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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
α
1000 N
i
j
Line of action
Point of application
magnitude
direction
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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 notationThe symbol representing the force bold face
or underlined letters
The magnitude of the force lightface (in the text book, + italic)
AAorA == 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 effectsExample 1: The body changes in motion
Example 2: The body deforms and produces (support) reactions The forces must be in balance
Not fixed, no (horizontal) support
Fa
F
Support Reactions
Fixed support
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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
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
ABBAR +=+=
RA
B
B
AR
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Resultant of Forces – Review on geometric laws
Law of Sines
Laws of Cosines
α
β
γ
cos2cos2cos2
222
222
222
accbaaccababbac
−+=
−+=
−+=
A
B
C
c
a
b
βγ
α
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Resultant of two concurrent forces
The magnitude of the resultant (R) is given by
The direction (relative to the direction of F1) can be given by the law of sines
φ
γ
cos2
cos2
212
22
12
212
22
12
FFFFRFFFFR
++=
−+=
RF φβ sinsin 2=
Pay attention to the angle and the sign of the last term !!!
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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
– The angle α between the R and the 900-lb force is given by
– The angle θ therefore is
lbRR
14133.141340cos)600)(900(2600900 0222
≈=++=
o836.153.1413
)40180sin(600
sin 00
=
−=
α
α
000 8.5035836.15 =+=θ
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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 FBand 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 !!
The direction of the components must be fixed (given)
FEDCHIGBAR
+=+=++=+= )(
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How to obtain the components of a force (arbitrary component directions)?
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
Parallel to u
oov
ou FF
110sin900
25sin45sin==
NF
NF
ov
o
u
405110sin
25sin900
677110sin
45sin900
0
0
==
==
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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
x y
z
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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 n-
direction (not x, y, nor z) Any vector can be represented as a multiplication of a
magnitude and a unit vector
nn AeeAA ==
nn BeeBB −=−=
A is in the positivedirection along n
B is in the negativedirection along n
AA
AAen ==
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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
x
y
Fy = Fy j
Fx = Fx i
i
j
F
θ
jiFFF yxyx FF +=+=
x
y
yx
y
x
FF
FFF
FFFF
1
22
tan
sincos
−=
+=
==
θ
θθ
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FF
FF
FFFF
FFFFFF
zz
yy
xx
zyx
zz
yy
xx
111
222
coscoscos
cos
coscos
−−− ===
++=
=
==
θθθ
θ
θθ
The rectangular components in 3D systems
FFFF
F
FFFF
zyxn
n
zyx
zyx
kjiFe
eFkji
FFFF
++==
=
++=
++=
x
y
z
Fy = Fy jFx = Fx i
Fz = Fz k
F
i
k
j
en
θz
θx θy
kjie zyxn θθθ coscoscos ++=
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Dot Products of two vectors
θθ coscos AB==•=• BAABBA
θ
A
B
It’s a scalar !!!Special cosines:
Cos 0o = 1Cos 30o = ½ √3Cos 45o = ½ √2Cos 60o = 0.5Cos 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)
nt
nnn
nnn
nnn
AAA
AAA
eeAA
eAeA
−=
•=
==•=
)(
cosθ (magnitude)
(the vectorial componentin the n direction)
The component along en
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)
x
y
Fy = Fy j
Fx = Fx i
i
j
F
θ
kF
jFiF
•=
=
−=•==•=
z
y
x
F
FFFFF
θ
θθ
sin
)90cos(cos
90-θAlso, in 3D,
jjFiiFjiFFF )()( •+•=+=+= yxyx FF
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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
The angle between two vectors (say between vectors Aand B)
zzyyxx BABABA ++=• BA
zzyyxx AAAAAAAA ++===• 22 0cosAA
++= −
ABBABABA zzyyxx1cosθ
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The rectangular components of arbitrary direction
znzynyxnx
nznynx
nzyx
nn
ttnn
zyx
zyx
FFFFFF
FFFF
FFFFF
θθθ coscoscos
)(
++=
•+•+•=
•++=•=
+=
++=
++=
ekejeiekji
eF
eeFkji
FFFF
kjie znynxnn θθθ coscoscos ++=
z
x
y
Fy = Fy jFx = Fx i
Fz = Fz k
F
i
k
j
en
θzn
θxn θyn
Fn
Ft
Can you show the following?
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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 self-
dot-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
θ
β
NFNFNFNF
FFFFFFFF
y
x
y
x
o
o
yx
yx
23832sin45038232cos450
39762sin45021162cos450
323062622890
)90cos(cos
)90cos(cos
'
''
====
====
=−=
=−=
−==
−==
β
θ
ββ
θθ
NN yx )238382()397211('' eejiF +=+=
Writing the F in Cartesian vector form:
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Example Problem 2-8 Find 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)
B
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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
oz
oy
ox
0.59831.53cos
7.46831.54cos
0.59831.53cos
1
1
1
==
==
==
−
−
−
θ
θ
θ
kNFFkNFFkNFF
zz
yy
xx
862.12cos
150.17cos862.12cos
==
====
θ
θθ
kN)862.12150.17862.12( kjiF ++=
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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)
– Remember, that any vector can be represented as a multiplication of its magnitude and a unit vector along its line of application
kjir 313 ++== AOAkjikji
kjirre A
688.0230.0688.036.4
313313
313222
++=++
=
++
++==
AOA
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Example Problem 8-2
The scalar component of F along OA
The vector component of F along OA
The vector component of F perpendicular to OA
The scalar component of F perpendicular to OA
OAeF •=OAF
kNFF
OA
OA
643.21688.0862.12230.0150.17688.0862.12)688.0230.0688.0()862.12150.17862.12(
=×+×+×=++•++= kjikji
kjieeFF OAOAOA
86.1497.486.14)688.0230.0688.0(6.21)(
++=++=•= kji
)218.122()86.1497.486.14()862.12150.17862.12(
kjikjikjiFFF OA
++−=++−++=−=t
kNF tt 50.12)2(18.12)2(|)218.122(||| 222 =−++−=−+−== kjiF
Check: kNFFF tOA 2550.12643.21 2222 ≈+=+=
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Resultants by rectangular components
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
∑ +== xxxx 21 FFFR
∑ +== yyyy 21 FFFR
jiRRR yxyx RR +=+=
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Resultants by rectangular components
The scalar components of the resultant
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
iiFFR 21 xxxxxx RFF =+=+= )( 21
jjFFR 21 yyyyyy RFF =+=+= )( 21
22yx RRR +=
RR
RR y
yx
x11 coscos −− == θθ
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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 2-37Determine the
components of F1 and F2in 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