Introduction to dynamic

Introduction to dynamic

Introduction

• Dynamic consist of i. Kinematics

analysis of the geometry of motion without concern for the forces causing the motion; which involving the quantity such as displacement, velocity, acceleration, and time.

ii. Kinetics the study of motion and the forces associated with motion; it

involves the determination of the motion resulting from given forces

Geometry of motion in kinematic

Displacement and Distance• Distance is a scalar quantity and displacement is a vector quantity.• To travel from point A to point B, say, there are 3 paths with different

distance, but there have the same displacement. Path 3 has longer distance than Path 1 or 2 but the entire three path have the same displacement.

• Displacement is the basic parameter values to determine velocity and acceleration

3

21

A B

Speed and velocity• Speed is a scalar quantity and velocity is a vector quantity.• Speed – change of distance per unit of time, SI unit is meters per second

(m/s).

• Velocity – rate of change of displacement with respect to time and has the same unit as speed.

• Velocity Resultant – combination of two or more velocity and can be determined using triangle law or parallel rectangular law to determine resultant vector.

t

s

ervalTime

ervalceDisvSpeed

int

inttan,

Acceleration• Acceleration

– change of velocity increment and decrement of velocity is known as de-acceleration. It is expressed in terms of displacement/time/time. SI units are .

or

• Ration velocity – difference between two velocities and can be determined by using

triangle and parallel rectangle method.

t

v

ervalTime

ervalVelocityaonAccelerati

int

int, v

dt

dswhere

ds

ds

dt

dva

Relationship of displacement, velocity and Time

Graph Displacement-time (s-t)

OL is the graph s-t for a body moves with constant velocity.

Therefore OA = constant acceleration

Gradient for OL,

Gradient for OL,

The gradient is the velocity of the body.

M

s

t

D

B

A

C

L

t (s)

s (m)

Graph velocity-time (v-t)

• AB = A body moves with constant velocity• PQ=constant acceleration• Distant = Area below the graph

– Area (1) = Area of ABCO– Area (2) = Area of OPQS

• Gradient, m = acceleration for the body– Graph 1,

– Graph 2,

sxsm )/(

0m

PR

QRm

Type of motion

• Type of motion– Rectilinear motion

• the particle or body moves in a straight lines and does not rotate about its center of mass

– Circular motion • a particle follows the path of a perfect circle.

– General plane motion • a particle may follow a path that is nether straight nor circular.

This also applies to a body have both rotating and rectilinear motion simultaneously

Rectilinear motion

• The three equation that relate the variables of time, displacement, velocity, and acceleration to each other are

where

atuvorasvv

atuvoratvv

attvs

2220

2

0

20

2

2

1

stimet

smonacceleratitconsa

smvelocityfinalv

smvelocityinitialuv

mntdisplacemes

,

/,tan

/,

/,)(

,

2

0

• Derivation of equation

20 2

1attvs

Rearranging equation ,dt

dva

atvv

atvv

dtadvor

dtadvtv

v

0

0

00

t

v

ervalTime

ervalVelocityaonAccelerati

int

int, v

dt

dswhere

ds

ds

dt

dva or

atvv 0

Rearranging equation dt

dsv

20

0

0

0

0

0

2

1attvs

dtatvds

dtatvds

atvvwheredtvds

ts

Using equation ds

dvva

sv

v

dsadvv00

asvv

22

20

2

asvv 220

2

asvv 220

2

To use these equations, the following points should bear in mind;

• Acceleration, although it may be in any direction, must be constant. Constant velocity is a special case in which acceleration is constant at zero.

• A free-failing body has an acceleration of

• Designate the direction that is to be positive. The direction of the initial velocity or displacement is often used as the positive direction.

• An object that decelerates or slows down in the positive direction is treated as having a negative acceleration (de-acceleration).

2/81.9 smga

Example 7.1A particle is moving on a straight line plane with a fixed point on a line. Displacement, s from point O is given by the equation, s = 6 + t –t2 . Calculate:-

(a) Displacement from point O when t = 0(b) Velocity at t = 0(c) Time when velocity is zero

Example 7.2

A bikers passing though bridge A with a velocity of 5 m/s. In 4 seconds, he reached bridge B with velocity 15 m/s. Calculate acceleration and displacement the bicycle.

Vertical motion (motion under gravitational effects)

• When an object falls to the earth, it will move very fast i.e., increased speed. – The increased motion which is free-fallings (air resistance is ignored)

close to the earth surface is known as gravitational acceleration.• Gravitational acceleration is same to all bodies, regardless of size or

chemical composition at any location.• Where motion upward is negative and motion downwards is positive.

s = maximum height

Example 7.3

• A particle is released freely from rest condition at the height of 150m. Determine the time taken to reach the earth and particle velocity at that moment.

Example 7.4

A stone is throwing up with an initial velocity of 24.5 m/s. Calculates:-a) Maximum peak the stone could reachb) How long will the stone to re-touch the earth surfacec) What when the velocity is equal to 4.9 m/sd) When it will be at 19.6 m from the earth surface.

Example 7.5

A stone is falling in a well. The water sprinkle heard at 2.56 seconds later. If the water level from the earth surface is 30m, what is sound velocity in the air?

Straight motion on slope plane

• Involved the effect of gravitational acceleration like vertical motion.• However, the calculated value of gravitational acceleration is the

component of gravitational acceleration which is parallel to the direction of motion.

• Acceleration component which is perpendicular to the motion is ignored because it does not gave any effect to the particle motion.

• Particles rolling from B to A, gravitational acceleration can be divided into component of rectangular, i.e., – At the direction of particle motion– At the direction parallel to the plane

asvv

atvv

attvs

2

2

1

20

2

0

20

singa

Example 7.6

A stone is rolling on the slope 30o from the horizontal plane. If it is rolls with the velocity of 5m/s , determine the displacement, distance taken, and velocity after 4 seconds.

g

30sing

sm /5

30

30cosg

sm

ga

s

st

smu

/905.4

)5.0)(81.9(

30sin

?

4

30

/5

(ii) Location of stone after 4s is at point C, i.e., 19.24m downward point A. To determine distance taken at the maximum peak, distance of AB need to be calculated (zero velocity)

)tan(55.2

)905.4(2

)5(0

2

?,2222

22

ABofceDism

a

uvs

sasuv

Distance taken = AB + BA + AC= 2.55+2.55+19.24= 24.34m

(iii) Velocity, atuv

sm /62.14

)4)(905.4(5

Newton Motion Law

• First Newton Law– states that a body in resting state will be in resting sate and a body

moving in constant velocity will move in straight line if no external force acting on it.

• Second Newton Law – states that momentum change rate will be proportionate to its

resultant force.

2

2

/

)/()()(

1

tan

smkgN

smaxkgmNF

kwheremaF

tconskwherekmaF

maF

The equation shows that:-Force, F should be acting continuously and constant acceleration produced constant acceleration, a.Force, F acting on the body is resultant force.Direction for the force, F and acceleration, a is the same.

• Third Newton Law – states that every single acting force will have another force which have

the same reaction but with different direction

T

N

W W

a b.

A body imposed an acting force, i.e., the weight of the object, W on the table surface and the table imposed a reaction force i.e., normal reaction, N upon the body

A body imposed a reaction force i.e., the weight of the object, W upon the tie rod and the tie rod imposed a reaction force i.e., tension upon the body.

• Application ( Example using Newton Law)– Two object connected by light tie – Constant Tension (T) along tie rod– Similar acceleration but tie rod in elastic– Use Second Newton Law

• Resultant acceleration with the statement that the body imposed the same or opposite direction

)()2..(....................

)()2.....(..........

)()()1....(..........

1

11

22

iiforamT

iforamgmT

iiandiforamTgm

maF

Newton ‘s Law Gravitational attraction

• Newton postulated the law governing the gravitational attraction between any two particles. Stated mathematically,

Where F = force of gravitation between the two particlesG = universal constant of gravitation; according to experimental evidence, G = 66.73 x 10-12 m3/kgs2.m1,m2=mass of each of the two particle

r = distance between the two particles

221

r

mmGF

• two particles or bodies have mutual attractive (gravitational) force acting between them.– In the case of a particle located at or near the surface of the earth,

however, the only gravitational force having any sizeable magnitude is that between the earth and particle.

• approximate expression for finding the weight, W of a particle having a mass m1 = m is developed– Assume the earth to be a non-rotating sphere of constant density and

having a mass m2 = Me , then if is the distance between the earth’s center and the particle, we have

• Letting, yields , so, yields so

21

r

MmGW e

2r

GMg e 2/807.9 smg

mgW maF

Example 7.7

Mass of 12 kg and 8 kg was attached by using inelastic light tie rod at its end passing at a light and smooth pulley and the system is released. Find the acceleration and tension at tie rod

a

a

T

a12 kg

8kg

Solution:-Assume that the acceleration for every mass is a, m/s and tension of the rod is T N.From 2nd Newton Law, maF

NT

sma

ag

agT

amgmT

aTg

amTgm

18.94

/962.1

204

)2()1(

)2....(..........88

)1(..........1212

2

22

11

m1g

m2g

Example 7.9

• A mass of 12kg now been placed on smooth horizontal plane. Determine its acceleration and tension.

NxT

smx

a

ag

aT

aTg

amT

amTgm

04.4792.312

/92.320

81.98

208

);2()1(

12

88

)2....(....................

)1....(..........

2

1

22

Solution

Friction force

• Friction force – the force that reacting when 2 surfaces rubbing each other and the

force reacting in different direction– Advantage of friction force enables a body to initiate motion and stop

N

F

WRough floor

CASE 1;

F < friction force

(The object will be in the resting condition and static).

Friction force = static force

mgWNwhereNFs

CASE 2;

F > friction force

(the object is moving in the direction of F and kinetic force exist)

NFk

• From the second Newton Law F= maF-Q = ma

Analysis involving kinetic friction• Direction of friction force is opposite to direction of object’s movement • The kinetic friction force value is proportionate to normal reaction force

upon rubbing surface.• The kinetic friction force is not depending upon the object’s velocity or net

body velocity (if both rubbing objects are moving with different velocity

a

Q FA body on smooth floor

• F and friction force from the contact surface i.e.,

• Since is proportionate to the normal reaction, N hence we can write;

μk is kinetic friction constant. The value is in the interval of 0 μk 1.0• From 3rd Newton Law, N=mg , therefore we can substitute the value of N

in (1),

• Therefore the equation in (a ) become

aF

N

W

Object on rough floor

forcefrictiontheisFwithaFFF gg ).......(..........

)1....(..........NFNF gg

mgF kg

mgFF k

Example 7.10

A force of 100N is acting on a body with a mass of 20 kg placed on the rough horizontal plane as shown below. Coefficient of kinetic friction on the contact surface is 0.3. Calculate the friction force acted on the body and its acceleration.

100NN

W

Fg

Solution

From 2nd Newton Law,

2/06.2

2086.58100

86.58

)81.920(3.0

sma

a

maFF

N

xx

mg

NF

g

g

Newton 2nd LawF=ma

Example 7.12

Determine the minimum force to restrain the object with a mass of 20kg from slipping down. The slope plane and static friction coefficient is given as θ = 35oand 0.2 respectively.

FN

W

cosWFg 35o

N

Wx

mgFg

14.32

)cos(2.0

0cos

;

mgN

maF

directionyIny

y

N

xF

maFmgF

directionxIn

g

64.144

14.325..112

)35cos81.920(2.035sin)81.9(20

sin

;

Solution:-

Fg=μN

Example 7.13

A block weighs 5 kg is pushing up on a slope plane of from horizontal plane. If the value of F=100N , friction coefficient between the block and plane,μ=0.3 and the block initially in the resting condition. Calculate the velocity of the block after 2 seconds and the distance of the block moving within the interval.

Solution

N

mgN

NFg

5.42

)30cos81.9(5

30cos

From 2nd Newton Law;

2/55.12

577.62

)3.0(5.4230sin)81.9(5100

sin

sma

a

ma

maFmgF

maF

g

sm

atuv

/1.25)2(55.120

The velocity after 2 seconds:-

The displacement after 2 seconds:-

m

atuts

1.25)2)(55.12(2

10

2

1

2

2

Example 7.14

Two wood blocks, A and B connected with a light and inelastic tie rod as shown below. It been pulled up from a horizontal surface with an acceleration of2 m/s2 by a constant force, F. Mass A and B is given as 3kg and 5kg respectively. If the friction coefficient between the block and surface is given as 0.2, calculate the F and tie rod tension, T.