Chapter 2&4_ Kinematics&Dynamic

7/29/2019 Chapter 2&4_ Kinematics&Dynamic

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Chapter 2:DESCRIBING MOTION/

Newtons Law of Motion


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Topics covered

Equation of motion in a straight line

Velocity-time graph

Newtons laws of motion

Coefficient of friction

Application of friction

Simple problems on mechanics involving Newtons

laws and friction


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Equation of motion in a

straight line

Initial Velocity (u);

Final Velocity (v),

Acceleration (a),

Distance Traveled (s),

Displacement (x) and

Time elapsed (t).

The equations which tell us the relationship between

these variables are as given below.

2

22

2

1

2

atuts

asuv

atuv


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Distance or displacement

Distance traveled and displacement are

different. When you traveled 50 km to the

East and then 20 km to the West, the total

distance you traveled is 70 km, but your

displacement is 30 km East.

Displacement, x = xf xi where xf is the final position

xi is the initial position.


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Motion: Distance, Velocity, and Acceleration

Distance

Final velocity

tvv

vts

2

0 2

0

2

1attvs

atvv 0 asvv 22

0

2


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Force

Any influence that can change momentum of an object eitherdirection, magnitude or both.

Force can be defined as a push or a pull.

Technically, force is something that can accelerate objects.

Example:

when you throw a baseball, you apply a force to the ball.

Force is measured by N (Newton).

A force that causes an object with a mass of 1 kg to accelerate

at 1 m/s2 is equivalent to 1 Newton. (F=ma)


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Net force

The sum of all forces acting on an object.

Example, in a tag of war, when one team is

pulling the tag with a force of 100 N and the

other with 80 N,

the net force would be 20 N at the direction of the

first team (100 N - 80 N = 20 N).


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Position-time Graph

shows the relationship

between time and position

From the graph:

find the average velocity of

the object at any given timeinterval

the instantaneous velocity

at any given time

Time (t) 0 1 2 3 4 5Position (s) 0 20 50 130 150 200


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Example:

A particles moves along the x axis. Its x coordinates varies

with time according to the expressionx= -4t + 2t2, wherex

is in meters and tis in second. The position-time graph for

this motion is shown in figure below.

determine the displacement of the particle in the time

intervals t = 0s to t = 1s and t = 1s to t= 3s

Calculate the average velocity in the time intervals t = 0

to t = 1s and t = 1s to t = 3s.

Find the instantaneous velocity of the particle at t = 2.5s


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distance versus time

-4

-2

0

2

4

6

8

10

12

14

16

18

0 1 2 3 4 5

t(s)

x(m)


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Answer (a)

The displacement can be calculated as

x = xf xi

= -2 0 = -2 m

For the second interval,at t = 1 s, x(m) = -2 m and at t = 3 s, x(m) = 6 m.

x = xf xi

= 6 (-2) = 8 m


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(b)

For the first time interval, the displacement, x, is -2 m so by

using the formula for average velocity:

For the second interval, the displacement is 8 m, time intervalt = 3 1 = 2 s.

sms

mV /2

1

2

sms

mV /4

2

8


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(c)

By drawing a tangential line at the graph at time t = 2.5s and

calculate its slope we find the instantaneous velocity v = 6 m/s

We can also differentiate the equation x = -4t + 2t2

smv

smtv

ttdt

d

dt

dxv

/6)5.21(4

/)1(4

)24( 2


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Velocity-time Graph

shows the relationship betweenvelocity and time.

Example, if a car moves at

constant velocity of 5 m/s for 10

seconds, you can draw a velocity-

time graph that looks like this: The area below the line

represents the displacement the

object traveled since it can be

calculated by xy, or (time *

velocity) which equals todisplacement


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Example

The velocity of a particlemoving along the x axis

varies in time according to

the expression v = (40-5t2)

m/s, where t is in second. Find the average

acceleration in the time

interval t = 0 to t = 2 s

velocity vs time

-50

-40

-30

-20

-10

0

10

20

30

40

50

0 1 2 3 4

t(s)

v(m/s)


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From the expression v = (40-5t2) we can calculate the valueof v when ti= 0 and tf= 2.

For ti= 0;

vi= (40 5t2)

= (40-5(0)2) = 40 m/s For tf= 2s;

vf= (40-5(2)2)

= (40 5(4))

= 40 20= 20 m/s Therefore the average acceleration in the time interval t = 1

to t = 2 s is:

smtt

vva

if

if/10

02

4020


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Inertia

Inertia is the resistance of an object to a

change in its state of motion.

The more of inertia that an object that rest

has, the more of an object resists changing in

state.

Inertia is a measure ofmass


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Newtons Laws of Motion

Newtons First Law of Motion

Newtons Second Law of Motion

Newtons Third Law of Motion


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1st Law

An object in motion will remain in motion,

with constant velocity,

and an object at rest will remain at rest unless

acted upon by an external force.

Sometimes called the Law of Inertia


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2nd Law

Newton's First Law deals with an object with no netforce.

Newton's Second Law talks about an object that hasnet force.

It states that when the net force acting on an objectis not zero, the object will accelerate at thedirection of the exerted force.

The acceleration is directly proportional to the net

force and inversely proportional to the mass.


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where:

F is the net force in N,

m is the mass of an object in

kg and

a is its acceleration in m/s2

we can say that force is

something that acceleratesan object.

F = ma


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Questions

How much net force is required to accelerate

a 1000 kg car at 5.00 m/s2? (5000N)

If you apply a net force of 1 N on 200g book,

what is the acceleration of the book? (5m/ s2)


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3rd Law

When one object applies a force on a second

object,

the second object applies a force on the first

that has an equal magnitude but opposite

direction.


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Newtons third Law

For example, when you push on a wall, the

wall will also push back on you with an equal

and opposite force.


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Question

What is the net force on 200 g ball when it hits

a wall with acceleration of 10 m/s2? (2N)


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Mass & Weight

Mass and weight are different in physics.

Example:

your mass doesn't change when you go to the

Moon,

but your weight does.

Mass shows the quantity, and

weight shows the size of gravity.


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How to calculate your weight?

If you know your mass, you can easily find

your weight because

W = mg

W is weight in Newton (N),

m is mass in kg, and

g is the acceleration of gravity in m/s2.

If mass is 70 kg on Earth,

weight is W=(70 kg)(9.8 m/s2) = 686 N.


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Question

Calculate the mass of an object that has a

weight of 115 N on the Moon? The gravity of

the Moon is 1/6 of g (which is 9.8 m/s2).

m=W/g

=115/[1/6(9.8)] = 70.4kg


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Exercises:

1. Determine the acceleration, when a12-N net force applied to a 3-kg object

and a 6-kg object?

2. A net force of 16 N causes a mass to

accelerate at a rate of 5 m/s2.

Determine the mass. 3. A object weighing 125N hangs from

a cable tied to two other cables

fastened to a support, as in the figure

below. The upper cables make angles

of 37.5o

and 53.0o

with the horizontal.Find the tension in the three cables.

370 530

object

TI T2

T3


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Answers:

1) Using F = ma

4 m/s2 and 2 m/s2

2)3.2 kg

3) Draw FBDT3

W

T1T2

-T3


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3)

From the first FBD, we have two forces acting on the object.

We know that W is 125 N so T3 must also be 125 N.

From the second FBD, we can construct the following table:

Forces x-Component y-Component

T1 -T1 cos 37.0o T1 sin 37.0

o

T2 T2 cos 53.0o T2 sin 53.0

o

T3 0 -125N


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For the x component:

Fx= -T1 cos 37.0o + T2 cos 53.0

o = 0

Fy= T1 sin 37.0o+ T2 sin 53.0

o + (-125 N) = 0

Try to solve it.. You should get the answer for: T1 = 75.1 N

T2 = 1.33(T1) = 1.33(75.1)= 99.9N


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Friction

Friction is the force that acts between two objects in contact

because ofaction-reaction.

When you slide your book on floor, it will come to stop

because of the force of friction.

There are two type of friction:

a. Static friction

b. Kinematics friction


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Normal force:

if a force F is applied to an object on a rough surface,

a frictional force of the same magnitude, -F, is produce in the

opposite direction of F.

The object is now under equilibrium.

N

F-F

W


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How to calculate force of friction?

where:

Ff:force of friction in N,

: coefficient of friction,

FN :normal force in N.

The value of depends on

surface you are dealing

with


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Example:

if you throw a 500 g book on floor where = 0.1, the forceof friction would be:

Ff= = (0.1)(0.5 * 9.8) = 0.49 N

Calculate the value of if the force of friction on a 300g

book was 0.5 N?

=0.5/(0.3*9.8)=0.17


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Application of Friction

Walking:

Static friction is the force that propels a person forward when walking.

On a slippery surface, i.e. almost no friction, a human could not stand

or walk without slipping.

Tug of Wars: The team with the highest static friction force has the advantage in tug

of wars. These contests are not won by pulling harder.


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Lets do some exercises

Simple problems on mechanics involving

Newtons Laws and Frictions
http://localhost/var/www/apps/conversion/tmp/scratch_4/Chap%203%20Newton's%20Law%20of%20motion-simple%20problem.dochttp://localhost/var/www/apps/conversion/tmp/scratch_4/Chap%203%20Newton's%20Law%20of%20motion-simple%20problem.dochttp://localhost/var/www/apps/conversion/tmp/scratch_4/Chap%203%20Newton's%20Law%20of%20motion-simple%20problem.dochttp://localhost/var/www/apps/conversion/tmp/scratch_4/Chap%203%20Newton's%20Law%20of%20motion-simple%20problem.doc

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