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Transcript
CHAPTER 4 FORCES AND MOTION Mechanics It is now time to take a closer look at the motion of objects and what makes them move the way they do. As part of our daily experiences we all know that to make an object move in a particular way you must apply the appropriate force i.e. a push or pull in the direction you wish the object to go. The harder you push or pull the faster it moves. These simple observations can be explained by Newton’s laws of motion which form the foundation of the study known as mechanics. They allow us to predict the motions of the planets, the path of a projectile and many more everyday events. Newton’s first law of motion Before Newton formulated his laws of motion most people held the Aristotelian view that an object remains at rest unless a force makes it move. In other words the ‘natural’ state of an object was that of rest, not motion. This seems reasonable when we consider some everyday examples. If the engine of a car doesn’t provide a force the car soon comes to a stop. If you push a book across the table it too will come to a stop. So it seems if you want to keep an object moving you need to keep applying a force to it. But let us perform a few more careful experiments. What if we push the book over a highly polished table? We find that it travels further. Now try it over a skating rink or an air track. It travels even further. We now imagine that by using even smoother surfaces the book will travel further and further. In the limit if we have an extremely smooth (frictionless) surface the book will continue moving without slowing down. We now conclude that you do not need a force to keep an object moving. We see that it is the presence of other forces (friction, air resistance etc.) that slows them down. This leads us to a formulation of Newton’s first law of motion NEWTON’S FIRST LAW: Consider an object on which no net force acts. If the body is at rest, it will remain at rest. If the body is moving with constant velocity it will continue to do so. Note that forces may still act on the object but the net (resultant) must be zero. Newton’s first law is sometimes called the law of inertia and it holds only in a special set of reference frames. A reference frame is quite simply a set of co-ordinate axes we use to make our measurements and observations. For example we can measure position relative to the Earth’s surface. We say the Earth is our reference frame. If we are moving along in a train and making measurements we can use the train as our reference frame. If the train is moving at a constant velocity we will find that Newton’s first law of motion is true. It is an inertial reference frame. If we tried to 19 do the same from a merry-go-round that was turning we would probably find that Newton’s first law was not true. This is because objects on the merry-go-round have a force acting on them, it keeps them moving in a circle. Strictly speaking the Earth is not an inertial reference frame but the acceleration of objects on the Earth’s surface is quite small. Newton’s second law of motion If a net force does act on an object then its effect is given by a mathematical formulation of Newton’s second law. NEWTON’S SECOND LAW: F = ma in other words the sum of all the forces acting on the object is equal to the mass of that object times the acceleration. Note that this is a vector equation and the direction of the acceleration is the same as the net force. The consequences of this law are very familiar from our everyday experiences. If we slide a book across a table we can do it quite easily and achieve a reasonable acceleration. Try pushing on a whole bookcase and the result is quite different. The larger the mass of the object the smaller the acceleration. We can now use Newton’s second law to carefully define force. Let us take the standard kilogram as our object. We place the object on a frictionless table and pull it along until, by trial and error, it has an acceleration of 1 m.s-2. This force is equal to 1 kg.m.s-2 and is given a special name called the newton (abbreviated N). It is the SI unit of force. Force is a vector quantity and so obeys the laws of addition and subtraction for vectors. If 2 or more forces act on a body we can find the net force by placing the tail of one force vector on the head of the next and so on. That forces behave as vectors is a fact that was verified by experiment. Newton’s second law is a way of expressing this fact, it does not prove it. We can also take components of forces. Using components we can write Newton’s second law as F x = ma x , F y = ma y , F z = maz Newton’s second law also allows us to define the mass of an object. Once we have a constant reproducible force the acceleration that force produces for an object is inversely proportional to the mass of that object. This means if the force produces an acceleration of 10 m.s-2 for the standard 1 kg mass and an acceleration of 5 m.s-2 for our test object then the mass of the test object is F = (1 kg) (10 m.s-2) = m (5 m.s-2) m = 2 kg 20 This is also referred to as the inertial mass of an object. It is the characteristic of the object that relates the force on the object to the resulting acceleration. Less precisely, we can think of it as a measure of the amount of matter in the object. This is different to the weight of an object which is a measure of the force due to gravity on that object. Newton’s second law shows that weight and mass are related but not equal. In order to use Newton’s second law to solve problems we have to realise that it applies to only one object at a time. We must find all the forces acting on that body and then we draw them in a free-body diagram. In it, the object is represented by a dot (make sure it is clearly visible) and each force by a vector with its tail on the dot. We only consider external forces i.e. forces exerted on the object by other objects. Internal forces in which parts of the object exert forces on other parts are not important here (we will see later that they cancel out). Let’s look at some examples of Newton’s second law and forces in action. Q. A student pushes a loaded sled of mass m = 240 kg for a distance d of 2.3 m over the frictionless surface of a frozen lake. In doing so he exerts a constant horizontal force F with magnitude F = 130 N. If the sled starts from rest, what is its final velocity? A. Our first step to solving any force problem is to draw a free-body diagram. This is shown below. The sled is represented by the dot and the force by an arrow. We also set up the relevant co-ordinate axes. In this case it is only a 1-dimensional problem so only the xaxis is drawn. We choose the force to be along the positive x-axis (we are free to do this since we are not told otherwise). Using Newton’s second law of motion we can find the acceleration of the sled. m F x Considering the x-component of F and the acceleration along the x-axis we obtain from Newton’s second law (written in component form) Fx = max F 130 N ax = x = m 240 kg = 0.542 m.s-2 From our equations of motion for constant acceleration we have v2 = v 02 + 2a(x - x0) 21 Since the sled starts from rest v0 = 0 and we identify that x - x0 = d. Our equation simplifies to v = 2ad (2)(0.542 m.s-2 )(2.3 m) = = 1.6 m.s-1 Q. In a 2-dimensional tug-of-war, Alex, Betty and Charles pull on an automobile tyre as shown in the diagram below. Alex pulls with a force FA (220 N) and Charles with a force FC (170 N). With what force must FB does Betty pull if the tyre remains stationary? y Alex Charles FA FC 47 o 137 x o Betty FB A. In this example we do need to use 2-dimensions to deal with the forces. Again we draw our free-body diagram as indicated above. We are free to choose any orientation for the xy co-ordinate system with respect to the actual forces. It is usually a good idea to align at least one force with one of the axes. Now we can apply Newton’s second law. Since the tyre is stationary the acceleration is zero so we can write F = FA + FB + FC = 0 You will find that the component form of this vector equation is more useful when solving problems. In terms of the x and y components we have F = FC cos FA cos 47.0 o = 0 F = FC sin FA sin 47.0 o FB = 0 x and y Note the use of minus signs to indicate a component is in the direction opposite to the positive x or y axes. Substituting values into the first equation we obtain 22 (170 N)( cos = (220 N) (cos 47.0 o) (220 N) (0.682) cos-1 = 28.0 o 170 N Substituting into the other equation we obtain FB = FC sin FA sin 47.0 o = (170 N) (sin 28.0 o) + (220 N)(sin 47.0 o) = 241 N Some particular forces There are many forces around us but there are some particular types that are very important and very common. We will now take a closer look at these. Weight The weight of an object is the force on that object due to the local gravitational field. If that object is near the Earth then the weight is due to the Earth’s gravitational field. From Newton’s second law we can write the weight of an object near the Earth’s surface as W = mg = m (9.8 m.s-2) Since weight is a force it has a direction. The direction is down towards the Earth’s centre. On the moon the same object would weigh less. The magnitude of the gravitational field on the moon is not as large as the Earth’s. The mass of the object, however, will be the same. It is a property of the object. In our everyday speech we tend to interchange the word weight and mass but in physics they have completely different meanings. Normal force Consider a book resting flat on a table. There is a weight force acting on the book. This, however, cannot be the only force since the net force must be zero, the book is stationary. There is a force exerted by the table on the book. This force is normal (perpendicular) to the surface of the table and is called the normal force (usually indicated by N). This is shown in the diagram below. y Normal force N N x W Weight W 23 We must remember that the normal force is always perpendicular to the surface and that the normal force is not always equal to the weight force. Friction If we try to slide the book over the table (or any surface) we find that the motion is resisted by a force. This force is called friction (usually indicated by f). It is parallel to the surface and opposite in direction to the intended motion. Friction is due to attraction (bonding) between the surface and the object. Sometimes we simply a problem by assuming there is no (very little) friction. Such a surface is called frictionless. Tension When an object is being pulled along by a rope the rope is under tension. There is a force exerted by the rope on the object. This force is called tension (usually indicated by T). The direction of the tension is along the rope, away from the object and toward the centre of the rope. This is indicated in the figure below. T T T T The rope is usually assumed to be massless. This means its mass is much less than the mass of the object. Any pulleys are also assumed to be massless and frictionless. This means its mass is also much smaller than the object and that the friction on the pulley axle is very small. Fundamental forces All the forces we have discussed so far and all forces so far known to physicists are manifestations of the four fundamental forces of nature. They are gravitational electromagnetic weak nuclear strong nuclear Most everyday forces (such as friction, tension, normal force) are due to electromagnetic forces. Newton’s third law of motion The most familiar of Newton’s laws of motion is probably his third and final law of motion. We are all aware that when we push on an object it pushes back. For instance to get out of a chair you may push down on that chair (arm rests or 24 somewhere) and it pushes back on you. Newton’s third law talks about the pair of forces acting between 2 bodies. NEWTON’S THIRD LAW: If a body A exerts a force FBA on a body B then that body exerts and equal and opposite force FAB on body A. Since force is a vector and has direction opposite means in the opposite direction. So the two forces have the same magnitude but are in the opposite direction. Mathematically we can write Newton’s third law as FAB = FBA Here the notation FAB means the force exerted on body A by body B. Sometimes you will hear people talk about action and reaction forces and they state Newton’s third law as “the reaction force is equal and opposite to the action force”. There is nothing wrong with this but it is not always clear which force is the action and which is the reaction. You may now think that if the two forces are equal and opposite they should cancel out and the net force will always be zero. If this is so then nothing will accelerate. The point to keep in mind here that the 2 forces are acting on different objects. In this case the forces do not cancel out. Let us take a closer look at a few examples of action-reaction pairs. Imagine a satellite orbiting around the Earth. The only force acting on the satellite is FSE the force exerted on the satellite by the gravitational attraction of the Earth. By Newton’s third law the satellite should exert an equal and opposite force on the Earth. This is the gravitational pull of the satellite on the Earth. Gravity is the mutual attraction of 2 objects. The force is equal in magnitude to that on the satellite but because the mass of the Earth is so large Newton’s second law tells us that the acceleration will be very small. FSE FES Now consider a book resting on a table. We can draw the forces acting on the book. There is the gravitational pull of the Earth on the book FBE(weight). Since the book is not accelerating this force is cancelled by the normal force of the table on the book FBT. However, these 2 forces do not form an action-reaction pair. They are equal and opposite but do not act on the same pair of objects. The ‘reaction’ to FBE is the 25 gravitational pull on the Earth due to the satellite FEB. The ‘reaction’ of the normal force due to the table is FTB, the force of the book on the table. FBT (normal from table) FBE FBT FEB FTB FBE (weight of book) Friction We are all familiar with the effects of friction. They are unavoidable in our everyday lives. However, the effects are not always to our disadvantage. Without friction we would not be able to walk, or write and objects would slide around. Friction provides the force that makes all these possible. There are different types of friction but in our case we will consider the friction between 2 dry solid surfaces (usually an object and some surface). Let us begin our investigation of friction with a few simple experiments. First we slide a book over the surface of a table. We observe that the book gradually slows down and eventually comes to rest. This is the first property of friction, it opposes motion. If we want to keep the book moving at a constant velocity we have to push it with a force that is equal and opposite to the frictional force. In our second experiment we gently push on the book. Eventhough we exert a force on the book it does not move. As we push even harder the book still does not move. The frictional force seems to adjust so as to be equal but opposite to the force we exert. What if we push really hard? The book does begin to move. At first it accelerates but as we reduce our push it moves with a constant velocity. So there is a maximum force that friction can exert but once the object moves then the frictional force is less. N N fs F W Book resting on table a N W W Frictional force equal but opposite to push F fk F Frictional force equal to maximum value v N fs F N fk F W W Book accelerates initially as F is greater than fk F is reduced until it is equal to fk 26 By performing such experiments carefully physicists have been able to formulate 2 expressions to account for the above behaviour. In the case where the object is not moving we call such friction the static frictional force and denote it by fs. The maximum value (magnitude) of this force is given by fs,max = sN where s is the static coefficient of friction (it has no units, it is dimensionless) and N is the normal force acting on the object. If the external force that is pushing is less than this maximum frictional force then the frictional force is equal to the external force i.e. fs sN Once the object is moving then the frictional force decreases. It is called the kinetic frictional force and is denoted by fk . The magnitude of this force is given by fk = kN where k is the coefficient of kinetic friction. Note that neither of the expressions for the frictional force say anything about direction. They are scalar expressions. The directions of fs and fk are always parallel to the surface and tend to oppose any other forces parallel to the surface or oppose any motion. These expressions are empirical (based on experimental data) and not fundamental equations like Newton’s second and third laws. Under some situations our expressions for frictional forces may not be accurate, at high speed for instance. They serve as approximations to what actually happens. The coefficients of friction depend on the nature of the 2 surfaces in contact. We usually state these as the coefficient of static friction between a rubber tyre and asphalt. Tables of such useful values exist. Friction is basically due to the attractive forces between the 2 surfaces. Q A 100 kg chest standing on a floor has a coefficient of static friction of 0.6. A boy approaches the chest and leans horizontally against it. (a) What is the frictional force exerted by the floor on the chest before the boy comes into contact with it? (b) What is the maximum force the boy can exert before the chest begins to slip? (c) What is the frictional force exerted by the floor when the exerts a force of 100 N on the chest? 27 A Drawing a free-body diagram we have N W (a) Initially the only forces acting on the chest are the weight and normal force. If there was friction too it would be the only force in the horizontal direction. This would result in an acceleration in the horizontal direction. However, the chest is at rest so there is no friction. (b) Now we add the push of the boy and the frictional force in the free body diagram Taking components in the horizontal direction we obtain F x N = fs F = 0 fs F and in the vertical direction we have W F y = N W = 0 From the first equation we obtain fs F so F is a maximum when fs is a maximum i.e. F = fs,max From the last equation we have N = W. Our expression for the static frictional force is fs,max = sN So we obtain F = fs,max = sW substituting we obtain F = 0.6 x (100 kg) (9.8 m.s-2) = 588 N (c) If the boy exerts a force of 100 N this is less than the maximum frictional force (588 N) so the block remains at rest and the frictional force is equal and opposite to the force the boy exerts. fs = 100 N 28 Q A block is at rest on an inclined plane as shown below. The coefficient of static friction between the block and incline is s. What is the maximum possible angle of inclination max of the surface for which the block will not slip? N fs W A In this problem we choose our co-ordinate axes to be parallel and perpendicular to the inclined surface. This will make the mathematics easier, the physics is always going to be the same. Taking components parallel to the surface and applying Newton’s second law fs W sin = 0 (taking up the incline as positive) taking components perpendicular to the incline we obtain N W cos = 0 (taking upwards as positive) The block begins to slip when fs = fs,max . Our expression for fs,max is fs,max = sN Substituting for fs and N we obtain fs,max = W sin s W cos rearranging we obtain sin s = = tan cos So max = tan-1 s 29 30