This document discusses force vectors and their representation. It covers key topics such as: - Scalars and vectors, and examples of each - Classifying forces as distributed or concentrated, and contact or body forces - Representing the direction and magnitude of a force vector using unit vectors - Performing vector addition and subtraction of forces using trigonometric relationships and Cartesian components - Solving examples of adding and subtracting forces to determine the resultant force vector

1. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-1
EML 3004C
CHAPTER TWO
Force Vectors

2. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-2
EML 3004C
2.1 Scalars and Vectors
 Force Vectors
 Scalars : A quantity represented be a number (positive or
negative)
Ex: Mass, Volume, Length
(in the book scalars are represented by italics)
 Vectors : A quantity which has both
A – magnitude (scalar)
B – direction (sense)
Ex: position, force, moment
Line of action
Sense

3. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-3
EML 3004C
Forces
 Classification of Forces
 Contact
1 – Contacting or surface forces (mechanical)
2 – Non-Contacting or body forces (gravitational, weight)
 Area
1 – Distributed Force, uniform and non-uniform
2 – Concentrated Force

4. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-4
EML 3004C
 Classification of Forces
 Force System
1 – Concurrent : all forces pass
through a point
2 – Coplanar : in the same plane
3 – Parallel : parallel line of action
4 – Collinear : common line of
action
 Three Types
1 – Free (direction, magnitude and
sense)
2 – Sliding
3 – Fixed

A
O
Origin
Forces

5. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-5
EML 3004C
2.2 Basic Vector Operations
Properties of Vectors
1 – Vector Addition
2 – Vector Subtraction
3 – Vector Multiplication

6. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-6
EML 3004C
Trigonometric Relations of a triangle


 sin
sin
sin
C
B
A


C
AB
B
A
C cos
2
2
2
2



A B
C



Phythagorean theorem is valid only for a right angled
triangle. For any triangle (not necessarily right angled)
Sine Law
Cosine Law

7. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-7
EML 3004C
Example
Ex: If the angle between F1 and F2 =60o and F1 = 54N
and F2 = 60N.
Find: the Resultant force and the angle .

R
F1
F2
120o
60o

8. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-8
EML 3004C
Example… contd
 
o
R
F
R
F
N
R
R
F
F
F
F
R
26
.
28
60
sin
sin
120
sin
sin
8
.
98
77
.
98
120
cos
54
60
2
54
60
cos
2
1
1
2
2
2
2
1
2
2
2
1
2



















R
F1
F2
120o
60o

9. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-9
EML 3004C
Properties of Parallelogram
A – sum of the three angles in a triangle is 180o
B – sum of the interior angles is 360o
C- opposite sides are equal
Parallelogram Law






 180


 

10. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-10
EML 3004C
Vector Addition
A
B
B
A
R 



R
A
B
A
B
C
B
A
R 


R
C
A
B

11. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-11
EML 3004C
Vector Subtraction
Vector Subtraction
B
A A
A - B
-B
 
B
A
B
A 




12. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-12
EML 3004C
2.3 Vector addition of forces
If we consider, only two forces at a time then the result
can be obtained using parallelogram law, and by using
law of sines and cosines of triangles.
Even if we have multiple (say 5) forces, we take two at
a time to resolve the resultant one by one.
Consider another example in Example 2.4 (a) page 24
Given 1, 2 and
F F 
To find: Resultant
Procedure: Use law of cosines to find the
magnitude, law of sines find the angles

13. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-13
EML 3004C
2.3 Vector addition of forces (example 2.4)
Given Resultant R 1 , 30
kN 
  
To find: 1 2
and
F F
Solution:
1. Draw the vector diagram
2. Use the sine law to find
1 2
and
F F
1
1
2
2
1000
653
30 130
1000
446
20 130
F
F N
Sin Sin
F
F N
Sin Sin
  
  

14. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-14
EML 3004C
2.4 Cartesian Coordinate Systems
A much more logical way to add/subtract/manipulate vectors is to
represent the vector in Cartesian coordinate system. Here we need to
find the components of the vector in x,y and z directions.
Simplification of Vector Analysis
i
j
k
x
y
z
x
y
A
z
n
en



x
y
z

15. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-15
EML 3004C
Vector Representation in terms unit vector
Suppose we know the magnitude of a vector in any
arbitrary orientation, how do we represent the vector?
Unit Vector : a vector with a unit magnitude
A
A
e
e
A
A
n
n




A
en
^

16. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-16
EML 3004C
2.5 Right-Handed Coordinate System
 Right-Handed System
If the thumb of the
right hand points in
the direction of the
positive z-axis when
the fingers are pointed
in the x-direction &
curled from the x-axis
to the y-axis. x
y
z

17. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-17
EML 3004C
Components of a Vector
 Cartesian (Rectangular) Components of a Vector
z
y
x A
A
A
A 


y
A
Ay
Ax
Az
z
x
y
A
Ay
Ax x
In 2-D y
x A
A
A 

In 3-D

18. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-18
EML 3004C
Cartesian Vectors
 Cartesian Unit Vectors
k
A
A
j
A
A
i
A
A
z
z
y
y
x
x
ˆ
ˆ
ˆ


 k
A
j
A
i
A
A x
y
x
ˆ
ˆ
ˆ 



y
A = Aen
Ay = Ayj
Ax = Axi
Az = Azk
z
x
n
i
j
k en

19. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-19
EML 3004C
Cartesian Vectors
 Magnitude of a Cartesian Vector
2
2
2
z
y
x A
A
A
A 


y
A
Ay
Ax
Az
z
x

20. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-20
EML 3004C
2.6. Addition and Subtraction of vectors
y
F
Fy
Fx
Fz
z
x



x
y
z
z
z
y
y
x
x
F
F
F
F
F
F



cos
cos
cos



F
F
F
F
F
F
z
z
y
y
x
x






cos
cos
cos

21. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-21
EML 3004C
Force Analysis
2 2 2
x y z
x y z
x y z
x y z
f x y z
f
F = F +F +F
ˆ ˆ ˆ
F = F i+F j+F k
ˆ ˆ ˆ
F = Fcosθ i+Fcosθ j+Fcosθ k
ˆ ˆ ˆ
F = F(cosθ i+cosθ j+cosθ k)
ˆ ˆ ˆ
Û = cosθ i+cosθ j+cosθ k
ˆ
F = FU
y
F
Fy
Fx
Fz
z
x



x
y
z

22. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-22
EML 3004C
Force Analysis
Ex:
y
z
x 10ft
6ft
8ft
F = 600
2 2 2
2 2 2
x x
y y
z z
d= x +y +z
= 6 +10 +8
=14.14ft
6
F =Fcosθ =600 =254lb
14.14
10
F =Fcosθ =600 =424lb
14.14
8
F =Fcosθ =600 =339lb
14.14
ˆ ˆ ˆ
F=255i+424j+339klb

23. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-23
EML 3004C
Summary of the Force Analysis
y
F
z
x
z
z
y
y
x
x
F
F
F
F
F
F



cos
cos
cos



2
2
2
z
y
x F
F
F
F
F 



F
Fx
x
1
cos


F
Fz
z
1
cos


2 2 2
Recall,
1
x y z
Cos Cos Cos
  
  
F
Fy
y
1
cos



ˆ
ˆ ˆ
ˆ
ˆ ˆ
cos cos cos
x y z
x y z
x y z
F F F F
F i F j F k
F i F j F k
  
  
  
  
x
F
Z
F
y
F

24. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-24
EML 3004C
2.6 Addition/Subtraction of Cartesion vectors
Since any vector in 3-D can be expressed as components in x,y,z
directions, we just need to add the corresponding components
since the components are scalars.
ˆ
ˆ ˆ
ˆ
ˆ ˆ
x y z
x y z
A A i A j A k
B B i B j B k
  
  
Then the addition
Then the subtraction
R A B
 
      ˆ
ˆ ˆ
x x y y z z
R A B i A B j A B k
     
R A B
  
      ˆ
ˆ ˆ
x x y y z z
R A B i A B j A B k
     

25. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-25
EML 3004C
2.6 ---Example 2-9, pg 41
Determine the magnitude and the coordinate direction angles of
the resultant force on the ring
Solution:
   
 
1 2
60 80 50 100 100
50 40 180 lb
R
F F F F
j k i j k
i j k
  
    
  

     
2 2 2
50 40 180
191.0 lb
R
F    


26. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-26
EML 3004C
2.6 ---Example 2-9…2
Unit vector and direction cosines
50 40 180
191.0 191.0 191.0
ˆ
ˆ ˆ
ˆ
ˆ ˆ
0.2617 0.2094 0.9422
R
R
F
R
F
u i j k
F
i j k
   
  
From these components, we can
determine the angles
0
0
0
cos 0.2617 =74.8
cos 0.2094 =102
cos 0.9422 =19.6
 
 
 

 


27. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-27
EML 3004C
2.4---Problem 2-27 (pg 35)
Four concurrent forces act on the plate.
Determine the magnitude of the resultant
force and its orientation measured
counterclockwise from the positive x axis.
Solution:
4
; 60 38cos30 + (50)
5
= 67.09 lb
3
; = 100 38 cos30 (50)
5
= 51 lb
Rx x Rx
Ry y Ry
F F F
F F F


    

    



Draw the free body diagram. Then resolve
forces in x and y directions.

28. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-28
EML 3004C
2.4---Problem 2-27…..2
2 2 2 2
1 1
Magnitude
= = 67.09 51 = 84.3 lb
Direction
51
tan = tan = 37.2
67.09
R Rx Ry
Ry
Rx
F F F
F
F
  
 
 
Find the magnitude and
direction.

29. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-29
EML 3004C
2.7. Position Vectors
y
z
x
r
( xa, ya, za)
(xb, yb, zb)
•Position vectors can be determined using the coordinates of
the end and beginning of the vector
     k
z
z
j
y
y
i
x
x
r
k
z
j
y
i
x
r
k
z
j
y
i
x
r
r
r
r
A
B
A
B
A
B
A
A
A
A
B
B
B
B
A
B
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ














A
r
B
r

30. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-30
EML 3004C
2.8---Problem 2-51, page 56
Determine the length of the connecting rod AB by first formulating
a Cartesian position vector form A to B and then determining its
magnitude.
 
 
2 2
Position vector:
r = 16-(-5sin30 ) i + (0-5cos30 )j
= 18.5i - 4.330j in
Magnitude:
18.5 4.330 19.0 in Ans
r
 
  
Solution

31. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-31
EML 3004C
Force is oriented along the vector AB (line AB)
Force Vector Along a Line
A force may be represented by a magnitude & a position
x
y
z
A
B
F
F
AB
AB
F
F
AB
AB
u
u
F
F
AB
AB




Unit vector along
the line AB

32. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-32
EML 3004C
2.6---Problem 2-36 (pg 47)
 
1 2
R 2
The two forces F and F acting at the end of the pipe have a resultant force
F 120 N. Determine the magnitude and direction angles of F .
i

 
 
 
 
   
 
1
2 2 2 2
1 2
2 2 2
2
F = 85 cos 30 sin45 i + cos30 cos45 sin 30 k
= 52.052i + 52.252j - 42.5k N
F = i + j + k N
F = 120i N
F =F +F
120i = 52.052i + 52.052j - 42.5k i + j + k
120i = 5
Force Vector
2.052+
x y z
R
R
x y z
x
j
F F F
F F F
F
     

   
2 2
i + 52.052+ j+ 42.5 k
y x
F F 
Solution:

33. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-33
EML 3004C
2.6---Problem 2-36…..2
2
2 2
2 2
2
Equating i, j, k component yeilds:
52.052 + 120 67.948N
52.052+ 0 52.052N
42.5 0 42.5 N
67.948i-52.0
x x
y y
x x
F F
F F
F F
F
 
  
  
  
2
2 2 2
2
52j+42.5k N
Magnitude of F
6.9487 ( 52.052) 42.5 = 95.56N = 95.6N Ans
F    

34. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-34
EML 3004C
2.6---Problem 2-36…..3
2
2
F
2
Coordinate direction angles:
F 67.948i-52.052j+42.05k
u
95.56
=0.7110i-0.5447j+0.4447k
cos 0.7110 =44.7 Ans
cos 0.5447 =
F
 
 
 
 
  123 Ans
cos 0.4447 =63.6 Ans
 

 

35. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-35
EML 3004C

2.9. Dot Product
Dot Product (Scalar Product) y

x
A
ˆ (1)cos
x
A A i A 
  
x y z
x y z
x x y y z z
A = A +A +A
B = B +B +B
A B = A B +A B +A B
i j k
i j k
A B=B A=ABcosθ
 
x
A
y
A
A
B

36. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-36
EML 3004C
2.9---Problem 2-76, pg 65
A force of F = 80 N is applied to the
handle of the wrench. Determine the
magnitudes of the components of the
force acting along the axis AB of the
wrench handle and perpendicular to
it.
F
U cos30 sin 45 cos30 cos45 j+sin30 k
=-0.6124i + 0.6124j + 0.58k
U j
ˆ
F = u 80(-0.6124i+0.6124j+0.5k)
= (-48.990i+48.99j+40k) N
F =F u ( 48.990i + 48.99j + 40k) (-j)
F
AB
AB AB
i
F
      
 

 
=49.0 N
Hand first rotates 45 in the xy
plane and 30 in that plane

37. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-37
EML 3004C
2.9---Problem 2-76…..2
2 2
.
2 2
Negative sign indicates that acts in the direction opposite to that of U
= 80 ( 49.0) 63.2N
Also, from problem 2-104 =128
cos 80cos128 4
AB AB
per AB
AB
F
F F F
F F


 
  

   
.
9.0N
sin 80cos128 = 63.2N
per
F F 
  

38. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-38
EML 3004C
Application of Dot Product
Component of a Vector along a line

A n
A
A
A  A cos
 A  ˆ
U
A  A ˆ
U  (A  ˆ
U) ˆ
U
A  A  A
A  A sin 

39. Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-39
EML 3004C
Application of dot product
Angle between two vectors

A
B
1
cos
cos
A B AB
AB
AB

 
 
 
  
 