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Outline:
Types and Properties of Forces
Representing Forces with Vectors
2D & 3D
Addition of Force Vectors
ENGR 1205 1 Chapter 4
a force represents the action of one body on another
it’s characterized by
i)
ii)
iii)
when working with particles we don’t need to worry
about point of application since there’s only one point
magnitude is in Newtons (N)
direction is defined by a
ENGR 1205 2 Chapter 4
Line of Action – the infinitely long straight line along
which the force vector lies (defined
Sense – pull/push, up/down, left/right,
tension/compression etc., defined by an arrowhead
e.g. opposite senses
ENGR 1205 3 Chapter 4
force is represented by a line segment along the line of
action
the length can be scaled to represent the magnitude
(make sure it’s clear whether scale does represent a
force’s magnitude)
e.g.
50N 100N
Typically, we the symbol “ F ” is used to represent a
force vector (but with multiple forces use other
meaningful letters or subscripts)
ENGR 1205 4 Chapter 4
forces acting on Rigid Bodies (RB) can be classified as
either external or internal forces
External Forces - the actions -
they are responsible for the externally observed
behaviour of the RB - they cause the RB to move, or
to stay at rest
e.g.
ENGR 1205 5 Chapter 4
the point of application for W is the centre of gravity
(CofG) [in Chapter 8 you will figure out how to locate
these]
W ↓ causes downward movement (the R forces in the case
of our truck example)
F’s point of application is where the rope is tied to the
truck, and F causes the
(whereas a jack would rotate it about an axis)
each external force acting on an RB can (if unopposed)
Internal Forces
e.g. glue, welds, rivets, intermolecular bonds, etc.
ENGR 1205 6 Chapter 4
Force vectors are “sliding vectors” and they obey the
principal of transmissibility
transmissibility means that the conditions of
equilibrium or motion
ENGR 1205 7 Chapter 4
F and F’ are said to be “equivalent”
they have the same effect on the RB (based on
experimental evidence)
with RBs, the point of application isn’t as important as
the line of action that a force is on
these are represented with “sliding vectors” (slide along
LOA)
So if we move W over on our truck example, reflecting a
change in the COG, is it the same system?
ENGR 1205 8 Chapter 4
Answer from prior slide: No because we changed the LOA
(ie: the truck will tip differently)
there are practical limits to the use of transmissibility and
equivalent forces
e.g.
Not really true, the internal forces and resultant
deformations (tension [vs] compression) are often very
different in real life
=
ENGR 1205 9 Chapter 4
=
= =
=
2 forces acting on a particle can be replaced by one
equivalent or resultant force, using the parallelogram
rule (where the diagonal is the resultant)
Two or more forces are said to be concurrent at point
Two forces applied at two different points can be
treated as concurrent at the point where their lines of
action intersect due to the Principle of
Transmissibility
ENGR 1205 10 Chapter 4
𝐑
Contact forces – produced by direct physical contact
Body forces – generated by position of a body with a
force field
Eg. gravity
Concentrated forces – the dimensions of the area of
application are very small compared with other
dimensions
Distributed forces – can be distributed over
An area
A volume
A line
ENGR 1205 11 Chapter 4
Every object exerts a force on every other
object.
This force is a body force called gravity and it
occurs in pairs.
Magnitude can be calculated using Newton’s
Law of Universal Gravitation (see chapter 1)
Most of the time we are analysing the
gravitational force between an object and the
earth
ENGR 1205 Chapter 4 12
𝑾𝑬𝑨𝑹𝑻𝑯 = 𝑭 = 𝒎𝒈
Normal Force
Occurs when two solid objects are in direct contact
Forces PERPENDICULAR to the contact surface
Direction is always towards the contact surface
Friction Force
occurs whenever there is a tendency for one object
to slide along another
Force is PARALLEL to the contact surface
Direction is always opposite (impending) motion
A normal force must occur if there is a friction force,
but not vice versa
ENGR 1205 Chapter 4 13
Tension Force
Occurs when a rope or cable attached to an object
is pulled taut
Force is along rope or cable
Direction is always away from object
Compressive Force
occurs when atoms in an object are pushed
together
Compression is an INTERNAL force (see chapter 10)
Shear Force
Occurs when atoms in an object slide against each
other
Shear is also an INTERNAL force (see chapter 10) ENGR 1205 Chapter 4 14
Force vectors can be represented:
Graphically (scale diagram- in 2D only)
By stating Magnitude and Direction
- direction in reference to some origin
- in 3D use space angles θx, θy, θz
By using Rectangular components
- a sum of vectors along perpendicular axes
- generally along x, y, z axes
ENGR 1205 Chapter 4 15
To define V
Draw scale diagram
State magnitude and angle
from x-axis
Write it as sum of Vx and Vy
ENGR 1205 Chapter 4 16
just as 2 force vectors can be
represented by 1 force vector, 1
force vector can be represented
by 2+ force vectors that
produce the same effect
these are called components of
the original force
the process is called “resolving
the force into components”
(usually x, y)
ENGR 1205 17 Chapter 4
Can resolve a force into 2 components
that are perpendicular to each other …
use the Cartesian system (x,y)
In this case, the parallelogram is a
proper rectangle
the x,y system is usually up/down/
left/right but it doesn’t have to be (it
can be tilted)
This feature is useful in
biomechanics where the reference
frame is the body or some part of
the body, instead of a “world”
frame of reference
ENGR 1205 18 Chapter 4
In two-dimensions:
We resolve forces into their
x & y components using “unit
vectors”.
“unit vectors” are vectors
with a magnitude of 1.
Unit vectors î & ĵ are along
the x & y axes such that:
ENGR 1205 19 Chapter 4
x x y yF F i and F F j
Unit vectors are any
vector, in any direction,
with a magnitude of 1.
We can write any vector
as a scalar magnitude
multiplied by a unit
vector along its line of
action.
ENGR 1205 20 Chapter 4
Describe vectors a, b and c using “vector
notation” (in terms of unit vectors i and j)
ENGR 1205 Chapter 4 21
Fx and Fy are the scalar components of
note that Fx and Fy can be positive or negative
(indicates direction)
another way of expressing Fx and Fy is
(note that these equations hold for all Ө , thereby
automatically dealing with the issue of sense/sign,
when taking Ө from the positive x axis)
ENGR 1205 Chapter 4 22
F
Determine the x & y components of the force vector F in
terms of angles θ, α, and β.
ENGR 1205 Chapter 4 23
The screw is subjected to
the force shown. Describe
the force in terms of
a) its magnitude and
direction
b) its scalar components
c) its vector notation
d) a unit vector along its
LOA
ENGR 1205 Chapter 4 24
ENGR 1205 Chapter 4 25
In three dimensions a Cartesian
system is made up of three
mutually perpendicular planes.
A 3-D cartesian system can be
left or right handed.
In a right-handed system you
can find the positive z-axis by
pointing the fingers of your
right hand in the positive x
direction and curling them into
the positive y direction. The
direction of your thumb is the
direction of the positive z-axis. ENGR 1205 26 Chapter 4
consider a force acting at origin O
in an x,y,z system
how can we define the direction
of ?
create plane OBAC with in it
the plane passes through the y
axis
find the orientation of plane
OBAC wrt the xy plane, using φ
(phi) wrt the positive x axis
Өy defines ’s direction within
the plane OBAC
F
F
F
ENGR 1205 27 Chapter 4
we can then break into
F
( )y h horizontal planeF and F
ENGR 1205 28 Chapter 4
but Fh can be resolved
into x and z components
the same way
ENGR 1205 29 Chapter 4
we have expressions for Fx, Fy, Fz
Fy is in terms of Өy (useful)
Fx, & Fz are in terms of Өy and φ (less useful)
We can repeat the same process using planes that go through
the other two axes (x & z) to get:
are called the direction cosines of F
cos , cos , cosx y z
ENGR 1205 30 Chapter 4
zzyyxx FFFFFF cos,cos,cos
For angles :
always measures these angles from the (+) side of an axis
angles should be between 0º and 180º
if θ<90º, the angle is on the positive side of the axis and
cos is (+)
if θ>90º, the angle is on the negative side of the axis and
cos is (-)
The direction of the force can be expressed using the
direction cosines or unit vectors.
where …
, ,x y z
ENGR 1205 31 Chapter 4
kFjFiFF zyxˆˆˆ
sub into
So,
Where u is the “unit vector” for
u has the same sense and is on
the same LOA as , but it’s of
unit length and u x = cos Өx,
u y = cos Өy, and u z = cos Өz
u x, u y, u z are not independent
u x2 + u y
2 + u z2 = 1
x y zF F i F j F k
ENGR 1205 32 Chapter 4
zzyyxx FFFFFF cos,cos,cos
F
F
u (Magnitude = 1)
= Fu
The magnitude of the force can be found:
Using
and,
therefore
and
22 2 2 2 2
2 2 2 2 2
22 2 2
2 2 2
y
x z
x y z
x y z
F OA OC CA OC F
OC OD DC F F
F F F F
F F F F
ENGR 1205 33 Chapter 4
From
we get
or
so if you know two of the angles, then the 3rd angle can
only be one of two values (Ө and 180-Ө)
knowing Fx, Fy, Fz, and
you can get Өx, Өy, Өz
22 2 2
x y z
22 2 2
x y z
2 2 2 2
x y z
F = F + F + F
= + +
1 = cos +cos +cos
yx zx y z
FF Fcos = , cos = , cos =
F F F
ENGR 1205 34 Chapter 4
2222
zyx uuuu
The tether of the balloon exerts an 800 N force F on the
hook at O. The vertical line AB intersects the x-z plane at
point A. The angle between the z axis and the line OA is
60o, and the angle between the line OA and F is 45o.
Express F in terms of components.
ENGR 1205 Chapter 4 35
O
y
z
x
B
F
B
A
ENGR 1205 Chapter 2 36
y
z
x
B
F
ANS: The force of the balloon on
the hook is 490i + 566j + 283k [N]
often, a force direction (not magnitude) will be
defined by two points M (x1,y1,z1) and N (x2,y2,z2).
The distance between M & N is d, and
dx = x2-x1, dy = y2-y1, dz = z2-z1,
so, if we take an origin at M
this vector points from M to N, not from N to M
x y zMN d i d j d k
ENGR 1205 37 Chapter 4
ENGR 1205 38 Chapter 4
u
we can get (the unit vector) by dividing by
for any force, , in the direction of :
Recall: and
ENGR 1205 39 Chapter 4
x y zd i d j d kF F F
d
MN
F
cosx xF F cos and ,x
adjadj d hyp d
hyp
u
u
u
Therefore
These equations make many problems easier to solve
when given:
We can use them to solve for Fx, Fy, Fz
for Өx, Өy, Өz recall that:
ENGR 1205 40 Chapter 4
, ,yx z
x y z
dd dF F i F F j F F k
d d d
2 1 2 1 2 1
2 2 2
, ,x y z
x y z
d x x d y y d z z
d d d d
cos , and cos , cos
x
yx x zx y z
dF
dF d dd
d d dF F
A 2-m tall man stands 4-m
from a wall and pulls on a
rope with a force of 70-N.
The rope is attached to the
wall at point A. Point A is 10-
m above the ground and 3-m
to the right of where the man
is standing. Represent the
man’s pulling force acting on
the support at A as a
Cartesian vector and
determine its direction
angles.
ENGR 1205 Chapter 4 41
y
z
x O
A
B
ENGR 1205 Chapter 2 42
y
z
x O
A
B
ENGR 1205 Chapter 2 43
ENGR 1205 Chapter 2 44
ANS: The pulling force is -22.3i + -59.4j + 29.7k
[Newtons] and it acts 109° from x-axis, 148° from
y-axis, and 65° from z-axis.
The resultant is the sum of concurrent forces
There are three methods to add vectors
Graphical Methods (tip-to-tail & scale diagram)
Geometric Method (trigonometry)
The use of Components
In 3D or with 3 or more forces it is better to use an analytic
solution by resolving all force vectors into x and y
components
𝑹 = 𝑷 + 𝑺 + 𝑸
P
Q
S
ENGR 1205 45 Chapter 4
and
also,
Therefore,
scalar components of R are the sums of the respective
scalar components of the given forces
PROCEDURE:
i) Break the given forces into components
ii) Get Rx, Ry, and Rz
iii) Write R in vector notation (or in 2D use trig to get
magnitude and angle)
ENGR 1205 46 Chapter 4
SQPR
kRjRiRR zyx
kSQPjSQPiSQP
kSjSiSkQjQiQkPjPiPR
zzzyyyxxx
zyxzyxzyx
zz
yy
xx
FR
FR
FR
The screw is subjected to two forces F1 and F2. Determine
the magnitude and direction of the resultant force.
a) Using trigonometry
b) Using components
ENGR 1205 Chapter 4 47
ENGR 1205 Chapter 4 48
ENGR 1205 Chapter 4 49
ENGR 1205 Chapter 4 50
Four forces act on bolt A as shown. Determine the
resultant of the forces on the bolt.
ENGR 1205 51 Chapter 4
ENGR 1205 Chapter 4 52
ENGR 1205 Chapter 4 53
ENGR 1205 Chapter 4 54
Determine the
magnitude and
direction of
resultant of the
two forces
shown knowing
P = 4 kN and
Q = 8 kN
ENGR 1205 55 Chapter 4
Determine the resultant of the three forces acting
at “O”. State in vector notation.
ENGR 1205 56 Chapter 4
ENGR 1205 Chapter 4 57
ENGR 1205 Chapter 4 58
Boom AB is held in the position shown by thrtee
cables. The tension in cable AC is 900N, the
tension in cable AD is 1200N, and the resultant of
the tensions exerted at point A of the boom is
directed along AB. Find the magnitude of the
tension in cable AE?
ENGR 1205 59 Chapter 4
ENGR 1205 Chapter 4 60
ENGR 1205 Chapter 4 61
ENGR 1205 Chapter 4 62
ENGR 1205 Chapter 4 63