6°
β = cos-1(-2/7) = 107°
γ = cos-1(-6/7) = 149°
34
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
17
Delta University
Faculty of Engineering
2.9 Dot Product
•
•
•
Dot product of vectors A and B is written as A·B (Read
A dot B)
Define the magnitudes of A and B and the angle
between their tails
A·B = AB cosθ where 0°≤ θ ≤180°
Referred to as scalar product of vectors as result is a
scalar
35
Lecture 3
Mechanics (1)
Delta University
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
2.9 Dot Product
Laws of Operation
1. Commutative law
A·B = B·A
2. Multiplication by a scalar
a(A·B) = (aA)·B = A·(aB) = (A·B)a
3. Distribution law
A·(B + D) = (A·B) + (A·D)
36
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
18
Delta University
Faculty of Engineering
2.9 Dot Product
Cartesian Vector Formulation
- Dot product of Cartesian unit vectors
i·i = (1)(1)cos0°
°=1
i·j = (1)(1)cos90°
°=0
- Similarly
i·i = 1
i·j = 0
j·j = 1
i·k = 0
k·k = 1
j·k = 0
37
Lecture 3
Mechanics (1)
Delta University
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
2.9 Dot Product
•
•
Cartesian Vector Formulation
Dot product of 2 vectors A and B
A·B = AxBx + AyBy + AzBz
Applications
The angle formed between two vectors or
intersecting lines.
θ = cos-1 [(A·B)/(AB)] 0°≤ θ ≤180°
The components of a vector parallel and
perpendicular to a line.
Aa = A cos θ = A·u
38
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
19
Delta University
Faculty of Engineering
Example
Determine the length of side BC of the triangular plate.
Solve the problem by finding the magnitude of rBC;
then check the result by first finding θ, rAB, and rAC and
then using the cosine law.
39
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Delta University
Faculty of Engineering
Solution:
A (0, 1, 1), B (0, 3, 4), C (3, 5, 0)
Side BC:
rBC = (xC − xB )i + (y C − y B ) j + (zC − zB )k
rBC = 3 i + 2 j - 4 k
rBC =
(xC − xB )2 + (y C − y B )2 + (zC − zB )2
rBC =
(3)2 + (2)2 + (− 4)2
= 5.39 m
( Ans.)
40
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
20
Delta University
Faculty of Engineering
Solution:
Side AB:
rAB = (xB − x A )i + (y B − y A ) j + (zB − zA )k
rAB = 0 i + 2 j + 3 k
rAB =
rAB =
(xB − x A )2 + (y B − y A )2 + (zB − zA )2
(0)2 + (2)2 + (3)2 = 3.6056
Side AC:
rAC = (xC − x A )i + (yC − y A ) j + (zC − zA )k
rAC = 3 i + 4 j - 1 k
rAC =
rAC =
(xC − x A )2 + (y C − y A )2 + (zC − zA )2
(3)2 + (4)2 + (− 1)2 = 5.099m
Lecture 3
Mechanics (1)
Delta University
41
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
Solution:
From Dot product:
cosθ =
5
rAB .rAC 0 + 4(2 ) + (− 1)(3)
=
=
= 0.27196
rAB rAC
3.6056 x 5.0999 78.385
θ = 74.219o
From triangle ABC:
rBC =
(rAB )2 + (rAC )2 − 2(rAB )(rAC )(cosθ )
rBC =
(5.099)2 + (3.6056)2 − 2(5.099)(3.6056)(cos 74.219o )
rBC = 5.39m
Lecture 3
( Αns.)
42
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
21
Delta University
Faculty of Engineering
Example 2.17
The frame is subjected to a horizontal force F = {300j} N.
Determine the components of this force parallel and
perpendicular to the member AB.
43
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Delta University
Faculty of Engineering
Solution
Since
r
r
r
r
r
rB
2i + 6 j + 3k
uB = r =
rB
(2)2 + (6)2 + (3)2
r
r
r
= 0.286i + 0.857 j + 0.429k
Thus
r
r
FAB = F cos θ
rr
r
r
r
r
= F .uB = 300 j ⋅ 0.286i + 0.857 j + 0.429k
(
)(
= (0)(0.286) + (300)(0.857) + (0)(0.429)
= 257.1N
)
44
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
22
Delta University
Faculty of Engineering
Solution
Since result is a positive scalar, FAB has the same sense of
direction as uB. Express in Cartesian form
r
r r
FAB = FAB u AB
(
r
r
r
= (257.1N ) 0.286i + 0.857 j + 0.429k
r
r
r
= {73.5i + 220 j + 110k }N
)
Perpendicular component
r
r r
r
r
r
r
r
r
r
F⊥ = F − FAB = 300 j − (73.5i + 220 j + 110k ) = {−73.5i + 80 j − 110k }N
45
Lecture 3
Mechanics (1)
Delta University
Prof.
M. Safwat
Eng.
Malek
Abuwarda
Faculty of Engineering
Solution
Magnitude can be determined from F┴ or from
Pythagorean Theorem,
r
F⊥ =
=
r2 r
F − FAB
2
(300N )2 − (257.1N )2
= 155N
46
Lecture 3
Mechanics (1)
Prof.
M. Safwat
Eng.
Malek
Abuwarda
23