Скачать презентацию 1 of 28 Boardworks Ltd 2009
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Скачать презентацию 1 of 28 Boardworks Ltd 2009
Скачать презентацию 1 of 28 Boardworks Ltd 2009
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What are we learning today? You will be a successful learner today if you can: 1. Describe the effects of balanced forces. 2. State Newton’s first Law. Success Criteria: Complete LO’s.
Isaac Newton was a great mathematician and scientist. His discoveries made major advances to our knowledge of the physical laws that govern the Universe. Isaac Newton 1643 - 1727 He invented a new branch of mathematics called calculus, that enabled mathematicians to solve many problems. He researched into gravity, light, rotating bodies, planetary orbits, and tidal motion. He had amazing powers of concentration, sometimes studying and solving problems continuously for 19 hours a day. In 1687, Newton published his most important discoveries in a famous book called “Principia Mathematica”.
• Youtube dog on skate board http: //www. youtube. com/watch? v=tgq. R 9 FGBk 6 E&feature=fvst
Newton’s First Law When a vehicle travels at a constant speed, the frictional forces exactly balance the driving force, so the resultant force is zero.
Newton’s First Law When a vehicle travels at a constant speed, the frictional forces exactly balance the driving force, so the resultant force is zero.
Applying Newton’s first law In everyday life, most things don’t move unless a force is applied. Why is this? What forces act on this car as it travels at a steady speed? Air resistance and friction acts against the direction of motion. Engine force from the car pushes the car forward. The car is travelling at a steady speed, so these forces (air resistance + friction and engine force) must be equal. 8 of 28 © Boardworks Ltd 2009
An object in space will continue at the same speed forever, unless it is acted upon by the gravity of a celestial body. 9 of 28 © Boardworks Ltd 2009
• Colorado Phet “Motion and Force”
• Colorado Phet “ Force”
Newton’s Laws of Motion Newton stated three laws concerning force and motion. Newton’s First Law If the forces acting on an object are balanced then the object will continue travelling at constant speed in a straight line or remain at rest. Examples: § If the resultant force acting on a stationary body is zero, the body will remain stationary. § If the resultant force acting on a moving body is zero, the body will continue to move at the same speed and in the same direction (i. e. with the same velocity).
Balanced forces Force is a vector. The vector sum of the forces on an object gives a net or resultant force. Look at the forces on this object: To see the vector sum of the forces, add the horizontal forces together. 3 N 3 N There is no ‘net’ or resultant force on the mass: the forces are balanced. 13 of 28 © Boardworks Ltd 2009
Balanced forces Force is a vector. The vector sum of all the forces on an object gives a net or resultant force. Look at the forces on this object: 4 N To see the vector sum of the forces, add the horizontal forces together first then add the vertical forces. 3 N 4 N 14 of 28 3 N Again there is no ‘net’ or resultant force on the mass: the forces are balanced. © Boardworks Ltd 2009
Relative motion 15 of 28 © Boardworks Ltd 2009
Newton’s first law: true or false? 16 of 28 © Boardworks Ltd 2009
What are we learning today? You will be a successful learner today if you can: 1. Describe the effects of unbalanced forces. 2. State Newton’s second Law. 3. Analyse skydiving. 4. Calculate tension in haulage systems. 5. Calculate forces in rocket motion. 6. Analyse forces in lifts. Success Criteria: Complete LO’s.
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There are 2 parts to Newton’s Second Law Part 1: The acceleration of a body is proportional to the resultant force and in the same direction as the force. Newton’s second law is represented by the equation: F = m x a Unbalanced Force mass acceleration § If the resultant or unbalanced force acting on a stationary body is not zero, the body will accelerate in the direction of the resultant force.
Demonstrating Newton’s Second Law
Using F = ma F = mxa m = F/a a = F/m
Unbalanced forces The forces on this 1 kg toy car are unbalanced: 3 N 18 N What is the resultant force? 18 N 3 N 15 N F = 18 - 3 = 15 N θ = 180° Force is a vector, resultant force is 15 N to the right There is a resultant force of 15 N at a direction of 180° from the horizontal. 22 of 28 © Boardworks Ltd 2009
Unbalanced forces The forces on this mass are unbalanced: What is the resultant force? Add the vectors together again, nose-to-tail. 8 N 3 N F 3 N 4 N θ 4 N 3 N F = √ 32 + 42 = 5 N 4 N θ = tan-1(4/3) = 53° There is a resultant force of 5 N at a direction of 53° from the horizontal. 23 of 28 © Boardworks Ltd 2009
Using F = ma connects the resultant force to the mass of the object and its acceleration: Force = mass x acceleration N kg m/s 2 Force is measured in Newtons (N). One Newton is the force required to accelerate a mass of 1 kg by 1 m/s 2. The formula, F = ma can be rearranged using a formula triangle. . .
Using F = ma If you know any two of these values, you can use the formula triangle to find the remaining value. . . m = F a = F F = mxa m a
Using F = ma Example 1 (General / NAT 4 level) A lorry with a mass of 3500 kg decelerates steadily at 3 m/s 2. What is the braking force applied? F = m x a F = 3500 x 3 F = 10500 N The braking force applied is 10500 N
Using F = ma Example 2 A force of 350 N acts on an object of mass 25 kg. If the object accelerates at 10 ms-2. What is the force of friction? F = mx a F = 25 x 10 F = 250 N But this is the unbalanced force 100 N 350 N So the force of friction is 100 N
Summarising Newton’s second law If there is a resultant force on an object, it accelerates in the direction of that force. Its acceleration is directly proportional to the force, and inversely proportional to the object’s mass. This can be stated in a vector equation: F = ma 28 of 28 © Boardworks Ltd 2009
Newton’s second law example 1 Haulage System A car pulls a caravan with an engine force of 4000 N. 1. Find the acceleration of the car and caravan. 4000 N 3000 kg 5000 kg The engine must accelerate both the car and the caravan: a = F/m = 4000 / (3000 kg + 5000 kg) = 0. 5 ms-2 29 of 28 © Boardworks Ltd 2009
Newton’s second law example 1 Haulage System Both the car and caravan are accelerating at 0. 5 ms-2. 2. Find the tension (force) in the tow-bar. ? 4000 N 3000 kg 5000 kg To answer this type of question consider putting your hand over the car and ask the question what force would cause the caravan to accelerate at 0. 5 ms-2. ? F (or tension)= ma = 5000 kg × 0. 5 ms-2 = 2500 N This is the tension in the tow-bar. 30 of 28 © Boardworks Ltd 2009
Newton’s second law example 2 Skydiving 31 of 28 © Boardworks Ltd 2009
Newton’s second law example 3 Rockets and Missiles A rocket of mass 2 x 106 kg is fired vertically into the air. The rocket motors provide a thrust of 50 x 106 N, and there is a drag force of 2 x 106 N. Find the acceleration of the rocket. F 1 a = F/m = unbalanced force / mass unbalanced force = F 1 – F 2 F 1 = 50 x 106 N F 2 = 2 x 106 N + mg F 2 = 2 x 106 N + 20 x 106 N = 22 x 106 N a = (50 x 106 – 22 x 106 ) 2 x 106 Acc =14 ms-2 32 of 28 Why would the acceleration of the rocket change? © Boardworks Ltd 2009
Newton’s second law example 4 Lift experiment Stand in a lift with a mass on a spring balance and record the reading. Note if the reading increases or decreases as the lift: accelerates upwards accelerates downwards decelerates upwards decelerates downwards 33 of 28 © Boardworks Ltd 2009
Newton’s second law example 4 Lifts What is the weight of the woman in this lift which is staionary? W = mg W = 60 x 10 (9. 8) W = 600 N 34 of 28 © Boardworks Ltd 2009
Newton’s second law example 4 Lifts What is the weight of the woman in this lift which is moving at a constant speed up or down? W = mg W = 60 x 10 (9. 8) W = 600 N 35 of 28 © Boardworks Ltd 2009
Newton’s second law example 4 Lifts The lift can accelerate and decelerate upwards. It can also accelerate and decelerate downwards. We now require to calculate the apparent weight of the women in each of these 4 situations. 36 of 28 © Boardworks Ltd 2009
Newton’s second law example 4 Lifts What is the apparent weight of the woman when the lift accelerates upwards at 1 ms-2? (speeding up on the way up) To answer this question we need to consider the 2 forces: weight and apparent weight As the lift accelerates it is the apparent weight which the lady feels and is displayed on the scales. Her weight, the force acting downwards, is always W = mg W = 60 x 10 (9. 8) W = 600 N weight 37 of 28 © Boardworks Ltd 2009
Newton’s second law example 4 Lifts If the lift accelerates upwards then there must be a greater upward force acting on the lady than downward force? So…the unbalanced force is? F 1– F 2 = ma apparent weight F 1 Apparent weight – weight = ma Apparent weight – 600 = 60 x 1 Apparent weight = 660 N She feels heavier! weight F 2 38 of 28 © Boardworks Ltd 2009
Newton’s second law example 4 Lifts What is the apparent weight of the woman when the lift deccelerates upwards at 1 ms-2? (slowing down on the way up) To answer this question we need to consider the 2 forces: weight and apparent weight As the lift decelerates upwards it is the apparent weight which the lady feels and is displayed on the scales. Her weight, the force acting downwards, is always W = mg W = 60 x 10 (9. 8) W = 600 N weight 39 of 28 © Boardworks Ltd 2009
Newton’s second law example 4 Lifts If the lift decelerates upwards then there must be a greater downward force acting on the lady than upward force? So…the unbalanced force is? F 2– F 1 = ma apparent weight F 1 Weight – apparent weight = ma 600 – apparent weight = 60 x 1 Apparent weight = 540 N She feels lighter! weight F 2 40 of 28 © Boardworks Ltd 2009
Newton’s second law example 4 Lifts What is the apparent weight of the woman when the lift accelerates downwards at 1 ms-2? (speeding up on the way down) To answer this question we need to consider the 2 forces: weight and apparent weight As the lift accelerates downwards it is the apparent weight which the lady feels and is displayed on the scales. Her weight, the force acting downwards, is always W = mg W = 60 x 10 (9. 8) W = 600 N weight 41 of 28 © Boardworks Ltd 2009
Newton’s second law example 4 Lifts If the lift accelerates downwards then there must be a greater downward force acting on the lady than upward force? So…the unbalanced force is? F 2– F 1 = ma apparent weight F 1 Weight – apparent weight = ma 600 – apparent weight = 60 x 1 Apparent weight = 540 N She feels lighter! weight F 2 42 of 28 © Boardworks Ltd 2009
Newton’s second law example 4 Lifts What is the apparent weight of the woman when the lift decelerates downwards at 1 ms-2? (slowing down on the way down) To answer this question we need to consider the 2 forces: weight and apparent weight As the lift decelerates downwards it is the apparent weight which the lady feels and is displayed on the scales. Her weight, the force acting downwards, is always W = mg W = 60 x 10 (9. 8) W = 600 N weight 43 of 28 © Boardworks Ltd 2009
Newton’s second law example 4 Lifts If the lift decelerates downwards then there must be a greater upward force acting on the lady than downward force? So…the unbalanced force is? F 1– F 2 = ma apparent weight F 1 Apparent weight – weight = ma Apparent weight - 600 = 60 x 1 Apparent weight = 660 N She feels heavier! weight F 2 44 of 28 © Boardworks Ltd 2009
Newton’s second law example 5 Components How would you calculate the acceleration of the vehicle shown below? What information do you need? 500 grams F 1– F 2 acceleration = force mass 50 grams Weight = mg = 50 x 10 -3 x 10 (9. 8) = 0. 5 N weight 45 of 28 acceleration = force = 0. 5 = 1 ms-2 mass 0. 5 © Boardworks Ltd 2009
Newton’s second law example 5 Components When this experiment was performed these results were recorded on the QED: 0. 69 ms-2, 0. 68 ms-2, 0. 72 ms-2, 0. 71 ms-2 and 0. 71 ms-2. Can you think why? Before we consider why lets calculate the average acceleration and the random uncertainty. 46 of 28 © Boardworks Ltd 2009
Newton’s second law example 5 Components Average acceleration = 0. 69 + 0. 68 + 0. 72 + 0. 71 Average acceleration = 0. 70 ms-2 Random uncertainty = largest value – smallest value number of values Random uncertainty = 0. 72– 0. 68 = 0. 04 = 0. 008 = 0. 01 5 5 Random uncertainty = + 0. 01 - Average acceleration = 0. 70 ms-2 47 of 28 + 0. 01 ms-2 © Boardworks Ltd 2009
Newton’s second law example 5 Components Now go ahead and try this experiment. Calculate theoretical value then obtain the experimental value with your mass and vehicles mass. ? grams F 1– F 2 acceleration = force mass ? grams Theoretical value Weight = mg = ? x 10 (9. 8) = ? N weight acceleration = force = ? ms-2 mass ? The experimental value was ? ms-2. 48 of 28 © Boardworks Ltd 2009
Newton’s second law example 5 Components The value obtained from the QED is less than the calculated value. Where have we gone wrong? F 1– F 2 This equation is OK acceleration = force mass This mass is OK But the force we use is too big. 49 of 28 © Boardworks Ltd 2009
Newton’s second law example 5 Components To understand why the experimental value is less than the calculated value we must now consider components of a force. A force can be split horizontal and vertical components. 50 of 28 © Boardworks Ltd 2009
Newton’s second law example 5 Components We must use the force which acts in the same direction to the one in which the vehicle accelerates. We must therefore find the horizontal component of force. Cos 45 = horizontal component of force 0. 5 Horizontal component of force = 0. 5 x cos 45 = 0. 35 N 51 of 28 © Boardworks Ltd 2009
Newton’s second law example 5 Components horizontal component of force: Cos 45 = horizontal component of force 0. 5 Horizontal component of force = 0. 5 x cos 45 = 0. 35 N acceleration = force = 0. 35 = 0. 7 ms-2 mass 0. 5 The value obtained experimentally 52 of 28 © Boardworks Ltd 2009
Newton’s second law example 6 Components A man pulls a garden roller of mass 100 kg with a force of 200 N acting at 300 to the horizontal. If there is a frictional force of 100 N between the roller and the ground, what is the acceleration of the roller? 100 N acceleration = unbalanced force = 200 - 100 = 1 ms-2 mass 100 Is this a good answer? 53 of 28 © Boardworks Ltd 2009
Newton’s second law example 6 Components No, it is not a good answer. When calculating the acceleration you must use only the bit of the force which acts in the direction the roller moves. You must calculate the……. . 100 N Horizontal component of force Cos 30 = horizontal component of force 200 Horizontal component of force = 200 x cos 30 = 173 N 54 of 28 © Boardworks Ltd 2009
Newton’s second law example 6 Components 100 N friction 173 N acceleration = unbalanced force = 173 - 100 = 0. 73 ms-2 mass 100 55 of 28 © Boardworks Ltd 2009
History lesson: Galileo (1564 – 1642) Galileo was a hugely influential Italian astronomer, physicist and philosopher. Galileo was born on 15 February 1564 near Pisa, the son of a musician. He began to study medicine at the University of Pisa but changed to philosophy and mathematics. In 1589, he became professor of mathematics at Pisa. In 1592, he moved to become mathematics professor at the University of Padua, a position he held until 1610. During this time he worked on a variety of experiments, including the speed at which different objects fall, mechanics and pendulums. In 1609, Galileo heard about the invention of the telescope in Holland. Without having seen an example, he constructed a superior version and made many astronomical discoveries. These included mountains and valleys on the surface of the moon, sunspots, the four largest moons of the planet Jupiter and the phases of the planet Venus. His work on astronomy made him famous and he was appointed court mathematician in Florence. In 1614, Galileo was accused of heresy for his support of the Copernican theory that the sun was at the centre of the solar system. This was revolutionary at a time when most people believed the Earth was in this central position. In 1616, he was forbidden by the church from teaching or advocating these theories. In 1632, he was again condemned for heresy after his book 'Dialogue Concerning the Two Chief World Systems' was published. This set out the arguments for and against the Copernican theory in the form of a discussion between two men. Galileo was summoned to appear before the Inquisition in Rome. He was convicted and sentenced to life imprisonment, later reduced to permanent house arrest at his villa in Arcetri, south of Florence. He was also forced to publicly withdraw his support for Copernican theory. Although he was now going blind he continued to write. In 1638, his 'Discourses Concerning Two New Sciences' was published with Galileo's ideas on the laws of motion and the principles of mechanics. Galileo died in Arcetri on 8 January 1642. 56 of 28 © Boardworks Ltd 2009
Galileo's Inclined Plane Experiment Galileo's use of the inclined plane to study the motion of objects is one of his most important contributions to science. As this video segment from NOVA illustrates, the inclined plane allowed Galileo to accurately measure the effect of gravity on falling objects and develop a universal law describing this effect. Galileo hypothesized that rolling balls down an inclined plane would cause them to accelerate uniformly. Dropping things vertically happened too quickly so he studied the motion on inclined planes. The speeds were very slow and this allowed him to analyse the motion Based on his book Two new Sciences Galileo. He measured the speed of a ball using water clocks and a swinging pendulum. He found that in the first unit of time it travelled 1 unit of time, in the second unit of time it travelled 3, in the next it travelled 5 and then 7 Galileo found that the velocity = g t and distance travelled was always proportional to 1/2 gt 2. He had found laws which apply to all falling bodies and this was a key used by Newton who established mechanics and gravity as a science later in the century. 57 of 28 © Boardworks Ltd 2009
Galileo In physics Galileo is remembered for discovering the laws of failing bodies and motions of projectiles; in astronomy he is known for pioneering telescopic observation and, through this, discovering sunspots, the irregular surface of the Moon, the satellites of Jupiter, and the phases of Venus. Most of all, he is remembered for championing Copernicus' Sun-centred universe and prompting the split that freed scientific knowledge from the restrictions of spiritual belief. Imprisoned in old-age for repeatedly publishing his views, his name has become immortalised as an emblem of resistance against the oppression of Truth. 58 of 28 © Boardworks Ltd 2009
Simple machine 59 of 28 © Boardworks Ltd 2009
Roads on hills 60 of 28 © Boardworks Ltd 2009
Wheel chair access 61 of 28 © Boardworks Ltd 2009
Inclined planes An inclined plane, commonly referred to as a ramp, is an even surface that is tilted at an angle. It helps reduce the force necessary to move an object by increasing the distance it must be moved. Picture a vertical wall two metres (m) tall. You would have to apply a lot of force to lift a 10 kilogram (kg) object over the wall. Now picture a 5 m ramp leading up to the top of the wall. It would be far easier to move the 10 kg object up the ramp than it would be to lift the object straight up. Ramps are used this way in many applications. They allow people with physical disabilities to move up to another floor in a building, for example, and they help to move objects into a truck. Some common applications of the use of the inclined plane are wheel chair access ramps, the ramps used in multilevel parking garages and road building: roads tend to wind up mountains instead of going straight up. 62 of 28 © Boardworks Ltd 2009
Newton’s second law example 7 Inclined planes It is easier to climb stairs to get to a second floor than to climb straight up a rope. It is easier to walk up a long, gentle hill than to climb up a short, steep hill. Here we examine 63 of 28 why © Boardworks Ltd 2009
Forces on Inclined Planes How difficult is it to keep this car stationary? 64 of 28 © Boardworks Ltd 2009
Forces on Inclined Planes How difficult is it to keep this car stationary? 65 of 28 © Boardworks Ltd 2009
Forces on Inclined Planes How difficult is it to keep this car stationary? 66 of 28 © Boardworks Ltd 2009
Forces on Inclined Planes Comment on the weight (force) acting down the incline ………. 67 of 28 © Boardworks Ltd 2009
Forces on Inclined planes Comment on the force acting into the surface ……… As 0, the angle of the slope increases; the force acting into the slope decreases. 68 of 28 © Boardworks Ltd 2009
Force on an inclined plane Perpendicular component Weight (mg) Parallel component When a mass sits on a slope we need to consider the components of its weight when studying its motion and to do this we use trigonometry Sin Cos = parallel component mg = perpendicular component mg 69 of 28 parallel component = mg sin perpendicular component = mgcos © Boardworks Ltd 2009
Forces on Inclined planes mgcos 90 mgcos 0 0 mgcos 45 As 0, the angle of the slope increases; the bit of weight (component of weight) acting into the slope (mgcos 0) decreases. 70 of 28 © Boardworks Ltd 2009
Forces on inclined planes mgsin 0 0 mgsin 0 mg sin 0 As the angle of the slope decreases, sin 0 decreases and the component acting down the slope decreases. 71 of 28 © Boardworks Ltd 2009
mgsin 0 & mgcos 0 • In this course we will concentrate only on the component acting down the slope: mgsin 0 72 of 28 © Boardworks Ltd 2009
Force on an inclined plane Components parallel and perpendicular to slope Sin = parallel component mg Cos = perpendicular component mg 73 of 28 parallel component = mg sin perpendicular component = mg cos © Boardworks Ltd 2009
Worked examples on inclined planes A mass of 10 kg is placed on a plane inclined at 300 to the horizontal. If there is a constant frictional force between the mass and the plane of 10 N, what is the acceleration of the mass when released? (4 ms- ) 10 N 2 mgsin 0 10 Kg 300 acceleration = unbalanced force = mgsin 30 - 10 = 4 ms-2 mass 10 74 of 28 © Boardworks Ltd 2009
Worked examples on inclined planes A mass of 10 kg is launched up a plane inclined at 300 to the horizontal with a force of 300 N. Calculate the frictional force between the mass and the plane when the acceleration of the mass is 10 ms-2. (150 N) 10 Kg 300 75 of 28 © Boardworks Ltd 2009
Worked examples on inclined planes A mass of 10 kg is launched up a plane inclined at 300 to the horizontal with a force of 300 N. Calculate the frictional force between the mass and the plane when the acceleration of the mass is 10 ms-2. (150 N) Friction (now acts Component of weight acting down slope 10 Kg mgsin 0 300 N downwards) 300 Force acting up the slope 76 of 28 © Boardworks Ltd 2009
Worked examples on inclined planes Friction (now acts Component of weight acting down slope 10 Kg mgsin 0 300 N downwards) 300 Force acting up the slope acceleration = unbalanced force = 300 –(mgsin 30 +friction) m 10 10 ms-2 = unbalanced force = 300 –(50 +friction) 10 10 10 = 300 –(50 +friction) 10 77 of 28 friction = 150 N © Boardworks Ltd 2009
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Action and reaction A football rests on flat ground. There are four key forces acting between the ball and the Earth. What are they? l The ball presses down on the ground. l The ground reacts to the ball. l Earth’s gravity acts on the ball. l The ball’s gravity acts on the Earth. 80 of 28 © Boardworks Ltd 2009
Newton’s third law For every action there is an equal and opposite reaction. Forces always come in pairs. List all the horizontal forces acting here, between the car, the caravan and the road: l l 81 of 28 The car pushes the road backwards. The road pushes the car forwards. The car pulls the caravan forwards. The caravan pulls the car backwards. © Boardworks Ltd 2009
Action and reaction “If every action has an equal and opposite reaction, how do unbalanced forces ever occur? ” Newton’s third law applies only to forces between objects. Pairs of action and reaction forces are of the same kind. This diagram shows a pair of frictional forces between a box and the ground: friction of the box on the ground, and friction of the ground on the box. 82 of 28 © Boardworks Ltd 2009
Balanced forces When considering the forces that cause acceleration, they must all act on the same object. These forces can be considered individually and can be of different kinds. For instance the pulling force on this box is not equal to the frictional force acting against it, so the box will start to move to the right. Free body diagrams only include the forces that act on one object. This makes it easy to work out whether the forces on it are balanced or not, and whether the object will accelerate. 83 of 28 © Boardworks Ltd 2009
Balanced forces: true or false? 84 of 28 © Boardworks Ltd 2009
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