em102 topic 2 - force vectors

8/17/2019 EM102 Topic 2 - Force Vectors

1/103

Statics, Fourteenth EditionR.C. Hibbeler

Copyright ©2016 by Pearson Education, Inc.All rights reserved.

EM102 Engineering Statics

Topic 2: Force Vectors

Scalars and vectors,

vector operations,

cartesian vectors,

position vectors,

equilibrium of a particle.

Dr. Elango Natarajan

Assistant Professor

Aug 2015


2/103



In-Class activities:• Reading Quiz

• Application of Adding Forces

•Resolution of a Vector Using

Cartesian Vector Notation (CVN)• Addition Using CVN

• Law of mechanics

•Resultant of two forces system

•Resultant of three forces system

•Resultant of multi forces system

Objective:

Students will be able to : a) Resolve a 2-D vector into

components.

b) Add 2-D vectors using

Cartesian vector notations.

FORCE VECTORS, VECTOR OPERATIONS &ADDITION COPLANAR FORCES


3/103



Scalars and vectors

Scalar is a quantity characterized by a positive or

negative number.

Example: Mass, Volume and Length

Vector is a quantity that has both a magnitude and a

direction.

Example: Position, force and moment.


4/103



Scalars Vectors

Examples: Mass, Volume Force, Velocity

Characteristics: It has a magnitude It has a magnitude

(positive or negative) and direction

Addition rule: Simple arithmetic Parallelogram law

Special Notation: None Bold font, a line, an

arrow or a “carrot”

In these PowerPoint presentations, a vector quantity is represented li ke this (in bold,

italics, and red).

SCALARS AND VECTORS(Section 2.1)


5/103



2. For vector addition, you have to use ______ law.A) Newton’s Second

B) the arithmetic

C) Pascal’s

D) the parallelogram

READING QUIZ

1. Which one of the following is a scalar quantity?

A) Force B) Position

C) Mass D) Velocity


6/103



There are three concurrentforces acting on the hook

due to the chains.

We need to decide if the

hook will fail (bend or break).

To do this, we need to know

the resultant or total forceacting on the hook as a

result of the three chains.

FR

APPLICATION OF VECTOR ADDITION


7/103



VECTOR OPERATIONS (Section 2.2)

Scalar Multiplicationand Division


8/103



FORCE SYSTEM

Before we proceed further with vectors, I would like

to introduce you the force systems.


9/103



Force

• Force is an action that can change motion.

• Force is what we call a push or a pull.

•

Force is any action that has the ability to change anobject’s motion.

• Forces can be used to increase the speed of an object,

decrease the speed of an object, or change the direction in

which an object is moving.


10/103



Force - Examples

• an object’s weight (self weight)

• tension in a rope

• friction

• attraction between an electron and proton


11/103



Force


12/103



Gravitational Force

Bodies don’t have to be in contact to exert forces on each other,

e.g., gravity.

Gravity is attraction between any two bodies.

It is weakest but most dominant.


13/103



Net force

When more than one force acts on a body, the net force (resultant

force) is the vector combination of all the forces, i.e., the “net effect.”

When forces act in the same line, we can just add or subtract their

magnitudes to find the net force.


14/103



Net force Examples

F1

F2

F3

Fnet


15/103



Net force - Example

2 kg

15 N 32 N

10 N

This is an example of collinear force system


16/103



Addition of Collinear force vectors


17/103



Inertia

• Inertia is a term used to measure the ability of an object to resist a

change in its state of motion.

• An object with a lot of inertia takes a lot of force to start or stop;

• An object with a small amount of inertia requires a small amount

of force to start or stop.

• The word “inertia” comes from the Latin word inertus, which can

be translated to mean “lazy”.


18/103



Inertia


19/103



Moment of a force

The Moment of a force is a measure of its tendency to cause a body to

rotate about a specific point or axis.

This is different from the tendency for a body to move, or translate, in

the direction of the force.

In order for a moment to develop, the force must act upon the body in

such a manner that the body would begin to twist.


20/103



Moment

This occurs every time a force is applied so that it does not pass

through the centroid of the body.

A moment is due to a force not having an equal and opposite force

directly along it's line of action.


21/103



Moment - More Examples


22/103



Couple

A couple is defined as two forces (coplaner) having the same

magnitude, line of action but in opposite sense.

It has pure rotation.


23/103



Force system and Static Equilibrium

A force system is a collection of forces acting at specified

locations

When all the forces that act upon an object are balanced, then the

object is said to be in a state of equilibrium.

The forces are considered to be balanced if the rightward forces are

balanced by the leftward forces and the upward forces are balanced

by the downward forces.


24/103



Static Equilibrium

This however does not necessarily mean that all the forces

are equal to each other.

If an object is at equilibrium, then the forces are balanced.

Balanced is the key word that is used to describe equilibrium

situations.

Thus, the net force is zero and the acceleration is 0 m/s2.


25/103



Types of force system

• Coplanar force systems have all the forces acting in in one plane.

• They may be concurrent, parallel, non-concurrent or non-parallel.

Concurrent coplanar


26/103



Resolution of a Vector Using

Cartesian Vector Notation (CVN)


27/103



VECTOR ADDITION USING EITHER THEPARALLELOGRAM LAW OR TRIANGLE

Parallelogram Law:

A and B are joined at their tails, Parallel lines are drawn from the head of each

vector intersect at a common point, thereby forming the adjacent sides of a

parallelogram.

The resultant R is a diagonal of the parallelogram.

R = A + B = B + A.


28/103



VECTOR ADDITION USING EITHER THEPARALLELOGRAM LAW OR TRIANGLE

Triangle method(always ‘tip to tail’):

A and B are joined in a “head – to- tail” fashion (by connectionthe head of A to the tail of B).

The resultant R extends from the tail of A to the head of B.


29/103



VECTOR SUBTRACTION USING EITHER THEPARALLELOGRAM LAW OR TRIANGLE

R' = A - B = A + (-B)


30/103



“Resolution” of a vector is breaking up a vector into components.

It is kind of like using the parallelogram law in reverse.

RESOLUTION OF A VECTOR


31/103



RESOLUTION OF A VECTOR - procedure

• If F in Fig. (a) is to be resolved into components acting along

the lines u and v , one starts at the head of F and extends a

line parallel to u until it intersects v .

• Likewise, a line parallel to v is drawn from the head of F to

the point of intersection with u.

• Two components F u and F

v are then drawn such that they

extend from the tail of F to the points of intersection as

shown in Fig. (b).


32/103



ADDITION OF A SYSTEM OF COPLANAR FORCES(Section 2.4)

• Each component of the vector is

shown as a magnitude and adirection.

• The directions are based on the x and y axes. We use the

“unit vectors” i and j to designate the x and y-axes.

• We ‘resolve’ vectors intocomponents using the x and y-axis

coordinate system.


33/103



For example,

F = Fx i + Fy j or F ' = F'x i + ( F'y ) j

The x and y-axis are always perpendicular to each other.

Together, they can be “set” at any inclination.


34/103



• Step 2 is to add all the x-

components together, followed by

adding all the y-componentstogether. These two totals are the

x and y-components of the

resultant vector.

• Step 1 is to resolve each forceinto its components.

ADDITION OF SEVERAL VECTORS

• Step 3 is to find the magnitude

and angle of the resultant

vector.


35/103



Break the three vectors into components, then add them.

F R = F 1 + F 2 + F 3

= F1x i + F1y j F2x i + F2y j + F3x i F3y j

= (F1x F2x + F3x) i + (F1y + F2y F3y) j

= (FRx) i + (FRy) j

An example of the process:


36/103



You can also represent a 2-D vector witha magnitude and angle.

1

tan

Ry

Rx

F

F

2 2

R Rx Ry F F F


37/103



1. Resolve F along x and y axes and write it in

vector form. F = { ___________ } N

A) 80 cos (30°) i – 80 sin (30°) j

B) 80 sin (30°) i + 80 cos (30°) j

C) 80 sin (30°) i – 80 cos (30°) j

D) 80 cos (30°) i + 80 sin (30°) j

2. Determine the magnitude of the resultant (F 1 + F 2) force in N

when F 1 = { 10 i + 20 j } N and F 2 = { 20 i + 20 j } N .

A) 30 N B) 40 N C) 50 N

D) 60 N E) 70 N

30°

xy

F = 80 N

ATTENTION QUIZ


38/103



Example 1

The link shown in Fig. is subjected to two forces F1 and

F2. Determine the magnitude and orientation of the

resultant force using Cartesian vector notation.


39/103



Example 1 - solution

• Resolving forces as shown:


40/103





41/103



Example 2 – solve yourself

Determine the

magnitude and

direction of theresultant force by

resolution of forces.


42/103




Answer:

F 1 = 1.41i-1.41j

F 2 = -3i-5.196jF R=-1.59i-6.606j

Magnitude of the resultant force is 6.79 kN

Direction of the resultant force is 76.47


43/103



Plan:

a) Resolve the forces into their x-y components. b) Add the respective components to get the resultant vector.

c) Find magnitude and angle from the resultant components.

EXAMPLE 3

Given: Three concurrent forces

acting on a tent post.

Find: The magnitude and

angle of the resultant

force by resolution of

forces.


44/103



F 1 = {0 i + 300 j } N

F 2 = { – 450 cos (45°) i + 450 sin (45°) j } N

= { – 318.2 i + 318.2 j } N

F 3 = { (3/5) 600 i + (4/5) 600 j } N

= { 360 i + 480 j } N

EXAMPLE 3 - solution


45/103



Summing up all the i and j components respectively, we get,

F R = { (0 – 318.2 + 360) i + (300 + 318.2 + 480) j } N

= { 41.80 i + 1098 j } N

x

y

FR Using magnitude and direction:

FR = ((41.80)2 + (1098)2)1/2 = 1099 N

= tan-1(1098/41.80) = 87.8°

EXAMPLE 3 - solution


46/103



Plan:

a) Resolve the forces into their x and y-components. b) Add the respective components to get the resultant vector.

c) Find magnitude and angle from the resultant components.

Exercise 4 – solve yourself

Given: Three concurrent

forces acting on a bracket.


angle of the resultant

force. Show the

resultant in a sketch.


47/103



F 1 = {850 (4/5) i 850 (3/5) j } N

= { 680 i 510 j } N

F 2 = {- 625 sin (30°) i 625 cos (30°) j } N

= {- 312.5 i 541.3 j } N

F 3 = {-750 sin (45°) i + 750 cos (45°) j } N

{- 530.3 i + 530.3 j } N

Exercise 4 – solution


48/103



Summing all the i and j components, respectively, we get,

F R = { (680 312.5 530.3) i + (510 541.3 + 530.3) j }N

= { 162.8 i 520.9 j } N

Now find the magnitude and angle,

FR = (( 162.8)2 + ( 520.9)2) ½ = 546 N

= tan – 1

( 520.9 / 162.8 ) = 72.6°

From the positive x-axis, = 253°

Exercise 4 – solution

FR

x

y

-520.9

-162.8


49/103





50/103



Exercise 5 - solution


51/103





52/103





53/103





54/103




Given: Three concurrent

forces acting on a

bracket


angle of the

resultant force.


55/103




• F 1 = [ 300 (5/13) i + 300 (12/13) j ] N

= [ 115.4 i + 276.9 j ] N

F 2 = [500 cos (30°) i + 500 sin (30°) j ] N= [ 433.0 i + 250 j ] N

F 3 = [ 600 cos (45°) i 600 sin (45°) j ] N

= [ 424.3 i 424.3 j ] N


56/103




Summing up all the i and j components respectively, weget

F R = [ (115.4 + 433.0 + 424.3)i + (276.9 + 250 - 424.3)] j

= [972.7 i + 102.7 j ] N


57/103



You’ll have more similar exercises in Tutorial class.


58/103



Further more…

Law of Mechanics

Solving problems on two forces system

Solving problems on three forces system

Solving problems on multi forces system

L f M h i


59/103



Law of Mechanics

Newton’s First Law of Motion

Second Law of Motion

Third Law of Motion

Gravitational Law of Attraction

Parallelogram Law of Forces

Sine Law

Cosine Law

Lami’s Theorem

Principle of Transmissibility of Force

Fi l f N


60/103



First law of Newton

“An object in motion tends to stay in motion; an object at rest tends to stay

at rest.”

Fi t L f N t


61/103



First Law of Newton

•A moving body will continue moving in the samedirection with the same speed until some net forceacts on it.

•A body at rest will remain at rest unless a net forceacts on it.

•Summing it up: It takes a net force to change abody’s velocity .

S d l f N t


62/103



Second law of Newton

• F net = m a

•The acceleration (a) an object is directly proportion

to the net force acting on it.

•Mass (m) is the constant of proportionality.

M f M l ti


63/103



More force – More acceleration

Thi d l f N t


64/103



Third law of Newton

“For every action there is an equal and oppositereaction.”

G it ti l l f tt ti


65/103



Gravitational law of attraction

This law states that two particles of mass m1 and m2

are attracted towards each other along the line

connecting them with a force whose magnitude F is

proportional to the product of their masses and

inversely proportional to the square of the distance(r ) between them.

G it ti l l f tt ti


66/103



Gravitational law of attraction

= (

)

where G is the universal constant of constant ofgravitation and its value is

(66.73±0.03) × 10-12 m3/kgs2

Parallelogram law of forces (two forces system)


67/103



Parallelogram law of forces (two forces system)

If two forces F1 and F2 acting at a point to be

represented in magnitude and direction by the two

adjacent sides of a parallelogram, then their

resultant is represented in magnitude and direction

by the diagonal of the parallelogram at that point.

Parallelogram law of forces


68/103



Parallelogram law of forces

Application of Parallelogram law


69/103






70/103




Sine law (three forces system)


71/103



Sine law (three forces system)

=

=

The law of sines relates two sides and the angles opposite

them. So any time you have two angles (and then can

easily figure out the third), it's easy to use the law of sines:

Angle-Side-Angle or Angle-Angle-Side.

Cosine law (three forces system)


72/103



Cosine law (three forces system)

The law of cosines relates 3 sides and an

angle. So if you know 3 sides, you can use it,

or if you know two sides and an angle, you

can find the third side. However, because of

the form of the equation, if you have anangle that's not between the two sides, you

get a quadratic equation in that case--a bit

messy.

So the law of cosines is best for Side-Side-

Side and Side-Angle-Side.

=

+

− 2

=

+

− 2

=

+

− 2

Lami’s theorem (three forces system)


73/103



Lami s theorem (three forces system)

If three forces acting at a point are in equilibrium,

each force will be proportional to the sine of the

angle between the other two forces.

Lami’s theorem


74/103



Lami s theorem

1

=

2

=

3

Principle of transmissibility of forces


75/103




The state of rest or motion of a rigid body is

unaltered if a force acting on the body is replaced by

another force of the same magnitude and direction

but acting anywhere on the body along the line of

action of the replaced force.



76/103




Two forces system


77/103



Two forces system

Two forces system


78/103



Two forces system

Two forces system


79/103



Two forces system

Exercise 1


80/103



Exercise 1

Determine the magnitude

and direction of the

resultant force.



81/103



Exercise 1 solution

Exercise 1- solution


82/103



Exercise 1 solution

Exercise 1- solution


83/103



Exercise 1 solution

Exercise 2 - do it yourself


84/103



Exercise 2 do it yourself



85/103



Exercise 2 solution



86/103



e c se so u o

Exercise 3 – do it yourself


87/103



y



88/103



• Resultant force FR = 629 N

• Angle of resultant force = 37.89

• Angle of resultant force from x axis = 30+37.89 = 67.89



89/103



y



90/103



Exercise 4 -solution


91/103





92/103



y



93/103





94/103





95/103



Exercise 6 -solution


96/103





97/103





98/103





99/103





100/103





101/103



Exercises to be done in tutorial classes


102/103



• Exercises on representation of forces in Cartesian vector.

• Exercises on two forces system – using Parallelogram law

•

Exercises on three forces system – using Parallelogram law orresolution forces

• Exercises on multi forces system – using resolution of forces.

TOPIC 2 FORCE VECTORS COMPLETED


103/103

em102 topic 2 - force vectors

Documents