* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Chapter 4 Newton`s Laws of Motion
Survey
Document related concepts
Coriolis force wikipedia , lookup
Hunting oscillation wikipedia , lookup
Frame of reference wikipedia , lookup
Equations of motion wikipedia , lookup
Electromagnetism wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Mechanics of planar particle motion wikipedia , lookup
Classical mechanics wikipedia , lookup
Fundamental interaction wikipedia , lookup
Inertial frame of reference wikipedia , lookup
Mass versus weight wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Fictitious force wikipedia , lookup
Centrifugal force wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Centripetal force wikipedia , lookup
Transcript
Chapter 4 Newton’s Laws of Motion 4.1 Forces and Interactions Fundamental forces. There are four types of fundamental forces: electromagnetic, weak, strong and gravitational. The first two had been successfully unified into electroweak theory and there are ongoing attempts to unify its with strong force. The task proved to be very difficult, but there are good reasons to believe that it is at least possible (for example, the so-called GUT theories). The situation is much worse with regards to gravitational force which is manifests itself not through exchange of participles like other forces, but through curvature of space-time. The string theory (perhaps the most promising candidate of unifying all forces) does describe a way of how to think about gravity (perturbatively), but it is too naive to expect that we will know the final answer any time soon. With regards to our everyday experience the (microscopic) forces and interactions manifest themselves trough (macroscopic) forces and interactions. For example, tension force from a rope, normal or friction forces from a surface or weight due to gravitational attraction. To these types of macroscopic forces we will refer as forces which are 3D vectors with direction and magnitude (representing its strength). Superposition of forces. There might be a number of different forces acting on a given object, and the total force (or net force) is given by ! ⃗ net = ⃗i = F ⃗1 +F ⃗2 +F ⃗ 3 + ... F F (4.1) i where the usual vector addition is used, i.e. " # ! ! ! (Fnetx , Fnety , Fnety ) = Fix , Fiy , Fiz . i 41 i i (4.2) 42 CHAPTER 4. NEWTON’S LAWS OF MOTION Strength of the net force is then Fnet = $ 2 2 2 Fnetx + Fnety + Fnetz %" #2 " #2 " #2 & & ! ! ! Fix + Fix + Fix . =' i i i (4.3) Example 4.1. Three professional wrestlers are fighting over a champion’s belt. The forces are on horizontal plane and have magnitudes and directions: F1 = 250 N and θ1 = 127◦ F2 = 50 N and θ2 = 0◦ F3 = 120 N and θ3 = 270◦ . (4.4) Find the components of the net force on the belt and it magnitude and direction. Step 1: Coordinate system. The 2D coordinate system was already chosen for us in the problem. Step 2: What is given? We know that the three wrestlers apply the following forces to the belt F⃗1 = (cos(127◦) · 250 N, sin(127◦) · 250 N) = (−150 N, 200 N) F⃗2 = (cos(0◦ ) · 50 N, sin(0◦ ) · 50 N) = (50 N, 0 N) F⃗3 = (cos(270◦) · 120 N, sin(270◦) · 120 N) = (0 N, −120 N) (4.5) Step 3: What do we have to find? From Eq. (4.1) the net force is ⃗ net = (−100 N, 80 N) F (4.6) CHAPTER 4. NEWTON’S LAWS OF MOTION and its magnitude and directions are $ (−100 N)2 + (80 N)2 = 128 N Fnet = 80 N θ = arctan + 180◦ = −39◦ + 180◦ = 141◦ . −100 N 43 (4.7) Note that θ had to be adjusted 180◦ due to the range where arctan function is defined. 4.2 Newton’s First Law First Law. A body acted on by no net force, i.e. ! ⃗i = 0 F (4.8) i has a constant velocity (which may be zero) and zero acceleration. (This tendency for a body to continue its motion is know as inertia and is extremely important concept in theory of general relativity.) Example 4.2. In the classic 1950 science-fiction film Rocketship X-M, a space-ship is moving in the vacuum of the outerspace, far from any star or planet, when it engine dies. As a result, the spaceship slows down and stops. What does Newton’s first law say about this scene? This is in a conflict with Newton’s first law. The spaceship should continue to move if there are no forces acting on it (no force from engine since it died and no significant gravitational force since all stars/planets are far away.) Example 4.3. You are driving a Maserati GranTurismo S on a straight testing track at a constant speed of 250 km/h. You pass a 1971 Volkswagen Beetle doing a constant speed 75 km/h. On which car is the net force greater? Since both cars move with constant velocities the net force on each car is zero. Inertial frames. It is important to note that the Newton’s first law is not obeyed in all reference frames (e.g. inside of an accelerating train). Those frames of references where it is obeyed is called inertial frame of reference (e.g. inside a train moving with constant velocity) and for this reason it is sometimes called the law of inertia. (Note that surface of Earth is not exactly an inertial reference frame (why?) but it is pretty close to being inertial.) For inertial reference frames one can easily go from one frame to another using Eqs. (3.62) and (3.63) which makes such frames particularly useful. CHAPTER 4. NEWTON’S LAWS OF MOTION 44 Examples. In which of the following situations is there zero net force on the body? An airplane flying due north as a steady 120 m/s and at a constant altitude? A car driving straight up a hill with a 3o slope at a s constant 90 km/h A hawk circling at a a constant 20 km/h at a constant height of 15 m above an open field? A box with slick, firctionless surface in the back truck as the truck accelerates on a a level road at 5 m/s2 . 4.3 Newton’s Second Law Second Law. If a net external force acts on a body, the body accelerates. The direction of acceleration is the same as the direction of the net force. The mass of the body times the acceleration vector of the body equals to the net force vector, i.e. ! ⃗ i = m⃗a F (4.9) i or ( ⃗ Fi ⃗a = i . (4.10) m Since Eq. (4.9) is a vector equation in 3D it is equivalent to three equations ! Fix = max i ! Fiy = may i ! Fiz = maz . (4.11) i Units. With the help of second law we can now relate the units of force to units of mass and acceleration. Since each equation must have the same units on both sides we see that N = kg · m/s2 . (4.12) So if a 1 kg object moves with acceleration 1 m/s2 then there must be a net fore of 1 N applied to it. In British system of units 1 lb = 4.448 N. (4.13) Example 4.4. A worker applies a constant horizontal force with magnitude 20 N to a box with mass m = 40 kg resting on a level floor with negligible CHAPTER 4. NEWTON’S LAWS OF MOTION 45 friction. What is the acceleration of the box? By using the second law (described by Eq. (4.10)) in 1D and converting the units (described by Eq. (4.12)) we get a= 4.4 20 N = 0.5 m/s2 . 40 kg (4.14) Mass and Weight Weight. Is a gravitational force acting on an object close to the surface of the Earth (to be precisely at the see level) ⃗ = m⃗g . w (4.15) It is a vector, but one often refers to the magnitude of the weight force as weight w = mg (4.16) (or even to mass itself just because one, i.e. m, is simply related to the other, i.e. w.) One should however be careful when weight is measured above or below the sea level (e.g. on the airplane), as the weight force can vary. Example 4.7. A 2.49 × 104 N Rolls-Royce Phantom traveling in the +x direction makes an emergency stop; the x-component of the net force acting on it is −1.83 × 104 N. What is its acceleration? The mass of Rolls-Royce Phantom is m= w 2.49 × 104 N = 2540 kg = g 9.8 m/s2 (4.17) CHAPTER 4. NEWTON’S LAWS OF MOTION 46 and the net force must only be along the direction of motion (call it x-axis) Fx = −1.83 × 104 N. (4.18) (Note that there are two more forces action on the car in vertical direction (weight and normal force), but they must balance each other or otherwise the car would be moving in vertical direction. More on this coming in the next section.) By plugging (4.17) and (4.18) into second law along x-axis (4.11) we get Fx = −7.20 m/s2 . (4.19) a= m 4.5 Newton’s Third Law Third Law. If a body A exerts a force on body B (an “action”), then body B exerts a force on body A (a “reaction”). These two forces have the same magnitude, but are opposite in direction. These two forces act on different bodies. ⃗ A on B = −F ⃗ B on A . F (4.20) Example 4.8. After your sport car breaks down, you start to push it to the nearest repair shop. While the car is starting to move, how does the force exert on the car compare to the force the car exerts on you? How do these forces compare when you are pushing the car along at a constant speed? The magnitude of forces in every action-reaction pair is always the same for any given setup. Thus the force exerted on the car has the same magnitude as the force car exerted on you. This is true when the car is moving with or without acceleration. On the other hand the force that needs to be applied to start moving is (usually) large than the force that needs to be applied to continue motion with constant speed. We will come back to why this is the case in next chapter in context of frictional forces. Example 4.9. An apple sits at rest on a table, in equilibrium (i.e. static). What forces act on the apple? What is the reaction force to each of the forces acting on the apple? What are the action-reaction pairs? CHAPTER 4. NEWTON’S LAWS OF MOTION 47 Apple experiences a gravitational attraction to the Earth (or weight) ⃗ ⃗ table on apple . These Fearth on apple as well as normal force from the table, F forces must balance each other, ⃗ earth on apple + F ⃗ table on apple = 0 or F ⃗ earth on apple = −F ⃗ table on apple . (4.21) F but they do not form an action-reaction pair. Since the Earth attracts apple, the apple must attract the Earth and this is the first action-reaction pair ⃗ earth on apple = −F ⃗ apple on earth . F (4.22) Since table repels the apple, the apple must also repel the earth and this the second action-reaction pair ⃗ table on apple = −F ⃗ apple on table . F (4.23) Example 4.10. A stonemason drags a marble block across a floor by pulling on a (weightless) rope attached to the block. The block is not necessarily in equilibrium. How are the various forces related? What are the action-reaction pairs? CHAPTER 4. NEWTON’S LAWS OF MOTION 48 There are three objects of interest: the block, the rope and the mason. The block is in contact with the rope and so it forms the first action-reaction pair: ⃗ Block on Rope = −F ⃗ Rope on Block . F The rope is in contact with the mason and so it forms the second actionreaction pair: ⃗ Rope on Manson = −F ⃗ Manson on Rope . F Note that if the rope is weightless accelerating, then FManson on Rope > FBlock on Rope . ⃗ Floor on Manson and F ⃗ Floor on Block How do you think the friction forces F Examples. You are driving a car on a country road when a mosquito splatters on the windshield. Which has the greater magnitude: the force that the car exerts on the mosquito or the force that the mosquito exerted on the car? Or are the magnitudes the same? If they are the different, how can you reconcile this fact with Newton’s third law? If they are equal, why is mosquito splattered and damaged while the car is undamaged? 4.6 Free-body diagram Free-body diagram. A useful tool in solving problems on Newton’s laws is to draw free-body diagram, where a chosen body appears by itself without of its surroundings, and with vectors drawn to show the magnitude and directions of all the forces that act on the body. the tricky part here is to include all of the forces that act on a chosen body, but not to include any other forces that act on other bodies.