Outline: Types and Properties of Forces Representing Forces...

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