2024 Mechanic Notes Part 1 - Newton's Laws and their application PDF

MECHANICS Mechanics is the branch of Physics dealing with the study of motion of objects when they are subjected to forces. There are 3 broad sections to the Mechanics which we study in our curriculum: Kinematics Newton's Laws and Application of Newton's Laws Momentum, Im...

MECHANICS Mechanics is the branch of Physics dealing with the study of motion of objects when they are subjected to forces. There are 3 broad sections to the Mechanics which we study in our curriculum: Kinematics Newton's Laws and Application of Newton's Laws Momentum, Impulse, Work, Energy and Power This module will include Newton's Laws and Application of Newton's Laws (this booklet), and Momentum and Impulse (following booklet). NEWTON'S LAWS AND APPLICATION OF NEWTON'S LAWS A: ALL ABOUT FORCE VECTORS Forces need to be described in terms of magnitude and direction, therefore they are VECTORS. A vector is a physical quantity that has both magnitude and direction. A scalar quantity is a physical quantity that has magnitude only. Vector quantities are represented by lines of exact length including an arrowhead to represent direction. E.g If the vectors lie along the same straight line we can indicate their directions by means of + and – signs. Usually left to right (W to E) is taken as positive (+) and right to left (E to W) as negative (-). The RESULTANT of a number of vectors is the single vector which has the same effect as the original vectors acting together. 1 Maximum vs. Minimum resultant Click on the following link: https://www.physicsclassroom.com/Physics-Interactives/Vectors-and-Projectiles/Vector- Addition/Vector-Addition-Interactive Each block represents 1 N of force. 1. Drag a BLUE arrow onto the canvas. Place it horizontally, close to the bottom of the grid. Make the length of the arrow 7 blocks. What is the magnitude and direction of the force represented by the blue arrow? __________________________________________________________________ 2. Drag a GREEN arrow onto the canvas. Place the tail of the green arrow adjacent to the head of the blue arrow. Make the length of the green arrow 3 blocks. What is the magnitude and direction of the force represented by the green arrow? __________________________________________________________________ 3. What is the angle between the two vectors? _______________________________ 4. Click on the tab on the right which says Show Resultant. What is the RESULTANT force of the blue and green arrow? __________________________________________________________________ 5. Now move the head of the green arrow so that it makes an acute angle of 45o from the tail of the blue arrow. Keep the length of the green arrow the same. 2 6. Click on Show Resultant. What is the resultant force of the blue and green arrow now? __________________________________________________________________ 7. Now move the head of the green arrow so that it makes an obtuse angle of 135 o from the tail of the blue arrow. Keep the length of the green arrow the same. 8. Click on Show Resultant. What is the resultant force of the blue and green arrow now? __________________________________________________________________ 9. Now move the head of the green arrow so that it is in the opposite direction to the blue arrow. Place the tail of the green arrow just above the head of the blue arrow. 10. What is the angle between the two vectors? _______________________________ 11. Click on Show Resultant. What is the resultant force of the blue and green arrow now? __________________________________________________________________ 12. When did the two force vectors produce the MAXIMUM resultant force? __________________________________________________________________ 13. When did the two force vectors produce the MINIMUM resultant force? __________________________________________________________________ 3 TO SUMMARISE: 1. The resultant vector is a maximum when the angle between the vectors is 0º (i.e. they are acting in the same direction) and the vectors are added. E.g. Resultant force = 6N + 2 N = 8 N right 2. The resultant of two vectors is a minimum when the angle between them is 180º (i.e. they are acting in opposite directions) and the vectors are subtracted. E.g. Resultant force = 6N + (-2 N) = 4 N right 3. How do you think the angle between the vectors (from 0o to 180o) affects the resultant? ____________________________________________________________________ ____________________________________________________________________ Multiple choice - Min/Max resultant 1. Which of the following values can NOT represent the resultant of 2 vectors of magnitude 3 N and 7 N? A 4N B 12 N C 5N D 10 N 2. The resultant of 2 forces of magnitude 13 N and 6 N, is 7 N. The angle between the forces is: A 0º B 60º C 90º D 180º 3. Two forces act on a point if the angle between them is increased from 30º to 150º. The magnitude of the resultant force will: A increase B decrease C increase then decrease D not change 4. Two forces, X and Y, have a resultant of 6 N. If the magnitude of X is 11 N then the magnitude of Y could be: A 2N B 4N C 6N D 18 N Bearings ° Bearings are used to describe the direction of a vector. All bearings are measured clockwise from north, with one revolution being equal to 360º. When giving direction, a BEARING (angle in degrees) must be given and NOT just a general compass direction. 4 Addition of Vectors Draw a rough diagram first. Only do an accurate scale diagram if asked to (rulers and protractors needed.) Note: All vector diagrams must have ARROWS ( NO arrows = NO marks) 1. Head to Tail method of vector addition This method is used when each vector starts where the previous one ended, i.e. the vectors act in 2 or more different directions one after the other. It is often used to represent successive displacements, and is the most common method for scale diagrams. 2. Parallelogram method of vector addition This method is used when 2 vectors act from the same point. In this case the vectors represent the sides of a parallelogram and the diagonal acting through the same starting point as the original vectors represents the resultant vector. i.e. all 3 vectors must act away from the same point. It is often used to represent force vectors. It is not normally used for a scale diagram, but is useful for working out forces and angles using trigonometry. Equilibrium of forces If an object is stationary then the forces acting on it are balanced. Such an object is said to be in equilibrium and it experiences zero resultant force. Example 1 Two boys push on a box from the same side with forces of 100 N and 200 N each. With what force must a third boy push in order to keep the box in equilibrium? Solution The third boy needs to push with a force of 300 N in the opposite direction to the first two boys. Note: The 300 N force is at 180º to the resultant force of the first two boys and is called the equilibrant force. The equilibrant of a number of vectors is the single vector which will balance the original vectors acting together. It is equal in magnitude but opposite in direction to the resultant. Example 2 Two forces act simultaneously on a body - one force of 5 N on a bearing of 45º and the other force of 10 N on a bearing of 135º. Determine: 2.1 the resultant force 2.2 the equilibrant force 5 Questions 1. Two forces of magnitude 10 N and 15 N act on an object. 1.1 What will be the magnitude of the maximum possible resultant? 1.2 What will be the magnitude of the minimum possible resultant? 1.3 Illustrate your answers using a diagram. 2. Two forces of 20 N West and 15 N North act on a block. Calculate the magnitude and direction of the resultant force. 3. Forces of 16 N and 12 N act on directions of 60º and 150º respectively. 3.1 What is the resultant force? 3.2 Determine the magnitude and direction of the single force that will balance the other two forces. 4.1 The diagrams show two forces of 5N and 12N which act at the same point. Find the resultant in each case. 4.2 When will two forces acting together have 4.2.1 a maximum resultant? 4.2.2 a minimum resultant? 4.3 For each of the diagrams shown above, give the magnitude of the force needed to balance the resultant. 5.1 Two forces X and Y can be replaced by a single force of 10 N. If X is a force of magnitude 4 N say whether each of the following could be possible values of Y. 5.1.1 24 N 5.1.2 6N 5.1.3 14 N 5.1.4 5N 5.2 Draw a rough vector diagram to explain each answer. 6. What happens to this resultant force if the tugs move closer to each other? Answers: 1.1 25 N 1.2 5 N 2. 25 N on a bearing of 306,87° 3.1 20 N on a bearing of 96,87° 3.2 20 N on a bearing of 276,87° 4.1 i) 17 N right ii) 7 N left iii) 13 N on a bearing of 22,62° 4.3a) 17 N left b) 7 N right c) 13 N on a bearing of 202,62° 6 B: TYPES OF FORCES 1. Weight (Fg) is the gravitational force the Earth exerts on any object on or near its surface. Weight can be calculated using the formula, Fg = m.g, where g is the acceleration due to gravity which has a value of 9,8 m.s-2 near the Earth’s surface. 2. The normal force (FN) is the perpendicular force exerted by a surface on an object in contact with it. 3. The frictional force (Ff) due to a surface is the force that opposes the motion of an object and acts parallel to the surface with which the object is in contact. 4. Tension – a pulling force, e.g. the force in a rope when it is pulled. 5. Compression – a pushing force, e.g. the force in a solid strut when pushed from both ends. Ff Force of table on box FN friction (normal) FN normal weight = mass x g Fg Gravitational force on Fg = mg weight box (weight) Fg NOTE:  For a body on a flat horizontal surface the normal force is equal in magnitude to the weight, therefore the resultant force in the first example is zero.  If the resultant of the 3 forces in the second example is zero then the block remains stationary. Rules for drawing free-body diagrams  A free-body diagram is used to represent the forces acting on a body.  The body is usually drawn as a point mass (large solid dot)  The forces are drawn as arrows in the direction of the force.  The forces all act away from the object.  The arrows must touch the circle representing the object. Body accelerating right  The relative lengths of the arrows should be accurate o If a body is accelerating then the length of the arrow in the direction of motion must be greater than the sum of the lengths of the arrows in the opposite direction. o If a body is moving at constant velocity then the length of the arrow in the direction of motion should be shown to be equal to the sum of the lengths of the arrows in the opposite direction to motion (see Newton’s 1st Law of Motion).  All bodies experience the force of gravity (weight) (Fg) down (except in space).  All bodies on a surface experience a normal force (FN) perpendicular to the surface.  All free-body diagrams must have a KEY, indicating the names of the forces labelled on the diagram. 7 Lifting and pushing bodies at an angle Note: The normal is the upward reaction force to the TOTAL downward force. Pushing: F Fy Total downward force = Fg + VC(Fy), ∴ FN > Fg Pulling: Fy F Total downward force = Fg - VC(Fy), ∴ FN < Fg Draw a free body diagram showing the vertical forces only: Pushing: Pulling 8 C: FINDING THE COMPONENTS OF FORCES (I.E. RESOLVING A FORCE INTO ITS COMPONENTS) The opposite of adding two forces together is finding the effect of a force in a certain direction. We call this resolving force into its components. We normally resolve into components that are at 90 degrees to each other. If we know the value of force R and the angle it makes with the horizontal (ϴ) we can find its vertical component R sinϴ and its horizontal component R cosϴ. Now practise 1. R is a force of 55 N which acts on a bearing of 60o. Draw a sketch of the force, then calculate the components in the North and East directions (i.e vertical and horizontal) 2. Draw a sketch of a force of 0,75 N acting at 40o to the horizontal and then calculate its vertical and horizontal components. 3. The rope is pulling the box of mass 15 kg but the box does not move. The rope exerts a force of 200 N and acts at an angle of 25o to the horizontal. 3.1 Draw a free body diagram showing all the forces acting on the block. 3.2 Calculate the vertical and horizontal components of the force exerted by the rope. 9 Exercise on components of forces (Ex. 7) 1 Two children pull ropes attached to a go-kart as shown in the diagram. 1.1 Calculate the component of the 30 N force which is acting forwards. 1.2 Calculate the component of the 24 N force which is acting forwards. 1.3 Calculate the resultant forward component of the two forces. 2 Consider the system of forces shown in the adjacent diagram and calculate: 2.1 the resultant horizontal force 2.2 the resultant vertical force 2.3 the overall resultant force 3 A flower pot of mass 8 kg is attached by cables to a wall as shown. 3.1 Draw a free-body diagram to show the forces acting on the pot. 3.2 Calculate the weight of the pot. 3.3 What is the vertical component of the tension in cable X? (Hint: Pot is stationary  net vertical force = 0) 3.4 What is the horizontal component of the tension in cable X? (Hint: Pot is stationary  net horizontal force = 0) 3.5 Calculate the magnitude of the tension in cable X and the angle θat which it is suspended from the ceiling. Resultant of multiple forces using components Resolve all of the forces into vertical and horizontal components. 1. Calculate the resultant (net) force of all the forces in the x-direction as the vector sum of all the components in the x-direction. 2. Calculate the resultant (net) force of all the forces in the y-direction as the vector sum of all the components in the y-direction. 3. Calculate the total net force by finding the vector sum of the resultant (net) x- component and the resultant (net) y-component. 10 Example 1 6N 8N Refer to the free body diagram opposite and calculate 1. The resultant horizontal force 25o 40o 2. The resultant vertical force 3. The overall resultant force Solution 5N 1. Let forces to the right be positive. Fx = (6cos40˚ – 8cos25˚) = 4,60 – 7,25 = -2,65 N 6sin40 Fx = 2,65 N to the left 8sin25 2. Let forces up be positive. 8cos25 6cos40 Fy = (8sin25˚ + 6sin40˚ - 5) = 3,38 + 3,86 - 5 = 2,24 N Fy = 2,24 N up 5N 3. FR2 = 2,242 + 2,652 FR = 3,47 N tanθ = 2,24 FR 2,65 2,24 N θ = 40,2o θ The resultant force is 3,47 N at 40,2o to horizontal 2,65 N If an object remains stationary (in equilibrium) then the resultant force acting on it must be zero. Resultant horizontal force = 0 Resultant vertical force = 0 The triangle rule for forces in equilibrium (an alternative to components) This is used when three forces act on the same object and keep it in equilibrium so that the object does not move or continues moving with a constant velocity. When three forces acting at a point are in equilibrium, they can be represented in both magnitude and direction by the three sides of a triangle taken in order. The three vectors drawn head to tail start and end at the same point i.e. they have a zero resultant. You are simply moving each vector so the three together make a closed triangle. The direction of each vector remains the same. 11 Practical problem solving techniques 1. Identify the point on which the forces are acting and the direction in which they are acting. When determining direction, remember the following: If the system is in equilibrium the resultant force must be zero and the three vectors should form a closed figure when joined head to tail in a vector diagram. Ask the question, “What would happen if that force were removed?” Example: If strut A were removed block B would move towards the wall. Therefore force in A must be pushing out from wall. 2. Using the forces you have identified, draw a separate vector diagram, ensuring that you maintain the correct angles between the vectors. In the case of 3 forces, one of the angles is usually a right angle. 3. Where strings or cables are involved and both act from the same height, then the longest string must represent the smallest force. Tension in strings and cables always acts away from the point of application. Worked example: If the OPEN sign has a weight of 100 N determine: a. The tension in A b. The compressional force in B Solution: 1. Identify the forces acting and fill them in on the original diagram. 1. Draw a suitable triangle of forces, joining the vectors head to tail. 12 3. Use the triangle rule of forces and trig to determine the unknown forces OR a scale diagram if it is not a right angled triangle. Examples using triangle of forces. 1. W=58, 7N, T=91, 4N 2. T=10, 8N, =33,7O 3. T1=104N, T2=60N 4. T=12,7N 13 D: THE WEIGHT OF OBJECTS ON A SLOPE When an object is on a horizontal surface its weight acts at right angles to the surface. The surface exerts an equal and opposite force on the object, this is called the reaction force or the normal. When an object is against a vertical surface, the weight of the object acts parallel to the surface. The object will slide along the surface as its weight is not supported. Weight = mg When an object is on a slope, part of the weight is supported by the slope but part of the weight acts parallel to the slope and makes the object slide. The slope has the effect of resolving the weight into two components. One component is perpendicular to the slope (let us call it W 1) This is equal and opposite to the normal or reaction force. The other component is parallel to the slope (let us call it W 2). This has to be balanced by a force such as friction. 14 Body on an inclined plane (refer to adjacent diagram) For a body on an inclined plane of slope angle θ the weight (Fg) can be resolved into two components; 1. The component of the weight which acts parallel to the slope is Fg.sinθ (mg.sinθ) 2. The component of the weight which acts perpendicular to the slope is Fg.cosθ (mg.cosθ) Weight (Fg), Fg = m.g If the body is at rest on a rough slope then the resultant (net) force is zero and we can conclude; Friction (Ff) up slope = Fg.sin θ = mg.sinθ down slope Normal (FN) = Fg.cosθ = mg.cosθ at 90o to slope Example The stationary crate in the adjacent diagram has a mass of 10 kg. The angle of the slope is 30o. Calculate 1 the frictional force between the slope and crate 2 the normal force Solution 1 Friction (Ff) = mgsinθ = (10 x 9,8)sin30˚ = 49 N up slope 2 FN = mgcosθ = (10 x 9,8)cos30˚ = 84,87 N perpendicular to slope Now practise 15 E: FRICTION Friction is a force that acts to oppose motion. It is caused by one surface dragging over another surface. The coefficient of friction symbolized by the Greek letter μ, is a dimensionless, scalar value which describes the ratio of the force of friction between a body and the surface and the force pressing them together (the normal - reaction force to the weight of the body on the surface). The general formula is: The coefficient of friction depends on the materials used; for example, ice on steel has a low coefficient of friction, while rubber on pavement has a high coefficient of friction. Coefficients of friction range from near zero to greater than one – under good conditions, a tyre on concrete may have a coefficient of friction of 1,7 The diagram below shows a wooden block which is pulled by a spring balance. Force to overcome static friction → Initially the block does not move but when the applied force is large enough the block starts to move. Let us call this force Ff. The reading on the spring balance gives a value which is known as STATIC FRICTION, Ffs (also known as limiting friction). The static friction force must be overcome by an applied force before an object can move, because it is the force that keeps an object at rest. Static friction is friction between two solid objects that are not moving relative to each other. For example, static friction can prevent an object from sliding down a sloped surface. The coefficient of static friction is typically denoted as μs. When the block is moving, the reading on the spring balance decreases slightly, showing that the force needed to keep the block moving is slightly less than the force of static friction. This force needed to sustain movement is known as KINETIC FRICTION, Ffk (also known as sliding or dynamic friction). Kinetic friction occurs when two objects are moving relative to each other and rub together. The coefficient of kinetic friction is typically denoted as μk, and is usually less than the coefficient of static friction for the same materials. If a small amount of force is applied to an object, the static friction has an equal magnitude in the opposite direction. As the force is increased, the static friction increases, until the force is great enough to JUST start the object moving. This value is known as maximum static friction, Ffsmax force of static friction ≤ (coefficient of static friction)(normal force) maximum force of static friction = (coefficient of static friction)(normal force) Ffs ≤ μs η , and Ffsmax = μs η Ffs = force of static friction μs = coefficient of static friction η = normal force Ffsmax = maximum force of static friction 16 Worked examples 1) A 5500 N force is applied to a sled full of firewood in a snow-covered forest. The skis of the sled have a coefficient of static friction μs = 0,75 with the snow. If the fully-loaded sled has a mass of 700 kg, what is the maximum force of static friction, and is the force applied enough to overcome it? Solution: On a flat surface, the normal force on an object is equal to its weight, ie η = mg. Using this, the maximum force of static friction can be found: Ffsmax = μs η Ffsmax = μs mg Ffsmax = (0,75)(700 kg)(9,8) Ffsmax = 5145 kg∙m.s-2 Ffsmax = 5145 N The maximum force of static friction is 5145 N, and therefore the applied force of 5500 N is enough to overcome it, and start moving the sled. 2) A person building a brick-making machine wants to measure the coefficient of static friction between brick and wood. To do this, she places a 2,00 kg brick on a flat piece of wood, and gradually applies more and more force until the brick moves. She finds that the brick moves when exactly 11,8 N of force is applied. What is the coefficient of static friction? Solution: The force that was applied was exactly the right amount to overcome static friction, so it is equal to Fsmax. On a flat surface, the normal force on an object is η = mg. The coefficient of static friction can be found by rearranging the formula for the maximum force of static friction: μs = 0,602 Now try these: 1. A force of 20 N is needed to start moving an object of mass 4 kg but it needs only 16 N to keep it moving at constant velocity. 1.1 What is the magnitude of the force of: a) kinetic friction b) static friction 1.2 What is the magnitude of the normal (reaction force) for this object? 17  1.3 Calculate μs and μK for the object. 2. A crate of mass 62 kg is being dragged across a level floor at constant velocity by a horizontal rope exerting a force of 240 N. Calculate the μK between the two surfaces. 3. A box of mass 20 kg rests on a flat rough surface. The slope of this surface is increased until the box just starts to move. The angle between the surface and the horizontal is 20˚. Calculate the coefficient of static friction. A little later on we will do Newton’s Second Law in conjunction with weight on a slope, and use the static and kinetic friction formula in examples which are a little more complex. 18 F: NEWTON’S LAWS A force is a push or pull which changes or tends to change the state of rest or motion of the body on which it acts. A force does not always cause motion – the force may be in equilibrium with an opposing force (e.g. friction) Galileo’s experiment: As the ball rolls down the slope the force of gravity accelerates it. On the flat there are no forces (friction having been reduced to a negligible force). On the upward slope the force of gravity slows it down (negative acceleration). The smaller the slope the less the decelerating force (i.e. gravity) so the further it rolls. The ball will tend to roll up to the height from which it began, thus if the slope is reduced to zero there will be no decelerating forces so the ball should roll on in a straight line forever. A stationary body will not begin to move unless acted on by a force. Once moving, it will not stop unless acted upon by an opposing force. NEWTON’S FIRST LAW: An object continues in a state of rest or uniform (moving with constant) velocity unless it is acted upon by a net or resultant force. This is sometimes called “The law of Inertia”. If a force is required to change the state of motion of a body, then it follows that the body has some property which resists change. This property is called inertia, and it depends on the mass of the object. Inertia is the property of an object that causes it to resist a change in its state of rest or uniform motion. 19 Demonstrations of Newton’s First Law: 1. Place a coin on a piece of card which is on top of a glass. Jerk the piece of card towards you. The coin falls into the glass. Explanation: The horizontal force acts only on the card. There is no horizontal force on the coin so it remains at rest due to its inertia. When the card is removed the force of gravity causes the coin to fall into the glass. 2. Stand a glass of water on a piece of paper. Quickly remove the piece of paper. The glass does not move. Explanation: The horizontal force acts only on the piece of paper, so the glass does not move, due to its inertia. If the paper is pulled slowly then the force is transmitted to the glass and both the glass and the piece of paper move. 3. When a car stops suddenly a parcel resting on the seat slides forwards. Explanation: The braking force when a car suddenly stops, acts only on the car. The parcel continues to move at the velocity the car was initially travelling at, due to its inertia, so it appears it is sliding forwards. 4. Drop a matchbox (end first) with the tray open. When the box hits the table, the tray slides into the box. Explanation: The cover of the matchbox hits the table first so a force acts to stop the movement of the cover. No force acts on the tray, so it continues to move, due to its inertia, until it also hits the table. 5. When a car goes around a corner, the parcels slide across the seat in the opposite direction. Explanation: When the driver turns the steering wheel to go around the corner, only the car experiences the change in direction. No forces act on the parcels, so the parcels continue moving in a straight line (i.e. the original direction of the car) due to their inertia, and so appear to slide across the seat. 6. When a bus accelerates forwards, the passengers are “jerked” backwards. Explanation: The force acts to increase the velocity of the bus. The passengers continue at constant velocity due to their inertia and appear to move backwards relative to the bus. 7. Space rockets can continue to move at constant velocity in space even though their engines are switched off. Explanation: In space, there are no friction forces to slow down the rocket, therefore, the rocket will continue to move with constant velocity, due to its inertia. 20 NEWTON’S SECOND LAW is concerned with the motion of the body once a resultant non- zero force has been applied. Mass is the amount of matter a body is made up of. Acceleration is the change in state of motion of a body. Acceleration is brought about by the application of a force. (Acceleration is the rate of change of velocity) Click on the following simulation: Intro to Newton 2 1. Keep the mass of both the top and bottom experiment at 1 kg, but pull the top mass with a force of 2 N and the bottom mass with a force of 10 N. Draw the pattern of the arrows in the top and bottom experiment, up to 1 s. What do you notice about the distances travelled in 1 s with the 2 N force compared to the 10 N force? Try to explain what is happening in terms of velocity and time 2. Now keep the applied force of both the top and bottom experiment at 10 N, but make the top mass 1 kg and the bottom mass 5 kg. Draw the pattern of the arrows in the top and bottom experiment, up to 1 s. What do you notice about the distances travelled in 1 s by the 1 kg mass compared to the 5 kg mass? Try to explain what is happening in terms of velocity and time 21 The TROLLEY EXPERIMENT shows how acceleration depends on two variables: - the force applied to the object - the mass of the object Test one variable at a time. Try to minimise friction by placing the trolley on a slight incline. Set up the apparatus as in the diagram Part 1 Create ticker tapes with the application of a force of 1 N, then 2 N, then 3 N, then 4 N. Mass of system must remain constant. (i.e. if one mass piece is over the edge, 3 mass pieces on trolley, if two mass pieces over the edge, two mass pieces on trolley) Cut each tape into 4 space lengths Each length represents the displacement covered in an equal time period i.e. velocity Stick tapes in order onto paper, thus producing a v vs. t bar graph Draw a line through the top of each tape, producing a line graph Find the gradient of this line (using cm, don’t worry about correct units) which is the acceleration Plot this calculated acceleration against force (F = 1, 2, 3 & 4), and write a conclusion. Part 2 Now keep force constant and vary mass by using one trolley and then two trolleys. Many errors occur with this experiment, so just observe the difference in acceleration Sketch a graph of a vs. m, then sketch a graph of a vs. 1 m Write a conclusion to your experiment. Newton’s Second Law: When a net force, Fnet, is applied to an object of mass, m, it accelerates in the direction of the net force. The acceleration, a, is directly proportional to the net force and inversely proportional to the mass. Fnet = ma Prove that N = kg·m·s-2 __________________________________________________________________________________________________ __________________________________________________________________ 22 The weight Fg, is the gravitational force the Earth exerts on any object on or near its surface. At a particular place, all objects fall with the same acceleration (± 9,8 m·s-2). So Fnet = ma can be rewritten as Fg = mg or Fg = m.9,8 The earth exerts a greater force on an object with bigger mass, but that object has greater inertia, and thus resists the force more, which results in the same acceleration for all objects. If the force applied to an object causes the object to move in a direction different to the direction of the force, the component of the original force which acts in the direction of movement, must be calculated before Fnet = ma may be used. Problems using Newton II 1 Calculate the acceleration of the frictionless block 12 N 5 kg 2 A force of 12 N is applied to a 5 kg block. When the block moves a frictional force of 4 N is exerted. Calculate the acceleration of the block. 4N 12 N 5 kg 3 A toy rocket has a mass of 30 kg. Its engine exerts a thrust of 480 N. Calculate the acceleration of the rocket. (Remember g = 9,8 m.s-2) Hint: 30 kg Fdown = Fup = Fnet= 4. A block of mass 8 kg is dragged across a rough horizontal surface, at a constant velocity of 2 m·s-1 by a horizontal force of 12 N. 4.1 What is the magnitude of the frictional force acting on the block? 4.2 If the same force of 12 N continued to drag the block across a smooth surface, where there was no friction, what would be the acceleration of the block? 4.3 At what velocity would the block be moving after it had travelled a distance of 7 m on the smooth surface? 5. A 60 kg block is pulled across a rough floor with a force of 62 N which acts at 60 o to the horizontal. A frictional force of 17 N acts on the block. Calculate the resultant force acting on the block. 23 6. For each of the following diagrams decide, by calculation, if an unbalanced (resultant) force acts. Calculate and describe the motion of the object using the phrases: at rest; moving with constant velocity; moving with constant positive or with constant negative acceleration. 6.1. Initial velocity = 0 m·s-1 6.2. Initial velocity = 0 m·s-1 6N 6N 2 kg 2 kg Frictionless surface 3 N friction 6.3. Vi = 0 m·s-1 6.4. Vi = 5 m·s-1 to the right 12 N 8N 60° 60° 2 kg 4N 2 kg friction Frictionless surface 6.5. Vi = 5 m·s-1 to the right 6.6. Vi = 5 m·s-1 to the right 8N 8N 60° 60° 2N 2 kg 5N 2 kg friction friction 6.7. Vi = 5 m·s-1 upwards 6.8. Vi = 5 m·s-1 upwards F = 19,6 N F = 39,6 N 2 kg 2 kg 24 6.9. Vi = 0 m·s-1 6.10. Vi = 5 m·s-1 upwards F = 9,6 N F = 9,6 N 2 kg 2 kg 7. A man of mass 75 kg is a passenger in a car moving at 18 m·s-1, and he is wearing a seatbelt. In a collision the car is brought to rest in 0,5 s. Determine the magnitude and direction of the force which the seatbelt exerts on the passenger. 8. A 3 kg block moves across a frictionless surface at a constant velocity. A force of 18 N is applied to the block in the opposite direction to its motion, and it slows the block down to half of its original speed while it covers a further 9 m. Calculate the time it takes for this to occur. Answers: 1. 2,4 m·s-2 right 2. 1,6 m·s-2 right 3. 6,2 m·s-2 up 4.1 12 N in opp dir to Fapp 4.2 1,5 m·s-2 in dir of Fapp 4.3 5 m·s-1 in dir of Fapp 5. 14 N in dir of motion 6.1 3 m·s-2 right 6.2 1,5 m·s-2 right 6.3 3 m·s-2 right 6.4 a = 0, v = 5 m·s-1 right 6.5 1 m·s-2 right 6.6 0,5 m·s-2 left 6.7 a = 0, v = 5 m·s-1 up 6.8 10 m·s-2 up 6.9 5 m·s-2 down 6.10 5 m·s-2 down 7. 2700 N backwards 8. 1s 25 Examples with friction/on an inclined plane 1. A block of wood of 5 kg rests on a flat horizontal table. A horizontally applied force of 30 N is just sufficient to make it slide. 1.1 Calculate μs between the block and the table. 1.2 The force of 30 N is maintained after it starts sliding, and the block now accelerates at 2,5 m.s-2. Calculate the μK between the block and the table. 2. The following block is pulled up a frictionless surface with a force of 7200 N. Calculate the acceleration of the block. 7200 N 25˚ 3. The 3 kg block in the diagram slides down a rough inclined plane. If the block experiences a frictional force of 15 N, calculate the acceleration of the block. 3 kg Ff = 15 N 50˚ 26 Using Newton’s Second Law with two-body problems The following are a few examples. More applications can be found on the Powerpoint. 27 28 Exercise 29 NEWTON’S THIRD LAW When object A exerts a force on object B, object B simultaneously exerts an oppositely directed force of equal magnitude on object A. This law involves pairs of forces (sometimes called an action/reaction pair) which always act on different objects and therefore they do NOT balance each other. Example 1 Walking When a person walks his foot exerts a force backwards on the floor (action force). The floor (earth) exerts an opposing reaction force against the foot, causing the body to move forwards. Action = foot on floor ; Reaction = floor on foot The reason the person moves forwards and the earth does not move backwards is that the force is big enough to overcome the inertia of the person, but not big enough to overcome the inertia of the earth. Example 2 Pressing against a wall If a boy, wearing roller skates, pushes forwards against a wall (action force) the wall will exert an opposing reaction force against the boy, who will move backwards, away from the wall. Action = boy on wall ; reaction = wall on boy. The wall does not move because the force is not big enough to overcome its inertia. Example 3 Space rockets The rapidly expanding gases which escape from the combustion chamber, experience a downward force, while the rocket experiences an upward force. Action = gases on chamber ; reaction = chamber on gases. The expanding gases exert forces on the sides of the combustion chamber, NOT on the surrounding air, therefore rockets can be propelled in a vacuum. An astronaut can propel himself in a certain direction by firing a small rocket in the opposite direction. Example 4 Rifles and recoil Since the bullet and rifle are different objects, action and reaction forces drive them in opposite directions during the explosion. The bullet has less inertia than the rifle and therefore moves with a greater speed. Example 5 Motion of a car The wheels of the car exert a force backward on the road and due to friction the road exerts a force forward on the car. Therefore, friction is the external force responsible for the motion of the car. Action = wheels on road ; reaction = road on wheels (Watch Colin Fairman’s video) 30 Note: lf one of the forces involved is the WEIGHT then this represents the force of the earth on an object and therefore, the reaction force is the force of the object on the earth. This concept is often tested in multiple choice questions. Example A book rests on a table. What force is the Newton reaction force to the weight of the book? A the force exerted by the book on the table B the force exerted by the table on the book C the force exerted by the book on the earth D the force exerted by the earth on the book Correct answer is “C”, since weight is “earth on book” so reaction is “book on earth” Problems involving free body diagrams These are the forces acting on one object. Problem 1 A boy sits on a stationary swing seat. The boy’s friend pulls the seat of the swing horizontally to the right as shown in the picture. She exerts a force of 300 N and the child has a mass of 40 kg. Draw a labelled diagram showing all the forces acting on the boy on the swing. (Treat the boy and the seat as one object.) 31 Problem 2 In the picture, a woman and her daughter are sitting side by side in a taxi. It accelerates to the right. The mother has twice the mass of her daughter. a) Give separate labelled diagrams showing the forces acting on the daughter and the mother. There is no need to be “exact”, but show the relative size of the forces acting on the mother and daughter. b) Explain, with diagrams, how the back rest of the seat is able to exert a force on the mother that has a different magnitude from the force it exerts on the daughter. Problem 3 The diagram shows the directions of three forces on a moving lorry. The lorry is travelling at a constant velocity. Which one of the following in the table is correct? Forward thrust Friction Air resistance A 3 500 kN 1 500 kN 2 500 kN B 3 500 kN 2 500 kN 2 500 kN C 3 500 kN 3 500 kN 3 500 kN D 3 500 kN 2 250 kN 1 250 kN Problem 4 A girl pulling a rope drags a box across a rough horizontal surface at a constant speed as shown. Which one of the following diagrams best represents all the forces acting on the box? 32 Solution to Problem 1 Solution to Problem 2 a) See figure below. b) The more we compress a spring, the greater the force it exerts. The springs in the seat are compressed more behind the mother than the child. See figure below. Solution to Problem 3 Option D Constant velocity → no resultant force sum of forces to the left = magnitude of force to the right Solution to Problem 4 Option C Constant velocity → no resultant force. The forces must be balanced. This is not possible for options A and D where there must be a resultant. For B to be correct the magnitude of the vertical component of the force with which the girl pulls must equal the magnitude of the weight. C is much more likely. 33 Conceptual questions involving Newton’s Laws To guide you in answering conceptual questions, always follow these steps: Step 1: Draw a sketch of the situation Step 2: Identify and label the forces acting on each object in the sketch Step 3: Identify the law that causes each object to react in the way that they do. Step 4: Using full sentences and referring to the name of the relevant law, explain how these laws link to the situation. Note: Newton’s First and Second Laws are linked – they concern one object and the forces that act on it. Newton’s Third Law is different because it defines the forces between two different objects. Example: Applying Newton’s Laws Ann is sitting in a seat in an aircraft. She has a safety belt on. When the aircraft accelerates on the runway for take-off, she feels as if something is pushing her backwards in the seat. a) Draw a labelled free body diagram to show all the forces acting on Ann as the aircraft accelerates on the runway. b) Using principles of physics, explain why Ann feels as if she is being pushed back in the seat. When the aircraft lands on the runway, it undergoes deceleration. The seat belt protects the passengers from moving forward. c) Draw a labelled free body diagram to show all the forces acting on Ann during the aircraft’s deceleration. d) Using principles of physics, explain why Ann has to wear a seat belt. 34 Solution: a) The free body diagram: b) There are no forces acting on Ann in a backwards direction. According to Newton’s Second Law (a  F), Ann experiences a net force forwards as a result of the acceleration of the aircraft. Her reaction force is an equal force backwards on the seat, as explained by Newton’s Third Law ( F1  F2 ). Her inertia causes her to maintain her rest position (Newton’s First Law), and this makes her feel as if she is being pushed back. c) The free body diagram: d) Ann is moving at the same velocity as the aircraft. When the aircraft decelerates, she continues to move forward at her original velocity due to her inertia (Newton’s First Law). To prevent her from injuries, she has to wear a seat belt. The seat belt exerts a force on her, which causes her to decelerate (Newton’s Second Law). She exerts a reaction force on the seat belt. Seat belts are manufactured from strong fabric to withstand reaction forces. 35