DYNAMICS & NEWTON’S LAW

DYNAMICS & NEWTON’S LAW

Topics:

Force

Newton’s Law of Motion

Application of Newton’s Law

1. Kinematics:is the branch of mechanics which studies a

motion that is observed based on physical quantities such as position, distance, displacement, speed, velocity and acceleration with no consider on what causes the motion (force).

Examples: Uniform rectilinear motion of a particle Accelerated uniform rectilinear motion Vertical motion Parabolic motion etc

2. Dynamics:is the branch of mechanics concernings

motion of object with regard to what causes the motion (force).

Examples: Motion of object on a smooth plane Motion of object on a rough plane Motion of object on an inclined plane Motion of object that connected by a pulley Lift etc

Forceis a quantity that can change the state of motions

(e.g., golf game), forms (e.g., bread batter making) and sizes (e.g.,spring) of an object.

force can be a push or a pull upon an object.

The first Newton’s Law

In the absence of external forces, an object at rest remains at rest and an object

in motion continues in motion with a constant velocity (that is, with a constant

speed in a straight line).

0F

V= 0

a = 0

At rest

GLB

The 2nd Newton’s Law

So, What will happen , if the resultant of Force is not ZERO ?

???0 F

mF

F aa = ?

a~F

a ~ 1/m m

Fa

• For example :

Sample problem

What is the acceleration resulted when a force resultant of 12 N is applied to an object which is 6 kg in mass?

Solution:

m.aF

m

Fa

22m/s6kg

12N

3. Newton’s Third Law

If two objects interact, the force F12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force F21 exerted by object 2 on object 1:

Faction = - F reaction

application

Normal force

Weight

On horisontal plane :

N = W

θ

ww cosθ

θw sinθ

N

On inclined plane :

N = W cosθ

W = mg

N

WN

w

w’

T

T

T = W

F

fk

Nf

Nf

ss

kk

There two kinds of frictional forceStatic frictional force: is a frictional force exerts on the

object which is at rest.Kinetic frictional force: is a frictional force works on the

moving object.

Nμf ss Nμf kk

fs = static frictional force (N)

fk = kinetic frictional force (N)

μs = coefficient of static friction

μk = coefficient of kinetic friction

μs > μk,

So … fs > fk

0 ≤ μ ≤ 1

μ = 0 (smooth)

μ > 0 (rough)

Friction force

F

fsmaks

Friction force

F

fs maks

If F < fsmaksat rest

If F = fsmaks Exactly will be move

If F > fsmaks Object is moving

m

fF

m

Fa k

fsmaks = μs N

Sample Problem Calculate the

acceleration of system , if Force applied is : (g=10m/s2 )a. 50 Nb. 52 Nc. 60 N d. 70 Ne. 61 N

12 kg

F

μS= 0,5 μk = 0,3

Sample problem

A car is 800 kg in mass, from rest it is accelerated with a constant acceleration, after 2 seconds the car travels at a distance of 20 m. Determine the force resulted by the car if during the motion it experiences the frictional force of 200 N!

Solution:

m.aF

m.afF Because a = constant, then the car moves in accelerated uniform rectilinear motion, therefore the acceleration (a) can be determined by;

2o at2

1tvs

2a(2s)21(0)(2s)20m

2m/s 4

2x20a

2m/s 01

2m/s kg.10 800N 200F

N 8000N 200F

N 2008

Sample problem:

Determine the normal force of cube A, B, and C, if m = 10 kg and g = 9.8 m/s2!

N=?

w

m

A

N=?

w

m

B

N=?

w

m

C

20 N 15 N

Cube A

Solution:

Because on cube A exerts no outside force,

N = w = m.g

= (10kg)(9.8m/s2) = 98 N

Cube B

Because on cube B works an outside force of 20 N,

N = ƩF = w + F = m.g + F

= (10kg)(9.8m/s2) + 20 N = 118 N

Cube C

Because on cube C works an outside force of 15 N,

N = ƩF = w - F = m.g - F

= (10kg)(9.8m/s2) - 15 N = 83 N

Problem 1

3 kg

7 kg

Calculate the acceleration of system and stress of string (percepatan system dan tegangan tali ) if flour is smooth ( jika lantai licin ) g = 10 m/s2

Answer

3 kg

7 kg

w=mg

= 30N

W2=70N

NBecause the floor is smooth , μ=0, fs =0

2

21

1

/3

73

30

sma

a

mm

wa

m

Fa

Problem

3 kg

7 kg

Calculate the acceleration of system and stress of string (percepatan system dan tegangan tali ) if flour is rough ( jika lantai kasar ) g = 10 m/s2

μs=0,4 μk=0,3

Answer

3 kg

7 kg

w=mg

= 30N

W2=70N

N

μs=0,4, fsmaks =0,4.70

= 28N

2

21

1

/9,0

10

9

73

2130

sma

a

mm

fwa

m

fFa

k

k

μs=0,4 μk=0,3

fk=μk.N

= 0,3.70=21N

Problem 2

Determine the acceleration (a) and tension of each string !

m1= 4 Kg

m2 = 6 Kg

There isn’t

Friction between string and pulley

Determine the acceleration (a) and tension of each string !

If the inclined plane is smooth. ( g = 10 m/s2)

3Kg

2Kg

30o

Answer

3Kg

2Kg

30oW1

T

W2

w2sin30o

w2cos30o

T2

21

2121

/4

520

51030

)23(.2030

)(30sin

sma

a

a

a

ammww

maFo

Tension of string (T)

3Kg

2Kg

30oW1

T

W2

w2sin30o

w2cos30o

T

NT

T

T

amwT

atau

NT

T

T

amTw

maF

18

810

4.2.20

30sin

18

1230

4.330

21

22

11