Charactteristics of forces; Vector to represent forces; Classification of forces; What is force system; Principles of forces; Resultant of forces; Components of forces; Solved numericals; examples; Solved problems; excercise;

1. Concurrent Force
Systems
Lecture-03
CL-101 ENGINEERING MECHANICS
B. Tech Semester-I
Prof. Samirsinh P Parmar
Mail: samirddu@gmail.com
Asst. Professor, Department of Civil Engineering,
Faculty of Technology,
Dharmsinh Desai University, Nadiad-387001
Gujarat, INDIA

2. Objectives
 To understand the basic characteristics of forces
 To understand the classification of force systems
 To understand some force principles
 To know how to obtain the resultant of forces in 2D and 3D systems
 To know how to obtain the components of forces in 2D and 3D
systems

3. Characteristics of forces
 Force: Vector with magnitude and direction
 Magnitude – a positive numerical value representing
the size or amount of the force
 Directions – the slope and the sense of a line segment
used to represent the force
– Described by angles or dimensions
– A negative sign usually represents opposite
direction
 Point of application
– A point where the force is applied
– A line of action = a straight line extending through
the point of application in the direction of the
force
 The force is a physical quantity that needs to be
represented using a mathematical quantity

4. Example

i
1000 N
j
Point of application
Line of action
magnitude
direction

5. Vector to represent Force
A vector is the mathematical representation that best
describes a force
A vector is characterized by its magnitude and
direction/sense
Math operations and manipulations of vectors can be used
in the force analysis

6. Free, sliding, and fixed vectors
 Vectors have magnitudes, slopes, and senses, and lines of applications
 A free vector
– The application line does not pass a certain point in space
 A sliding vector
– The application line passes a certain point in space
 A fixed vector
– The application line passes a certain point in space
– The application point of the vector is fixed

7. Vector/force notation
The symbol representing the force  bold face or underlined
letters
The magnitude of the force  lightface (in the text book, +
italic)
A  A or A  A

8. Classification of forces
 Based on the characteristic of the interacting bodies:
– Contacting vs. Non-contacting forces
 Surface force (contacting force)
– Examples:
» Pushing/pulling force
» Frictions
 Body force (non-contacting force)
– Examples:
» Gravitational force
» Electromagnetic force

9. Classification of forces
 Based on the area (or volume) over
which the force is acting
– Distributed vs. Concentrated forces
 Distributed force
– The application area is relatively large
compare to the whole loaded body
– Uniform vs. Non-uniform
 Concentrated force
– The application area is relatively small
compare to the whole loaded body

10. What is a force system?
 A number of forces (in 2D or 3D system) that is
treated as a group:
 A concurrent force system
– All of the action lines intersect at a common
point
 A coplanar force system
– All of the forces lie in the same plane
 A parallel force system
– All of the action lines are parallel
 A collinear force system
– All of the forces share a common line of action

11. The external and internal effects
 A force exerted on the body has two effects:
– External effects
» Change of motion
» Resisting forces (reactions)
– Internal effects
» The tendency of the body to deform  develop strain, stresses
– If the force system does not produce change of motion
» The forces are said to be in balance
» The body is said to be in (mechanical) equilibrium

12. External and internal effects
Example 1: The body changes in motion
Example 2: The body deforms and produces
(support) reactions  The forces must be in
balance
F
Not fixed, no (horizontal) support
a
F
Fixed support
Support Reactions

13. Principle for force systems
 Two or more force systems are equivalent when their applications to a body
produce the same external effect
 Transmissibility
 Reduction =
– A process to create a simpler equivalent system
– to reduce the number of forces by obtaining the “resultant” of the forces
 Resolution =
– The opposite of reduction
– to find “the components” of a force vector 
“breaking up” the resultant forces

14. Principle of Transmissibility
 Many times, the rigid body assumption is taken  only the external effects are the
interest
 The external effect of a force on a rigid body is the same for all points of application
of the force along its line of action



15. Resultant of Forces – Review on vector addition
 Vector addition
 Triangle method (head-to-tail
method)
– Note: the tail of the first vector
and the head of the last vector
become the tail and head of the
resultant  principle of the force
polygon/triangle
 Parallelogram method
– Note: the resultant is the diagonal
of the parallelogram formed by
the vectors being summed
R  A B  B  A
R
A
B
B
A
R

16. Resultant of Forces – Review on geometric laws
Law of Sines
Laws of Cosines
c2
 a2
b2
 2abcos
b2
 a2
 c2
 2accos
a2
 b2
 c2
 2accos
A
B
C
c
a
b




17. Resultant of two concurrent forces
 The magnitude of the resultant (R) is given by
1 2 1 2
R2
 F2
 F2
 2F F cos
1 2 1 2
 The direction (relative to the direction of F1) can be given by the law
of sines
R2
 F2
 F2
 2F F cos
R
sin  
F2 sin
Pay attention to the angle
and the sign of the last
term !!!

18. Resultant of three concurrent forces and more
 Basically it is a repetition of finding resultant of two forces
 The sequence of the addition process is arbitrary
 The “force polygons” may be different
 The final resultant has to be the same

19. Resultant of more than two forces
 The polygon method becomes tedious when dealing with three and more forces
 It’s getting worse when we deal with 3D cases
 It is preferable to use “rectangular-component” method

20. Example Problem 2-1
 Determine:
– The resultant force (R)
– The angle  between the R and the x-axis
 Answer:
– The magnitude of R is given by
R2
 9002
 6002
 2(900)(600)cos400
R 1413.3 1413lb
– The angle  between the R and the 900-lb
force is given by
sin 
sin(1800
 400
)
600 1413.3
 15.836o
– The angle  therefore is
 15.8360
 350
 50.80

21. Example Problem 2-2
Determine
– The resultant R
– The angle between the R
and the x-axis

22. Another example
 If the resultant of the force system is zero, determine
– The force FB
– The angle between the FB and the x-axis

23. Force components

24. Resolution of a force into components
 The components of a resultant force are
not unique !!
 The direction of the components must be
fixed (given)
R  A B  (G  I)  H
 C  D  E  F

25.  Steps:
– Draw lines parallel to u and v crossing
the tip of the R
– Together with the original u and v
lines, these two lines produce the
parallelogram
– The sides of the parallelogram
represent the components of R
– Use law of sines to determine the
magnitudes of the components
Parallel to v
How to obtain the components of a force
(arbitrary component directions)?
Parallel to u
900
sin110o
Fv
sin 25o
Fu
sin 45o


9 0 0 sin 4 5 o
s i n 11 0 0
9 0 0 sin 2 5 0
s i n 11 0 o
 4 0 5 N
Fv 
 6 7 7 N
Fu 

26. Example Problem 2-5
 Determine the components of F =
100 kN along the bars AB and AC
 Hints:
– Construct the force
triangle/parallelogram
– Determine the angles   
– Utilize the law of sines

27. Another example
Determine the magnitude of the components of R in the
directions along u and v, when R = 1500 N

28. Rectangular components of a force
 What and Why rectangular components?
– Rectangular components  all of the components are perpendicular to each other (mutually
perpendicular)
– Why? One of the angle is 90o ==> simple
 Utilization of unit vectors
 Rectangular components in 2D and 3D
 Utilization of the Cartesian c.s.
Arbitrary rectangular

29. The Cartesian coordinate system
 The Cartesian coordinate axes
are arranged following the
right-hand system (shown on
the right)
 The setting of the system is
arbitrary, but the results of the
analysis must be independent of
the chosen system x
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y
z

30. Unit vectors
 A dimensionless vector of unit magnitude
 The very basic coordinate system used to specify coordinates in the space is the
Cartesian c.s.
 The unit vectors along the Cartesian coordinate axis x, y and z are i, j, k,
respectively
 The symbol en will be used to indicate a unit vector in some n- direction (not x, y,
nor z)
 Any vector can be represented as a multiplication of a magnitude and a
unit vector
A  Aen  Aen
B   Ben  Ben
A is in the positive
direction along n
B is in the negative
direction along n
en

A

A
A A
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31. The rectangular components of a force in 2D system
 While the components must be perpendicular to each other, the directions do not have to be
parallel or perpendicular to the horizontal or vertical directions
x
y
Fy = Fy j
Fx = Fx i
i
j
F

F  Fx  Fy  Fxi  Fy j
F  F 2
 F 2
x y
  tan1 Fy
Fx
Fx  F cos
Fy  F sin

32. F
F
F
F
F
F z
z
y
y
x
x
z
y
x
z z
y y
F 2
 F 2
 F2
F 
F  F cos
F  F cos
Fx  F cosx
1
1
1
  cos
  cos
  cos
The rectangular components in 3D systems
F
n
F
F
Fxi  Fy j Fzk

e 
 Fxi  Fy j Fzk
F  Fen
F  Fx  Fy  Fz
x
y
z
Fy = Fy j
Fx = Fx i
Fz = Fz k
F
i
k
j
en
z
x y
en  cosxi  cosy j coszk

33. Dot Products of two vectors
AB  B A  A B cos  ABcos

A
B
It’s a scalar !!!
Special cosines:
Cos 0o = 1
Cos 30o = ½ √3
Cos 45o = ½ √2
Cos 60o = 0.5
Cos 90o = 0

34.  The dot product can be used to obtain the rectangular components of a force (a vector in
general)
At  A An
The component along et
Remember, en and et are perpendicular
An  (Aen )en
An  Anen
An  Aen  Acosn (magnitude)
(the vectorial component
in the n direction)
The component along en
Dot products and rectangular components

35. Cartesian rectangular components
 The dot product is particularly useful when the unit vectors
are of the Cartesian system (the i, j, k)
x
y
Fy = Fy j
Fx = Fx i
i
j
F

Fx  Fi  F cos
Fy  F j  F cos(90 )
 F sin
Also, in 3D,
Fz  Fk
90-
F  Fx  Fy  Fxi  Fy j  (Fi)i  (F j)j
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36. More usage of dot products …
 Dot products of two vectors written in Cartesian system
 The magnitude of a vector (could be a force vector), here A is the
vector magnitude
A A  A2
cos0  A2
 A A  A A  A A
x x y y z z
 The angle between two vectors (say between vectors A
and B)
AB  Ax Bx  Ay By  Az Bz



 AB
1 Ax Bx  Ay By  Az Bz 
  cos

37. The rectangular components of arbitrary direction
 Fx cosxn  Fy cosyn  Fz coszn
n
 (Fxi  Fy j Fzk)e
Fn  Fen
F  Fx  Fy  Fz
 Fxi  Fy j Fzk
F  Fnen  Ftet
en  cosxni  cosyn j cosznk
z
 Fxi en  Fy jen  Fzk en x
y
Fy = Fy j
Fx = Fx i
Fz = Fz k
F
i
k
j
en
zn
xn
yn
Ft
Fn
Can you show the following?

38. Summarizing ….
 The components of a force resultant are not unique
 Graphical methods (triangular or parallelogram methods) combined with law of
sinus and law of cosines can be used to obtain components in arbitrary direction
 Rectangular components are components of a force (vector) that perpendicular to each
other
 The dot product can be used to
– obtain rectangular components of a force vector
– obtain the magnitude of a force vector (by performing self- dot-product)
– Obtain the angle between two (force) vectors

39. Example Problem 2-6
 Find the x and y scalar components of the force
 Find the x’ and y’ scalar components of the force
 Express the force F in Cartesian vector form for the xy- and x’y’- axes

40. Example Problem 2-6


Fx  F cos Fy  F cos(90 )
Fx'  F cos Fy'  F cos(90  )
  90  28  62o
  62 30  32o
Fx  450cos62  211N
Fy  450sin 62  397N
Fx'  450cos32  382N
Fy  450sin32  238N
F  (211i  397j)N  (382ex'  238ey'
)N
Writing the F in Cartesian vector form:
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41. Example Problem 2-8
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 Find the angles x, y, and z
x is the angle between OB and x axis and so
on ..)
 The x, y, and x scalar components of
the force.
 The rectangular component Fn of the force
along line OA
 The rectangular component of the force
perpendicular to line OA (say Ft)
B

42. Example Problem 2-8
 To find the angles:
– Find the length of the
diagonal OB, say d
– d = 5.831 m
– Use cosines to get the
angles
 The scalar components in the
x, y, and z directions:
B
z
y
o
x
3
5.831
4
5.831
3
5.831
 59.0o
 46.7o
 59.0
1
  cos1
  cos1
  cos
Fx  F cosx 12.862kN
Fy  F cosy 17.150kN
Fz  F cosz 12.862kN
F  (12.862i 17.150j12.862k)kN
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43. Example Problem 2-8
 To find the rectangular component Fn of
the force along line OA:
– Needs the unit vector along OA
– Method 1 : Follow the method
described in the book
– Method 2: utilize the vector
position of A (basically vector OA)
– Remember, that any vector can be
represented as a multiplication of
its magnitude and a unit vector
along its line of application
OA  rA  3i 1j 3k r
e

3i 1j 3k
 0.688i  0.230j 0.688k
4.36


rA 3i 1j3k
32
12
32
A
OA

44. Example Problem 8-2
FOA  FeOA
 The scalar component of F along OA
FOA  (12.862i 17.150j12.862k)(0.688i  0.230j 0.688k)
FOA 12.8620.68817.1500.23012.8620.688  21.643kN
 The vector component of F along OA
FOA  (FeOA )eOA  21.6(0.688i  0.230 j  0.688k)
14.86i  4.97j14.86k
 The vector component of F perpendicular to OA
Ft  F  FOA  (12.862i 17.150j12.862k)  (14.86i  4.97j14.86k)
 (2i 12.18j 2k)
 The scalar component of F perpendicular to OA
F | F || (2i 12.18j 2k) | (2)2
12.182
 (2)2
12.50kN
t t
OA t
F 2
 F 2
 21.6432
12.502
 25kN
Check: F 

45.  The Cartesian rectangular components of forces can be utilized to obtain the
resultant of the forces
x
y
F1
F2
F1x
F2x
F2y
F1y
•Adding the x vector components, we obtain the x vector
component of the resultant
•Adding the y vector components, we obtain the y
vector component of the resultant
•The resultant can be obtained by performing the
vector addition of these two vector components
Rx  Fx  F1x  F2x
Ry  Fy  F1y  F2 y
R  Rx  Ry  Rxi  Ry j
Resultants by rectangular components

46. Resultants by rectangular components
 The magnitude of the resultant
 The angles formed by the resultant and the Cartesian axes
 All of the above results can be easily extended for 3D system
 The scalar components of the resultant
Rx  F1x  F2x  (F1x  F2x )i  Rxi
Ry  F1y  F2 y  (F1y  F2 y )j  Ry j
R  R2
 R2
x y
R
R R
R y
y
x
x
1
1
  cos
  cos

47. HW Problem
 Determine the non-rectangular components of R

48. HW Problem
Determine the
components of F1 and F2
in x-y and x’-y’ systems

49. HW Problem
 Express the cable tension in Cartesian
form
 Determine the magnitude of the
rectangular component of the cable force
 Determine the angle  between cables
AD and BD
Typo in the problem!!!
B(4.9,-7.6,0)
C(-7.6,-4.6,0)
Don’t worry if you don’t get the solution in the back of the
book

50. HW Problem
e
 Determine the scalar components
 Express the force in Cartesian vector form
 Determine the angle  between th force and line AB

51. HW problems
 Given: F1 = 500 lb, F2 = 300
lb, F3 = 200 lb
 Determine the resultant
 Express the resultant in the
Cartesian format
 Find the angles formed by the
resultant and the coordinate
axes

52. HW Problem
 Given T1 and T2 are 650 lb,
 Determine P so that the resultant of T1, T2 and P is zero