Everybody remains in a state of rest or continues to move in a uniform motion, in a straight line, unless acting on by an external force.

Transcription

1 NEWTON S LAWS OF MOTION Newton s First Law Everybody remains in a state of rest or continues to move in a uniform motion, in a straight line, unless acting on by an external force. Inertia (Newton s 1 st Law can be used to explain inertia ) This is the laziness or tendency of matter to resist changes in its motion. Inertia makes difficult for an object to start or stop moving, change direction or accelerate. Example: A passenger standing on a bus observes the effects of inertia, especially when the bus moves off and stops suddenly. Newton s Second Law When a force acts on a body, the rate of change on momentum is proportional to the applied force, and takes place in the direction on the force. force = mass acceleration F = ma 1N is defined as the force needed to give a mass of 1kg an acceleration of 1m/s² Momentum (Newton s 1 st and nd Laws can be used to explain Momentum) momentum = mass velocity p = m v (units of momentum = kgm/s) 1

2 Note: Momentum is a vector quantity. When a force acts on a body, it causes the velocity to change, and hence the momentum changes. We can determine the force applied by measuring the rate of change of momentum. F = m v t where: v = change in velocity = v u (final velocity initial velocity) Therefore: F = m(v u) t Impulse We can calculate the impulse, which is equal to change of momentum. It is determined by multiplying the force by time. Ft = mv mu unit Ns Newton s Third Law If body A exerts a force on body B, then body B exerts an equal but opposite force. NB: When an object is at rest on a flat level surface the forces acting are the weight (W) and the reaction to its weight acting perpendicular (Normal Reaction).

3 Terminal Velocity (Free Fall) When a fallen object has gained a velocity, a friction force (FR) opposes its weight (W), the resultant downward force (F) can be calculated by F = W F R We can use this equation to calculate the acceleration of the object. Since we know that F = ma And We can therefore state F = W F R Hence ma = W F R a = W F R m As the velocity of an object increases so does the magnitude of the frictional force, this continues until the friction force reaches the same value as the weight. Since it acts in an opposite direction to the frictional force, the resultant force is zero, W = F R W F R = 0 F R = 0 At this point the acceleration is also zero so the object is falling at a constant speed. This constant speed is called terminal velocity, and it is the maximum speed for particular objects, free falling through liquids. 3

4 Example: A marble falling through a viscous fluid. Stage A: When the marble is first released from rest, there s no frictional force (FR = 0). Therefore F = W W Stage B: Velocity increases as the marble falls hence frictional force increases, but is less than the weight (FR < W). The F which acts downwards is now W FR (F = W FR) FR W Stage C: Terminal velocity is reached when the frictional force is equal to the weight of the marble (W = FR). Therefore the resultant force is equal to zero, no acceleration (F = 0). FR W 4

5 Example 1 A woman of 70kg jumps out of an aeroplane, and before she could open her parachute she was falling at an acceleration of 7m/s². Calculate the frictional drag, if g = 10N/kg. Example A body of mass 7 kg rests on the floor of a lift. Calculate the force of R exerted on the body by the floor when the lift has: a. An upward acceleration m/s b. A downwards acceleration of 3 m/s c. A constant movement of velocity Example 3 A 600N Physics student stands on a bath scale in an elevator. What is the scale reading if the elevator ha an acceleration of: a. 1.8 m/s upwards b. 1.8 m/s downwards c. 10 m/s downwards Example 4 A block of mass kg rests on the floor of a lift which has an acceleration of 5m/s upwards. Find the reaction between the block and the lift. Example 5 A box of 60kg is pulled by a 400N force acting at an angle of 30 0 to the horizontal. The friction between the surface and the box is 00N. Calculate: a. the acceleration of the box b. the normal reaction acting on the box 5

6 Inclined Plane R F x Ѳ Ѳ W F y Parallel to Plane: Perpendicular to Plane: F x = W sinθ F y = W cosθ Consider an object on an incline plane, its weight acts vertically downwards. There is a normal reaction to the weight and its value is R. Assuming that the object is motionless on the slope, the fore that keeps it in contact with the surface is WcosѲ and the force that will cause it to move down the slope is WsinѲ. Example 1 An object slides down a slope which makes an angle of 30 0 to the horizontal. If the mass of the object is 1kg, calculate: a. the acceleration of the object down the slop assuming the slope is frictionless b. the normal reaction between the slope and the object Example A body of mass 5 kg is pulled up a smooth plane inclined at 30 0 to the horizontal by a force of 40N acting parallel t the plane. Calculate: a. the acceleration of the body b. the force exerted on the body on the plane 6

7 Example 3 A body of mass 10 kg is pulled up a smooth plane inclined at 60 0 to the horizontal by a force of 100 N acting parallel t the plane. Calculate: a. the acceleration of the body b. the force exerted on the body on the ground Example 4 A car of mass 900 kg accelerates up an inclined plane of angle 15 0 at a speed of 15m/s. If the car experiences a constant resistance to its motion of 150 N, calculate the driving force of the engine in the car. Example 5 A bag of cement of mass 50 kg is placed on a ramp and allowed to slide down it. Given that the ramp is at 30 0 to the horizontal, calculate: a. the component which acts down the slide b. the component which acts on the surface of the ramp c. the frictional force, if the acceleration of the bag down the slope is m/s 7

8 MOTION Definitions Displacement is the distance moved in a particular direction Velocity is the rate of change of displacement Acceleration is the rate of change of velocity Speed is the rate of change of distance Distance Time Graphs gradient = change in distance time taken gradient = Δd Δt = velocity Velocity Time Graphs gradient = change in velocity time taken gradient = Δv Δt = acceleration acceleration = v u t Area under the curve is the distance travelled 8

9 Equations for Motion Motion for uniform acceleration acceleration = rate of change of velocity time a = v u t at = v u v = u + at eq (i) average velocity = v + u displacement(s) = average velocity time From equation 1 we get that v + u ( ) t eq (ii) t = v u a When we substitute into equation, we get that v + u ( u ) (v ) a (v + u)(v u) a v u a v = u + as eq (iii) Using equation 1 and equation, we get that: 9

10 v + u ( ) t eq (ii) ut + vt But we know that v = u + at eq (i) Hence we can say that ut + (u + at)t ut + ut + at ut + at ut + ½ at eq (iv) Example 1 An object travelling at a speed at 10 m/s accelerated for 8 seconds to reach a speed of 50 m/s. Calculate: a. the acceleration of the object assuming it to be uniformed b. the distance travelled by the object after 8 seconds Example A car travelling at 0 m/s decelerates to a speed of m/s for 8 seconds. Calculate: a. the acceleration of the car after 8 seconds b. the distance the car travels after 8 seconds c. the total distance the car travels when it comes to rest 10

11 Example 3 A car travelling at 10 m/s accelerates uniformly at 1 m/s to reach a speed of 15 m/s. Calculate: a. the time it takes to reach 15 m/s b. the distance the car travelled during that time c. the velocity reached at 100 m from the point where the acceleration began Example 4 An object is thrown vertically upwards with an initial speed of 35 m/s. taking g as 10 m/s, calculate: a. the maximum height reached b. the time taken to reach the maximum height c. the time taken for the displacement to be 40 m d. velocity of the object after 5 seconds 11