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Transcript
Newton’s Third Law of Motion SOMNATH DATTA 656 “Snehalata, 13th Main, 4th Stage, T.K. Layout, Mysore 5700009, India E.mail: [email protected] http://sites.google.com/site/physicsforpleasure 1 A Slide Show on Newton’s Third Law of Motion 1. Newton’s first and second laws tell us that a body accelerates due to external forces. The net external force equals the mass of the body times its acceleration. These laws do not tell us where this force comes from. Does the force that drives this car come from the engine? 2. If so, why is this car stuck on wet clay road when the engine is running in full throttle? The force that pushes the car forward does not come from the engine. This force comes from the road due to friction between the tyre and the road, and is called traction. 3. Does the force come from his muscles? Physics Education • July − September 2009 213 4. If so, why is he slipping? No runner is his able to run on a slippery track in spite of his muscle power. The force that drives the runner forward comes from the ground when there is sufficient friction between his feet and the ground. 5. This car requires 40,000 N to speed up at the rate of 10m/s2. This sprinter requires 300 N to speed up at the rate of 5m/s2. How is the road able to judge the individual force requirements for the car and the sprinter? 6. Answer to this comes from Newton’s Third Law of Motion: • • Nothing in the universe can act without being acted upon. To every action there is an equal and opposite reaction. The first statement is somewhat philosophical. The second one is vague. In exact terms what the first statement means is that whenever any object A offers a force F, it itself experiences the same force back on itself in the reverse direction. 214 Physics Education • July − September 2009 7. The second statement can be interpreted as follows. Assume two objects A and B. A offers a force on B directed from A to B. We denote this force by FAB. The 3rd Law says B will simultaneously offer a force FBA on A which is directed from B to A. These two forces are equal and opposite, even if the objects A and B are grossly unequal in size and mass. Written as vectors, FBA = −FAB. (1) We can refer to any one of the pair of forces, say the force FAB, as the action force and the second one, i.e. FBA, as the reaction force. Then Eq. (1) is a mathematical expression of the second statement. In the particular example shown these two forces are repulsive forces. 8. Here they are attractive forces. A pulls B with a force FAB. As a result A is itself being pulled towards B with a force FBA, and FBA = −FAB. 9. A runner can run only if the ground kicks him forward. To get this kick the runner must kick the ground backward. It is possible to kick a solid and firm object like football or grass with sufficient force, but it is difficult to kick a loose object like air or slippery road with sufficient force. Therefore the runner is unable to run on a slippery road. Physics Education • July − September 2009 215 10. It is possible to get the necessary force of 40,000 N from a dry ground because the tyre, powered by the engine, can press against the road with this force due to friction between the two surfaces (of the tyre and the ground.) A frictionless surface on the other hand will not allow the tyre to rub with a large force and because of this the car will not be able to accelerate. 11. There is a common misconception about the Third Law especially among the beginners. They think that this law holds only when the interacting bodies are in equilibrium since the forces of action and reaction will then cancel each other. This is entirely wrong. Here we show two bodies A and B. 12. But the force FAB is acting on B only. Therefore B is not in equilibrium. 13. The reaction force FBA is acting on A only. Therefore A is not in equilibrium. 216 Physics Education • July − September 2009 14. Now see the complete picture. The forces are equal and opposite. But they do not cancel each other. Because they are acting on different bodies. 15. We see that the third law holds between any two bodies, whether at rest or in motion. 16. Let us illustrate the Third Law with a few more examples. This stone of mass 1 kg is falling to the ground in a parabolic trajectory with acceleration g equal to 9.81 m/s2, like any other stone. The gravitational pull on the stone is m times g, that is, 9.81 N. Since the earth is pulling the stone with force 9.81 N downward, the stone is at the same time pulling the earth with a force of the same magnitude of 9.81 N, but directed upward. 17. Therefore the earth is also accelerating toward the stone! However, since the mass of the earth is very LARGE, its acceleration (toward the stone) is very small. Physics Education • July − September 2009 217 18. Another example: A football player kicks a football. Therefore according to the 3rd Law, the football kicks the player back. 19. The kick is an impulsive force, lasting for a very short time, say, 101 of a second. Taking the mass of the ball to be ¼ kg, we have calculated the kick force to be 250 N. 20. The 3rd Law tells us that the ball also kicks the player back in the opposite direction. However, since the mass of the player is 100 kg, i.e., 400 times the mass of the ball, his acceleration is 400 times less and lasts exactly for 101 second. 21. Here we see the complete picture. The mutual kick forces are equal and opposite. But they do not cancel each other, because they act on different bodies. 218 Physics Education • July − September 2009 22. Let us try to understand how this horse is able to pull the cart. 23. The cart requires a force of 2500 N to accelerate. But there is a backward frictional drag on the wheels. Therefore the horse must pull the cart with a force of 3000 N. 24. The horse needs 1500 N to accelerate. According to the 3rd Law, the cart is pulling horse back with force 3000 N. Therefore the ground must provide the necessary traction force of 4500 N in the forward direction. 25. How does the horse get this 4500 N traction force? By pressing against the ground with this force, according to the 3rd Law. Also since the wheels are experiencing frictional drag of 500 N backward, the ground is at the same time dragged forward with 500 N, again due to the 3rd Law. The resulting acceleration of the ground is very small, since the mass of the earth is very LARGE. Physics Education • July − September 2009 219 26. Now see the complete picture. For every force there is an equal and opposite reaction force. But the forces on any one of the three bodies are not in equilibrium. Therefore, the cart, the horse and the ground are all accelerating (though in different degrees). 27. As final example we consider the motion of the earth and the moon due to the gravitational attraction between the two. Note the distance between the centres of the earth and the moon, and the masses of the earth and the moon. The gravitational force with which the earth is pulling the moon is 20.6 × 1019 N. 28. As a result the moon is accelerating toward the earth and is moving in a circle. 29. The earth is experiencing a force of exactly the same magnitude of 20.6 × 1019 N but directed toward the moon. However, the mass of the earth is about 83 times the mass of the moon. Therefore the earth is undergoing a much smaller acceleration of 3.44 × 10−5m/s2. As a result the centre of the earth is also moving in a circle, a much smaller circle. 220 Physics Education • July − September 2009 30. Both the moon and the earth are accelerating toward each other according to the Third Law of Motion. The centre of each is moving in a circle, one circle being much larger than the other because one mass is much smaller than the other. 2 Free Body Diagrams Newton’s first law of motion is difficult to believe. Why should anyone believe that an object keeps moving forever along a straight line in the absence of external interferences, when our everyday encounters seem to contradict the statement? Similarly Newton’s third law of motion is rarely believed. It is easy to “imagine” equal and opposite action-reaction forces when, for example, a stone is resting on a table. But how to trust the same statement when the table breaks and the stone crashes down? The confusion is understandable, and persists. The reason − our apathy to explain the theory properly with effective drawings. For understanding concepts and principles of mechanics drawing freebody diagrams is an essential first step. Without this Newton’s laws of motion can never be appreciated. • A car and the ground in Slide 10; • A falling stone and the earth in Slides 1617; • A football player, football and the ground in Slides 19- 21; • A cart, a horse pulling it, and the ground in Slides 23-26; • The earth and the moon in Slides 27-29. 2.2 The Golden Rules for Understanding Newton’s Laws of Motion For understanding Newton’s laws of motion properly, especially their applications to extended objects, it is essential that the following golden rules be kept in mind. 1. 2.1 What is a Freebody Diagram? 2. A freebody diagram (FBD) is a pictorial formulation of the mechanics problem through figures in which each object in a system of interacting bodies is shown separately and the forces acting on this body (and on this body alone) are clearly delineated and also shown separately. The slides presented in the previous section have shown many examples of FBDs. Some of the important ones are the following 3. • A runner and the ground in Slide 9 (The gravity force has been suppressed); Physics Education • July − September 2009 Clearly identify the “system” whose motion you are going to analyze. Sometimes it may be useful to draw a boundary around this system in order to identify it and isolate it from all neighbouring systems. All the forces that are coming from outside this system you have just identified − by this I mean the forces that are coming from outside the boundary you have drawn - are to be called external forces. Forces of “interaction” among objects inside the system that is, forces coming from objects that are confined within the boundary you have drawn − are all internal forces. Therefore draw a freebody diagram (FBD) of all the external forces. You must remember that FBDs are a must for understanding mechanics. It is impossible 221 to appreciate the principles of mechanics, the forces and how they guide the motion of objects, without drawing FBDs. Add up vectorially all the external forces acting on the system. In some cases this vector sum may be zero. In that case a central point of the object, called the Centre of Mass, to be abbreviated as CM, will be either stationary or moving with constant velocity (i.e., constant speed along a straight line.) 4. We shall refer to the motion of the CM as the bulk translatory motion of the system. If the vector sum of the external forces is non-zero, then that vector sum Ftotal will be called the Resultant Force on the system. The bulk translatory motion of the system is determined by Newton’s second law of motion, which can now be written as M r = Ftotal (2) where M is the total mass of all the component parts of the system, and r is the radius vector of the CM of the system, so that r is the acceleration of the CM. We have illustrated Rules 1,2,3 in Figures 1. In the Top Box we have shown three objects A, B and C interacting with one another, and forming a system. We assume that this system of three objects is not subjected to any “external force”. Individually the objects A, B and C are moving and accelerating only due to the forces of interaction among themselves. In frame (a) we have depicted each one of them as a single system, by drawing boundaries (with dotted lines) around each. The forces of interaction are (F1,F′1), (F2,F′2), (F3,F′3). The pair of forces within parentheses are the “equal and opposite” action-reaction pairs. For example if F′1 is the action force, then its counterpart F1 is the corresponding (equal and opposite) reaction force. Similarly if F3 is the action force, then its counterpart F′3 is the corresponding (equal and opposite) reaction force. Note also that we have drawn “shooting” lines on each one of the three objects indicating that the objects are all moving objects having velocities va, vb, vc respectively. Each one of the three systems A,B,C is subjected to the following external forces: (1) external forces F1,F2 on A; (2) external forces F′1,F′3 on B; (3) external forces F′2,F3 on C. In frame (b) we have clubbed A and B forming a single system AB. Now it is an interaction between two systems namely, AB and C. The only external force on AB is F′, and the only external force on C is F = −F′. In frame (c) we have taken A, B and C together as a single system ABC. There are no external forces on this system ABC. We shall now illustrate Rule 4 in the form of the following equations of motion. System A Net external force on the system Ftotal = F1 + F2 Eqn of Motion ma ra = F1 + F2 B Ftotal = F′1 + F′3 mb rb = F′1 + F′3 C Ftotal = F′2 + F3 mc rc = F′2 + F3 AB Ftotal = −F = −(F′2+F3) = F2+ F′3 (ma+mb) rab = F′=−F ABC Ftotal = 0 (ma+mb+mc) rabc =0 222 (3) Physics Education • July − September 2009 Figure 1. Forces of interaction between systems of material objects. In the above table ma,mb,mc represent the masses of the systems A, B, C respectively, and ra, rb, rc, rab, rabc represent the locations of CMs of the systems A, B, C, AB, ABC respectively. In the Bottom Box of Figure 1 we have shown a stone falling under the force of gravity. The bulk motion of the stone is determined by the gravity force mg. We had shown an FBD of this object in slide 16 of the previous section. However, we have repeated the FBD in frame (a). The stone is composed of a large number of atoms, of the order of, say, 1025. These atoms are pulling or pushing each other. Yet, they are all internal forces and, therefore, have no effect on the bulk motion of the stone. Now, “imagine” a boundary line Σ dividing the stone into parts A and B, which we shall refer to as A-stone and B-stone respectively. Then there are action-reaction forces (F,F′) Physics Education • July − September 2009 between the two parts, which are always equal and opposite. We have shown the FBD of both the parts in frame (b). The CM of A-stone is moving under the effect of (1) the gravity force mag, and (2) the force F on it coming from B. These two forces cause the falling of the Astone under gravity, along with rotational motion of the CM.1 In the same manner the CM of the B-stone is moving under the effect of (1) the gravity force mbg, and (2) the reaction force F′ = −F on it coming from A. These two forces together cause the falling under gravity along with rotation of the CM of B. We shall now apply Rule 4 to the falling stone and its two parts and obtain the following equations of motion. 1. There can also be oscillatory motion of the CMs of A and B, due to vibration induced by the forces F and F′ respectively. 223 System Whole stone A-stone B-stone Net external force on the system Ftotal = Mg Ftotal =mag+ F Ftotal = mbg − F In the above table (M = ma + mb, r), (ma, ra), (mb, rb) represent (mass, location of the CM) of the whole stone, A-stone, B-stone respectively. The reader should note here an important point. The only external force on the whole stone is the gravitational force. Hence the point C (representing the CM of the stone) falls in a parabolic path like an ordinary projectile. On the other hand A-stone and B-stone are subjected to extra forces F and −F respectively, besides the force of gravity. Their CMs Ca and Cb follow “nonparabolic” paths. Eqn of Motion M r = Mg ma r a = mag+F r mb b= mbg+F′ = mbg−F (4) 4 Every Real Force has a Parent In Figures 2-5 we have cited a number of instances in which the third law presents us with a paradox. We have illustrated graphically each situation by identifying the forces with FBDs, and in two cases listing them in a tabular form. This, we hope, will facilitate clarification of the Second and the Third Laws of Motion. We have presented in Tables 1 and 2 the “Forces at Work” corresponding to two of the examples cited, namely, (a) Diving, and (b) Tug of War. In the right column of each table you will see the forces on the subsystems. The action-reaction forces on these subsystems taken together add up to zero, so that the net force on the “Total” system is nil. In writing the vectors in Table 2 we have taken the forces T1,T2,F1,F2 to be in the direction of the positive X axis (i.e., rightward), and the forces N1,N2,M1g,M2g to be in the direction of the positive Z axis (i.e., upward). Newton’s first law of motion gives a qualitative meaning of force. A force is something which makes a particle, a system, deviate from its “most natural” motion, namely, moving in a straight line with constant velocity. The second law pinpoints the concept further by painting force in the colour of a vector, a directed straight line with a specified length − equal to the rate of change of momentum of the object. The third law refines the concept of force further by making the grand statement, “force is an interaction between two objects in which each one is equal.” The earth pulls a mosquito with the same force with which the mosquito pulls the earth. A proton pulls an electron with the same force with which the electron (about two thousand times lighter) pulls the proton. These statements are as profound as the shock and disbelief they generate in our minds. Yet the statements are true. However, the reader must not miss a crucial assumption in the statements of the laws of motion. Velocity, momentum, acceleration are frame dependent quantities. They have no meaning unless you have specified a frame of reference. What frame? The frame in which the statements are valid. That reads like the old joke: which came first, the egg or the chicken? However, the assumption of an inertial frame is not an absurdity. The concept of a family of inertial frames (the name of the frames in which the laws of motion hold) lies at the foundation of classical mechanics. 224 Physics Education • July − September 2009 3 Further Examples of FBDs Figure 2: Free body diagrams of diving. The action-reaction pair of forces are shown with the same roman bold letter appearing twice, once with prime and once again without it. The forces at play are the tension forces (T,T′) between the board and the left pillar; the compressive (pressure) forces (P,P′) between the board and the left pillar; the force R′ with which the diver kicks the board and the kickback reaction force R he receives, the gravitational forces (G1,G′1), (G2,G′2), (G3,G′3), (G4,G′4). Drawings on Left show Normal Diving. Drawings on Right show accident in diving. The board gets uprooted from one of the supporting pillars, and the diver tumbles into the water. In this case the pair of forces T,T′ disappear, and they disappear simultaneously. Physics Education • July − September 2009 225 Figure 3. We have now written each pair of action reaction forces with the same symbol (even though they have opposite signs.) For example (T1,T1) are action-reaction forces of tension. Top : Tug of War. Note that the tension forces T1,T2 are not equal in magnitude, because the rope has considerable mass m and it is accelerating. The contestant on the right is pushing the ground with friction force F1, and a downward force N1, and at the same time pulling the earth upward with a gravitational force M1g. The earth is repaying him back with the same force in opposite direction. Bottom : Boy jumping off a boat. In Figure (b) we have shown the forces acting on the “boat + boy” system. Figures (c) and (d) show the FBDs of the boat and the boy separately. We have removed the earth from our “system”. 226 Physics Education • July − September 2009 Figure 4. Top: FBD of an unfortunate man who is sinking into a quicksand. He sinks because the gravitational force downward is larger in magnitude than the upward reaction force provided by the ground. We have not shown the action-reaction counterpart of this gravitational force (pulling the earth upward.) Bottom : Action-reaction forces between a balloon and the escaping gas. We have marked out a rectangular column of the gas which is pushed downward with a gas-pressure force of magnitude F1. This causes an upward reaction force of the same magnitude acting on the balloon. The balloon accelerates upward because this reaction force F1 is larger than the weight of the balloon (same as the downward force of gravity on it. The lateral pressure forces F2, F3, F4, F5 are balanced causing no sidewise acceleration of the gas. Their reaction counterparts are similarly balanced causing no sidewise acceleration of the balloon. Physics Education • July − September 2009 227 Figure 5. A locomotive steam engine hauling a train consisting of only two coaches. We assume that the train is moving with constant speed, so that the net force on the train is zero. (a) General picture. The train is moving from right to left as indicated by the arrow at the front and by the direction in which the Driving Wheels are turning. (b) FBD of the engine and the coaches. Note that the driving wheels, connected to the piston (reciprocating inside the steam chamber), shown as a box) by the connecting rod, are pressing the rails backward thereby getting the traction force Ft, directed forward. It is this traction force that moves the train. The other wheels are subjected to rolling frictions Fr1,Fr2,Fr3,Fr4, directed backward, and trying to slow down the train. T1,T2 are tensions in the chains that connect the engine to coach 1, and coach 1 to coach 2 respectively. (c) FBD of the whole train. (d) FBD of the rails. We have suppressed the gravity forces. In the boxes we have indicated the force equalities. When the train accelerates, the equality symbol “=” becomes “greater than” symbol “>”. 228 Physics Education • July − September 2009 Table 1. Forces at work in diving System Diver Diving board Left pillar Right pillar Earth Total External force on the system R+G1 T+G2+P+R′ T′+G4 P′+G3 G′1+G′2+G′3+G′4 0 (5) Table 2. Forces at work in tug of war System Player-right Player-left Rope Earth Total Net external force on the system −T1+F1+N1−M1g T2−F2+N2−M2g T1 −T2 −F1+F2−N1−N2+M1g+M2g 0 Anyone who applies Newton’s laws of motion must be standing in an inertial frame. The space capsule in which the modern day heroes − the astronauts − are floating is the best example of an inertial frame. In contrast, the earth is not. What are the forces acting on a missile shot out from an earth station? Along with the main force, the force of gravity, there are other forces, the centrifugal force and the Coriolis force. These “other forces” are not forces of interaction. And Newton’s third law of motion does not apply to them. Imagine yourself sitting in a fast accelerating train (better still imagine yourself sitting inside a jet plane as it accelerates along the runway before taking off.) You feel as if someone is pushing you backward, against the chair in which you are sitting. Let us refer to such forces as force type l. Think of another sensation. You are on a rotating platform of a merry −go-round − your first stop on entering a carnival park. If you are sitting on a chair on the go-round, you feel that someone is trying to throw you outward, away Physics Education • July − September 2009 (6) from the centre, but the chair is saving you from such disaster. Even stranger will be your feeling if you try to walk on the platform. Besides the outward “throw out” force, called the centrifugal force, you will also feel a “transverse force” (i.e. perpendicular to the direction you are walking in) making you feel dizzy. This sideward force is called Coriolis force. Let us refer to these two forces as force types 2 and 3) The force types 1,2 and 3, are not forces of action-reaction. They completely disappear when the platform you are on (train, aeroplane, merry-go-round) comes to a stop. People club such forces under fancy names : fictitious forces, pseudo-forces. The pseudo forces are clearly distinct from the real forces. All real forces have parents. They originate from some material source, and reach out to other material objects altering their motions in accordance with Newton’s laws. In contrast, the pseudo-forces do not have parents, i.e., they do not originate from material sources. We mistake them to be forces 229 because we are always thinking our frames to be inertial even when they are not. Physicists do not feel shy of applying Newton’s laws of motion even when sitting in noninertial frames. They make up for their errors by inventing the fictitious pseudo-forces. References 5 Mach’s Principle 1. Pseudo-forces are not as untouchable, bereft of respectability, as their category may indicate. They have a trail of glorious history. It was Ernst Mach’s deep insight that brought pseudoforces on par with real forces by suggesting that the pseudo-forces originate from the mighty galaxies of an accelerating universe (when your frame is accelerating with respect to an inertial frame, the universe is also accelerating backward with respect to your frame.) Albert Einstein was so influenced by this profound message that he removed all distinctions between inertial and non-inertial frames and proposed a theory of gravitation in which pseudo-forces are no different from gravitational forces. Acknowledgment The author gratefully acknowledges the help provided by Michael Murphy and Professor A.V. Gopala Rao help in installing Linux for preparation of the manuscript. 230 There is one full chapter (18 pages long) on FBDs, with a wide variety of examples and illustrated drawings in the following book which the author found in the “References” of the article “Friction” in Wikipedia. http://en.wikipedia.org/wiki/Friction. Andy Ruina, Rudra Pratap, Introduction to Statics and Dynamics, Pre-print for Oxford Univ Press, January 2002. Books on engineering mechanics usually contain many examples of FBDs. See for example the following book. 2. F.P. Beer, E.R. Johnston, Vector Mechanics for Engineers, McGraw Hill, N.Y. (1962). Rolling friction and traction forces are well explained in 3. A.N. Mateev, Mechanics and Theory of Relativity, Mir Publishers, Moscow (1989). The drawings in this article were all drawn by the author. In particular the drawings appearing in Section 1 (Slide Show) and the drawings appearing in Fig. 2 were published in this journal in 1986. 4. S. Datta, A Film Strip on Newton’s Third Law of Motion, Physics Education, Vol 2, No. 4 (January-March 1986), pp 30-40. The drawings appearing in Figures 3 and 4 were part of a Teacher’s Guide handwritten by the author in December 1989 at Kavaratti, capital of Lakshadweep islands, as part of a teacher training programme organized by the Regional College of Education (NCERT), Mysore. Physics Education • July − September 2009