Newton s Laws of Motion

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1 Chapter 4 Newton s Laws of Moton 4.1 Forces and Interactons Fundamental forces. There are four types of fundamental forces: electromagnetc, weak, strong and gravtatonal. The frst two had been successfully unfed nto electroweak theory and there are ongong attempts to unfy ts wth strong force. The task proved to be very dffcult, but there are good reasons to beleve that t s at least possble (for example, the so-called GUT theores). The stuaton s much worse wth regards to gravtatonal force whch s manfests tself not through exchange of partcples lke other forces, but through curvature of space-tme. The strng theory (perhaps the most promsng canddate of unfyng all forces) does descrbe a way of how to thnk about gravty (perturbatvely), but t s too nave to expect that we wll know the fnal answer any tme soon. Wth regards to our everyday experence the (mcroscopc) forces and nteractons manfest themselves trough (macroscopc) forces and nteractons. For example, tenson force from a rope, normal or frcton forces from a surface or weght due to gravtatonal attracton. To these types of macroscopc forces we wll refer as forces whch are 3D vectors wth drecton and magntude (representng ts strength). Superposton of forces. There mght be a number of dfferent forces actng on a gven object, and the total force (or net force) s gven by F net = F = F 1 + F 2 + F (4.1) where the usual vector addton s used,.e. ( (F netx,f nety,f nety )= F x, 41 F y, F z ). (4.2)

2 CHAPTER 4. NEWTON S LAWS OF MOTION 42 Strength of the net force s then ( ) 2 ( ) 2 ( ) 2 F net = Fnetx 2 + Fnety 2 + Fnetz 2 = F x + F x + F x. (4.3) Example 4.1. Three professonal wrestlers are fghtng over a champon s belt. The forces are on horzontal plane and have magntudes and drectons: F 1 = 250N and θ 1 =127 F 2 = 50N and θ 2 =0 F 3 = 120N and θ 3 =270. (4.4) Fnd the components of the net force on the belt and t magntude and drecton. Step 1: Coordnate system. The 2D coordnate system was already chosen for us n the problem. Step 2: What s gven? We know that the three wrestlers apply the followng forces to the belt F 1 = (cos(127 ) 250 N, sn(127 ) 250 N) =( 150 N, 200 N) F 2 = (cos(0 ) 50 N, sn(0 ) 50 N) =(50N, 0 N) F 3 = (cos(270 ) 120 N, sn(270 ) 120 N) =(0N, 120 N) (4.5) Step 3: What do we have to fnd? From Eq. (4.1) thenetforces F net =( 100 N, 80 N) (4.6)

3 CHAPTER 4. NEWTON S LAWS OF MOTION 43 and ts magntude and drectons are F net = ( 100 N) 2 +(80N) 2 =128N θ = arctan 80 N 100 N +180 = =141. (4.7) Note that θ had to be adjusted 180 due to the range where arctan functon s defned. 4.2 Newton s Frst Law Frst Law. Abodyactedonbynonetforce,.e. F =0 (4.8) has a constant velocty (whch may be zero) and zero acceleraton. (Ths tendency for a body to contnue ts moton s know as nerta and s extremely mportant concept n theory of general relatvty.) Example 4.2. In the classc 1950 scence-fcton flm Rocketshp X-M, aspace-shpsmovngnthevacuumoftheouterspace,farfrom any star or planet, when t engne des. As a result, the spaceshp slows down and stops. What does Newton s frst law say about ths scene? Ths s n a conflct wth Newton s frst law. The spaceshp should contnue to move f there are no forces actng on t (no force from engne snce t ded and no sgnfcant gravtatonal force snce all stars/planets are far away.) Example 4.3. You are drvng a Maserat GranTursmo S on a straght testng track at a constant speed of 250 km/h. You pass a 1971 Volkswagen Beetle dong a constant speed 75 km/h. On whch car s the net force greater? Snce both cars move wth constant veloctes the net force on eachcars zero. Inertal frames. It s mportant to note that the Newton s frst law s not obeyed n all reference frames (e.g. nsde of an acceleratng tran). Those frames of references where t s obeyed s called nertal frame of reference (e.g. nsde a tran movng wth constant velocty) and for ths reason t s sometmes called the law of nerta. (Note that surface of Earth s not exactly an nertal reference frame (why?) but t s pretty close to beng nertal.) For nertal reference frames one can easly go from one frame toanother usng Eqs. (3.62) and(3.63) whchmakessuchframespartcularlyuseful.

4 CHAPTER 4. NEWTON S LAWS OF MOTION 44 Examples. In whch of the followng stuatons s there zero net force on the body? An arplane flyng due north as a steady 120 m/s and at a constant alttude? Acardrvngstraghtupahllwtha3 o slope at a s constant 90 km/h Ahawkcrclngataaconstant20 km/h at a constant heght of 15 mabove an open feld? A box wth slck,frctonless surface n the back truck as the truck accelerates on a a level road at 5 m/s Newton s Second Law Second Law. If a net external force acts on a body, the body accelerates. The drecton of acceleraton s the same as the drecton of the net force. The mass of the body tmes the acceleraton vector of the body equals to the net force vector,.e. F = m a (4.9) or F a = m. (4.10) Snce Eq. (4.9) savectorequatonn3dtsequvalenttothreeequatons F x = ma x F y = ma y F z = ma z. (4.11) Unts. Wth the help of second law we can now relate the unts of force to unts of mass and acceleraton. Snce each equaton must have the same unts on both sdes we see that N = kg m/s 2. (4.12) So f a 1 kg object moves wth acceleraton 1 m/s 2 then there must be a net fore of 1 Nappledtot.InBrtshsystemofunts 1 lb =4.448 N. (4.13) Example 4.4. Aworkerapples aconstanthorzontalforcewthmagntude 20 Ntoaboxwthmassm =40kg restng on a level floor wth neglgble

5 CHAPTER 4. NEWTON S LAWS OF MOTION 45 frcton. What s the acceleraton of the box? By usng the second law (descrbed by Eq. (4.10)) n 1D and convertng the unts (descrbed by Eq. (4.12)) we get 4.4 Mass and Weght a = 20 N 40 kg =0.5 m/s2. (4.14) Weght. Is a gravtatonal force actng on an object close to the surface of the Earth (to be precsely at the see level) w = m g. (4.15) It s a vector, but one often refers to the magntude of the weght force as weght w = mg (4.16) (or even to mass tself just because one,.e. m, ssmplyrelatedtotheother,.e. w.) One should however be careful when weght s measured above or below the sea level (e.g. on the arplane), as the weght force canvary. Example 4.7. A NRolls-RoycePhantomtravelngnthe+x drecton makes an emergency stop; the x-component of the net force actng on t s N. What s ts acceleraton? The mass of Rolls-Royce Phantom s m = w g = N 9.8 m/s 2 =2540kg (4.17)

6 CHAPTER 4. NEWTON S LAWS OF MOTION 46 and the net force must only be along the drecton of moton (call t x-axs) F x = N. (4.18) (Note that there are two more forces acton on the car n vertcal drecton (weght and normal force), but they must balance each other or otherwse the car would be movng n vertcal drecton. More on ths comng n the next secton.) By pluggng (4.17) and(4.18) nto second law along x-axs (4.11) weget 4.5 Newton s Thrd Law a = F x m = 7.20 m/s2. (4.19) Thrd Law. If a body A exerts a force on body B (an acton ), then body B exerts a force on body A (a reacton ). These two forces have the same magntude, but are opposte n drecton. These two forces act on dfferent bodes. F A on B = F B on A. (4.20) Example 4.8. After your sport car breaks down, you start to push t to the nearest repar shop. Whle the car s startng to move, how doestheforce exert on the car compare to the force the car exerts on you? How do these forces compare when you are pushng the car along at a constant speed? The magntude of forces n every acton-reacton par s always the same for any gven setup. Thus the force exerted on the car has the same magntude as the force car exerted on you. Ths s true when the car smovng wth or wthout acceleraton. On the other hand the force that needs to be appled to start movng s (usually) large than the force that needstobe appled to contnue moton wth constant speed. We wll come back to why ths s the case n next chapter n context of frctonal forces. Example 4.9. An apple sts at rest on a table, n equlbrum (.e. statc). What forces act on the apple? What s the reacton force to each oftheforces actng on the apple? What are the acton-reacton pars?

7 CHAPTER 4. NEWTON S LAWS OF MOTION 47 Apple experences a gravtatonal attracton to the Earth (or weght) F earth on apple as well as normal force from the table, F table on apple. These forces must balance each other, F earth on apple + F table on apple =0 or F earth on apple = F table on apple. (4.21) but they do not form an acton-reacton par. Snce the Earth attracts apple, the apple must attract the Earth and ths s the frst acton-reacton par F earth on apple = F apple on earth. (4.22) Snce table repels the apple, the apple must also repel the earth and ths the second acton-reacton par F table on apple = F apple on table. (4.23) Example Astonemasondragsamarbleblockacrossafloorby pullng on a (weghtless) rope attached to the block. The block s not necessarly n equlbrum. How are the varous forces related? What are the acton-reacton pars?

8 CHAPTER 4. NEWTON S LAWS OF MOTION 48 There are three objects of nterest: the block, the rope and the mason. The block s n contact wth the rope and so t forms the frst acton-reacton par: F Block on Rope = F Rope on Block. The rope s n contact wth the mason and so t forms the second actonreacton par: F Rope on Manson = F Manson on Rope. Note that f the rope s weghtless acceleratng, then F Manson on Rope >F Block on Rope. How do you thnk the frcton forces F Floor on Manson and F Floor on Block Examples. You are drvng a car on a country road when a mosquto splatters on the wndsheld. Whch has the greater magntude: the force that the car exerts on the mosquto or the force that the mosquto exerted on the car? Or are the magntudes the same? If they are the dfferent, how can you reconcle ths fact wth Newton s thrd law? If they are equal, why s mosquto splattered and damaged whle the car s undamaged? 4.6 Free-body dagram Free-body dagram. AusefultoolnsolvngproblemsonNewton slaws s to draw free-body dagram, where a chosen body appears by tself wthout of ts surroundngs, and wth vectors drawn to show the magntude and drectons of all the forces that act on the body. the trcky part here s to nclude all of the forces that act on a chosen body, but not to nclude any other forces that act on other bodes.

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