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Chapter 3  forces
 

Chapter 3 forces

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    Chapter 3  forces Chapter 3 forces Presentation Transcript

    • Unit: 3 FORCE Effects of forces
    • Learning objectives
      • At the end of this unit you should be able to :
      • Understand the meaning of effects of forces;
      • Know various types of effects of forces.
    • Force
        • The environment affects a
        • body by exerting forces on it.
        • Force simply as a push and
        • pull.
    • Force
      • Force is the effect of pull and push.
      Boy exerts a pull on the rope Boy exerts a push on the boat
    • SI unit of force
      • SI Unit of force
        • newton , N
        • 1 N = 1 kg x 1 ms -1
    • Examples of Forces
        • Friction
        • Tension
        • Weight
        • Electric force
        • Magnetic force
    • Effects of Force
      • Force can
      • change the shape and size of an object
    • Effects of Force
      • Force can
        • change the motion of the body by:
        • Moving a stationary body.
        • Increasing speed or accelerating a body;
        • Decreasing speed or decelerating a body;(i.e.stop a body)
      • Force can
      • Change the direction of motion of an object.
      Effects of Force
    • Newton’s law of Motion 1st Law & 2nd Law
    • Learning objectives
      • At the end of this unit you should be able to :
      • quote and use the Newton’s three laws of motion.
    • Assumption
      • To simplify Newton’s laws of motion, assume that any object being considered:
      • has a constant mass;
      • is not rotating;
      • is in surroundings which are not
      • accelerating.
    • Newton’s Laws of Motion Newton’s First Law of Motion Newton’s Second law of Motion Newton’s Third Law of Motion
    • The First Law
      • An object is at rest will remain at rest, and moving at constant speed in a straight line will continue to do so unless an unbalanced external force is applied to it.
    • Simple illustration of first law
      • To Newton,
        • ‘ uniform motion in a straight line’ meant constant velocity .
        • The ‘state of rest’ meant zero velocity .
      • The velocity of an object remains constant if the net force on it is zero.
    • The Second Law
      • The acceleration of an object is directly proportional to and in the same direction as the resultant force acting on it, and inversely proportional to the mass of the object.
    • 2nd Law
      • Thus, a  F and a  1/m
      • Hence,  F/m  F  ma
      • F = ma
        • If m = 1 kg, a = 1 ms -2
        • F = 1 kg x 1 ms -2 = 1 N
    • 2nd Law
      • Since F = ma = m(v-u)/t
      • as (mass x velocity) = momentum and
      • (mv - mu) / t is the rate of change of momentum
      • Therefore, 2rd Law can be stated as
      • The rate of change of the momentum of an object is equal to the unbalanced force applied to it. The direction of momentum takes place in the direction of the force .
    • 2nd Law F = ma 1 Newton is a force which causes a mass of 1 kilogram to have an acceleration of 1 ms -2 .
    • Simple illustration of 2nd Law
      • The second law is really a description of how a body responds mechanically to its environment.
      • The influence of the environment is the net force, F, the body’s response is the acceleration, a, and the strength of the response is the inversely proportional to the mass, m.
    • 3rd Law
      • For every action, there is an equal and opposite reaction.
    • Simple illustration of 3rd Law 1. Newton’s third law is a statement about the forces rather than about motion. 2. Action and reaction refer to forces. 3. This law plays important role in the development of the methods used to analyze the motion of extended bodies that rotate as they move through space.
    • Example
      • What force is required to accelerate a 2000 kg car from 36 km h -1 to 72 km h -1 in 5 s ?
      • < Solution>
      • v - u = 72 - 36 = 36 km h -1
      • =(36x1000)/3600 = 10 ms -1
      • Since a = (v-u)/t = 10/5 = 2 ms -1
      • F = ma = 2000 x 2 = 4000N
    • Example
      • A car of mass 1000 kg travelling at 20 ms -1 is brought to rest in 5 s. Find
      • (a) the average deceleration,
      • (b) the average braking force,
      • (c) the distance travelled before stopping.
      • <Solution>
      • (a) a = (v-u) / t = (0-20)/5 = -4 ms - 2 ,
      • deceleration = 4 ms -2
      • F = ma, the braking force = 1000 x 4
      • = 4000 N
      • (c) s = ut + 1/2 at 2 =(20 x 5)-(0.5 x 4 x 25)
      • = 50 m
    • GCE O-Level Past Examination Paper Science (Physics)
    • Nov 1991 1. A horizontal force of 5 N was applied to a block of mass 2 kg resting on a frictionless table. What was the acceleration of the block? A 0.4 ms -2 B 2.5 ms -2 C 10 ms -2 D 25 ms -2 B Hint: F =ma
    • GCE O Nov 1994 2. A horizontal force of 5 N was applied to a block of mass 2 kg, resting on a frictionless table. What was the acceleration of the block ? A 0.4 ms -2 B 2.5 ms -2 C 10.0 ms -2 D 25.0 ms -2 B
    • Nov 1998 3. A horizontal force of 8 N is applied to to a block of mass 2 kg, resting on a frictionless table. What is the acceleration of the block? A 0.25 ms -2 B 4.0 ms -2 C 6.0 ms -2 D 16 ms -2 B Hint: F = ma
    • Nov 1990 4. A ball of mass 0.4 kg was thrown vertically into the air. It reached a maximum height of 3.0 m. (g = 10N/kg) a) What was the force on it at the highest point? [1] b) What was its acceleration at the highest point? [1] c) After it had fallen 1 m from its highest point, it’s acceleration was 8 m/s 2 . Why was it not 10m/s 2 .? [2] (a). At highest point, v = 0  a = 0  force = 0 N (b). Acceleration = 0 ms -2 (c ). due to air resistance, acceleration will gradually decreased from 10 ms -2 and hence after it had fallen to I m, a = 8 ms -2 .