Chapter 8 Applications of Newton s Second Law

Transcription

1 81 Force Laws 2 Chapter 8 Applications of Newton s Second Law 811 Hooke s Law Principle of Equivalence: Gravitational Force near the Surface of the Earth Electric Chare and Coulomb s Law 7 Example 81 Coulomb s Law and the Universal Law of Gravitation 8 83 Constraint Forces Contact Forces 10 Example 82 Normal Component of the Contact Force and Weiht Kinetic and Static Friction Free-body Force Diaram System Modelin Tension in a Rope Definition of Tension in a Rope 17 Example 83 Tension in a Massive Rope 19 Example 84 Tension in a Suspended Rope Continuous Systems and Newton s Second Law as a Differential Equations Dra Forces in Fluids 26 Example 85 Dra Force at Low Speeds 28 Example 86 Dra Forces at Hih Speeds Worked Examples 32 Example 87 Staircase 32 Example 88 Cart Movin on a Track 34 Example 89 Pulleys and Ropes Constraint Conditions 38 Example 810 Acceleratin Wede 42 Example 811: Capstan 45 Example 812 Free Fall with Air Dra 48

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3 Chapter 8 Applications of Newton s Second Law Those who are in love with practice without knowlede are like the sailor who ets into a ship without rudder or compass and who never can be certain whether he is oin Practice must always be founded on sound theory 1 81 Force Laws Leonardo da Vinci There are forces that don't chane appreciably from one instant to another, which we refer to as constant in time, and forces that don't chane appreciably from one point to another, which we refer to as constant in space The ravitational force on an object near the surface of the earth is an example of a force that is constant in space There are forces that depend on the confiuration of a system When a mass is attached to one end of a sprin, the sprin force actin on the object increases in strenth whether the sprin is extended or compressed There are forces that spread out in space such that their influence becomes less with distance Common examples are the ravitational and electrical forces The ravitational force between two objects falls off as the inverse square of the distance separatin the objects provided the objects are of a small dimension compared to the distance between them More complicated arranements of attractin and repellin interactions ive rise to forces that fall off with other powers of r : constant, 1/ r, 1/ r 2, 1 / r 3,, A force may remain constant in manitude but chane direction; for example the ravitational force actin on a planet underoin circular motion about a star is directed towards the center of the circle This type of attractive central force is called a centripetal force A force law describes the relationship between the force and some measurable property of the objects involved We shall see that some interactions are describable by force laws and other interactions cannot be so simply described 811 Hooke s Law In order to stretch or compress a sprin from its equilibrium lenth, a force must be exerted on the sprin Consider an object of mass m that is lyin on a horizontal surface Attach one end of a sprin to the object and fix the other end of the sprin to a wall Let l 0 denote the equilibrium lenth of the sprin (neither stretched or compressed) Assume 1 Notebooks of Leonardo da Vinci Complete, tr Jean Paul Richter, 1888, Vol1-2

4 that the contact surface is smooth and hence frictionless in order to consider only the effect of the sprin force If the object is pulled to stretch the sprin or pushed to compress the sprin, then by Newton s Third Law the force of the sprin on the object is equal and opposite to the force that the object exerts on the sprin We shall refer to the force of the sprin on the object as the sprin force and experimentally determine a relationship between that force and the amount of stretch or compress of the sprin Choose a coordinate system with the oriin located at the point of contact of the sprin and the object when the sprin-object system is in the equilibrium confiuration Choose the î unit vector to point in the direction the object moves when the sprin is bein stretched Choose the coordinate function x to denote the position of the object with respect to the oriin (Fiure 81) wall l 0 î m equilibrium confiuration+ frictionless surface l 0 x = 0 x î m stretched: x > 0 x = 0 x î m compressed: x < 0 x = 0 Fiure 81 Sprin attached to a wall and an object Initially stretch the sprin until the object is at position x Then release the object and measure the acceleration of the object the instant the object is released The manitude of the sprin force actin on the object is F = m a Now repeat the experiment for a rane of stretches (or compressions) Experiments show that for each sprin, there is a rane of maximum values x max > 0 for stretchin and minimum values x min < 0 for compressin such that the manitude of the measured force is proportional to the stretched or compressed lenth and is iven by the formula F = k x, (811) -3

5 where the sprin constant k has units N m 1 The free-body force diaram is shown in Fiure 82 x î F = F x î = kx î x = 0 Fiure 82 Sprin force actin on object The constant k is equal to the neative of the slope of the raph of the force vs the compression or stretch (Fiure 83) slope = -k F x x min x max x Fiure 83 Plot of x -component of the sprin force F x vs x The direction of the acceleration is always towards the equilibrium position whether the sprin is stretched or compressed This type of force is called a restorin force Let F x denote the x -component of the sprin force Then F x = kx (812) Now perform similar experiments on other sprins For a rane of stretched lenths, each sprin exhibits the same proportionality between force and stretched lenth, althouh the sprin constant may differ for each sprin It would be extremely impractical to experimentally determine whether this proportionality holds for all sprins, and because a modest samplin of sprins has confirmed the relation, we shall infer that all ideal sprins will produce a restorin force, which is linearly proportional to the stretched (or compressed) lenth This experimental relation reardin force and stretched (or compressed) lenths for a finite set of sprins has now been inductively eneralized into the above mathematical model for ideal sprins, a force law known as a Hooke s Law This inductive step, referred to as Newtonian induction, is the critical step that makes physics a predictive science Suppose a sprin, attached to an object of mass m, is 8-4

6 stretched by an amount Δx Use the force law to predict the manitude of the force between the rubber band and the object, F = k Δx, without havin to experimentally measure the acceleration Now use Newton s Second Law to predict the manitude of the acceleration of the object F k Δx a = = (813) m m Carry out the experiment, and measure the acceleration within some error bounds If the manitude of the predicted acceleration disarees with the measured result, then the model for the force law needs modification The ability to adjust, correct or even reject models based on new experimental results enables a description of forces between objects to cover larer and larer experimental domains Many real sprins have been wound such that a force of manitude F 0 must be applied before the sprin beins to stretch The value of F is referred to as the pre-tension of the 0 sprin Under these circumstances, Hooke s law must be modified to account for this pretension, F x = F 0 kx, x > 0 (814) F x = +F 1 kx, x < 0 Note the value of the pre-tension F0 sprin and F 1 may differ for compressin or stretchin a 82 Fundamental Laws of Nature Force laws are mathematical models of physical processes They arise from observation and experimentation, and they have limited ranes of applicability Does the linear force law for the sprin hold for all sprins? Each sprin will most likely have a different rane of linear behavior So the model for stretchin sprins still lacks a universal character As such, there should be some hesitation to eneralize this observation to all sprins unless some property of the sprin, universal to all sprins, is responsible for the force law Perhaps sprins are made up of very small components, which when pulled apart tend to contract back toether This would suest that there is some type of force that contracts sprin molecules when they are pulled apart What holds molecules toether? Can we find some fundamental property of the interaction between atoms that will suffice to explain the macroscopic force law? This search for fundamental forces is a central task of physics In the case of sprins, this could lead into an investiation of the composition and structural properties of the atoms that compose the steel in the sprin We would investiate the eometric properties of the lattice of atoms and determine whether there is some fundamental property of the atoms that create this lattice Then we ask how stable is 8-5

7 this lattice under deformations This may lead to an investiation into the electron confiurations associated with each atom and how they overlap to form bonds between atoms These particles carry chares, which obey Coulomb s Law, but also the Laws of Quantum Mechanics So in order to arrive at a satisfactory explanation of the elastic restorin properties of the sprin, we need models that describe the fundamental physics that underline Hooke s Law 821 Universal Law of Gravitation At points sinificantly far away from the surface of Earth, the ravitational force is no loner constant with respect to the distance to the center of Earth Newton s Universal Law of Gravitation describes the ravitational force between two objects with masses, m 1 and m 2 This force points alon the line connectin the objects, is attractive, and its manitude is proportional to the inverse square of the distance, r 1,2, between the two point-like objects (Fiure 84a) The force on object 2 due to the ravitational interaction between the two objects is iven by G m 1 m = G 2 F1,2 2 r ˆ 1,2, (821) where r = r r is a vector directed from object 1 to object 2, r 1, ,2 = r 1,2, and rˆ = 1,2 r 1,2 / r 1,2 is a unit vector directed from object 1 to object 2 (Fiure 84b) The constant of proportionality in SI units is G = N m 2 k 2 r 1,2 Fiure 84 (a) Gravitational force between two point-like objects Coordinate system for the two-body problem Fiure 84 (b) 822 Principle of Equivalence: The Principle of Equivalence states that the mass that appears in the Universal Law of Gravity is identical to the inertial mass that is determined with respect to the standard 8-6

8 kiloram From this point on, the equivalence of inertial and ravitational mass will be assumed and the mass will be denoted by the symbol m 823 Gravitational Force near the Surface of the Earth Near the surface of Earth, the ravitational interaction between an object and Earth is mutually attractive and has a manitude of where is a positive constant F G = m (822) earth,object The International Committee on Weihts and Measures has adopted as a standard value for the acceleration of an object freely fallin in a vacuum = m s 2 The actual value of varies as a function of elevation and latitude If φ is the latitude and h the elevation in meters then the acceleration of ravity in SI units is = ( cos(2φ) cos 2 (2φ) h) m s 2 (823) This is known as Helmert s equation The strenth of the ravitational force on the standard kiloram at 42 latitude is N k 1, and the acceleration due to ravity at sea level is therefore = m s 2 for all objects At the equator, = 978 m s 2 and at the poles = 983 m s 2 This difference is primarily due to the earth s rotation, which introduces an apparent (fictitious) repulsive force that affects the determination of as iven in Equation (822) and also flattens the spherical shape of Earth (the distance from the center of Earth is larer at the equator than it is at the poles by about 265 km ) Both the manitude and the direction of the ravitational force also show variations that depend on local features to an extent that's useful in prospectin for oil, investiatin the water table, naviatin submered submarines, and as well as many other practical uses Such variations in can be measured with a sensitive sprin balance Local variations have been much studied over the past two decades in attempts to discover a proposed fifth force which would fall off faster than the ravitational force that falls off as the inverse square of the distance between the objects 824 Electric Chare and Coulomb s Law Matter has properties other than mass Matter can also carry one of two types of observed electric chare, positive and neative Like chares repel, and opposite chares attract each other The unit of chare in the SI system of units is called the coulomb [C] The smallest unit of free chare known in nature is the chare of an electron or proton, which has a manitude of e = C (824) 8-7

9 It has been shown experimentally that chare carried by ordinary objects is quantized in interal multiples of the manitude of this free chare The electron carries one unit of neative chare ( q e = e ) and the proton carries one unit of positive chare ( q p = +e ) In an isolated system, the chare stays constant; in a closed system, an amount of unbalanced chare can neither be created nor destroyed Chare can only be transferred from one object to another r 1, 2 Consider two point-like objects with chares q 1 and q 2, separated by a distance in vacuum By experimental observation, the two objects repel each other if they are both positively or neatively chared (Fiure 84a) They attract each other if they are oppositely chared (Fiure 85b) The force exerted on object 2 due to the interaction between objects 1 and 2 is iven by Coulomb's Law, q 1 q F E = k 2 ˆ (825) r 1, 2 e 2 1, 2 r1, 2 where ˆ = / is a unit vector directed from object 1 to object 2, and in SI units, r 1,2 r 1,2 r 1,2 k e = N m 2 C 2, as illustrated in the Fiure 85a This law was derived empirically by Charles Auustin de Coulomb in the late 18 th century Fiure 85 (a) and 85 (b) Coulomb interaction between two chares Example 81 Coulomb s Law and the Universal Law of Gravitation Show that both Coulomb s Law and the Universal Law of Gravitation satisfy Newton s Third Law Solution: To see this, interchane 1 and 2 in the Universal Law of Gravitation to find the force on object 1 due to the interaction between the objects The only quantity to chane sin is the unit vector 8-8

10 Then ˆ = ˆ (826) r 2,1 r 1, 2 m 2 m 1 m 1 m 2 F G = G rˆ = G ˆ = F G (827) 2,1 2 2,1 2 r 1, 2 1, 2 r 2,1 r 1, 2 Coulomb s Law also satisfies Newton s Third Law since the only quantity to chane sin is the unit vector, just as in the case of the Universal Law of Gravitation 83 Constraint Forces Knowlede of all the external and internal forces actin on each of the objects in a system and applyin Newton s Second Law to each of the objects determine a set of equations of motion These equations of motion are not necessarily independent due to the fact that the motion of the objects may be limited by equations of constraint In addition there are forces of constraint that are determined by their effect on the motion of the objects and are not known beforehand or describable by some force law For example: an object slidin down an inclined plane is constrained to move alon the surface of the inclined plane (Fiure 86a) and the surface exerts a contact force on the object; an object that slides down the surface of a sphere until it falls off experiences a contact force until it loses contact with the surface (Fiure 86b); as particles in a sealed vessel are constrained to remain inside the vessel and therefore the wall must exert force on the as molecules to keep them inside the vessel (86c); and a bead constrained to slide outward alon a rotatin rod is actin on by time dependent forces of the rod on the bead (Fiure 86d) We shall develop methods to determine these constraint forces althouh there are many examples in which the constraint forces cannot be determined inclined plane (a) (b)' bead rotatin rod (c) (d)' 8-9

11 Fiure 86 Constrained motions: (a) particle slidin down inclined plane, (b) particles slidin down surface of sphere, (c) as molecules in a sealed vessel, and (d) bead slidin on a rotatin rod 831 Contact Forces Pushin, liftin and pullin are contact forces that we experience in the everyday world Rest your hand on a table; the atoms that form the molecules that make up the table and your hand are in contact with each other If you press harder, the atoms are also pressed closer toether The electrons in the atoms bein to repel each other and your hand is pushed in the opposite direction by the table Accordin to Newton s Third Law, the force of your hand on the table is equal in manitude and opposite in direction to the force of the table on your hand Clearly, if you push harder the force increases Try it If you push your hand straiht down on the table, the table pushes back in a direction perpendicular (normal) to the surface Slide your hand ently forward alon the surface of the table You barely feel the table pushin upward, but you do feel the friction actin as a resistive force to the motion of your hand This force acts tanential to the surface and opposite to the motion of your hand Push downward and forward Try to estimate the manitude of the force actin on your hand The force of the table actin on your hand, F C C, is called the contact force This force has both a normal component to the surface, C N, called the normal force, and a tanential component to the surface, C f, called the friction force (Fiure 86) " C N C C f Fiure 86 Normal and tanential components of the contact force The contact force, written in terms of its component forces, is therefore C= C + C " N+ f (831) Any force can be decomposed into component vectors so the normal component, N, and the tanential component, f, are not independent forces but the vector components of the contact force, perpendicular and parallel to the surface of contact The contact force is a distributed force actin over all the points of contact between your hand and the surface 8-10

12 For most applications we shall treat the contact force as actin at sinle point but precaution must be taken when the distributed nature of the contact force plays a key role in constrainin the motion of a riid body In Fiure 87, the forces actin on your hand are shown These forces include the contact force, C, of the table actin on your hand, the force of your forearm, F forearm, actin on your hand (which is drawn at an anle indicatin that you are pushin down on your hand as well as forward), and the ravitational interaction, F, between the earth and your hand C F forearm F Fiure 87 Forces on hand when movin towards the left One point to keep in mind is that the manitudes of the two components of the contact force depend on how hard you push or pull your hand and in what direction, a characteristic of constraint forces, in which the components are not specified by a force law but dependent on the particular motion of the hand Example 82 Normal Component of the Contact Force and Weiht Hold a block in your hand such that your hand is at rest (Fiure 88) You can feel the weiht of the block aainst your palm But what exactly do we mean by weiht? N F = m Fiure 88 Block restin in hand Fiure 89 Forces on block 8-11

13 There are two forces actin on the block as shown in Fiure 89 One force is the ravitational force between the earth and the block, and is denoted by F = m The other force actin on the block is the contact force between your hand and the block Because our hand is at rest, this contact force on the block points perpendicular to the surface, and hence has only a normal component, N Let N denote the manitude of the normal force Because the object is at rest in your hand, the vertical acceleration is zero Therefore Newton s Second Law states that N+ F = 0 (832) Choose the positive direction to be upwards and then in terms of vertical components we have that N m = 0 (833) which can be solved for the manitude of the normal force N = m (834) When we talk about the weiht of the block, we often are referrin to the effect the block has on a scale or on the feelin we have when we hold the block These effects are actually effects of the normal force We say that a block feels lihter if there is an additional force holdin the block up For example, you can rest the block in your hand, but use your other hand to apply a force upwards on the block to make it feel lihter in your supportin hand The word weiht, is often used to describe the ravitational force that Earth exerts on an object We shall always refer to this force as the ravitational force instead of weiht When you jump in the air, you feel weihtless because there is no normal force actin on you, even thouh Earth is still exertin a ravitational force on you; clearly, when you jump, you do not turn ravity off This example may also ive rise to a misconception that the normal force is always equal to the mass of the object times the manitude of the ravitational acceleration at the surface of the earth The normal force and the ravitational force are two completely different forces In this particular example, the normal force is equal in manitude to the ravitational force and directed in the opposite direction because the object is at rest The normal force and the ravitational force do not form a Third Law interaction pair of forces In this example, our system is just the block and the normal force and ravitational force are external forces actin on the block Let s redefine our system as the block, your hand, and Earth Then the normal force and ravitational force are now internal forces in the system and we can now identify the various interaction pairs of forces We explicitly introduce our interaction pair notation to enable us to identify these interaction pairs: for example, let F E denote,b the ravitational force on the block due to the interaction with Earth The ravitational 8-12

14 force on Earth due to the interaction with the block is denoted by F B, and these two,e forces form an interaction pair By Newton s Third Law, F = F E Note that these,b B,E two forces are actin on different objects, the block and Earth The contact force on the block due to the interaction between the hand and the block is then denoted by N H,B The force of the block on the hand, which we denote by N B,H, satisfies N B, H = N H,B Because we are includin your hand as part of the system, there are two additional forces actin on the hand There is the ravitational force on your hand F, satisfyin E,H F E,H = F H,E, where F H,E is the ravitational force on Earth due to your hand Finally there is the force of your forearm holdin your hand up, which we denote F F,H Because we are not includin the forearm in our system, this force is an external force to the system The forces actin on your hand are shown in diaram on your hand is shown in Fiure 810, and the just the interaction pairin of forces actin on Earth is shown in Fiure 811 (we are not representin all other external forces actin on the Earth) F F,H F H,E F B,E F E,H N B,H Fiure 810 Free-body force diaram on hand Fiure 811 Gravitational forces on earth due to object and hand 832 Kinetic and Static Friction When a block is pulled alon a horizontal surface or slidin down an inclined plane there is a lateral force resistin the motion If the block is at rest on the inclined plane, there is still a lateral force resistin the motion This resistive force is known as dry friction, and there are two distinuishin types when surfaces are in contact with each other The first type is when the two objects are movin relative to each other; the friction in that case is called kinetic friction or slidin friction When the two surfaces are non-movin but there is still a lateral force as in the example of the block at rest on an inclined plane, the force is called, static friction 8-13

15 Leonardo da Vinci was the first to record the results of measurements on kinetic friction over a twenty-year period between and about 1515 Based on his measurements, the force of kinetic friction, f k, between two surfaces, he identified two key properties of kinetic friction The manitude of kinetic friction is proportional to the normal force between the two surfaces, f k = µ k N, (835) where µ k is called the coefficient of kinetic friction The second result is rather surprisin in that the manitude of the force is independent of the contact surface Consider two blocks of the same mass, but different surface areas The force necessary to move the blocks at a constant speed is the same The block in Fiure 812a has twice the contact area as the block shown in Fiure 812b, but when the same external force is applied to either block, the blocks move at constant speed These results of da Vinci were rediscovered by Guillaume Amontons and published in 1699 The third property that kinetic friction is independent of the speed of movin objects (for ordinary slidin speeds) was discovered by Charles Auustin Coulomb F f k F f k (a) Fiure 812 (a) and (b): kinetic friction is independent of the contact area k The kinetic friction on surface 2 movin relative to surface 1 is denoted by, The direction of the force is always opposed to the relative direction of motion of surface 2 relative to the surface 1 When one surface is at rest relative to our choice of reference frame we will denote the friction force on the movin object by f k The second type of dry friction, static friction occurs when two surfaces are static relative to each other Because the static friction force between two surfaces forms a third s law interaction pair, will use the notation to denote the static friction force on surface f 1,2 2 due to the interaction between surfaces 1 and 2 Push your hand forward alon a surface; as you increase your pushin force, the frictional force feels stroner and stroner Try this Your hand will at first stick until you push hard enouh, then your hand slides forward The manitude of the static frictional force, f s, depends on how hard you push If you rest your hand on a table without pushin horizontally, the static friction is zero As you increase your push, the static friction increases until you push hard enouh that your hand slips and starts to slide alon the surface Thus the manitude of static (b) f 1,2 8-14

16 friction can vary from zero to some maximum value, ( f ), when the pushed object s max beins to slip, 0 f ( f ) (836) s s max Is there a mathematical model for the manitude of the maximum value of static friction between two surfaces? Throuh experimentation, we find that this manitude is, like kinetic friction, proportional to the manitude of the normal force ( f ) = µ N (837) s max s Here the constant of proportionality is µ s, the coefficient of static friction This constant is slihtly reater than the constant µ k associated with kinetic friction, µ s > µ k This small difference accounts for the slippin and catchin of chalk on a blackboard, finernails on lass, or a violin bow on a strin The direction of static friction on an object is always opposed to the direction of the applied force (as lon as the two surfaces are not acceleratin) In Fiure 813a, an external force, F, is applied the left and the static friction, f s, is shown pointin to the riht opposin the external force In Fiure 813b, the external force, F, is directed to the riht and the static friction, f s, is now pointin to the left F (a) f s f s (b) F Fiure 813 (a) and (b): External forces and the direction of static friction Althouh the force law for the maximum manitude of static friction resembles the force law for slidin friction, there are important differences: 1 The direction and manitude of static friction on an object always depends on the direction and manitude of the applied forces actin on the object, where the manitude of kinetic friction for a slidin object is fixed 2 The manitude of static friction has a maximum possible value If the manitude of the applied force alon the direction of the contact surface exceeds the manitude of the maximum value of static friction, then the object will start to slip (and be subject to kinetic friction) We call this the just slippin condition 8-15

17 84 Free-body Force Diaram 841 System When we try to describe forces actin on a collection of objects we must first take care to specifically define the collection of objects that we are interested in, which define our system Often the system is a sinle isolated object but it can consist of multiple objects Because force is a vector, the force actin on the system is a vector sum of the individual forces actin on the system F = F +F + (841) 1 2 A free-body force diaram is a representation of the sum of all the forces that act on a sinle system We denote the system by a lare circular dot, a point (Later on in the course we shall see that the point represents the center of mass of the system) We represent each force that acts on the system by an arrow (indicatin the direction of that force) We draw the arrow at the point representin the system For example, the forces that reularly appear in free-body diaram are contact forces, tension, ravitation, friction, pressure forces, sprin forces, electric and manetic forces, which we shall introduce below Sometimes we will draw the arrow representin the actual point in the system where the force is actin When we do that, we will not represent the system by a point in the free-body diaram Suppose we choose a Cartesian coordinate system, then we can resolve the force into its component vectors F =F î + F ĵ+ x F z ˆk (842) Each one of the component vectors is itself a vector sum of the individual component vectors from each contributin force We can use the free-body force diaram to make these vector decompositions of the individual forces For example, the x - component of the force is F x = F 1,x + F 2,x + (843) y 845 Modelin One of the most central and yet most difficult tasks in analyzin a physical interaction is developin a physical model A physical model for the interaction consists of a description of the forces actin on all the objects The difficulty arises in decidin which forces to include For example in describin almost all planetary motions, the Universal Law of Gravitation was the only force law that was needed There were anomalies, for example the small shift in Mercury s orbit These anomalies are interestin because they may lead to new physics Einstein corrected Newton s Law of Gravitation by introducin General Relativity and one of the first successful predictions of the new theory was the perihelion precession of Mercury s orbit On the other hand, the anomalies may simply -16

18 be due to the complications introduced by forces that are well understood but complicated to model When objects are in motion there is always some type of friction present Air friction is often nelected because the mathematical models for air resistance are fairly complicated even thouh the force of air resistance substantially chanes the motion Static or kinetic friction between surfaces is sometimes inored but not always The mathematical description of the friction between surfaces has a simple expression so it can be included without makin the description mathematically intractable A ood way to start thinkin about the problem is to make a simple model, excludin complications that are small order effects Then we can check the predictions of the model Once we are satisfied that we are on the riht track, we can include more complicated effects 85 Tension in a Rope 851 Definition of Tension in a Rope Let s return to our example of the very liht rope (object 2 with m 2 0 ) that is attached to a block (object 1) at the point B, and pulled by an applied force at point A, F A,2 (Fiure 818a) 1 B A 2 F A,2 Choose a coordinate system with the Fiure 818a Massless rope pullin a block ĵ -unit vector pointin upward in the normal direction to the surface, and the î -unit vector pointin in the positive x -direction, (Fiure 818b) The force diarams for the system consistin of the rope and block is shown in Fiure 819, and for the rope and block separately in Fiure 820, where F 2,1 is the force on the block (object 1) due to the rope (object 2), and F 1,2 is the force on the rope due to the block f 1 m 1 N ĵ î B A F A,2 2 m 2 0 +x Fiure 818b Forces actin on system consistin of block and rope 8-17

19 The forces on the rope and the block must each sum to zero Because the rope is not acceleratin, Newton s Second Law applied to the rope requires that F A,2 F 1,2 = m 2 a (where we are usin manitudes for all the forces) f N 1 F 2,1 F 1,2 B m 1 ĵ 2 î m 2 Fiure 819 Separate force diarams for rope and block Because we are assumin the mass of the rope is neliible therefore 0 A F A,2 F A,2 F 1,2 = 0; (massless rope) (851) If we consider the case that the rope is very liht, then the forces actin at the ends of the rope are nearly horizontal Then if the rope-block system is movin at constant speed or at rest, Newton s Second Law is now F A,2 F 1,2 = 0; (constant speed or at rest) (852) Newton s Second Law applied to the block in the +î -direction requires that F 2,1 f = 0 Newton s Third Law, applied to the block-rope interaction pair requires that F 1,2 = F 2,1 Therefore F A,2 = F 1,2 = F 2,1 = f (853) Thus the applied pullin force is transmitted throuh the rope to the block since it has the same manitude as the force of the rope on the block In addition, the applied pullin force is also equal to the friction force on the block How do we define tension at some point in a rope? Suppose make an imainary slice of the rope at a point P, a distance x P from point B, where the rope is attached to the block The imainary slice divides the rope into two sections, labeled L (left) and R (riht), as shown in Fiure 820 B P A L R x P imainary slice Fiure 820 Imainary slice throuh the rope 8-18

20 There is now a Third Law pair of forces actin between the left and riht sections of the rope Denote the force actin on the left section by F R,L (x P ), and the force actin on the riht section by F L,R (x P ) Newton s Third Law requires that the forces in this interaction pair are equal in manitude and opposite in direction F R,L (x P ) = F L,R (x P ) (854) The force diaram for the left and riht sections are shown in Fiure 821 where the force on the left section of the rope due to the block-rope interaction (We had previously denoted that force by F 1,2 ) Now denote the force on the riht section of the rope side due to the pullin force at the point A by F A,R, (which we had previously denoted by F A,2 ) F 1,L is F 1,L B P A F P R,L (x P ) F L,R (x P ) L R F A,R x P Fiure 821 Force diaram for the left and riht sections of rope The tension T (x P ) at a point P in rope lyin a distance x from one the left end of the rope, is the manitude of the action -reaction pair of forces actin at the point P, T (x P ) = (x P ) = (x P ) (855) F R,L F L,R For a rope of neliible mass, under tension, as in the above case, (even if the rope is acceleratin) the sum of the horizontal forces applied to the left section and the riht section of the rope are zero, and therefore the tension is uniform and is equal to the applied pullin force, T = F A,R (856) Example 83 Tension in a Massive Rope B 1 2 A F A,R Fiure 822a Massive rope pullin a block 8-19

21 Consider a block of mass m 1 that is lyin on a horizontal surface The coefficient of kinetic friction between the block and the surface is µ k A uniform rope of mass m 2 and lenth d is attached to the block The rope is pulled from the side opposite the block with an applied force of manitude F A,2 = F A,2 Because the rope is now massive, the pullin force makes an anle φ with respect to the horizontal in order to balance the ravitational force on the rope, (Fiure 822a) Determine the tension in the rope as a function of distance x from the block Solution: In the followin analysis, we shall assume that the anle φ is very small and depict the pullin and tension forces as essentially actin in the horizontal direction even thouh there must be some small vertical component to balance the ravitational forces The key point to realize is that the rope is now massive and we must take in to account the inertia of the rope when applyin Newton s Second Law Consider an imainary slice throuh the rope at a distance x from the block (Fiure 822b), dividin the rope into two sections The riht section has lenth d x and mass m R = (m 2 / d)(d x) The left section has lenth x and mass m L = (m 2 / d)(x) B L R A imainary slice + x O x d x Fiure 822b Imainary slice throuh the rope The free body force diarams for the two sections of the rope are shown in Fiure 822c, where T (x) is the tension in the rope at a distance x from the block, and F 1,L = F 1,L F 1,2 is the manitude of the force on the left-section of the rope due to the rope-block interaction ĵ î F 1,L B T (x) T (x) A L R m L x m R F A, R Fiure 822c Force diaram for the left and riht sections of rope Apply Newton s Second Law to the riht section of the rope yieldin 8-20

22 F A,R T (x) = m R a R = m 2 (d x)a d R, (857) where a R is the x -component of the acceleration of the riht section of the rope Apply Newton s Second Law to the left slice of the rope yieldin T (x) F 1,L = m L a L = (m 2 / d)x a L, (858) where a L is the x -component of the acceleration of the left piece of the rope 1 N F L,1 ĵ î f k m1 Fiure 823 Force diaram on slidin block The force diaram on the block is shown in Fiure 823 Newton s Second Law on the block in the +î -direction is F L,1 f k = m 1 a 1 and in the +ĵ -direction is N m 1 = 0 The kinetic friction force actin on the block is f k = µ k N = µ k m 1 Newton s Second Law on the block in the +î -direction becomes F L,1 µ k m 1 = m 1 a 1, (859) Newton s Third Law for the block-rope interaction is iven by F L,1 = F 1,L Eq (858) then becomes T (x) (µ k m 1 + m 1 a 1 ) = (m 2 / d)xa L (8510) Because the rope and block move toether, the accelerations are equal which we denote by the symbol a a 1 = a L Then Eq (8510) becomes T (x) = µ k m 1 + (m 1 + (m 2 / d)x)a (8511) This result is not unexpected because the tension is acceleratin both the block and the left section and is opposed by the frictional force Alternatively, the force diaram on the system consistin of the rope and block is shown in Fiure

23 f k 1 N B 2 A F A, R m 1 m 2 Fiure 824 Force diaram on block-rope system Newton s Second Law becomes F A,R µ k m 1 = (m 2 + m 1 )a (8512) Solve Eq (8512) for F A,R and substitute into Eq (857), and solve for the tension yieldin Eq (8511) Example 84 Tension in a Suspended Rope A uniform rope of mass M and lenth L is suspended from a ceilin (Fiure 825) The manitude of the acceleration due to ravity is (a) Find the tension in the rope at the upper end where the rope is fixed to the ceilin (b) Find the tension in the rope as a function of the distance from the ceilin (c) Find an equation for the rate of chane of the tension with respect to distance from the ceilin in terms of M, L, and L rope of mass M ĵ Fiure 825 Rope suspended from ceilin Fiure 826 Coordinate system for suspended rope Solution: (a) Bein by choosin a coordinate system with the oriin at the ceilin and the positive y -direction pointin downward (Fiure 826) In order to find the tension at the upper end of the rope, choose as a system the entire rope The forces actin on the rope are the force at y = 0 holdin the rope up, T ( y = 0), and the ravitational force on the entire rope The free-body force diaram is shown in Fiure y 8-22

24 T ( y = 0) ĵ M Fiure 827 Force diaram on rope Because the acceleration is zero, Newton s Second Law on the rope is M T ( y = 0) = 0 Therefore the tension at the upper end is T ( y = 0) = M (b) Recall that the tension at a point is the manitude of the action-reaction pair of forces actin at that point Make an imainary slice in the rope a distance y from the ceilin separatin the rope into an upper sement 1, and lower sement 2 (Fiure 828a) Choose the upper sement as a system with mass m 1 = ( M / L) y The forces actin on the upper sement are the ravitational force, the force T ( y = 0) holdin the rope up, and the tension T ( y) at the point y, that is pullin the upper sement down The free-body force diaram is shown in Fiure 828b (a) (b) T ( y = 0) y y 1 ĵ 1 ĵ L y 2 T ( y) m 1 Fiure 828 (a) Imainary slice separates rope into two pieces (b) Free-body force diaram on upper piece of rope Apply Newton s Second Law to the upper sement: m 1 + T ( y) T ( y = 0) = 0 Therefore the tension at a distance y from the ceilin is T ( y) = T ( y = 0) m 1 Because m 1 = ( M / L) y is the mass of the sement piece and M is the tension at the upper end, Newton s Second Law becomes 8-23

25 T ( y) = M(1 y / L) (8513) As a check, we note that when y = L, the tension T ( y = L) = 0, which is what we expect because there is no force actin at the lower end of the rope (c) Differentiate Eq (8513) with respect to y yieldin dt = ( M / L) (8514) dy The rate that the tension is chanin at a constant rate with respect to distance from the top of the rope 852 Continuous Systems and Newton s Second Law as a Differential Equations We can determine the tension at a distance y from the ceilin in Example 84, by an alternative method, a technique that will eneralize to many types of continuous systems Choose a coordinate system with the oriin at the ceilin and the positive y - direction pointin downward as in Fiure 825 Consider as the system a small element of the rope between the points y and y + Δy This small element has lenth Δy, The small element has mass Δm = ( M / L)Δy and is shown in Fiure 829 ĵ m y y y + y y = L Fiure 829 Small mass element of the rope The forces actin on the small element are the tension, T ( y) at y directed upward, the tension T ( y + Δy) at y + Δy directed downward, and the ravitational force Δm directed downward The tension T ( y + Δy) is equal to the tension T ( y) plus a small difference ΔT, T ( y + Δy) = T ( y) + ΔT (8515) 8-24

26 The small difference in eneral can be positive, zero, or neative The free body force diaram is shown in Fiure 830 T ( y) m ĵ y y y + y T ( y) + T Fiure 830 Free body force diaram on small mass element Now apply Newton s Second Law to the small element Δm + T ( y) (T ( y) + ΔT ) = 0 (8516) The difference in the tension is then ΔT = Δm We now substitute our result for the mass of the element Δm = ( M / L)Δy, and find that that ΔT = ( M / L)Δy (8517) Divide throuh by Δy, yieldin ΔT / Δy = ( M / L) Now take the limit in which the lenth of the small element oes to zero, Δy 0, ΔT lim = ( M / L) (8518) Δy 0 Δy Recall that the left hand side of Eq (8518) is the definition of the derivative of the tension with respect to y, and so we arrive at Eq (8514), dt = ( M / L) dy We can solve the differential equation, Eq (8514), by a technique called separation of variables We rewrite the equation as dt = ( M / L)dy and interate both sides Our interal will be a definite interal in which we interate a dummy interation variable y from y = 0 to y = y and the correspondin T from T = T ( y = 0) to T = T ( y) : T =T ( y) y = y dt = ( M / L) y =0 dy (8519) T '=T ( y=0) 8-25

27 After interation and substitution of the limits, we have that T ( y) T ( y = 0) = ( M / L)y (8520) Us the fact that tension at the top of the rope is T ( y = 0) = M and find that T ( y) = M(1 y / L) in areement with our earlier result, Eq (8513) 86 Dra Forces in Fluids When a solid object moves throuh a fluid it will experience a resistive force, called the dra force, opposin its motion The fluid may be a liquid or a as This force is a very complicated force that depends on both the properties of the object and the properties of the fluid The force depends on the speed, size, and shape of the object It also depends on the density, viscosity and compressibility of the fluid For objects movin in air, the air dra is still quite complicated but for rapidly movin objects the resistive force is rouhly proportional to the square of the speed v, the cross-sectional area A of the object in a plane perpendicular to the motion, the density ρ of the air, and independent of the viscosity of the air Traditional the manitude of the air dra for rapidly movin objects is written as Table 81 Dra Coefficients Shape Dra coefficient Sphere 047 Half-sphere 042 Cone 050 Cube 105 F dra = 2 1 CD Aρv 2 (861) The coefficient C D is called the dra coefficient, a dimensionless number that is a property of the object Table 81 lists the dra coefficient for some simple shapes, (each of these objects has a Reynolds number of order 10 4 ) Anled cube Lon cylinder Short cylinder Streamlined body Streamlined half-body The above model for air dra does not extend to all fluids An object dropped in oil, molasses, honey, or water will fall at different rates due to the different viscosities of the fluid For very low speeds, the dra force depends linearly on the speed and is also proportional to the viscosity η of the fluid For the special case of a sphere of radius R, 8-26

28 the dra force law can be exactly deduced from the principles of fluid mechanics and is iven by F dra = 6πηRv (sphere) (862) This force law is known as Stokes Law The coefficient of viscosity η has SI units of [N m 2 s] = [Pa s] = [k m 1 s 1 ] ; a cs unit called the poise is often encountered Some typical coefficients of viscosity are listed in Table 82 Table 82: Coefficients of viscosity fluid Temperature, 0 C Coefficient of viscosity η ; [k m 1 s 1 ] Acetone Air Benzene Blood 37 (3 4) 10 3 Castor oil Corn Syrup Ethanol Glycerol Methanol Motor oil (SAE 10W) Olive Oil Water Water Water Water This law can be applied to the motion of slow movin objects in a fluid, for example: very small water droplets fallin in a ravitational field, rains of sand settlin in water, or the sedimentation rate of molecules in a fluid In the later case, If we model a molecule as a sphere of radius R, the mass of the molecule is proportional to R 3 and the dra force is proportion to R, therefore different sized molecules will have different rates of acceleration This is the basis for the desin of measurin devices that separate molecules of different molecular weihts In many physical situations the force on an object will be modeled as dependin on the object s velocity We have already seen static and kinetic friction between surfaces modeled as bein independent of the surfaces relative velocity Common experience (swimmin, throwin a Frisbee) tells us that the frictional force between an object and a 8-27

29 fluid can be a complicated function of velocity Indeed, these complicated relations are an important part of such topics as aircraft desin Example 85 Dra Force at Low Speeds marble: mass m and radius R olive oil h Fiure 831 Example 85 A spherical marble of radius R and mass m is released from rest and falls under the influence of ravity throuh a jar of olive oil of viscosity η The marble is released from rest just below the surface of the olive oil, a heiht h from the bottom of the jar The ravitational acceleration is (Fiure 831) Nelect any force due to the buoyancy of the olive oil (i) Determine the velocity of the marble as a function of time, (ii) what is the maximum possible velocity v = v(t = ) (terminal velocity), that the marble can obtain, (iii) determine an expression for the viscosity of olive oil η in terms of, m, R, and v =, (iv) determine an expression for the position of the marble from just below the v surface of the olive oil as a function of time Solution: Choose positive y -direction downwards with the oriin at the initial position of the marble as shown in Fiure 832(a) ĵ O + y v(t) y(t) h (a) Fiure 832 (a) Coordinate system for marble; (b) free body force diaram on marble There are two forces actin on the marble: the ravitational force, and the dra force which is iven by Eq (862) The free body diaram is shown in the Fiure 832(b) Newton s Second Law is then dv m 6πηRv = m, (863) dt O + y F dra m ĵ (b) 8-28

30 where v is the y -component of the velocity of the marble Let γ = 6πηR / m ; the SI units γ are [s 1 ] Then Eq (863) becomes dv γ v =, (864) dt Suppose the object has an initial y -component of velocity v(t = 0) = 0 We shall solve Eq (863) usin the method of separation of variables The differential equation may be rewritten as dv = γ dt (865) (v / γ ) The interal version of Eq (865) is then v =v(t ) v =0 t =t dv = γ dt v / γ (866) t =0 Interatin both sides of Eq (866) yields ln v(t) / γ / γ = γ t (867) Recall that e ln x = x, therefore upon exponentiation of Eq (867) yields v(t) / γ γ t = e (868) / γ Thus the y -component of the velocity as a function of time is iven by γ t m (6πηR/m)t v(t) = (1 e ) = (1 e ) (869) γ 6πηR A plot of v(t) vs t is shown in Fiure 831 with parameters R = m, 1 1 η = k m 1 s, m = k, and / γ = 187 m s 8-29

31 v(t) 5 [m s 1 ] 1 [s] 2 [s] t Fiure 833 Plot of y -component of the velocity v(t) vs t for marble fallin throuh oil with / γ = 187 m s 1 For lare values of t, the term e terminal velocity (6πηR/m)t approaches zero, and the marble reaches a m v = v(t = ) = (8610) 6πηR The coefficient of viscosity can then be determined from the terminal velocity by the condition that m η = (8611) 6π Rv ter Let ρ m denote the density of the marble The mass of the spherical marble is m = (4 / 3)ρ m R 3 The terminal velocity is then 2ρ R 2 m v = (8612) 9η The terminal velocity depends on the square of the radius of the marble, indicatin that larer marbles will reach faster terminal speeds The position of the marble as a function of time is iven by the interal expression t =t y(t) y(t = 0) = v(t )dt, t =0 (8613) which after substitution of Eq (869) and interation usin the initial condition that y(t = 0) = 0, becomes y(t) = t + (e 1) γ t (8614) γ γ