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EML 3004C
CHAPTER TWO
Force Vectors
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-1
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
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-2
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
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-3
EML 3004C
Forces
 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
O
3 – Fixed
A

Origin
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-4
EML 3004C
2.2 Basic Vector Operations
Properties of Vectors
1 – Vector Addition
2 – Vector Subtraction
3 – Vector Multiplication
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-5
EML 3004C
Trigonometric Relations of a triangle
Phythagorean theorem is valid only for a right angled
triangle. For any triangle (not necessarily right angled)
A
B
C


sin  sin  sin 
Sine Law
C 2  A2  B 2  2 AB cos C Cosine Law
A

B


C
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-6
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
120o

60o
F2
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-7
EML 3004C
Example… contd
R 2  F12  F22  2 F1 F2 cos 
R  60  54  2  60  54  cos120 
2
2
2
R  98.77  98.8 N
sin  sin 120

F1
R
F1
sin   sin 60
R
  28.26o
Namas Chandra
Introduction to Mechanical engineering
R
F1
120o

60o
F2
Hibbler
Chapter 2-8
EML 3004C
Parallelogram Law
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

    180
 
Namas Chandra
Introduction to Mechanical engineering



Hibbler
Chapter 2-9
EML 3004C
Vector Addition
B
B
R
A
A
C
A
B
R  A B  B  A
R
R  A B C
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-10
EML 3004C
Vector Subtraction
Vector Subtraction
 
A B  A  B
A-B
A
A
B
Namas Chandra
Introduction to Mechanical engineering
-B
Hibbler
Chapter 2-11
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
F1, F2 and 
To find: Resultant
Procedure: Use law of cosines to find the
magnitude, law of sines find the angles
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-12
EML 3004C
2.3 Vector addition of forces (example 2.4)
Given Resultant R  1kN ,  30
To find:
F1 and F2
Solution:
1. Draw the vector diagram
2. Use the sine law to find
F1 and F2
F1
1000

 F1  653N
Sin30
Sin130
F2
1000

 F2  446 N
Sin20
Sin130
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-13
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
z
z
en
 z
k
n
A
 y
j
y
 x
y
i
x
Namas Chandra
Introduction to Mechanical engineering
x
Hibbler
Chapter 2-14
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 en

A
en 
A
Namas Chandra
Introduction to Mechanical engineering
e^n
A
Hibbler
Chapter 2-15
EML 3004C
2.5 Right-Handed Coordinate System
z
 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.
Namas Chandra
Introduction to Mechanical engineering
y
x
Hibbler
Chapter 2-16
EML 3004C
Components of a Vector
 Cartesian (Rectangular) Components of a Vector
In 3-D A  Ax  Ay  Az
z
Az
y
A
Ay
A
Ay
Ax
Ax
Namas Chandra
Introduction to Mechanical engineering
x
x
y
In 2-D
A  Ax  Ay
Hibbler
Chapter 2-17
EML 3004C
Cartesian Vectors
 Cartesian Unit Vectors
Ax  Ax iˆ
 A  Ax iˆ  Ay ˆj  Ax kˆ
Ay  Ay ˆj
Az  Az kˆ
z
A z = A zk
n
k
A = Aen
en
y
i
Ax = Axi
Namas Chandra
Introduction to Mechanical engineering
x
j
Ay = Ayj
Hibbler
Chapter 2-18
EML 3004C
Cartesian Vectors
 Magnitude of a Cartesian Vector
A  Ax2  Ay2  Az2
z
Az
A
Ay
Ax
y
x
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-19
EML 3004C
2.6. Addition and Subtraction of vectors
Fx  F cos x
Fx
cos  x 
F
Fy
cos  y 
F
F
cos  z  z
F
Fy  F cos y
Fz  F cos  z
z
Fz
 z
F
 y
Fx
 x
Fy
y
x
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-20
EML 3004C
Force Analysis
2
x
2
y
z
2
z
F = F +F +F
Fz
F = Fx ˆi+Fy ˆj+Fz kˆ
 z
F = Fcosθ x ˆi+Fcosθ y ˆj+Fcosθ z kˆ
ˆ
F = F(cosθ x ˆi+cosθ y ˆj+cosθ z k)
Û f = cosθ x ˆi+cosθ y ˆj+cosθ z kˆ
ˆ
F = FU
f
Namas Chandra
Introduction to Mechanical engineering
F
 y
Fx
Fy
y
 x
x
Hibbler
Chapter 2-21
EML 3004C
Force Analysis
Ex:
d= x 2 +y 2 +z 2
= 62 +102 +82
=14.14ft
6
Fx =Fcosθ x =600
=254lb
14.14
10
Fy =Fcosθ y =600
=424lb
14.14
8
Fz =Fcosθ z =600
=339lb
14.14
ˆ
ˆ
ˆ
F=255i+424j+339klb
Namas Chandra
Introduction to Mechanical engineering
z
F = 600
8ft
y
6ft
x
10ft
Hibbler
Chapter 2-22
EML 3004C
Summary of the Force Analysis
F  Fx  Fy  Fz
z
 Fx iˆ  Fy ˆj  Fz kˆ
 F cos x iˆ  F cos y ˆj  F cos z kˆ
FZ
Fx  F cos  x
F
Fy  F cos  y
Fz  F cos  z
 x  cos 1
Fx
F
 y  cos
Fy
Fx
1
Fy
F
x
y
F  F  Fx  Fy  Fz
2
2
2
Recall,
Cos 2 x  Cos 2 y  Cos 2 z  1
Namas Chandra
Introduction to Mechanical engineering
Fz
 z  cos
F
1
Hibbler
Chapter 2-23
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.
A  Axiˆ  Ay ˆj  Az kˆ
B  Bxiˆ  By ˆj  Bz kˆ
Then the addition
R  AB
R   Ax  Bx  iˆ   Ay  By  ˆj   Az  Bz  kˆ
Then the subtraction
R  A  B
R   Ax  Bx  iˆ   Ay  By  ˆj   Az  Bz  kˆ
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-24
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:
FR   F  F1 F2
  60 j  80k    50i  100 j  100k 
 50i  40 j  180k  lb
FR 
 50    40   180 
2
2
2
 191.0 lb
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-25
EML 3004C
2.6 ---Example 2-9…2
Unit vector and direction cosines
uFR 
FR
FR
50 ˆ
40 ˆ
180 ˆ
 191.0
i  191.0
j  191.0
k
 0.2617iˆ  0.2094 ˆj  0.9422kˆ
From these components, we can
determine the angles
cos   0.2617
 =74.80
cos   0.2094
 =1020
cos   0.9422
 =19.60
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-26
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:
Draw the free body diagram. Then resolve
forces in x and y directions.

 FRx   Fx ;

  FRy   Fy ;
Namas Chandra
Introduction to Mechanical engineering
4
FRx  60  38cos30+ (50)
5
= 67.09 lb 
3
FRy = 100  38 cos30  (50)
5
= 51 lb 
Hibbler
Chapter 2-27
EML 3004C
2.4---Problem 2-27…..2
Find the magnitude and
direction.
Magnitude
FR = FRx2  FRy2 = 67.092  512 = 84.3 lb
Direction
  tan
1
FRy
FRx
= tan
Namas Chandra
Introduction to Mechanical engineering
1
51
= 37.2
67.09
Hibbler
Chapter 2-28
EML 3004C
2.7. Position Vectors
z
r  rB  rA
(xb, yb, zb)
r
rB  xB iˆ  y B ˆj  z B kˆ
rA  x Aiˆ  y A ˆj  z A kˆ
( xa, ya, za)
rB
y
rA
x
r  xB  x A iˆ   y B  y A  ˆj  z B  z A kˆ
•Position vectors can be determined using the coordinates of
the end and beginning of the vector
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-29
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.
Solution
Position vector:
r = 16-(-5sin30) i + (0-5cos30)j
= 18.5i - 4.330j in
Magnitude:
r
18.52  4.3302  19.0 in
Namas Chandra
Introduction to Mechanical engineering
Ans
Hibbler
Chapter 2-30
EML 3004C
Force Vector Along a Line
A force may be represented by a magnitude & a position
Force F is oriented along the vector AB (line AB)
B
F  F u AB
u AB 

z
F
Unit vector along
the line AB
A
AB
AB
F  F
Namas Chandra
Introduction to Mechanical engineering
AB
AB
y
x
Hibbler
Chapter 2-31
EML 3004C
2.6---Problem 2-36 (pg 47)
The two forces F1 and F2 acting at the end of the pipe have a resultant force
FR  120i N. Determine the magnitude and direction angles of F2 .
Solution:
Force Vector
F1 = 85 cos 30sin45i + cos30cos45 j  sin 30k
= 52.052i + 52.252j - 42.5k N
F2 =  F2 x i + F2 y j + F2 z k N
FR = 120i N
FR =F1 +F2
120i = 52.052i + 52.052j - 42.5k   F2 xi + F2 y j + F2 z k
120i =  52.052+F2 x  i +  52.052+F2 y  j+  F2 x  42.5  k
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-32
EML 3004C
2.6---Problem 2-36…..2
Equating i, j, k component yeilds:
52.052 + F2 x  120
Fx  67.948N
52.052+ F2 y  0
F2 y  52.052N
F2 x  42.5  0
F2 x  42.5 N
F2  67.948i-52.052j+42.5k N
Magnitude of F2
F2  6.9487 2  (52.052) 2  42.52 = 95.56N = 95.6N
Namas Chandra
Introduction to Mechanical engineering
Ans
Hibbler
Chapter 2-33
EML 3004C
2.6---Problem 2-36…..3
Coordinate direction angles:
u F2 
F2 67.948i-52.052j+42.05k

F2
95.56
=0.7110i-0.5447j+0.4447k
cos  0.7110
 =44.7 Ans
cos  0.5447
 =123 Ans
cos  0.4447
 =63.6 Ans
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-34
EML 3004C
2.9. Dot Product
Dot Product (Scalar Product)
y
A  B=B  A=ABcosθ
A
Ax  A  iˆ  A(1) cos
Ay

Ax
x
A = A x i +A y j +A z k
B

B = Bx i +B y j +Bz k
A B = A x Bx +A y B y +A z Bz
A
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-35
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.
U F  cos 30 sin 45i  cos 30 cos 45 j+sin30k
=-0.6124i + 0.6124j + 0.58k
U AB   j
Hand first rotates 45 in the xy
plane and 30 in that plane
F = Fuˆ F  80(-0.6124i+0.6124j+0.5k)
= (-48.990i+48.99j+40k) N
FAB =F u AB  ( 48.990i + 48.99j + 40k) (-j)
=49.0 N
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-36
EML 3004C
2.9---Problem 2-76…..2
Negative sign indicates that FAB acts in the direction opposite to that of U AB
2
Fper .  F 2  FAB
= 802  (49.0) 2  63.2N
Also, from problem 2-104  =128
FAB  F cos  80cos128  49.0N
Fper .  F sin   80cos128 = 63.2N
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-37
EML 3004C
Application of Dot Product
Component of a Vector along a line
A
 A cos 
ˆ
 AU
A
ˆ  (A  U
ˆ )U
ˆ
A U
A
A
A A
A
 A sin 
A

An
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-38
EML 3004C
Application of dot product
Angle between two vectors
B
A  B  AB cos
 AB 
  cos 

 AB 
1

A
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-39