NEWTON’S FIRST LAWDefinition- An object at rest will remain at rest, or if it is moving, it will continue to move with constant velocity, unless acted upon by a resultant force.sometimes called the “law of inertia”. Inertia of an object is the object resistance to changes from its state of rest or motion.In a balanced applied force (Fnet=0), unmoving body will remains at rest, and a body already in motion remains in motion with a constant velocity (constant speed and direction)

note: net force Fnet, is the vector sum ∑Fi, or resultant of all the forces acting on the object system

Balanced and Unbalanced Force

Balanced Force (zero net force)the net force is zero when forces of equal magnitude act in opposite directionExample :

Fnet= F1-F2=0

Unbalanced Force (nonzero net force)

- Only one force equal to the net force were acting. Net force always produces an acceleration

Eg: a)

b)

QUIZ 2

What is the net force acting on the object shown above?

15 N 15 N

10 Nx

y

a. 40 N

b. 0 N

c. 10 N down

d. 10 N up

Definition: The rate of change of momentum of an object is directly proportional to the resultant force acting on the object and is in the same direction as the resultant force.

Resultant force, F = d/dt (mv)Where m is the mass and v is velocity

We can express the equation;

where the net force is equal to the product of the mass times the acceleration

SI unif of force: newton (N) or kgms-2

NEWTON’S SECOND LAW

amFnet

When the object is drop, it falls towards the earth. Force is acting on the object, its weight (w) is the net force, acceleration due to gravity (g). Therefore equation for weight in terms of mass is a form of Newton second law;

Example 1:

By using Newton’s Second Law, Shows that where F is measured in Newton m in kg and a in ms-2

maFnet

mgw

30°C20°C

35

a

Example 2 :

One boxes with the mass of 5 kg is act by the two force as shown in Figure above. Calculate the acceleration of the boxes.

Figure above shows the relationship among force, acceleration and mass (assuming no friction)

The Second Law in Component FormA force may be expressed in component notation in 2 dimension :

ymaxmyaxamyFxF yxyxyx ˆˆa)ˆˆ(ˆˆ

xmaFx and

amFnet

yy maF

Newton’s Third Law

Definition: Every action has a reaction that is equal in magnitude but opposite in direction

In symbol notation, Newton’s third law is

2112 FF

12F

21F

: the force exert on object 1 by object 2

:the equal and opposite force exert on object 2 by object 1 (minus indicates opposite direction)

Force pair of the third law is that the action-reaction forces do not act on the same object.

N

W=mg

W= -N

Fb Fa

A B

Magnitude normal force N and weight is equal but different direction.

FORCE FROM NEWTON’S LAW

Frictional Forces are divided to 2 conditions:

a)Static Friction- exist when the object is static/before motion occurs

b)Kinetic Friction- exist when the object is moving.

Static Friction-is the resistance to motion that occurs between contacting surfaces

N, normal reaction

F, pulling force

W weight force

Fg, friction force

Static friction increase if F increase at:a) F=Fg, object start to move, Fs is max, called as limited friction forceb) F> Fg, object accelerate. Fg depends on the nature of the two surface in contact.

The ratio of the magnitude of limiting static friction Fg to the normal reaction R is known as coefficient of static friction,µs limiting static friction Fg=µsR R=mg

c) Fg>F, reaction of friction force when motion occur and is called kinetic friction. Fg=µsR

Depends on,- normal reaction force for that object- surface area of contact- velocity of the movement an object

Fg- friction force

R-normal reaction

µ- coefficient of friction

Determination Coefficient of Static Friction

θ mg cos θ

mg sin θ

Friction

mg

R

The normal reaction R is not necessarily always equals to the weight, mg of the objectFigure above shows the box at rest on a rough plane inclined at an angle θ to the horizontal.The weight mg of the box may be resolved into two orthogonal directions. One component mg sin θ Is parallel to the inclined plane and another component mg cos θ is perpendicular to the inclined plane, since theBox is start to move;

R= mg cos θ and friction, Fg =mg sin θ

From the same figure,Inclined planeFg= mg sin θ

R= mg cos θTherefore coefficient static friction (µ);

Fg=µR

µ=Fg/R =mg sinθ/mg cos θ= tan θ

Eg:

100N

One object with mass 20 kg is located on the table is pulled by horizontal force 100N. Coefficient friction between object and table is 0.4.a) Is the object moving?b) What is the acceleration?