2 Understanding motion. Written by. Dr Janet Taylor

Transcription

1 Module Understanding motion UNDERSTANDING MOTION Written by Dr Janet Taylor Revised by Lucy George 00

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3 Introduction Module The Nature of Physics Module Understanding Motion Module Understanding motion. In our real world, motion is fundamental everything moves. We walk, run, jump, breathe. Inside us, blood flows. Inside plants, fluids circulate. In the heavens, stars and planets rotate and orbit. Within the atom, electrons are constantly on the move. Even things that appear to be at rest are moving, relative to another object. Motion is observed everywhere. As easy as motion is to recognise it is hard to describe. This module is concerned with our understanding of motion what it is, how it is measured and represented and how different forces and energies relate to different motions. Before starting this module consider the following scenario. Many of you, as drivers, would have been faced with the dilemma of whether or not you should speed through an orange light or stop? Lateness for appointments, stress levels, red light cameras and intuition aside, how can we decide if we should risk it? (There are certain calculations we can make to judge more accurately if we will make it through.) We will come back to this challenge at the end of this module. Viewing the physics video will also help you with this challenge.. Linear motion.. How to measure motion For an object to be in motion it must be changing position relative to another object. Speed is a measure of how fast something is moving or changing position. It is described as a rate or a quantity divided by time. In this case the quantity is the distance moved. So if you are late for work and running to catch a bus, you might run 00 metres in 0 seconds. Your average speed would be calculated as follows: Average speed distance time interval 00 metres seconds 5 metres/second This is written as 5 m/s or 5 ms If you miss the bus and catch a taxi to the next town 40 kilometres away, it might take 35 minutes. Your average speed would then be:

4 . TPP760 Preparatory physics Average speed 40 kilometres minutes 40 km hour 68.6 km/hour (rounded to one decimal place) This is written as 68.6 km/h or 68.6 km h. In SI units this would be converted to ms by the following calculation ms (rounded to decimal place). (Note that if you have a number of calculations keep them as accurate as possible before finally rounding.) However if you glance at the speedometer of the taxi during the above trip, you would notice that it changes. On a straight run it could read 00 km/hr, while on the corners it could read 0 km/hr. The speedometer measures the speed of a car at any one instant. This is termed instantaneous speed. Distance and speed are quantities that tell you something about the size or magnitude of the change in position, and how fast that position has changed. The problem is that distance and speed don t tell you anything about the direction of that change in position. To take this into account two further measures were developed: displacement and velocity. Displacement is a measure which incorporates the magnitude and direction of the change in position. Velocity is a measure which incorporates how fast the change in position takes place and the direction of that change... How to represent motion You can represent velocity diagrammatically by drawing a line whose length represents the speed (or magnitude of velocity) and its direction the direction of the velocity. Such lines are called vectors. Scalars are quantities like speed or distance which have no direction i.e. Scalar (magnitude only) distance (e.g. 0 m) speed (e.g. 0 ms ) Vector (magnitude and direction) displacement (e.g. 0 m east) velocity (e.g. 0 ms north)

5 Module Understanding motion.3 Later you might like to add to this table as we investigate other scalar and vector quantities. Example: Let s consider the situation discussed previously. Imagine you are running for the bus at 6.7 metres/second with a wind blowing at.7 metres/second in the same direction. Such a situation could be represented by scaled vectors so that cm represents metre per second. Scale: cm : ms.7 m/s.7cm 6.7 m/s 6.7 cm To find your velocity: Draw an arrow to represent each individual velocity. The arrow points in the direction of the velocity and its length represents the size of the velocity. Redraw the second arrow so that its tail is at the tip of the first. Draw a vector from the tail of the first arrow to the tip of the second arrow If you write: V y to be the vector representing your velocity. V w to the vector representing the velocity of the wind. V wy to be the vector representing your final velocity. thus V y + V w V wy Consider if you were running into the wind. It is harder running when the wind is against you because V wy V y V w This can be represented using vectors. V y V w In the olympics wind assistance is considered when deciding whether new records are allowed.

6 .4 TPP760 Preparatory physics Example: Consider the situation of running for the bus with the wind blowing perpendicular to your direction of running (as shown) what would be your resultant speed relative to the ground and what would be your direction (relative to the ground means as observed by someone who is stationary, since the earth itself rotates on its own axis and also revolves around the sun). You could represent it using the following vectors. V w wind velocity your resultant velocity V y your velocity To calculate the resultant velocity we need to calculate two things: (i) (ii) magnitude direction. Firstly draw a vector diagram by completing the parallelogram below (which in this case is a rectangle)..7 r.7 m/s σ 6.7 m/s Then using Pythagoras Theorem r r r 5.8 r 7. ms The resultant velocity has a magnitude of 7.ms.

7 Module Understanding motion.5 To calculate the direction of the velocity, use the vector triangle and the tan ratio..7 tan So you are running at 7. metres per second at an angle of.9 to your original direction. This velocity is called the resultant velocity. The wind has increased your speed but at the same time changed your direction. Vectors are a very useful tool to help understand and predict behaviours of moving objects. They can be used to represent many different quantities, if these quantities have magnitude and direction components. You will return to the study of these later. Activity. Question Question : A dog searching for a bone walks 3.5 m south, then 8. m at an angle of 30 north of east and finally 5 m west. Find the dog s resultant displacement vector by drawing the vectors accurately to scale. A motor boat is travelling due east at 4 ms across a river that is flowing north at 6 ms. (i) Use both graphical (by scaled drawing) and mathematical methods to find the resultant velocity of the boat. (ii) If the river is 00 m wide, how long does it take the boat to reach the other side? (iii) How far downstream is the boat when it reaches the other shore? Question 3: A plane is headed directly east at 340 km/hr when the wind is from the south at 45 km/hr. What is the plane s velocity with respect to the ground?

8 .6 TPP760 Preparatory physics..3 Acceleration Previously we examined instantaneous velocity where, while driving along in a taxi, the speedometer variously changed from 60 to 80 then back to 60 km/hr (for example). If you calculate how fast this velocity is changing then you are calculating acceleration. Acceleration change in velocity time interval (change in time) For instance, if you were driving in the above taxi along a straight road and increased your velocity from 60 to 80 km/hr in a two-second period then Acceleration 80 km/hr 60 km/hr second change in velocity change in time 0 km/hour/second (0 km per hour per second) or (.8 ms in SI units) This is written as 0 km/h/s or in SI units as.8 ms On the other hand, if you decrease your velocity from 80 to 60 km/hr: Acceleration km/hour/second (or.8 ms ) Acceleration can be considered as the rate at which velocity changes. Just as velocity is dependent on the magnitude and direction of the change in position change, so acceleration is dependent on the magnitude and direction of the velocity change. If you ride on a merry-go-round you can feel the changes that are taking place as you swing around. The speed (or the magnitude of the velocity) of the merry-go-round will be constant but the velocity will be changing because of the changing direction, so therefore there is an acceleration. (As long as one component is changing, be it the magnitude or the direction of the velocity.) Suppose that a taxi does not change its speed (the magnitude of the velocity) but turns a corner. Is this acceleration? Yes, because although the speed is not changing the direction is thus a change in velocity has resulted this is defined as acceleration. Acceleration has magnitude and direction components and can be represented by a vector. It is a vector quantity.

9 Module Understanding motion.7 Activity. Question : Question : A car s velocity increases from 3.0 ms to 3.5 ms over a 4 second period. What is its acceleration? The handbrake gives way on your car and it rolls down the hill reaching a speed of 0 ms in the first 3 seconds. What is its acceleration? A road train slows down from 50 ms to 0 ms in 5 seconds in order to round a corner. What is its acceleration?..4 Motion in a straight line Let us consider a simple form of motion motion in a straight line where the object accelerates uniformly (i.e. acceleration is constant). There are many examples of this in real life. a train constantly accelerating along a straight track a dragster speeding along a drag strip an athlete running the 00 m (as long as we assume that the acceleration is constant) fruit falling from a tree. In all of these situations an object is falling or travelling in a straight line with constant acceleration (at least for part of the time). It would be handy for us, as well as scientists in general, to have a simple set of equations that describe how far and how fast an object is moving or falling. To derive such equations the first step is to define the variables. Let a constant acceleration u the initial velocity of an object moving in a straight line v the velocity after time, t t time Question 3: s the distance travelled in time, t We know from section..3 acceleration change in velocity change in time v u a (using SI units this would be ms ) t

10 .8 TPP760 Preparatory physics Making v the subject of the formula above, v u + at (Equation ) This is often referred to as the st equation of motion. Also from section.. distance travelled average velocity (average velocity is used when velocity change in time changes over the period of time) u + v s - t Making s the subject of the formula above, s u + v t s u + u + at t substituting for v u + at (from equation ) s ut + ut + at s ut + -- at (Equation ) This is referred to as the nd equation of motion. Finally consider the equations u + v t s and a v u t we rearrange both equations to make t the subject of the formula s t and t u + v v u a equating for t we have, s v u u + v a (v + u) (v u) as(cross-multiply) v u as This is referred to as the 3rd equation of motion. v u + as (Equation 3)

11 Module Understanding motion.9 The three equations for motion in a straight line for uniform or constant acceleration are v u + at s ut + -- at v u + as In situations of bodies falling on or near the surface of the earth, the acceleration due to gravity is called g with an approximate magnitude of 9.8 ms, and directed towards the centre of the earth. Example: A hammer is dropped from the roof of a building. (a) What would its velocity be after 4s? (b) How far does it fall during this time. Solution: (a) Firstly set out your given information u (initial velocity) 0 (dropped) a (acceleration due to gravity) 9.8 ms t (time) 4 s Since u, a, t are known, and we want to find its final velocity v after 4 s, we need to think about the concept involved here. Basically a hammer is dropped. As it drops it gains speed because of gravity accelerating it. g 9.8 ms - u 0 t 4s An image of what s going on is always helpful. v? As we visualise the event, the actual concept here is acceleration.

12 .0 TPP760 Preparatory physics We know that acceleration change in velocity time v u or a (a g in this case) t v v or v 39. ms in the direction of the acceleration. We could also just refer to the first equation of motion where v u + at v (relating the variables we are interested in.) 39. ms A word of advice: Students tend to look at physics problems by firstly asking themselves which formula do I use? Instead I would advise you to firstly: Set out your given information Draw a picture or vector diagram labelled with the given data Think about what basic concept(s) is involved here Use a formula(s) that describes that concept(s) in a nutshell (that s what a formula does) v u As you can see I chose to use a in the example above because it describes our t central concept of acceleration. Remember the 3 equations of motion are simplified versions describing the concepts of linear v u motion. The first equation of motion v u + at came from simplifying a t So I hope you will not make the mistake of solving physics problems by asking which formula do I use? but instead think about the concepts involved first. With experience and practice, you will learn to use simple basic formulas with ease. Remember formulas are concepts in a nutshell. (b) This question asks how far the hammer falls during this time. We want to find its displacement or change in position.

13 Module Understanding motion. u 0 s? t 4 s v 39. ms g 9.8 ms Since there is a change in velocity due to acceleration, we know that average velocity displacement time displacement average velocity time s u + v time m in the direction of the acceleration. OR we could have used the equation of motion s ut + -- gt where u 0, o s m o or v u + as or as v (39.) 0 + ( 9.8 s) s v a Home experiment s 78.4m Try this simple experiment for yourself m You will need a small ball, a stopwatch or a watch from which you can count seconds, and possibly an assistant. Throw a ball straight up into the air to an approximate known height (e.g. the height of your house). Measure the time it takes for the ball to reach the ground after it has reached its maximum height. Using the equations of motion, assuming that the acceleration due to gravity is 9.8 ms, calculate the velocity of the ball when it reaches the ground.

14 . TPP760 Preparatory physics What would you expect the velocity to be? Detail accurately what methods you used to make the measurements in this experiment and how you calculated the velocity. What did you calculate the velocity to be? Were your predicted and observed velocities different? How can you explain this, and could the experiment be improved? Did you refer to any books or other sources of information to complete this experiment. If so list them here. Well, how did you go with this first home experiment? Before starting the experiment you could have calculated, using Newton s Equations of Motion, what you might expect the time a ball was in the air if it was thrown straight up.

15 Module Understanding motion.3 Height of ball Time of flight t (time to reach maximum height time to fall down to starting point) Final velocity (v) 0 m u 0 v u + at, u 0, s ut + --at, u 0 a 9.8 ms a 9.8 ms t.4 s t s C t v? t time it takes to drop through a displacement of 0m 0 t t t.43 s time of flight t.86 s v v 3.9 ms After falling back to its starting point, its final velocity 3.9 ms 0 m u 0 s 0 ut + --at t v u + at v ms t s t t.0 s time of flight 4.04 s C v? Then you could have conducted the experiment calculating the velocity of the ball by measuring the maximum height of the throw and the time the ball takes to reach this maximum height distance velocity time If your calculations in the experiment do not match your theoretical prediction, then you have to look at your methods and conditions on the day to determine what other factors could cause your results to differ from the theoretical. Maybe it was a windy day, or maybe you could not estimate the height accurately enough. A range of factors could be involved but you were there so only you know what they could be.

16 .4 TPP760 Preparatory physics Activity.3 Question : Question : Question 3: Question 4: A bus moving at 5 ms slows down at a rate of ms. How far will it go before it stops? A driver of a car travelling at 60 km/hour sees a child crossing the road 00 m away. Calculate the acceleration required by the car to stop before hitting the child (in SI units). Is your answer possible? Explain. What velocity would a bungi jumper reach two seconds after jumping from a platform? How far would she have fallen during this time? A cat falls from a branch of a tree and is in the air for 0.5 seconds before touching the ground. How high was the branch from the ground? Question 5: What is the acceleration of a car that reaches a speed at 8 ms from rest after travelling 40 m? The motorist having reached this speed maintains it through a built up area where the maximum speed is 60 km h. Is the motorist above or below this speed limit and by how much? Question 6: A lift is travelling downward at 4.4 ms and is accelerated upward at.5 ms for two seconds. What is its final velocity? (remember that velocity includes direction.) Section review In this section we have examined concepts of motion including velocity and acceleration. We have also discussed how these concepts can be represented using vectors and how problems involving motion can be solved using Newton s equations of motion. You should now assess yourself as to whether you have met the module objectives. You should now be able to: explain the difference between speed and velocity use vectors and scalars to represent velocity and speed explain the meaning of acceleration and its relationship with velocity use equations of motion. In your study of this section you should have been making notes of the main points and listing those concepts that were difficult to understand. Have you been able to revise those concepts that were difficult to understand? Have you sought help? Use the following post-test to see how you are going so far.

17 Module Understanding motion.5 Post-test. Question : Complete this summary of the previous section. The rate of change of displacement with time is called. Velocity is a quantity i.e. its and must be specified. The magnitude of the velocity of a body is called its. Speed is a quantity. If we divide total displacement by time taken the result is the velocity in the direction of displacement. The velocity of a moving object at any particular time is called velocity. The SI unit of velocity is. If velocity is changing with time, the rate at which it changes is called the of the body. The SI unit of acceleration is. If the velocity changes by a constant amount every second, the acceleration is said to be. Gravitational acceleration at any particular point on the earth s surface is an example of acceleration. If an object undergoes a displacement s is time t, starting with an initial velocity, u, and acceleration a in the direction of the displacement and after time t the velocity is v, three equations of motion can be written.. 3. If the body starts from rest these equations become.. 3.

18 .6 TPP760 Preparatory physics Question : Can a body have a constant speed and a changing velocity? Can a body have a constant velocity and a changing speed? Explain your answers. Question 3: If a ball is thrown straight up into the air, describe the changes in its velocity and acceleration.

19 Module Understanding motion.7. Motion and forces In our world pigs don t fly or if they did we would probably look for supporting wires or some special effects. Throughout our lives we have been observing nature and have accepted how objects behave on our planet. In the past many observers have watched and recorded how objects move. Isaac Newton (64 77), the famous scientist and mathematician, synthesised the work of many others to formulate these observations into three laws: Newton s First Law of Motion (Law of Inertia) Newton s Second Law of Motion Newton s Third Law of Motion.. Newton s First Law of Motion (Law of Inertia) Have you ever slipped on a wet floor, or worse still, tried to stop when ice skating? Or have you ever tried the magicians s trick of pulling a tablecloth out from under a set of cups and saucers without dropping the crockery? These are all examples of the operation of Newton s First Law of Motion, often called the Law of Inertia. Newton s First Law says: An object will continue moving in a straight line at a constant speed and a stationary object will remain at rest unless acted upon by an unbalanced force. Home experiment Try this experiment at home. You will need a glass of water, a dollar coin and a square of shiny cardboard bigger than the top of the glass. Place the cardboard over the top of the glass and the coin in the centre of the cardboard and then try to remove the cardboard. Hint: Try flicking the cardboard out with your finger or ruler. What do you think will happen to the coin? Describe in detail the methods you used to perform this experiment. Describe what did happen to the coin. What is your explanation of this? Could this experiment be improved? Did you refer to any books or other people to help you in your explanations? If so list what or who you referred to.

20 .8 TPP760 Preparatory physics There are many other examples of the operation of Newton s first law in our everyday world around us. You don t slip on a dry floor, but can very easily slip on a wet floor, because the force of friction is much less on the wet floor than the dry floor. When driving your car, if you take your foot off the accelerator it does not automatically come to a stop. It coasts along for a while. If you want it to stop you have to apply the brakes, that is, apply another force. The force then is what we have to apply to cause a change in motion or a change in velocity. It is a vector quantity because it has a magnitude and a direction and is measured in a unit called a Newton. A force of magnitude of Newton (N) causes a mass of one kilogram to accelerate at a rate of one metre per second per second. The concept of a force is important throughout physics and will be returned to a number of times throughout these modules. What is Inertia? Newton s first law of motion is often called the Law of Inertia but what is inertia? Have you ever tried to push a loaded shopping trolley uphill or to push a bogged car out of a hole. These objects are said to possess inertia a tendency to remain in the same state of motion (in this case still). Their inertia is dependent on their mass, so that mass is often used as a measure of inertia. Mass is often confused with weight. Mass is the measure of the amount of matter in an object; it depends only on the type and number of atoms in it. In SI units it is measured in kilograms. Weight on the other hand is not a fixed property of an object but depends on the amount of gravitational force acting on it, hence it varies with location. Weight is directly proportional to the acceleration due to gravity. In SI units weight mass acceleration due to gravity. Note that the term weight is often used incorrectly by many of us, when we get a measure on the scales that is really our mass, not weight, as weight is a force and is measured in Newtons. Note mass is a scalar quantity, while weight is a vector. Activity.4 Question : What is your weight in Newtons? Question : What would it be on the moon which has a gravitational force / 6 that of earth? Question 3: What would it be on Mars where the acceleration due to gravity is 8.3 ms? Question 4: What would it be in outer space?

21 .. Newton s Second Law of Motion Module Understanding motion.9 We all know it takes a greater force to push a full shopping trolley than an empty one. This is an illustration of Newton s Second Law which says that: The acceleration of a body is directly proportional to the force acting on it and inversely proportional to the mass of the body. In mathematical terms F ma where F is the force in Newtons Weight is measured in Newtons and mass is measured in kilograms in the SI System. What forces act on a body? m is mass in kilograms a is acceleration in ms Consider your study book sitting on your desk. What forces are acting on it? Gravity would be one, but if it were the only force then the book would be accelerating downward through the desk. If this is not happening then the desk must also be pushing upward on the book. There are often many forces acting on an object, but the object will stay still if the sum of all those forces is zero. Remember that force is a vector quantity with magnitude and direction. So it follows that if an object is made to move then it must have a set of unbalanced forces acting on it. What we see is the affect of resultant force. For example if we push our study book across the desk there are another two forces acting on it; the force we exert and the friction of the desk s surface. If we represent this diagrammatically we have: Upward force exerted by desk Backward force exerted by friction between book and desk Forward force e by hand Table s surface Actual diagram Downward force exerted by gravity A free body diagram The first law tells us when a force is acting. The second law tells what the force does when it acts. So if we know the forces acting on a system of known mass we can predict its future motion. This is intuitively accepted. For example the harder you throw a ball the further it goes and the greater the force the greater the acceleration. Newton s second law defines the balance between resultant force and mass in producing an acceleration.

22 .0 TPP760 Preparatory physics However Newton s second law does not imply that every time a force acts motion results. For example you can push against a wall without moving it; your study book on a desk is under the influence of gravity yet does not move down through the desk towards the centre of the earth. In both cases the wall and the desk are exerting their own forces that balance the one that acts on them. It is only the net or unbalanced forces that actually give rise to acceleration. Before we consider how these unbalanced and balanced forces act on an object, it is important to examine Newton s third law...3 Newton s Third Law of Motion Previously you looked at the situation of pushing a wall and it pushing back, or the study book on the desk not falling down towards the centre of the earth. Similarly if you touch a person that person cannot help but touch you back. These are all examples of Newton s third law. For every action there is an equal and opposite reaction. Newton s third law tells us that whenever a force is applied to an object, that object simultaneously exerts an equal and opposite force forces always act simultaneously in pairs. How does this tie in with Newton s other laws? If you catch a ball, the first law says you have to exert a force so that the ball will stop, the third law says the ball will exert an equal and opposite force on our hand, and the second law says that a resultant force on your hand will cause your hand to move. So how can Newton s three laws help us to understand and analyse the forces working on moving and non-moving bodies?..4 Forces acting horizontally and vertically Consider again your study book sitting on your desk. Upward force exerted by desk Backward force exerted by friction between book and desk Forward force exerted by hand Table s surface Actual diagram Downward force exerted by gravity A free body diagram

23 Module Understanding motion. The forces acting on it would be as shown downward force (F w ) due to gravity (known as weight mg) and an upward force (F n ) termed the normal force. As the book is stationary (in the vertical direction) these forces are in equilibrium so that sum of forces in the vertical direction 0 i.e. F 0 ( is the symbol for summation) or mg + F n 0 Remembering that forces are vectors with direction, F n and weight have equal magnitude but opposite directions i.e. F n mg (the negative sign here indicates direction only). Now what happens if you push the book across the desk. If the desk is highly polished it will move easily, if on the other hand it has a rough surface it will be more difficult to move. Two more forces are involved the force you exert by pushing the book (F p ) and the frictional frictional force (F f ) between the book and the surface. This means that there are four forces acting on the book. F n F f F p table s surface F w Since the book is being pushed forward, there must be a net force in the forward direction. Summing horizontal forces, F F net F p + F f F net (in this case, F p > F f for an acceleration to occur) Experimentally it has been found that the frictional force increases as you first try to move an object. This force is called the static frictional force. Once the object begins to move, the frictional force acting is called the kinetic frictional force. This is much smaller than static friction. Friction is dependent on the nature of the two surfaces independent of the size (measured as contact area) of the two surfaces in contact. The frictional force is actually proportional to the normal force pushing the two surfaces in contact with each other.

24 . TPP760 Preparatory physics The result can be summarised by the equation: F f s F n where F f is the magnitude of the frictional force Once the book moves, F k k F n where F k is the force of kinetic friction k is the coefficient of kinetic friction. The coefficient of friction being a ratio of the magnitude of two forces is a unit-less number. It depends on the types of materials in contact and the condition of the surfaces (polished, rough) and often other variables such as temperature. Typically the coefficient of static friction ( s ) ranges from 0.0 for smooth surfaces to.5 for rough surfaces while the values of k are much smaller. Note that if F p F k i.e. net force 0 there is no acceleration, and the book moves forward with constant velocity. Example: F n is the magnitude of the normal force, exerted by the desk on the book s is coefficient of static friction which depends on the nature of the surfaces in contact. A person presses a.6 kg book against a vertical brick wall. If the static coefficient of friction between the book and the wall is 0.35, how hard would the person have to push to allow the book to slide down at a constant speed. mg F f becomes F n F push F p F w Actual Solution: F frictional Free body diagram mg If the book is moving down at a constant speed it has no acceleration. There is no net force and the vertical forces are in equilibrium, with the gravitational force dragging it down equal to the frictional force holding it up. i.e. Gravitational force mg N Frictional force, F f 5.68 N.

25 Module Understanding motion.3 The force exerted on the book by the wall is the normal force, and is calculated from the equation: F f s F n where F f 5.68 N F f F n ---- s 0.35 s F n F n is force exerted by the wall and is unknown. F n 44.8 N The force that the person exerts on the book to hold it against the wall has a magnitude of 44.8 N. Activity.5 Question : The coefficient of kinetic friction between rubber car tyres and a wet road is If the driver of a 800 kg car travelling at 30 ms applies the brakes and skids to a stop: (a) (b) (c) What is the size and direction of the force of friction that the road exerts on the car? What is the resulting acceleration of the car? How long will it take for the car to come to rest, and how far will it skid? To drag a 50 N box along the footpath at a constant speed a horizontal force at 36 N is exerted. What is the coefficient of friction between the footpath and the box?..5 Forces at any angle Question : Have you ever taken a dog for a walk? If the dog sits down, then you exert a force on the lead to make the dog stand again. This force is made up of two components: a horizontal component that is pulling the dog forward and a vertical component pulling the dog up. Of course there are other forces acting on the dog but we will ignore these for the time being. upward component forward component

26 .4 TPP760 Preparatory physics It is possible to resolve the force on the lead into these two components. The process of transforming a vector into its horizontal and vertical components is called vector resolution. If the force used to pull the dog is 0 N and is exerted at an angle at 35 to the horizontal, then this can be represented by the following vector diagram. 0 N 35 Using trigonometry, to resolve the force in the lead, the components will be: F sin F F cos Vertical component of F F sin 0 sin N Horizontal component of F F cos 0 cos N. In general, any vector can be resolved into its horizontal and vertical components. Remember that the vector to be resolved is always the hypotenuse in a right angled triangle. The technique of resolution of vectors into their components is useful we want to sum up vectors in a certain direction. Consider the following examples. You will need to refresh your knowledge of basic coordinate geometry and trigonometry before progressing further. You will also need a pair of compass and protractor for your geometrical constructions. Example: Consider two forces acting on a body B, one of 30 N at 30 to the horizontal and the second of 0 N at 40 to the horizontal. What would be the resultant force on the body? Solution:. This problem could be solved graphically by the addition of vector components. To solve the problem graphically, you need to accurately construct a parallelogram (figure.) using the force vectors as the sides. Another way is to draw a vector triangle, (figure.) by joining the vectors tip to tail and making sure that the directions are maintained. The following diagrams illustrate the graphical solution, drawn correctly to a scale of inch: 0 N 0 N 30 N C 40º 30º B Given Figure.: D 0 N B Resultant 69º 30 N A

27 Module Understanding motion.5 B Resultant 69º 30º 30 N 0 N The resultant force on B is 30 N and directed at 69 to the horizontal. Figure.. The problem could also be solved by resolving each vector into its horizontal and vertical components and then adding each new component to give the resultant. This is shown as follows: Force Horizontal Component (x) Vertical Component (y) 30 N 30 cos N 30 sin 30 5 N 0 N 0 cos N 0 sin N 40º 30º The horizontal component for the resultant vector, R x is the sum of all the horizontal components, ( 5.3) 0.66 (Note: negative sign denotes that it is in the opposite direction, i.e. to the left) The vertical component for the resultant vector, R y is We will now add the resultant components by using Pythagoras Theorem. This can be represented diagrammatically as: R x R θ 0.66 R y 7.86 Using Pythagoras rule, the magnitude of the resultant vector is: R R x + R y

28 .6 TPP760 Preparatory physics Using trigonometry, the direction of the resultant vector is given by: Thus the resultant force would be 30 N at The problem could also be solved by straight calculation Firstly from Figure., ABˆ D B C Applying the Cosine Rule, where tan R y R x 69. C a b + c bc Cos  b 0 N (0)(30) cos 70 a 70º A c 30 N B a 9.8 N The resultant force is approximately 30 N To find the direction of the resultant force, we can apply the cosine rule or the sine rule. The sine rule is easier in this case. sinbˆ Using b sin Bˆ sinâ a b sin  a 0 sin Bˆ 39 The direction of the resultant is at an angle of ( ) 69 to the horizontal. As you can see, it is much easier to use the graphical method.

29 Module Understanding motion.7 Note: Some of you may already be familiar with the sine and cosine rules. If not, you could use the calculation shown earlier where Pythagoras Theorem is employed. Example: Four forces act on a body at point A as in the following diagram. Use algebra and trigonometry to find the resultant force acting on the body. (Hint: Use the direction of the 80 N force as the frame of reference, and measure other angles anticlockwise.) 0 N N 00 A N 60 N Solution: Table of component vectors. Force (N) Horizontal Component (x) Vertical Component (y) cos sin cos sin cos sin cos sin Resultant R x 95 N R y 7N Note that R x is directly opposite in direction to the 80 N force since it is negative. See figure.3. Using figure.4, the magnitude of R R x + R y R y 7 N 8.6 N Figure.3 R x - 95 N

30 .8 TPP760 Preparatory physics Direction of R, tan tan tan Figure.4 since x-component is negative and y-component is positive the angle must be in the nd quadrant, so 43 Resultant force is 9 N at 43 R y ---- R x R y Resultant, R R x θ original reference line Try to solve this graphically and check your answer with the solution above. (Hint: It is much easier to use graph paper when solving by drawing). Activity.6 Question : Two soccer players kick a ball at exactly the same time. One player s foot exerts a force at 65 N north. The other players foot exerts a force of 83 N east. What is the magnitude and direction of the resultant force on the ball? Question : Two 5 N forces act concurrently on point A. Find the magnitude of their resultant when the angle between them is 35. Previously we examined the nature of the forces that acted on a book on your desk, with the technique of resolution of vectors into components we can consider further situations where several forces act on a body. Example: A 40 kg block of stone is being pulled at a steady speed by rope at an angle of 30 to the horizontal and the tension in the rope being 00 N. Find: (a) (b) the frictional force the force exerted by the earth on the block.

31 Module Understanding motion.9 Solution: Let F f represent the frictional force, and F n represent the force exerted by the earth normal to the plane F n F f F n 00 N 30 o F f mg mg (a) In the horizontal direction The frictional force must be equal to the horizontal component of the pulling force. Therefore F f 00 cos N (b) In the vertical direction The downward force (weight) must be equal to the sum of the upward forces. Therefore mg F n + F sin F n + 00 sin 30 F n 34 N The normal reaction of the earth is 34 N Example: A sign over a bookshop weighs 9 N and the angle of the supporting wire is 5 to the horizontal. What is the force in the rope? F 5 Bookshop 9 N F 5 F cos 5 F sin 5 mg 9 N

32 .30 TPP760 Preparatory physics Solution: In the vertical direction, when forces are in equilibrium mg F sin so 9 F sin 5 F sin N Force in the rope is 94.7 N..6 A special force gravity Gravity is the most obvious force in our daily lives and we have already talked about it a great deal. If a stone is released and falls to earth, then its motion is due to a force acting on it. The force acting on it is called its weight and is due to something called gravity. The true nature of gravity is still under investigation. Let s have a look at some further situations where gravity plays an important role. On a hill (an inclined plane) gravity will still pull straight down, not at an angle downhill. However a component of the weight will act parallel to the inclined surface. mg sin θ B θ θ α θ mg mg cos θ α + A and + B Bˆ Â The component of weight pulling the body down the hill is mg sin. Example: A truck weighing 56 N is resting on a plane inclined at 0 to the horizontal. Draw a diagram indicating the relevant forces. Find the magnitudes of the parallel and perpendicular components of the weight.

33 Module Understanding motion.3 Solution: Force parallel to the inclined plane F p is mg sin. F p θ 0 0 mg F n F p 56 sin 0 9 N Force perpendicular to the inclined plane F n is mg cos. F n 56 cos 0 58 N Activity.7 Question : A car weighing. 0 4 N is parked on a 35 slope. (a) Find the force tending to cause the car to roll down the hill. (b) What is the force the car exerts perpendicular to the hill? Question : A well-oiled, friction-less wagon with a mass of 75 kg is steadily pulled uphill, using a force of only 0 N. What is the slope of the hill? Question 3: A force of 89 N is needed to push a trolley up an incline of 35 to the horizontal. Find the weight of the trolley along the slope if friction is neglected. Question 4: A child of mass 0 kg is standing on a plank inclined at 0 to the horizontal. How much force does the child exert normally (perpendicularly) against the plank?

34 .3 TPP760 Preparatory physics Home experiment Measurement of coefficient of static friction. In this experiment you will need some small objects e.g. a coin, ice block, small ball etc., a stiff flat surface that you can tilt up and down, and a protractor. Put a coin on the cover of a book, then slowly tilt the cover of the book until the coin begins to slide down the cover of the book. What angle did you think the coin would begin to move? Measure your angle with a protractor and determine what angle the coin actually began to move at. Draw a diagram indicating the actual situation. When motion is just ready to start friction force is at a maximum and the forces parallel or perpendicular to the book cover are balanced. What information would you use to determine the coefficient of static friction. If you said we would use the relationship F f F n where s is the coefficient of static friction, F f is the frictional force which is equal to W sin and F n is the normal force which is equal to W cos, you would have been correct. Calculate the coefficient of static friction between the book and the coin. Determine the coefficient of static friction for at least 5 objects you have around the home. (Describe the method and characteristics of each object). What combination has the greatest friction?

35 Module Understanding motion.33 What combination has the least friction? Are there any limitations to using this technique to measure the coefficient of static friction? How would these limitations effect your answer? Activity.8 Question : Question : Suppose a large block is at rest on an inclined plane. The angle of the inclined plane is slowly increased until at 4 to the horizontal the block begins to slide. What is the coefficient of static friction between the block and the incline? When a force of 500 N pushes a 5 kg box up an incline of 40 the resultant force is 8 N. (a) (b) (c) (d) Draw a diagram indicating the relevant forces. Write an equation to determine the frictional force (hint: balance all forces parallel to inclined plane). Write an equation to determine the normal. (hint: balance all forces perpendicular to the inclined plane). Calculate the coefficient of kinetic friction. Question 3: At a tourist resort in the Southern Alps of Australia a horse pulls a sled up a steep snow-covered slope of 8. The sled has a mass of 300 kg and the frictional force between the sled and the snow is 350 N. What is the coefficient of friction between the snow and the sled?

36 .34 TPP760 Preparatory physics..7 Falling bodies If you ever get a chance to go to one of the science centres situated in most Australian capital cities they traditionally have a display of a feather and some solid object, usually a coin, falling in a vacuum. Observation of this tells us that the feather and the coin will land at the same time. Gravity, the attraction between two objects, the feather and the earth, say, is the only force acting on the feather because in a vacuum, air resistance (a frictional force) does not exist. We say the feather would be in free fall. A heavier object, such as the coin, is attracted to the earth with more force than the lighter object (feather). But they do not travel with different accelerations because the acceleration of an object depends on force and mass (See Newton s second law and the relationship F ma). So double the force working on double the mass results in the same acceleration as half the force on half the mass.the feather and coin will arrive at the bottom of the tube at the same time. The acceleration will always be a constant, g 9.8 ms. However, not many of us experience seeing objects falling in a vacuum. What happens in the real situation of an object falling through our atmosphere. In air, feather and coins behave very differently. Home experiment You will need a feather or a small piece of paper about the size of a feather and a coin. Drop the feather and coin from the same height. What would you expect to happen? Describe your methods used. Describe what happened. Give an explanation for your result. If you have referred to any books, list them.

37 Module Understanding motion.35 Are there any limitations to the design of this experiment and how could you improve the design. Did you notice that the feather dropped in air, accelerates very briefly then floats to the ground at constant speed. This happens because of the effects on the feather of air resistance which quickly increases until it equals the weight of the feather, so that the feather then has net force of zero acting on it. If there is no net force acting on the feather then the feather must no longer be accelerating and the speed of the feather is constant. This speed is called the terminal speed or if direction is considered, terminal velocity. For a feather it is about 0.0 ms. From your observations what factors affect the amount of air resistance on a feather or other object. If you said the size (i.e. cross-sectional area) and the speed of the falling object you would have been correct. The size is important because it determines the amount of air the object must fall through in falling, while the speed is important because at greater speeds the object collides with more air molecules increasing the impact forces. Consider the situation of a parachutist or sky diver. They can control their rate of descent by changing the cross-sectional area of their parachute or the position of their limbs and body. height wt Drag or air resistance time Two parachutists of the same weight must land at the same time. The first parachutist will leave the plane and slow down his fall by extending his arms and legs. Terminal velocity is reached at about 3 ms and the parachutist opens his chute to slow down his descent further. (Terminal velocity about 5 ms ). The second parachutist must travel faster than the first in order to catch up and reduces her surface area by bringing arms and legs closer to the body. The terminal velocity of the nd parachute is about 34 ms. Both parachutists can land at the same time despite the fact that they left the plane at different times.

38 .36 TPP760 Preparatory physics Activity.9 Question : A 0.50 kg ball falls from the window of a tall building. (a) (b) (c) At one point air resistance acting on the ball is 0.8 N. What is the acceleration of the ball? At another stage of its journey the ball is falling with constant velocity. What is the magnitude of the air resistance now? If there were no air resistance what would be the acceleration of the ball? Question : Two balls of equal size, but different mass, (one solid and one a shell) are dropped at the same time from a hot air balloon. Both experience air resistance as they fall. (a) (b) Which ball reaches terminal velocity first? Do both hit the ground at the same time? Explain your answer. Question 3: Why is it that a tennis ball dropped from the top of a 50 storey building will hit the ground with no greater velocity than if it was dropped from a 0 storey building...8 Newton s Law of Universal Gravitation If we drop a stone it falls to the earth. The ancient Greeks explained this by saying that every object had a home and would try to return home. For a stone, its home was the earth, and that was where it returned. As knowledge increased, it was realised that if an object was to move a force was necessary. If a stone was released and fell to the earth then it was due to a force acting on it, and the force was called weight (see section..) and due to something called gravity. It is reported that Newton was sitting in his garden when an apple fell from a tree. Newton asked himself what would happen if you took the apple 00 metres up and dropped it would it fall back to earth because of gravity? What if you took the apple to 000 metres or metres would it drop back? Newton reasoned that the apple would fall back to earth from these distances and then asked the big question what if you take the apple to km (the moon-earth distance) would it still fall back to earth. He reasoned that the apple would fall back to earth and the force on the moon that kept it in orbit, was the same type of force that would act on the apple earth s gravity. Twenty years later after much thought, and cross checking, Newton published his Law of Universal Gravitation which states: Between any two objects in the universe there is an attractive force (gravity) that is proportional to the masses of the objects and inversely proportional to the square of the distance between them.