Chapter 2:
Concurrent force systems
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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
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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


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Example
direction
j
i
1000 N
α
magnitude
Point of application
Line of action
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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
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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


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Vector/force notation
symbol representing the force  bold face
or underlined letters
 The
magnitude of the force  lightface (in the
text book, + italic)
 The
A= A
or
A= A
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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
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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
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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
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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
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External and internal effects
 Example
1: The body changes in motion
a
F
Not fixed, no (horizontal) support
 Example
2: The body deforms and produces
(support) reactions  The forces must be in
balance
F
Fixed support
Support Reactions
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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
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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


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Resultant of Forces –
Review on vector addition

Vector addition
B
R = A+B =B+A


A
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
R
A
B
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Resultant of Forces – Review on geometric laws
 Law
of Sines
A
α
 Laws
c
of Cosines
γ
c 2 = a 2 + b 2 − 2ab cos γ
β
b 2 = a 2 + c 2 − 2ac cos β
a = b + c − 2ac cos α
2
2
2
b
B
C
a
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Resultant of two concurrent forces

The magnitude of the resultant (R) is given by
R 2 = F12 + F22 − 2 F1 F2 cos γ
Pay attention to the angle
and the sign of the last
term !!!
R 2 = F12 + F22 + 2 F1 F2 cos φ

The direction (relative to the direction of F1) can be given by the law
of sines
F2 sin φ
sin β =
R
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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
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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
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Example Problem 2-1


Determine:
– The resultant force (R)
– The angle θ between the R and the x-axis
Answer:
– The magnitude of R is given by
R 2 = 900 2 + 600 2 + 2(900)(600) cos 400
R = 1413.3 ≈ 1413lb
– The angle α between the R and the 900-lb
force is given by
sin α sin(1800 − 400 )
=
600
1413.3
α = 15.836o
– The angle θ therefore is
θ = 15.8360 + 350 = 50.80
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Example Problem 2-2
 Determine
– The resultant R
– The angle between the R
and the x-axis
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Another example

If the resultant of the force
system is zero, determine
– The force FB
– The angle between the FB
and the x-axis
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Force components
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Resolution of a force into components

The components of a resultant force are
not unique !!
R = A + B = (G + I ) + H
=C+D = E+F

The direction of the components must be
fixed (given)
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How to obtain the components of a force
(arbitrary component directions)?
Parallel to u
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
Parallel to v – Use law of sines to determine the
magnitudes of the components

Fu
Fv
900
=
=
sin 45o sin 25o sin 110 o
900 sin 45o
Fu =
= 677 N
sin 110 0
900 sin 250
Fv =
= 405 N
sin 110 o
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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 α, β, γ
– Utilize the law of sines
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Another example
 Determine
the magnitude of the components of
R in the directions along u and v, when R =
1500 N
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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
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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
z
x
y
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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 ndirection (not x, y, nor z)
Any vector can be represented as a multiplication of a
magnitude and a unit vector
A is in the positive
direction along n
A = A e n = Ae n
B is in the negative
direction along n
B = − B e n = − Be n
A A
=
en =
A A
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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
F = Fx + Fy = Fx i + Fy j
Fx = F cos θ
y
Fy = F sin θ
F
Fy = Fy j
j
θ
i
F = Fx + Fy
2
Fx = Fx i
θ = tan
x
−1
2
Fy
Fx
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The rectangular components in 3D systems
z
en
k
F = Fx + Fy + Fz
Fz = Fz k
= Fx i + Fy j + Fz k
F = Fe n
F
F Fx i + Fy j + Fz k
en = =
F
F
θz
Fx = F cos θ x
θx
Fy = F cos θ y
θy
Fy = Fy j
Fx = Fx i
Fz = F cos θ z
j
i
F = Fx + Fy + F
2
Fx
θ x = cos
F
−1
2
2
z
θ y = cos
x
−1
Fy
F
e n = cos θ x i + cos θ y j + cos θ z k
Fz
θ z = cos
F
−1
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y
Dot Products of two vectors
A • B = B • A = A B cos θ = AB cos θ
A
It’s a scalar !!!
Special cosines:
θ
B
Cos 0o = 1
Cos 30o = ½ √3
Cos 45o = ½ √2
Cos 60o = 0.5
Cos 90o = 0
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Dot products and rectangular components

The dot product can be used to obtain the rectangular
components of a force (a vector in general)
An = A • e n = A cos θ n
A n = Ane n
(magnitude)
(the vectorial component
in the n direction)
A n = ( A • e n )e n
The component along en
At = A − A n
The component along et
Remember, en and et are perpendicular
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Cartesian rectangular components

The dot product is particularly useful when the unit vectors
are of the Cartesian system (the i, j, k)
Fx = F • i = F cos θ
y
Fy = F • j = F cos(90 − θ )
F
Fy = Fy j
= F sin θ
90-θ
j
θ
i
Fx = Fx i
Also, in 3D,
x
Fz = F • k
F = Fx + Fy = Fx i + Fy j = (F • i )i + (F • j) j
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More usage of dot products …
 Dot
products of two vectors written in Cartesian system
A • B = Ax Bx + Ay B y + Az Bz
 The
magnitude of a vector (could be a force vector),
here A is the vector magnitude
A • A = A2 cos 0 = A2 = Ax Ax + Ay Ay + Az Az
 The angle between two vectors (say between vectors A
and B)
 Ax Bx + Ay B y + Az Bz
θ = cos 
AB

−1



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The rectangular components of arbitrary direction
z
k
Fz = Fz k
en
F = Fx + Fy + Fz
F
= Fx i + Fy j + Fz k
F = Fne n + Ft et
Ft
θzn
Fn
Fx = Fx i
θxn
Fy = Fy j
θyn
Fn = F • e n
j y
= ( Fx i + Fy j + Fz k ) • e n
= Fx i • e n + Fy j • e n + Fz k • e n x
i
= Fx cos θ xn + Fy cos θ yn + Fz cos θ zn
Can you show the following?
e n = cos θ xn i + cos θ yn j + cos θ znk
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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 selfdot-product)
– Obtain the angle between two (force) vectors
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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
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Example Problem 2-6
Fx = F cos θ
Fy = F cos(90 − θ )
Fx ' = F cos β
Fy ' = F cos(90 − β )
θ = 90 − 28 = 62o
β = 62 − 30 = 32
β
o
Fx = 450 cos 62 = 211N
Fy = 450 sin 62 = 397 N
θ
Fx ' = 450 cos 32 = 382 N
Fy = 450 sin 32 = 238 N
Writing the F in Cartesian vector form:
F = (211i + 397 j) N = (382e x ' + 238e y ' ) N
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Example Problem 2-8
B
the angles θx, θy, and θz
(θx 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)
 Find
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Example Problem 2-8
B

To find the angles:
– Find the length of the
3
o
−1
θ
=
=
cos
59
.
0
x
diagonal OB, say d
5.831
4
– d = 5.831 m
θ y = cos −1
= 46.7 o
5.831
– Use cosines to get the
3
θ z = cos −1
= 59.0o
angles
5.831

The scalar components in the
x, y, and z directions:
Fx = F cos θ x = 12.862kN
Fy = F cos θ y = 17.150kN
Fz = F cos θ z = 12.862kN
F = (12.862i + 17.150 j + 12.862k )kN
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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)
OA = rA = 3i + 1j + 3k
– Remember, that any vector can
be represented as a multiplication
of its magnitude and a unit vector
along its line of application
rA
3i + 1j + 3k
eOA =
=
rA
32 + 12 + 32
3i + 1j + 3k
=
= 0.688i + 0.230 j + 0.688k
4.36
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Example Problem 8-2

FOA = F • eOA
The scalar component of F along OA
FOA = (12.862i + 17.150 j + 12.862k ) • (0.688i + 0.230 j + 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 • e OA )e OA = 21.6(0.688i + 0.230 j + 0.688k )
= 14.86i + 4.97 j + 14.86k

The vector component of F perpendicular to OA
Ft = F − FOA = (12.862i + 17.150 j + 12.862k ) − (14.86i + 4.97 j + 14.86k )
= (−2i + 12.18 j + 2k )

The scalar component of F perpendicular to OA
Ft =| Ft |=| (−2i + 12.18 j − 2k ) |= (−2) 2 + 12.182 + (−2) 2 = 12.50kN
Check:
2
F = FOA
+ Ft 2 = 21.6432 + 12.50 2 ≈ 25kN
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Resultants by rectangular components
 The
Cartesian rectangular components of forces can be
utilized to obtain the resultant of the forces
•Adding the x vector components, we obtain the x
vector component of the resultant
y
R x = ∑ Fx = F1 x + F2 x
F1
F1y
•Adding the y vector components, we obtain the y
vector component of the resultant
F2x
F1x
x
R y = ∑ Fy = F1 y + F2 y
•The resultant can be obtained by performing the
vector addition of these two vector components
F2
F2y
R = R x + R y = Rx i + R y j
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Resultants by rectangular components

The scalar components of the resultant
R x = F1 x + F2 x = ( F1x + F2 x )i = Rx i
R y = F1 y + F2 y = ( F1 y + F2 y ) j = R y j
R= R +R

The magnitude of the resultant

The angles formed by the resultant and the Cartesian axes
Rx
θ x = cos
R
−1

θ y = cos
2
x
−1
2
y
Ry
R
All of the above results can be easily extended for 3D system
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Please do example problems 2-9, 2-10, and 2-11
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HW Problem 2-20

Determine the non-rectangular components of R
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HW Problem 2-37
 Determine
the
components of F1 and F2
in x-y and x’-y’ systems
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HW Problem 2-44



Express the cable tension in Cartesian
form
Determine the magnitude of the
rectangular component of the cable force
Determine the angle α 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
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HW Problem 2-46



Determine the scalar components
Express the force in Cartesian
vector form
Determine the angle α between the
force and line AB
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HW problems 2-55
 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
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HW Problem 2-49


Given T1 and T2 are 650 lb,
Determine P so that the resultant of T1, T2 and P is
zero
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