STATICS
Assist. Prof. Dr. Cenk Üstündağ
2
Force Vectors
Chapter Objectives
• Parallelogram Law
• Cartesian vector form
• Dot product and angle between 2 vectors
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
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. Position, 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 sines
– Magnitude of the two components can be determined by
the law of sines
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
100 N 2  150 N 2  2100 N 150 N  cos115
10000  22500  30000 0.4226   212.6 N  213 N
FR 

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 
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  173 N  173 N 
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: F  240 N 
2x
F2 y  100 N  100 N 
Cartesian Vector Notation: F1   100i  173 jN
F2  240i  100 jN
Solution
Scalar Notation
F1x  200 sin 30 N  100 N  100 N 
F1 y  200 cos 30 N  173 N  173 N 
Hence, from the slope triangle, we have:
5
  tan  
12
1
Cartesian Vector Notation
F1  100i 173jN
F2  240i 100j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 
236.8 N 2  582.8 N 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.