Transcript Force Vector 1 - UniMAP Portal

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Engineering Mechanics
STATIC
FORCE VECTORS
Chapter Outline
1. Scalars and Vectors
2. Vector Operations
3. Vector Addition of Forces
4. Addition of a System of Coplanar Forces
5. Cartesian Vectors
6. Addition and Subtraction of Cartesian Vectors
7. Position Vectors
8. Force Vector Directed along a Line
9. Dot Product
2.1 Scalars and Vectors
 Scalar
– A quantity characterized by a positive or
negative number
– Indicated by letters in italic such as A
e.g. Mass, volume and length
2.1 Scalars and Vectors
 Vector
– A quantity that has magnitude and direction
e.g.force and moment

– Represent by a letter with an arrow over it, A

– Magnitude is designated as A
– In this subject, vector is presented as A and its
magnitude (positive quantity) as A
2.2 Vector Operations
 Multiplication and Division of a Vector by a
Scalar
- Product of vector A and scalar a = aA
- Magnitude = aA
- Law of multiplication applies e.g. A/a = ( 1/a )A,
a≠0
2.2 Vector Operations
 Vector Addition
- Addition of two vectors A and B gives a
resultant vector R by the parallelogram law
- Result R can be found by triangle construction
- Communicative e.g. R = A + B = B + A
- Special case: Vectors A and B are collinear
(both have the same line of action)
2.2 Vector Operations
 Vector Subtraction
- Special case of addition
e.g. R’ = A – B = A + ( - B )
- Rules of Vector Addition Applies
2.3 Vector Addition of Forces
Finding a Resultant Force
 Parallelogram law is carried out to find the
resultant force
 Resultant,
FR = ( F1 + F2 )
2.3 Vector Addition of Forces
Procedure for Analysis
 Parallelogram Law
Make a sketch using the parallelogram law
 2 components forces add to form the
resultant force
 Resultant force is shown by the diagonal of
the parallelogram
 The components is shown by the sides of
the parallelogram

2.3 Vector Addition of Forces
Procedure for Analysis
 Trigonometry
Redraw half portion of the parallelogram
 Magnitude of the resultant force can be
determined by the law of cosines
 Direction if the resultant force can be
determined by the law of sine


Magnitude of the two components can be
determined by the law of sine
Example 2.1
The screw eye is subjected to two forces, F1 and
F2. Determine the magnitude and direction of the
resultant force.
Solution
Parallelogram Law
Unknown: magnitude of FR and angle θ
Solution
Trigonometry
Law of Cosines
2
2
FR  100 N   150 N   2100 N 150 N  cos115
 10000  22500  30000 0.4226  212.6 N  213N
Law of Sines
150 N 212.6 N

sin  sin 115
150 N
0.9063
sin  
212.6 N
  39.8
Solution
Trigonometry
Direction Φ of FR measured from the horizontal
  39.8  15
 54.8 
Exercise 1:
 Determine magnitude of the resultant force acting on
the screw eye and its direction measured clockwise
from the x- axis
Exercise 2:
 Two forces act on the hook. Determine the magnitude
of the resultant force.
2.4 Addition of a System of Coplanar
Forces
 When a force resolved into two components
along the x and y axes, the components are
called rectangular components.
 Can represent in scalar notation or cartesan
vector notation.
2.4 Addition of a System of Coplanar
Forces
 Scalar Notation
 x and y axes are designated positive and
negative
 Components of forces expressed as
algebraic scalars
F  Fx  Fy
Fx  F cos  and Fy  F sin 
2.4 Addition of a System of Coplanar
Forces
 Cartesian Vector Notation
 Cartesian unit vectors i and j are used to
designate the x and y directions
 Unit vectors i and j have dimensionless
magnitude of unity ( = 1 )
 Magnitude is always a positive quantity,
represented by scalars Fx and Fy
F  Fx i  Fy j
2.4 Addition of a System of Coplanar
Forces
 Coplanar Force Resultants
To determine resultant of several coplanar forces:
 Resolve force into x and y components
 Addition of the respective components using scalar
algebra
 Resultant force is found using the parallelogram
law

Cartesian vector notation:
F1  F1x i  F1 y j
F2   F2 x i  F2 y j
F3  F3 x i  F3 y j
2.4 Addition of a System of Coplanar
Forces
 Coplanar Force Resultants
 Vector resultant is therefore
FR  F1  F2  F3
 FRx i  FRy  j

If scalar notation are used
FRx  F1x  F2 x  F3 x
FRy  F1 y  F2 y  F3 y
2.4 Addition of a System of Coplanar
Forces
 Coplanar Force Resultants
 In all cases we have
FRx   Fx
FRy   Fy

* Take note of sign conventions
Magnitude of FR can be found by Pythagorean
Theorem
FR  F  F
2
Rx
2
Ry
and   tan
-1
FRy
FRx
Example 2.5
Determine x and y components of F1 and F2
acting on the boom. Express each force as a
Cartesian vector.
Solution
Scalar Notation
F1x  200 sin 30 N  100 N  100 N 
F1 y  200 cos 30 N  173N  173N 
Hence, from the slope triangle, we have
5
  tan  
 12 
1
Solution
By similar triangles we have
 12 
F2 x  260   240 N
 13 
5
F2 y  260   100 N
 13 
Scalar Notation:
F2 x  240 N 
F2 y  100 N  100 N 
F1   100i  173 jN
Cartesian Vector Notation:
F2  240i  100 jN
Solution
Scalar Notation
F1x  200 sin 30 N  100 N  100 N 
F1 y  200 cos 30 N  173N  173N 
Hence, from the slope triangle, we have:
5
  tan 1  
 12 
Cartesian Vector Notation
F1   100i  173 jN
F2  240i  100 jN
Example 2.6
The link is subjected to two forces F1 and F2.
Determine the magnitude and orientation of the
resultant force.
Solution I
Scalar Notation:
FRx  Fx :
FRx  600 cos 30 N  400 sin 45 N
 236.8 N 
FRy  Fy :
FRy  600 sin 30 N  400 cos 45 N
 582.8 N 
Solution I
Resultant Force
FR 
236.8N 2  582.8N 2
 629 N
From vector addition, direction angle θ is
 582.8 N 
  tan 

 236.8 N 
 67.9
1
Solution II
Cartesian Vector Notation
F1 = { 600cos30°i + 600sin30°j } N
F2 = { -400sin45°i + 400cos45°j } N
Thus,
FR = F1 + F2
= (600cos30ºN - 400sin45ºN)i
+ (600sin30ºN + 400cos45ºN)j
= {236.8i + 582.8j}N
The magnitude and direction of FR are
determined in the same manner as before.
Exercise 1:
 Determine the magnitude and direction of the
resultant force.
Exercise 2:
 Determine the magnitude of the resultant force and
its direction θ measured counterclockwise from the
positive x-axis.