NEWTON’S FIRST LAW

1. NEWTON’S FIRST LAW • Definition- 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

2. Balanced and Unbalanced Force Balanced Force (zero net force) the net force is zero when forces of equal magnitude act in opposite direction Example : Fnet= F1-F2=0

3. Unbalanced Force (nonzero net force) - Only one force equal to the net force were acting. Net force always produces an acceleration Eg: a)

4. b)

5. y 15 N 15 N x 10 N QUIZ 2 What is the net force acting on the object shown above? • 40 N • 0 N • 10 N down • 10 N up

6. NEWTON’S SECOND LAW 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 samedirection 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

7. 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

8. 5 3 a 20°C 30°C 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.

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

10. The Second Law in Component Form • A force may be expressed in component notation in 2 dimension : and

11. 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 : 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)

12. N Fb Fa A B W= -N W=mg Force pair of the third law is that the action-reaction forces do not act on the same object. Magnitude normal force N and weight is equal but different direction.

13. FORCE FROM NEWTON’S LAW Frictional Forces are dividedto 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

14. N, normal reaction Fg, friction force F, pulling force W weight force Fg- friction force R-normal reaction µ- coefficient of friction • Static friction increase if F increase at: • F=Fg, object start to move, Fs is max, called as limited friction force • 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

15. Determination Coefficient of Static Friction R Friction mg sin θ θ mg cos θ mg The normal reaction R is not necessarily always equals to the weight, mg of the object Figure 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 cosθ is perpendicular to the inclined plane, since the Box is start to move; R= mg cosθ and friction, Fg =mg sin θ

16. From the same figure, Inclined plane Fg= mg sin θ R= mg cosθ Therefore coefficient static friction (µ); Fg=µR µ=Fg/R =mg sinθ/mg cosθ= tan θ

17. 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. • Is the object moving? • What is the acceleration?

18. END OF CHAPTER 2