Laws of Motion PDF

CHAPTER FIVE LAWS OF MOTION 5.1 INTRODUCTION In the preceding Chapter, our concern was to describe the motion of a particle in space quantitativel...

CHAPTER FIVE LAWS OF MOTION 5.1 INTRODUCTION In the preceding Chapter, our concern was to describe the motion of a particle in space quantitatively. We saw that 5.1 Introduction uniform motion needs the concept of velocity alone whereas 5.2 Aristotle’s fallacy non-uniform motion requires the concept of acceleration in 5.3 The law of inertia addition. So far, we have not asked the question as to what 5.4 Newton’s first law of motion governs the motion of bodies. In this chapter, we turn to this 5.5 Newton’s second law of basic question. motion Let us first guess the answer based on our common 5.6 Newton’s third law of motion experience. To move a football at rest, someone must kick it. To throw a stone upwards, one has to give it an upward 5.7 Conservation of momentum push. A breeze causes the branches of a tree to swing; a 5.8 Equilibrium of a particle strong wind can even move heavy objects. A boat moves in a 5.9 Common forces in mechanics flowing river without anyone rowing it. Clearly, some external 5.10 Circular motion agency is needed to provide force to move a body from rest. 5.11 Solving problems in Likewise, an external force is needed also to retard or stop mechanics motion. You can stop a ball rolling down an inclined plane by Summary applying a force against the direction of its motion. Points to ponder In these examples, the external agency of force (hands, Exercises wind, stream, etc) is in contact with the object. This is not Additional exercises always necessary. A stone released from the top of a building accelerates downward due to the gravitational pull of the earth. A bar magnet can attract an iron nail from a distance. This shows that external agencies (e.g. gravitational and magnetic forces ) can exert force on a body even from a distance. In short, a force is required to put a stationary body in motion or stop a moving body, and some external agency is needed to provide this force. The external agency may or may not be in contact with the body. So far so good. But what if a body is moving uniformly (e.g. a skater moving straight with constant speed on a horizontal ice slab) ? Is an external force required to keep a body in uniform motion? 90 PHYSICS 5.2 ARISTOTLE’S FALLACY true law of nature for forces and motion, one has The question posed above appears to be simple. to imagine a world in which uniform motion is However, it took ages to answer it. Indeed, the possible with no frictional forces opposing. This correct answer to this question given by Galileo is what Galileo did. in the seventeenth century was the foundation 5.3 THE LAW OF INERTIA of Newtonian mechanics, which signalled the Galileo studied motion of objects on an inclined birth of modern science. plane. Objects (i) moving down an inclined plane The Greek thinker, Aristotle (384 B.C– 322 accelerate, while those (ii) moving up retard. B.C.), held the view that if a body is moving, (iii) Motion on a horizontal plane is an something external is required to keep it moving. intermediate situation. Galileo concluded that According to this view, for example, an arrow an object moving on a frictionless horizontal shot from a bow keeps flying since the air behind plane must neither have acceleration nor the arrow keeps pushing it. The view was part of retardation, i.e. it should move with constant an elaborate framework of ideas developed by velocity (Fig. 5.1(a)). Aristotle on the motion of bodies in the universe. Most of the Aristotelian ideas on motion are now known to be wrong and need not concern us. For our purpose here, the Aristotelian law of motion may be phrased thus: An external force is required to keep a body in motion. (i) (ii) (iii) Aristotelian law of motion is flawed, as we shall Fig. 5.1(a) see. However, it is a natural view that anyone Another experiment by Galileo leading to the would hold from common experience. Even a same conclusion involves a double inclined plane. small child playing with a simple (non-electric) A ball released from rest on one of the planes rolls toy-car on a floor knows intuitively that it needs down and climbs up the other. If the planes are to constantly drag the string attached to the toy- smooth, the final height of the ball is nearly the car with some force to keep it going. If it releases same as the initial height (a little less but never the string, it comes to rest. This experience is greater). In the ideal situation, when friction is common to most terrestrial motion. External absent, the final height of the ball is the same forces seem to be needed to keep bodies in as its initial height. motion. Left to themselves, all bodies eventually If the slope of the second plane is decreased come to rest. and the experiment repeated, the ball will still What is the flaw in Aristotle’s argument? The reach the same height, but in doing so, it will answer is: a moving toy car comes to rest because travel a longer distance. In the limiting case, when the external force of friction on the car by the floor the slope of the second plane is zero (i.e. is a opposes its motion. To counter this force, the child horizontal) the ball travels an infinite distance. has to apply an external force on the car in the In other words, its motion never ceases. This is, direction of motion. When the car is in uniform of course, an idealised situation (Fig. 5.1(b)). motion, there is no net external force acting on it: the force by the child cancels the force ( friction) by the floor. The corollary is: if there were no friction, the child would not be required to apply any force to keep the toy car in uniform motion. The opposing forces such as friction (solids) and viscous forces (for fluids) are always present in the natural world. This explains why forces by external agencies are necessary to overcome the frictional forces to keep bodies in uniform motion. Now we understand where Aristotle Fig. 5.1(b) The law of inertia was inferred by Galileo went wrong. He coded this practical experience from observations of motion of a ball on a in the form of a basic argument. To get at the double inclined plane. LAWS OF MOTION 91 In practice, the ball does come to a stop after accomplished almost single-handedly by Isaac moving a finite distance on the horizontal plane, Newton, one of the greatest scientists of all times. because of the opposing force of friction which Newton built on Galileo’s ideas and laid the can never be totally eliminated. However, if there foundation of mechanics in terms of three laws were no friction, the ball would continue to move of motion that go by his name. Galileo’s law of with a constant velocity on the horizontal plane. inertia was his starting point which he Galileo thus, arrived at a new insight on formulated as the first law of motion: motion that had eluded Aristotle and those who Every body continues to be in its state followed him. The state of rest and the state of of rest or of uniform motion in a straight uniform linear motion (motion with constant line unless compelled by some external velocity) are equivalent. In both cases, there is force to act otherwise. Ideas on Motion in Ancient Indian Science Ancient Indian thinkers had arrived at an elaborate system of ideas on motion. Force, the cause of motion, was thought to be of different kinds : force due to continuous pressure (nodan), as the force of wind on a sailing vessel; impact (abhighat), as when a potter’s rod strikes the wheel; persistent tendency (sanskara) to move in a straight line(vega) or restoration of shape in an elastic body; transmitted force by a string, rod, etc. The notion of (vega) in the Vaisesika theory of motion perhaps comes closest to the concept of inertia. Vega, the tendency to move in a straight line, was thought to be opposed by contact with objects including atmosphere, a parallel to the ideas of friction and air resistance. It was correctly summarised that the different kinds of motion (translational, rotational and vibrational) of an extended body arise from only the translational motion of its constituent particles. A falling leaf in the wind may have downward motion as a whole (patan) and also rotational and vibrational motion (bhraman, spandan), but each particle of the leaf at an instant only has a definite (small) displacement. There was considerable focus in Indian thought on measurement of motion and units of length and time. It was known that the position of a particle in space can be indicated by distance measured along three axes. Bhaskara (1150 A.D.) had introduced the concept of ‘instantaneous motion’ (tatkaliki gati), which anticipated the modern notion of instantaneous velocity using Differential Calculus. The difference between a wave and a current (of water) was clearly understood; a current is a motion of particles of water under gravity and fluidity while a wave results from the transmission of vibrations of water particles. no net force acting on the body. It is incorrect to The state of rest or uniform linear motion both assume that a net force is needed to keep a body imply zero acceleration. The first law of motion can, in uniform motion. To maintain a body in therefore, be simply expressed as: uniform motion, we need to apply an external If the net external force on a body is zero, its force to ecounter the frictional force, so that acceleration is zero. Acceleration can be non the two forces sum up to zero net external zero only if there is a net external force on force. the body. To summarise, if the net external force is zero, a body at rest continues to remain at rest and a Two kinds of situations are encountered in the body in motion continues to move with a uniform application of this law in practice. In some velocity. This property of the body is called examples, we know that the net external force inertia. Inertia means ‘resistance to change’. on the object is zero. In that case we can A body does not change its state of rest or conclude that the acceleration of the object is uniform motion, unless an external force zero. For example, a spaceship out in compels it to change that state. interstellar space, far from all other objects and with all its rockets turned off, has no net 5.4 NEWTON’S FIRST LAW OF MOTION external force acting on it. Its acceleration, Galileo’s simple, but revolutionary ideas according to the first law, must be zero. If it is dethroned Aristotelian mechanics. A new in motion, it must continue to move with a mechanics had to be developed. This task was uniform velocity. 92 PHYSICS Galileo Galilei (1564 - 1642) Galileo Galilei, born in Pisa, Italy in 1564 was a key figure in the scientific revolution in Europe about four centuries ago. Galileo proposed the concept of acceleration. From experiments on motion of bodies on inclined planes or falling freely, he contradicted the Aristotelian notion that a force was required to keep a body in motion, and that heavier bodies fall faster than lighter bodies under gravity. He thus arrived at the law of inertia that was the starting point of the subsequent epochal work of Isaac Newton. Galileo’s discoveries in astronomy were equally revolutionary. In 1609, he designed his own telescope (invented earlier in Holland) and used it to make a number of startling observations : mountains and depressions on the surface of the moon; dark spots on the sun; the moons of Jupiter and the phases of Venus. He concluded that the Milky Way derived its luminosity because of a large number of stars not visible to the naked eye. In his masterpiece of scientific reasoning : Dialogue on the Two Chief World Systems, Galileo advocated the heliocentric theory of the solar system proposed by Copernicus, which eventually got universal acceptance. With Galileo came a turning point in the very method of scientific inquiry. Science was no longer merely observations of nature and inferences from them. Science meant devising and doing experiments to verify or refute theories. Science meant measurement of quantities and a search for mathematical relations between them. Not undeservedly, many regard Galileo as the father of modern science. More often, however, we do not know all the normal force R must be equal and opposite to the forces to begin with. In that case, if we know weight W ”. that an object is unaccelerated (i.e. it is either at rest or in uniform linear motion), we can infer from the first law that the net external force on the object must be zero. Gravity is everywhere. For terrestrial phenomena, in particular, every object experiences gravitational force due to the earth. Also objects in motion generally experience friction, viscous drag, etc. If then, on earth, an object is at rest or in uniform linear motion, it is not because there are no forces acting on it, but Fig. 5.2 (a) a book at rest on the table, and (b) a car because the various external forces cancel out moving with uniform velocity. The net force i.e. add up to zero net external force. is zero in each case. Consider a book at rest on a horizontal surface Consider the motion of a car starting from Fig. (5.2(a)). It is subject to two external forces : rest, picking up speed and then moving on a the force due to gravity (i.e. its weight W) acting smooth straight road with uniform speed (Fig. downward and the upward force on the book by (5.2(b)). When the car is stationary, there is no the table, the normal force R. R is a self-adjusting net force acting on it. During pick-up, it force. This is an example of the kind of situation mentioned above. The forces are not quite known accelerates. This must happen due to a net fully but the state of motion is known. We observe external force. Note, it has to be an external force. the book to be at rest. Therefore, we conclude The acceleration of the car cannot be accounted from the first law that the magnitude of R equals for by any internal force. This might sound that of W. A statement often encountered is : surprising, but it is true. The only conceivable “Since W = R, forces cancel and, therefore, the book external force along the road is the force of is at rest”. This is incorrect reasoning. The correct friction. It is the frictional force that accelerates statement is : “Since the book is observed to be at the car as a whole. (You will learn about friction rest, the net external force on it must be zero, in section 5.9). When the car moves with according to the first law. This implies that the constant velocity, there is no net external force. LAWS OF MOTION 93 The property of inertia contained in the First It relates the net external force to the law is evident in many situations. Suppose we acceleration of the body. are standing in a stationary bus and the driver Momentum starts the bus suddenly. We get thrown Momentum of a body is defined to be the product backward with a jerk. Why ? Our feet are in touch of its mass m and velocity v, and is denoted with the floor. If there were no friction, we would by p: remain where we were, while the floor of the bus p=mv (5.1) would simply slip forward under our feet and the back of the bus would hit us. However, Momentum is clearly a vector quantity. The following common experiences indicate the fortunately, there is some friction between the importance of this quantity for considering the feet and the floor. If the start is not too sudden, effect of force on motion. i.e. if the acceleration is moderate, the frictional force would be enough to accelerate our feet Suppose a light-weight vehicle (say a small along with the bus. But our body is not strictly car) and a heavy weight vehicle (say a loaded a rigid body. It is deformable, i.e. it allows some truck) are parked on a horizontal road. We all relative displacement between different parts. know that a much greater force is needed to What this means is that while our feet go with push the truck than the car to bring them to the bus, the rest of the body remains where it is the same speed in same time. Similarly, a due to inertia. Relative to the bus, therefore, we greater opposing force is needed to stop a are thrown backward. As soon as that happens, heavy body than a light body in the same time, however, the muscular forces on the rest of the if they are moving with the same speed. body (by the feet) come into play to move the body If two stones, one light and the other heavy, along with the bus. A similar thing happens are dropped from the top of a building, a when the bus suddenly stops. Our feet stop due person on the ground will find it easier to catch to the friction which does not allow relative the light stone than the heavy stone. The motion between the feet and the floor of the bus. mass of a body is thus an important But the rest of the body continues to move parameter that determines the effect of force forward due to inertia. We are thrown forward. on its motion. The restoring muscular forces again come into Speed is another important parameter to play and bring the body to rest. consider. A bullet fired by a gun can easily t pierce human tissue before it stops, resulting Example 5.1 An astronaut accidentally in casualty. The same bullet fired with gets separated out of his small spaceship moderate speed will not cause much damage. accelerating in inter stellar space at a Thus for a given mass, the greater the speed, constant rate of 100 m s-2. What is the the greater is the opposing force needed to stop acceleration of the astronaut the instant after the body in a certain time. Taken together, he is outside the spaceship ? (Assume that the product of mass and velocity, that is there are no nearby stars to exert momentum, is evidently a relevant variable gravitational force on him.) of motion. The greater the change in the momentum in a given time, the greater is the Answer Since there are no nearby stars to exert force that needs to be applied. gravitational force on him and the small A seasoned cricketer catches a cricket ball spaceship exerts negligible gravitational coming in with great speed far more easily attraction on him, the net force acting on the than a novice, who can hurt his hands in the astronaut, once he is out of the spaceship, is act. One reason is that the cricketer allows a zero. By the first law of motion the acceleration longer time for his hands to stop the ball. As of the astronaut is zero. t you may have noticed, he draws in the hands 5.5 NEWTON’S SECOND LAW OF MOTION backward in the act of catching the ball (Fig. The first law refers to the simple case when the 5.3). The novice, on the other hand, keeps net external force on a body is zero. The second his hands fixed and tries to catch the ball law of motion refers to the general situation when almost instantly. He needs to provide a much there is a net external force acting on the body. greater force to stop the ball instantly, and 94 PHYSICS this hurts.The conclusion is clear: force not This force is provided by our hand through only depends on the change in momentum, the string. Experience suggests that our hand but also on how fast the change is brought needs to exert a greater force if the stone is about. The same change in momentum rotated at greater speed or in a circle of brought about in a shorter time needs a smaller radius, or both. This corresponds to greater applied force. In short, the greater the greater acceleration or equivalently a greater rate of change of momentum, the greater is rate of change in momentum vector. This the force. suggests that the greater the rate of change in momentum vector the greater is the force applied. Fig. 5.3 Force not only depends on the change in momentum but also on how fast the change is brought about. A seasoned cricketer draws Fig. 5.4 Force is necessary for changing the direction in his hands during a catch, allowing greater of momentum, even if its magnitude is time for the ball to stop and hence requires a constant. We can feel this while rotating a smaller force. stone in a horizontal circle with uniform speed by means of a string. Observations confirm that the product of mass and velocity (i.e. momentum) is basic to These qualitative observations lead to the the effect of force on motion. Suppose a fixed second law of motion expressed by Newton as force is applied for a certain interval of time follows : on two bodies of different masses, initially at The rate of change of momentum of a body is rest, the lighter body picks up a greater speed directly proportional to the applied force and than the heavier body. However, at the end of takes place in the direction in which the force the time interval, observations show that each acts. body acquires the same momentum. Thus Thus, if under the action of a force F for time the same force for the same time causes interval ∆t, the velocity of a body of mass m the same change in momentum for changes from v to v + ∆v i.e. its initial momentum different bodies. This is a crucial clue to the p = m v changes by ∆p = m ∆v. According to the second law of motion. In the preceding observations, the vector Second Law, character of momentum has not been evident. ∆p ∆p F∝ or F = k In the examples so far, momentum and change ∆t ∆t in momentum both have the same direction. where k is a constant of proportionality. Taking But this is not always the case. Suppose a stone is rotated with uniform speed in a ∆p the limit ∆t → 0, the term becomes the horizontal plane by means of a string, the ∆t magnitude of momentum is fixed, but its derivative or differential co-efficient of p with direction changes (Fig. 5.4). A force is needed dp respect to t, denoted by. Thus to cause this change in momentum vector. dt LAWS OF MOTION 95 dp on the particle and a stands for acceleration F =k (5.2) of the particle. It turns out, however, that the dt For a body of fixed mass m, law in the same form applies to a rigid body or, even more generally, to a system of particles. dp d dv In that case, F refers to the total external force = (m v ) = m = ma (5.3) dt dt dt on the system and a refers to the acceleration of the system as a whole. More precisely, a is i.e the Second Law can also be written as the acceleration of the centre of mass of the F = kma (5.4) system about which we shall study in detail in which shows that force is proportional to the chapter 7. Any internal forces in the system product of mass m and acceleration a. are not to be included in F. The unit of force has not been defined so far. In fact, we use Eq. (5.4) to define the unit of force. We, therefore, have the liberty to choose any constant value for k. For simplicity, we choose k = 1. The second law then is dp F= = ma (5.5) dt In SI unit force is one that causes an acceleration of 1 m s-2 to a mass of 1 kg. This unit is known as newton : 1 N = 1 kg m s-2. Fig. 5.5 Acceleration at an instant is determined by Let us note at this stage some important points the force at that instant. The moment after a about the second law : stone is dropped out of an accelerated train, 1. In the second law, F = 0 implies a = 0. The it has no horizontal acceleration or force, if second law is obviously consistent with the air resistance is neglected. The stone carries no memory of its acceleration with the train first law. a moment ago. 2. The second law of motion is a vector law. It is equivalent to three equations, one for each 4. The second law of motion is a local relation component of the vectors : which means that force F at a point in space dp x (location of the particle) at a certain instant Fx = = ma x of time is related to a at that point at that dt instant. Acceleration here and now is dp y Fy = = ma y determined by the force here and now, not by dt any history of the motion of the particle dp (See Fig. 5.5). Fz = z =m a z (5.6) dt Example 5.2 A bullet of mass 0.04 kg t This means that if a force is not parallel to the velocity of the body, but makes some angle moving with a speed of 90 m s-1 enters a with it, it changes only the component of heavy wooden block and is stopped after a velocity along the direction of force. The distance of 60 cm. What is the average component of velocity normal to the force resistive force exerted by the block on the remains unchanged. For example, in the bullet? motion of a projectile under the vertical gravitational force, the horizontal component Answer The retardation ‘a’ of the bullet of velocity remains unchanged (Fig. 5.5). (assumed constant) is given by 3. The second law of motion given by Eq. (5.5) is applicable to a single point particle. The force –u2 – 90 × 90 a= = m s −2 = – 6750 m s −2 F in the law stands for the net external force 2s 2 × 0.6 96 PHYSICS The retarding force, by the second law of ordinary forces. Newtonian mechanics has no motion, is such distinction. Impulsive force is like any other = 0.04 kg × 6750 m s-2 = 270 N force – except that it is large and acts for a short time. The actual resistive force, and therefore, retardation of the bullet may not be uniform. The t Example 5.4 A batsman hits back a ball answer therefore, only indicates the average straight in the direction of the bowler without resistive force. t changing its initial speed of 12 m s–1. t Example 5.3 The motion of a particle of If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball. (Assume 1 2 linear motion of the ball) mass m is described by y = ut + gt. Find 2 the force acting on the particle. Answer Change in momentum Answer We know = 0.15 × 12–(–0.15×12) 1 2 = 3.6 N s, y = ut + gt 2 Impulse = 3.6 N s, Now, in the direction from the batsman to the bowler. dy This is an example where the force on the ball v= = u + gt by the batsman and the time of contact of the dt ball and the bat are difficult to know, but the dv impulse is readily calculated. t acceleration, a = =g dt 5.6 NEWTON’S THIRD LAW OF MOTION Then the force is given by Eq. (5.5) F = ma = mg t The second law relates the external force on a Thus the given equation describes the motion body to its acceleration. What is the origin of the of a particle under acceleration due to gravity external force on the body ? What agency and y is the position coordinate in the direction provides the external force ? The simple answer of g. in Newtonian mechanics is that the external force on a body always arises due to some other Impulse body. Consider a pair of bodies A and B. B gives We sometimes encounter examples where a large rise to an external force on A. A natural question force acts for a very short duration producing a is: Does A in turn give rise to an external force finite change in momentum of the body. For on B ? In some examples, the answer seems example, when a ball hits a wall and bounces clear. If you press a coiled spring, the spring is back, the force on the ball by the wall acts for a compressed by the force of your hand. The very short time when the two are in contact, yet compressed spring in turn exerts a force on your the force is large enough to reverse the momentum hand and you can feel it. But what if the bodies of the ball. Often, in these situations, the force are not in contact ? The earth pulls a stone and the time duration are difficult to ascertain downwards due to gravity. Does the stone exert separately. However, the product of force and time, a force on the earth ? The answer is not obvious which is the change in momentum of the body since we hardly see the effect of the stone on the remains a measurable quantity. This product is earth. The answer according to Newton is: Yes, called impulse: the stone does exert an equal and opposite force Impulse = Force × time duration on the earth. We do not notice it since the earth = Change in momentum (5.7) is very massive and the effect of a small force on A large force acting for a short time to produce a its motion is negligible. finite change in momentum is called an impulsive Thus, according to Newtonian mechanics, force. In the history of science, impulsive forces force never occurs singly in nature. Force is the were put in a conceptually different category from mutual interaction between two bodies. Forces LAWS OF MOTION 97 always occur in pairs. Further, the mutual forces comes before reaction i.e action is the cause between two bodies are always equal and and reaction the effect. There is no cause- opposite. This idea was expressed by Newton in effect relation implied in the third law. The the form of the third law of motion. force on A by B and the force on B by A act To every action, there is always an equal and at the same instant. By the same reasoning, opposite reaction. any one of them may be called action and the other reaction. Newton’s wording of the third law is so crisp and 3. Action and reaction forces act on different beautiful that it has become a part of common bodies, not on the same body. Consider a pair language. For the same reason perhaps, of bodies A and B. According to the third law, misconceptions about the third law abound. Let us note some important points about the third FAB = – FBA (5.8) law, particularly in regard to the usage of the (force on A by B) = – (force on B by A) terms : action and reaction. 1. The terms action and reaction in the third law Thus if we are considering the motion of any mean nothing else but ‘force’. Using different one body (A or B), only one of the two forces is terms for the same physical concept relevant. It is an error to add up the two forces can sometimes be confusing. A simple and claim that the net force is zero. and clear way of stating the third law is as However, if you are considering the system follows : of two bodies as a whole, FAB and FBA are internal forces of the system (A + B). They add Forces always occur in pairs. Force on a up to give a null force. Internal forces in a body A by B is equal and opposite to the body or a system of particles thus cancel away force on the body B by A. in pairs. This is an important fact that 2. The terms action and reaction in the third law enables the second law to be applicable to a may give a wrong impression that action body or a system of particles (See Chapter 7). Isaac Newton (1642 – 1727) Isaac Newton was born in Woolsthorpe, England in 1642, the year Galileo died. His extraordinary mathematical ability and mechanical aptitude remained hidden from others in his school life. In 1662, he went to Cambridge for undergraduate studies. A plague epidemic in 1665 forced the university town to close and Newton had to return to his mother’s farm. There in two years of solitude, his dormant creativity blossomed in a deluge of fundamental discoveries in mathematics and physics : binomial theorem for negative and fractional exponents, the beginning of calculus, the inverse square law of gravitation, the spectrum of white light, and so on. Returning to Cambridge, he pursued his investigations in optics and devised a reflecting telescope. In 1684, encouraged by his friend Edmund Halley, Newton embarked on writing what was to be one of the greatest scientific works ever published : The Principia Mathematica. In it, he enunciated the three laws of motion and the universal law of gravitation, which explained all the three Kepler’s laws of planetary motion. The book was packed with a host of path-breaking achievements : basic principles of fluid mechanics, mathematics of wave motion, calculation of masses of the earth, the sun and other planets, explanation of the precession of equinoxes, theory of tides, etc. In 1704, Newton brought out another masterpiece Opticks that summarized his work on light and colour. The scientific revolution triggered by Copernicus and steered vigorously ahead by Kepler and Galileo was brought to a grand completion by Newton. Newtonian mechanics unified terrestrial and celestial phenomena. The same mathematical equation governed the fall of an apple to the ground and the motion of the moon around the earth. The age of reason had dawned. 98 PHYSICS magnitude of force cannot be ascertained since Example 5.5 Two identical billiard balls t the small time taken for the collision has not strike a rigid wall with the same speed but been specified in the problem. at different angles, and get reflected without any change in speed, as shown in Fig. 5.6. Case (b) What is (i) the direction of the force on the wall due to each ball? (ii) the ratio of the ( px ) initial ( ) = m u cos 30
, p y initial = − m u sin 30
magnitudes of impulses imparted to the balls by the wall ? ( px ) = – m u cos 30
, py ( ) = − m u sin 30
final final Note, while px changes sign after collision, py does not. Therefore, x-component of impulse = –2 m u cos 30° y-component of impulse = 0 The direction of impulse (and force) is the same as in (a) and is normal to the wall along the negative x direction. As before, using Newton’s third law, the force on the wall due to the ball is normal to the wall along the positive x direction. The ratio of the magnitudes of the impulses Fig. 5.6 imparted to the balls in (a) and (b) is Answer An instinctive answer to (i) might be 2 that the force on the wall in case (a) is normal to ( 2 m u/ 2 m u cos 30
= ) 3 ≈ 1.2 t the wall, while that in case (b) is inclined at 30° to the normal. This answer is wrong. The force 5.7 CONSERVATION OF MOMENTUM on the wall is normal to the wall in both cases. The second and third laws of motion lead to How to find the force on the wall? The trick is an important consequence: the law of to consider the force (or impulse) on the ball conservation of momentum. Take a familiar due to the wall using the second law, and then example. A bullet is fired from a gun. If the force use the third law to answer (i). Let u be the speed on the bullet by the gun is F, the force on the gun of each ball before and after collision with the by the bullet is – F, according to the third law. wall, and m the mass of each ball. Choose the x The two forces act for a common interval of time and y axes as shown in the figure, and consider ∆t. According to the second law, F ∆t is the the change in momentum of the ball in each change in momentum of the bullet and – F ∆t is case : the change in momentum of the gun. Since Case (a) initially, both are at rest, the change in momentum equals the final momentum for each. ( px ) initial = mu (p ) y initial =0 Thus if pb is the momentum of the bullet after firing and pg is the recoil momentum of the gun, ( px ) final = −mu (p ) y final =0 pg = – pb i.e. pb + pg = 0. That is, the total momentum of the (bullet + gun) system is Impulse is the change in momentum vector. conserved. Therefore, Thus in an isolated system (i.e. a system with x-component of impulse = – 2 m u no external force), mutual forces between pairs y-component of impulse = 0 of particles in the system can cause momentum Impulse and force are in the same direction. change in individual particles, but since the Clearly, from above, the force on the ball due to mutual forces for each pair are equal and the wall is normal to the wall, along the negative opposite, the momentum changes cancel in pairs x-direction. Using Newton’s third law of motion, and the total momentum remains unchanged. the force on the wall due to the ball is normal to This fact is known as the law of conservation the wall along the positive x-direction. The of momentum : LAWS OF MOTION 99 The total momentum of an isolated system of interacting particles is conserved. An important example of the application of the law of conservation of momentum is the collision of two bodies. Consider two bodies A and B, with initial momenta pA and pB. The bodies collide, get apart, with final momenta p′ A and p′ B respectively. By the Second Law Fig. 5.7 Equilibrium under concurrent forces. FAB ∆t = p′A − p A and In other words, the resultant of any two forces say F1 and F2, obtained by the parallelogram FBA ∆t = p′B − p B law of forces must be equal and opposite to the (where we have taken a common interval of time third force, F3. As seen in Fig. 5.7, the three forces in equilibrium can be represented by the for both forces i.e. the time for which the two sides of a triangle with the vector arrows taken bodies are in contact.) in the same sense. The result can be Since FAB = − FBA by the third law, generalised to any number of forces. A particle is in equilibrium under the action of forces F1, p′A − p A = −( p′B − p B ) F2,... Fn if they can be represented by the sides of a closed n-sided polygon with arrows directed i.e. p′A + p′B = p A + p B (5.9) in the same sense. which shows that the total final momentum of Equation (5.11) implies that the isolated system equals its initial momentum. F1x + F2x + F3x = 0 Notice that this is true whether the collision is F1y + F2y + F3y = 0 elastic or inelastic. In elastic collisions, there is a second condition that the total initial kinetic F1z + F2z + F3z = 0 (5.12) energy of the system equals the total final kinetic where F1x, F1y and F1z are the components of F1 energy (See Chapter 6). along x, y and z directions respectively. Example 5.6 See Fig. 5.8 A mass of 6 kg t 5.8 EQUILIBRIUM OF A PARTICLE is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the Equilibrium of a particle in mechanics refers to horizontal direction is applied at the mid- the situation when the net external force on the point P of the rope, as shown. What is the particle is zero.* According to the first law, this angle the rope makes with the vertical in means that, the particle is either at rest or in equilibrium ? (Take g = 10 m s-2). Neglect uniform motion. the mass of the rope. If two forces F1 and F2, act on a particle, equilibrium requires F1 = − F2 (5.10) i.e. the two forces on the particle must be equal and opposite. Equilibrium under three concurrent forces F1, F2 and F3 requires that the vector sum of the three forces is zero. (a) (b) (c) F1 + F2 + F3 = 0 (5.11) Fig. 5.8 * Equilibrium of a body requires not only translational equilibrium (zero net external force) but also rotational equilibrium (zero net external torque), as we shall see in Chapter 7. 100 PHYSICS Answer Figures 5.8(b) and 5.8(c) are known as other types of supports), there are mutual free-body diagrams. Figure 5.8(b) is the free-body contact forces (for each pair of bodies) satisfying diagram of W and Fig. 5.8(c) is the free-body the third law. The component of contact force diagram of point P. normal to the surfaces in contact is called Consider the equilibrium of the weight W. normal reaction. The component parallel to the Clearly,T2 = 6 × 10 = 60 N. surfaces in contact is called friction. Contact Consider the equilibrium of the point P under forces arise also when solids are in contact with the action of three forces - the tensions T1 and fluids. For example, for a solid immersed in a T2, and the horizontal force 50 N. The horizontal fluid, there is an upward bouyant force equal to and vertical components of the resultant force the weight of the fluid displaced. The viscous must vanish separately : force, air resistance, etc are also examples of contact forces (Fig. 5.9). T1 cos θ = T2 = 60 N Two other common forces are tension in a T1 sin θ = 50 N string and the force due to spring. When a spring which gives that is compressed or extended by an external force, 5 5 a restoring force is generated. This force is tanθ = or θ = tan −1 = 40
usually proportional to the compression or 6 6 elongation (for small displacements). The spring Note the answer does not depend on the length force F is written as F = – k x where x is the of the rope (assumed massless) nor on the point displacement and k is the force constant. The at which the horizontal force is applied. t negative sign denotes that the force is opposite 5.9 COMMON FORCES IN MECHANICS to the displacement from the unstretched state. For an inextensible string, the force constant is In mechanics, we encounter several kinds of very high. The restoring force in a string is called forces. The gravitational force is, of course, tension. It is customary to use a constant tension pervasive. Every object on the earth experiences the force of gravity due to the earth. Gravity also T throughout the string. This assumption is true governs the motion of celestial bodies. The for a string of negligible mass. gravitational force can act at a distance without In Chapter 1, we learnt that there are four the need of any intervening medium. fundamental forces in nature. Of these, the weak All the other forces common in mechanics are and strong forces appear in domains that do not contact forces.* As the name suggests, a contact concern us here. Only the gravitational and force on an object arises due to contact with some electrical forces are relevant in the context of other object: solid or fluid. When bodies are in mechanics. The different contact forces of contact (e.g. a book resting on a table, a system mechanics mentioned above fundamentally arise of rigid bodies connected by rods, hinges and from electrical forces. This may seem surprising Fig. 5.9 Some examples of contact forces in mechanics. * We are not considering, for simplicity, charged and magnetic bodies. For these, besides gravity, there are electrical and magnetic non-contact forces. LAWS OF MOTION 101 since we are talking of uncharged and non- exist by itself. When there is no applied force, magnetic bodies in mechanics. At the microscopic there is no static friction. It comes into play the level, all bodies are made of charged constituents moment there is an applied force. As the applied (nuclei and electrons) and the various contact force F increases, fs also increases, remaining forces arising due to elasticity of bodies, molecular equal and opposite to the applied force (up to a collisions and impacts, etc. can ultimately be certain limit), keeping the body at rest. Hence, it traced to the electrical forces between the charged is called static friction. Static friction opposes constituents of different bodies. The detailed impending motion. The term impending motion means motion that would take place (but does microscopic origin of these forces is, however, not actually take place) under the applied force, complex and not useful for handling problems in if friction were absent. mechanics at the macroscopic scale. This is why We know from experience that as the applied they are treated as different types of forces with force exceeds a certain limit, the body begins to their characteristic properties determined move. It is found experimentally that the limiting empirically. value of static friction f s ( ) max is independent of 5.9.1 Friction the area of contact and varies with the normal Let us return to the example of a body of mass m force(N) approximately as : at rest on a horizontal table. The force of gravity (f ) s max = µs N (5.13) (mg) is cancelled by the normal reaction force where µ s is a constant of proportionality (N) of the table. Now suppose a force F is applied depending only on the nature of the surfaces in horizontally to the body. We know from contact. The constant µs is called the coefficient experience that a small applied force may not of static friction. The law of static friction may be enough to move the body. But if the applied thus be written as force F were the only external force on the body, f s ≤ µs N (5.14) it must move with acceleration F/m, however small. Clearly, the body remains at rest because ( ) If the applied force F exceeds f s max the body some other force comes into play in the begins to slide on the surface. It is found horizontal direction and opposes the applied experimentally that when relative motion has force F, resulting in zero net force on the body. started, the frictional force decreases from the This force fs parallel to the surface of the body in contact with the table is known as frictional ( ) static maximum value f s max. Frictional force force, or simply friction (Fig. 5.10(a)). The that opposes relative motion between surfaces subscript stands for static friction to distinguish in contact is called kinetic or sliding friction and it from kinetic friction fk that we consider later is denoted by fk. Kinetic friction, like static friction, is found to be independent of the area (Fig. 5.10(b)). Note that static friction does not of contact. Further, it is nearly independent of the velocity. It satisfies a law similar to that for static friction: fk = µk N (5.15) where µk′ the coefficient of kinetic friction, depends only on the surfaces in contact. As mentioned above, experiments show that µk is Fig. 5.10 Static and sliding friction: (a) Impending less than µs. When relative motion has begun, motion of the body is opposed by static the acceleration of the body according to the friction. When external force exceeds the second law is ( F – fk )/m. For a body moving with maximum limit of static friction, the body constant velocity, F = fk. If the applied force on begins to move. (b) Once the body is in the body is removed, its acceleration is – fk /m motion, it is subject to sliding or kinetic friction and it eventually comes to a stop. which opposes relative motion between the The laws of friction given above do not have two surfaces in contact. Kinetic friction is the status of fundamental laws like those for usually less than the maximum value of static gravitational, electric and magnetic forces. They friction. are empirical relations that are only 102 PHYSICS approximately true. Yet they are very useful in Answer The forces acting on a block of mass m practical calculations in mechanics. at rest on an inclined plane are (i) the weight Thus, when two bodies are in contact, each mg acting vertically downwards (ii) the normal experiences a contact force by the other. Friction, force N of the plane on the block, and (iii) the by definition, is the component of the contact force static frictional force fs opposing the impending parallel to the surfaces in contact, which opposes motion. In equilibrium, the resultant of these impending or actual relative motion between the forces must be zero. Resolving the weight mg two surfaces. Note that it is not motion, but along the two directions shown, we have relative motion that the frictional force opposes. m g sin θ = fs , m g cos θ = N Consider a box lying in the compartment of a train As θ increases, the self-adjusting frictional force that is accelerating. If the box is stationary relative to the train, it is in fact accelerating along fs increases until at θ = θmax, fs achieves its with the train. What forces cause the acceleration of the box? Clearly, the only conceivable force in maximum value, f s ( ) max = µs N. the horizontal direction is the force of friction. If Therefore, there were no friction, the floor of the train would slip by and the box would remain at its initial tan θmax = µs or θmax = tan–1 µs position due to inertia (and hit the back side of When θ becomes just a little more than θmax , the train). This impending relative motion is there is a small net force on the block and it opposed by the static friction fs. Static friction begins to slide. Note that θmax depends only on provides the same acceleration to the box as that µs and is independent of the mass of the block. of the train, keeping it stationary relative to the train. For θmax = 15°, µs = tan 15° Example 5.7 Determine the maximum t = 0.27 t acceleration of the train in which a box t lying on its floor will remain stationary, Example 5.9 What is the acceleration of given that the co-efficient of static friction the block and trolley system shown in a between the box and the train’s floor is Fig. 5.12(a), if the coefficient of kinetic friction 0.15. between the trolley and the surface is 0.04? What is the tension in the string? (Take g = Answer Since the acceleration of the box is due 10 m s-2). Neglect the mass of the string. to the static friction, ma = fs ≤ µs N = µs m g i.e. a ≤ µs g ∴ amax = µs g = 0.15 x 10 m s–2 = 1.5 m s–2 t t Example 5.8 See Fig. 5.11. A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle θ = 15° with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ? (a) (b) (c) Fig. 5.11 Fig. 5.12 LAWS OF MOTION 103 Answer As the string is inextensible, and the is the reason why discovery of the wheel has pully is smooth, the 3 kg block and the 20 kg been a major milestone in human history. trolley both have same magnitude of Rolling friction again has a complex origin, acceleration. Applying second law to motion of though somewhat different from that of static the block (Fig. 5.12(b)), and sliding friction. During rolling, the surfaces 30 – T = 3a in contact get momentarily deformed a little, and Apply the second law to motion of the trolley (Fig. this results in a finite area (not a point) of the 5.12(c)), body being in contact with the surface. The net effect is that the component of the contact force T – fk = 20 a. parallel to the surface opposes motion. Now fk = µk N, We often regard friction as something Here µk = 0.04, undesirable. In many situations, like in a N = 20 x 10 machine with different moving parts, friction = 200 N. does have a negative role. It opposes relative Thus the equation for the motion of the trolley is motion and thereby dissipates power in the form T – 0.04 x 200 = 20 a Or T – 8 = 20a. of heat, etc. Lubricants are a way of reducing 22 kinetic friction in a machine. Another way is to These equations give a = m s –2 = 0.96 m s-2 use ball bearings between two moving parts of a 23 and T = 27.1 N. t machine [Fig. 5.13(a)]. Since the rolling friction between ball bearings and the surfaces in Rolling friction contact is very small, power dissipation is A body like a ring or a sphere rolling without reduced. A thin cushion of air maintained slipping over a horizontal plane will suffer no between solid surfaces in relative motion is another effective way of reducing friction (Fig. friction, in principle. At every instant, there is 5.13(a)). just one point of contact between the body and In many practical situations, however, friction the plane and this point has no motion relative is critically needed. Kinetic friction that to the plane. In this ideal situation, kinetic or dissipates power is nevertheless important for static friction is zero and the body should quickly stopping relative motion. It is made use continue to roll with constant velocity. We know, of by brakes in machines and automobiles. in practice, this will not happen and some Similarly, static friction is important in daily resistance to motion (rolling friction) does occur, life. We are able to walk because of friction. It i.e. to keep the body rolling, some applied force is impossible for a car to move on a very slippery is needed. For the same weight, rolling friction road. On an ordinary road, the friction between is much smaller (even by 2 or 3 orders of the tyres and the road provides the necessary magnitude) than static or sliding friction. This external force to accelerate the car. Fig. 5.13 Some ways of reducing friction. (a) Ball bearings placed between moving parts of a machine. (b) Compressed cushion of air between surfaces in relative motion. 104 PHYSICS 5.10 CIRCULAR MOTION is the static friction that provides the centripetal acceleration. Static friction opposes the We have seen in Chapter 4 that acceleration of impending motion of the car moving away from a body moving in a circle of radius R with uniform the circle. Using equation (5.14) & (5.16) we get speed v is v2/R directed towards the centre. the result According to the second law, the force f providing c this acceleration is : mv 2 f ≤ µs N = mv 2 R fc = (5.16) R µs RN v2 ≤ = µs Rg [∵ N = mg] where m is the mass of the body. This force m directed forwards the centre is called the which is independent of the mass of the car. centripetal force. For a stone rotated in a circle This shows that for a given value of µs and R, by a string, the centripetal force is provided by there is a maximum speed of circular motion of the tension in the string. The centripetal force the car possible, namely for motion of a planet around the sun is the v max = µs Rg (5.18) (a) (b) Fig. 5.14 Circular motion of a car on (a) a level road, (b) a banked road. gravitational force on the planet due to the sun. Motion of a car on a banked road For a car taking a circular turn on a horizontal We can reduce the contribution of friction to the road, the centripetal force is the force of friction. circular motion of the car if the road is banked The circular motion of a car on a flat and (Fig. 5.14(b)). Since there is no acceleration along banked road give interesting application of the the vertical direction, the net force along this laws of motion. direction must be zero. Hence, Motion of a car on a level road N cos θ = mg + f sin θ (5.19a) Three forces act on the car (Fig. 5.14(a): (i) The weight of the car, mg The centripetal force is provided by the horizontal (ii) Normal reaction, N components of N and f. (iii) Frictional force, f mv 2 As there is no acceleration in the vertical N sin θ + f cos θ = (5.19b) R direction N – mg = 0 But f ≤ µs N N = mg (5.17) Thus to obtain v we put The centripetal force required for circular motion max is along the surface of the road, and is provided f = µs N. by the component of the contact force between Then Eqs. (5.19a) and (5.19b) become road and the car tyres along the surface. This by definition is the frictional force. Note that it N cos θ = mg + µs N sin θ (5.20a) LAWS OF MOTION 105 N sin θ + µs N cos θ = mv2/R t turn. t (5.20b) From Eq. (5.20a), we obtain Example 5.11 A circular racetrack of mg radius 300 m is banked at an angle of 15°. N= If the coefficient of friction between the cosθ – µs sinθ wheels of a race-car and the road is 0.2, Substituting value of N in Eq. (5.20b), we get what is the (a) optimum speed of the race- mg (sinθ + µs cosθ ) mv max 2 car to avoid wear and tear on its tyres, and = (b) maximum permissible speed to avoid cosθ – µs sinθ R slipping ? 1  µ + tanθ  2 Answer On a banked road, the horizontal or v max =  Rg s  (5.21) component of the normal force and the frictional  1 – µs tanθ  force contribute to provide centripetal force to Comparing this with Eq. (5.18) we see that keep the car moving on a circular turn without maximum possible speed of a car on a banked slipping. At the optimum speed, the normal road is greater than that on a flat road. reaction’s component is enough to provide the needed centripetal force, and the frictional force is not needed. The optimum speed vo is given by For µs = 0 in Eq. (5.21 ), Eq. (5.22): vo = ( R g tan θ ) ½ (5.22) vO = (R g tan θ)1/2 At this speed, frictional force is not needed at all Here R = 300 m, θ = 15°, g = 9.8 m s-2; we to provide the necessary centripetal force. have Driving at this speed on a banked road will cause vO = 28.1 m s-1. little wear and tear of the tyres. The same equation also tells you that for v < vo, frictional The maximum permissible speed vmax is given by force will be up the slope and that a car can be Eq. (5.21): parked only if tan θ ≤ µs. t 1/2 µ + tan θ vmax = Rg s = 38.1 m s −1 t 1 − µs tan θ Example 5.10 A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the 5.11 SOLVING PROBLEMS IN MECHANICS speed. The co-efficient of static friction The three laws of motion that you have learnt in between the tyres and the road is 0.1. Will this chapter are the foundation of mechanics. the cyclist slip while taking the turn ? You should now be able to handle a large variety of problems in mechanics. A typical problem in Answer On an unbanked road, frictional force mechanics usually does not merely involve a alone can provide the centripetal force needed single body under the action of given forces. to keep the cyclist moving on a circular turn More often, we will need to consider an assembly without slipping. If the speed is too large, or if of different bodies exerting forces on each other. the turn is too sharp (i.e. of too small a radius) Besides, each body in the assembly experiences or both, the frictional force is not sufficient to the force of gravity. When trying to solve a provide the necessary centripetal force, and the problem of this type, it is useful to remember cyclist slips. The condition for the cyclist not to the fact that we can choose any part of the slip is given by Eq. (5.18) : assembly and apply the laws of motion to that v2 ≤ µs R g part provided we include all forces on the chosen Now, R = 3 m, g = 9.8 m s-2, µs = 0.1. That is, part due to the remaining parts of the assembly. µs R g = 2.94 m2 s-2. v = 18 km/h = 5 m s-1; i.e., We may call the chosen part of the assembly as v2 = 25 m2 s-2. The condition is not obeyed. the system and the remaining part of the The cyclist will slip while taking the circular 106 PHYSICS assembly (plus any other agencies of forces) as equal to 2 × 10 = 20 N; and the normal force the environment. We have followed the same R of the floor on the block. By the First Law, method in solved examples. To handle a typical the net force on the block must be zero i.e., problem in mechanics systematically, one R = 20 N. Using third law the action of the should use the following steps : block (i.e. the force exerted on the floor by (i) Draw a diagram showing schematically the the block) is equal to 20 N and directed various parts of the assembly of bodies, the vertically downwards. links, supports, etc. (b) The system (block + cylinder) accelerates (ii) Choose a convenient part of the assembly downwards with 0.1 m s-2. The free-body as one system. diagram of the system shows two forces on (iii) Draw a separate diagram which shows this the system : the force of gravity due to the system and all the forces on the system by earth (270 N); and the normal force R ′ by the the remaining part of the assembly. Include floor. Note, the free-body diagram of the also the forces on the system by other system does not show the internal forces agencies. Do not include the forces on the between the block and the cylinder. Applying environment by the system. A diagram of the second law to the system, this type is known as ‘a free-body diagram’. 270 – R′ = 27 × 0.1N (Note this does not imply that the system ie. R′ = 267.3 N under consideration is without a net force). (iv) In a free-body diagram, include information about forces (their magnitudes and directions) that are either given or you are sure of (e.g., the direction of tension in a string along its length). The rest should be treated as unknowns to be determined using laws of motion. (v) If necessary, follow the same procedure for another choice of the system. In doing so, employ Newton’s third law. That is, if in the free-body diagram of A, the force on A due to B is shown as F, then in the free-body diagram of B, the force on B due to A should be shown as –F. The following example illustrates the above t procedure : Example 5.12 See (Fig. 5.15) A wooden Fig. 5.15 block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg By the third law, the action of the system on is placed on top of the block, the floor yields the floor is equal to 267.3 N vertically downward. steadily and the block and the cylinder Action-reaction pairs together go down with an acceleration of For (a): (i) the force of gravity (20 N) on the block 0.1 m s–2. What is the action of the block by the earth (say, action); the force of on the floor (a) before and (b) after the floor gravity on the earth by the block yields ? Take g = 10 m s–2. Identify the (reaction) equal to 20 N directed action-reaction pairs in the problem. upwards (not shown in the figure). (ii) the force on the floor by the block Answer (action); the force on the block by the (a) The block is at rest on the floor. Its free-body floor (reaction). diagram shows two forces on the block, the For (b): (i) the force of gravity (270 N) on the force of gravitational attraction by the earth system by the earth (say, action); the LAWS OF MOTION 107 force of gravity on the earth by the happen to be equal and opposite can never system (reaction), equal to 270 N, constitute an action-reaction pair. The force of directed upwards (not shown in the gravity on the mass in (a) or (b) and the normal figure). force on the mass by the floor are not action- (ii) the force on the floor by the system reaction pairs. These forces happen to be equal (action); the force on the system by the and opposite for (a) since the mass is at rest. floor (reaction). In addition, for (b), the They are not so for case (b), as seen already. force on the block by the cylinder and The weight of the system is 270 N, while the the force on the cylinder by the block normal force R′ is 267.3 N. t also constitute an action-reaction pair. The practice of drawing free-body diagrams is The important thing to remember is that an of great help in solving problems in mechanics. action-reaction pair consists of mutual forces It allows you to clearly define your system and which are always equal and opposite between consider all forces on the system due to objects two bodies. Two forces on the same body which that are not part of the system itself. A number of exercises in this and subsequent chapters will help you cultivate this practice. SUMMARY 1. Aristotle’s view that a force is necessary to keep a body in uniform motion is wrong. A force is necessary in practice to counter the opposing force of friction. 2. Galileo extrapolated simple observations on motion of bodies on inclined planes, and arrived at the law of inertia. Newton’s first law of motion is the same law rephrased thus: “Everybody continues to be in its state of rest or of uniform motion in a straight line, unless compelled by some external force to act otherwise”. In simple terms, the First Law is “If external force on a body is zero, its acceleration is zero”. 3. Momentum (p ) of a body is the product of its mass (m) and velocity (v) : p = mv 4. Newton’s second law of motion : The rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which the force acts. Thus dp F =k =kma dt where F is the net external force on the body and a its acceleration. We set the constant of proportionality k = 1 in SI units. Then dp F= = ma dt The SI unit of force is newton : 1 N = 1 kg m s-2. (a) The second law is consistent with the First Law (F = 0 implies a = 0) (b) It is a vector equation (c) It is applicable to a particle, and also to a body or a system of particles, provided F is the total external force on the system and a is the acceleration of the system as a whole. (d) F at a point at a certain instant determines a at the same point at that instant. That is the Second Law is a local law; a at an instant does not depend on the history of motion. 5. Impulse is the product of force and time which equals change in momentum. The notion of impulse is useful when a large force acts for a short time to produce a measurable change in momentum. Since the time of action of the force is very short, one can assume that there is no appreciable change in the position of the body during the action of the impulsive force. 108 PHYSICS 6. Newton’s third law of motion: To every action, there is always an equal and opposite reaction In simple terms, the law can be stated thus : Forces in nature always occur between pairs of bodies. Force on a body A by body B is equal and opposite to the force on the body B by A. Action and reaction forces are simultaneous forces. There is no cause-effect relation between action and reaction. Any of the two mutual forces can be called action and the other reaction. Action and reaction act on different bodies and so they cannot be cancelled out. The internal action and reaction forces between different parts of a body do, however, sum to zero. 7. Law of Conservation of Momentum The total momentum of an isolated system of particles is conserved. The law follows from the second and third law of motion. 8. Friction Frictional force opposes (impending or actual) relative motion between two surfaces in contact. It is the component of the contact force along the common tangent to the surface in contact. Static friction fs opposes impending relative motion; kinetic friction fk opposes actual relative motion. They are independent of the area of contact and satisfy the following approximate laws : ( ) f s ≤ fs max = µs R f =µ R k k µs (co-efficient of static friction) and µk (co-efficient of kinetic friction) are constants characteristic of the pair of surfaces in contact. It is found experimentally that µk is less than µs. POINTS TO PONDER 1. Force is not always in the direction of motion. Depending on the situation, F may be along v, opposite to v, normal to v or may make some other angle with v. In every case, it is parallel to acceleration. 2. If v = 0 at an instant, i.e. if a body is momentarily at rest, it does not mean that force or acceleration are necessarily zero at that instant. For example, when a ball thrown upward reaches its maximum height, v = 0 but the force continues to be its weight mg and the acceleration is not zero but g. 3. Force on a body at a given time is determined by the situation at the location of the body at that time. Force is not ‘carried’ by the body from its earlier history of motion. The moment after a stone is released out of an accelerated train, there is no horizontal force (or acceleration) on the stone, if the effects of the surrounding air are neglected. The stone then has only the vertical force of gravity. LAWS OF MOTION 109 4. In the second law of motion F = m a, F stands for the net force due to all material agencies external to the body. a is the effect of the force. ma should not be regarded as yet another force, besides F. 5. The centripetal force should not be regarded as yet another kind of force. It is simply a name given to the force that provides inward radial acceleration to a body in circular motion. We should always look for some material force like tension, gravitational force, electrical force, friction, etc as the centripetal force in any circular motion. 6. Static friction is a self-adjusting force up to its limit µs N (fs ≤ µs N). Do not put fs= µs N without being sure that the maximum value of static friction is coming into play. 7. The familiar equation mg = R for a body on a table is true only if the body is in equilibrium. The two forces mg and R can be different (e.g. a body in an accelerated lift). The equality of mg and R has no connection with the third law. 8. The terms ‘action’ and ‘reaction’ in the third Law of Motion simply stand for simultaneous mutual forces between a pair of bodies. Unlike their meaning in ordinary language, action does not precede or cause reaction. Action and reaction act on different bodies. 9. The different terms like ‘friction’, ‘normal reaction’ ‘tension’, ‘air resistance’, ‘viscous drag’, ‘thrust’, ‘buoyancy’ ‘weight’ ‘centripetal force’ all stand for ‘force’ in different contexts. For clarity, every force and its equivalent terms encountered in mechanics should be reduced to the phrase ‘force on A by B’. 10. For applying the second law of motion, there is no conceptual distinction between inanimate and animate objects. An animate object such as a human also requires an external force to accelerate. For example, without the external force of friction, we cannot walk on the ground. 11. The objective concept of force in physics should not be confused with the subjective concept of the ‘feeling of force’. On a merry-go-around, all parts of our body are subject to an inward force, but we have a feeling of being pushed outward – the direction of impending motion. EXERCISES (For simplicity in numerical calculations, take g = 10 m s-2) 5.1 Give the magnitude and direction of the net force acting on (a) a drop of rain falling down with a constant speed, (b) a cork of mass 10 g floating on water, (c) a kite skillfully held stationary in the sky, (d) a car moving with a constant velocity of 30 km/h on a rough road, (e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields. 5.2 A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, (a) during its upward motion, (b) during its downward motion, (c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance. 5.3 Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg, (a) just after it is dropped from the window of a stationary train, (b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h, (c ) just after it is dropped from the window of a train accelerating with 1 m s-2, 110 PHYSICS (d) lying on the floor of a train which is accelerating with 1 m s-2, the stone being at rest relative to the train. Neglect air resistance throughout. 5.4 One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is : mv2 mv2 (i) T, (ii) T − , (iii) T + , (iv) 0 l l T is the tension in the string. [Choose the correct alternative]. 5.5 A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s-1. How long does the body take to stop ? 5.6 A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s-1 to 3.5 m s-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force ? 5.7 A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body. 5.8 The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle ? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg. 5.9 A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s-2. Calculate the i