Physics 101: Chapter 4 Newton's Laws

Physics 101: Chapter 4 Newton's Laws Textbook sections 4.1-4.12 Concepts of Mass and Force Newton s Three Laws Gravity Normal Force Friction Tension But first, let s review last lecture.. UB, Phy101: Chapter 4, Pg 1

Summary of Lecture 4 1. 2D Kinematics 2. Projectile Motion x = x 0 + v 0 t v = v 0x y = y 0 + v 0y t - 1/2 gt 2 v y = v 0y -gt v y2 = v 0y2-2g y UB, Phy101: Chapter 4, Pg 2

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Newton s s First Law The motion of an object does not change unless it is acted upon by a net force. If v=0, it remains 0 If v is some value, it stays at that value Another way to say the same thing: No net force velocity is constant acceleration is zero DEMO UB, Phy101: Chapter 4, Pg 5

Mass or Inertia Mass (m) is the property of an object that measures how hard it is to change its motion. Units: [M] = kg UB, Phy101: Chapter 4, Pg 6

Newton s s Second Law This law tells us how motion changes when a force is applied Demo In words. acceleration = (net force)/mass r r F in symbols: a = = M r F M net r Ftot = M alternate way to write it r r F = M a net : UB, Phy101: Chapter 4, Pg 7

Newton s s Second Law r F net = r M a Units: [F] = [M] [a] [F] = kg-m/s 2 1 Newton (N) 1 kg-m/s 2 A vector equation: F net,x = Ma x F net,y = Ma y UB, Phy101: Chapter 4, Pg 8

Chapter 4, Pre-Flight Questions An airplane is flying from Willard airport to O'Hare. Many forces act on the plane, including weight (gravity), drag (air resistance), the trust of the engine, and the lift of the wings. At some point during its trip the velocity of the plane is measured to be constant (which means its altitude is also constant). At this time, the total force on the plane: 1. is pointing upward 2. is pointing downward 3. is pointing forward 4. is pointing backward 5. is zero correct lift drag thrust weight UB, Phy101: Chapter 4, Pg 9

Chapter 4, Pre-Flight Questions (great answers) Newton's first law states that if no net force acts on an object, then the velocity of the object remains unchanged. Since at some point during the trip, the velocity is constant, then the total force on the plane must be zero, according to Newton's first law. lift ΣF= ma = m0 = 0 drag thrust weight the upward and downward forces on the plane would have to be equal for it to fly at a constant altitude, and the same is true of the forward and backward forces for the plane to go at a constant speed. UB, Phy101: Chapter 4, Pg 10

Chapter 4, Pre-Flight (common misconceptions) The total force on the plane is pointing upward otherwise the plane wouldn't be in the air. With the force of gravity, there is a constant acceleration downwards. Therefore the total force is only downwards The plane is still traveling forward, so the net force on the plane must be forward. Because we all love physics All other forces would be at equilibrium at constant velocity and altitude, except since the plane is still moving forward it still faces air resistance UB, Phy101: Chapter 4, Pg 11

Chapter 4, Pre-Flight Question You are watching an old episode of Vampires from Outer Space, your favorite Sci-Fi TV show, when you see the following scene: A starship is shown cruising through space with a constant velocity, it's engines turned on full blast. As the starship nears the space station it wants to visit, the captain turns the engines off and the ship is shown gliding to a stop. Looking at this through the eyes of a physicist, briefly explain what things are wrong with this scene. First, if the engines are on full blast, the starship is accelerating. Second, once the engines are turned off, there isn't a force in space acting to slow the star ship down. Third, vampires don't come from outer space. Fourth, I don't have any favorite Sci-Fi shows. First of all, if I am watching this particular show, it's way too late to be thinking about physics (obviously the show comes on at a rather late hour). Neglecting that fact, I, as a physicist, would notice that the ship would not be slowing down at all if this were a real situation. In space there are no opposing forces that would stop the craft. It would keep right on going into oblivion. Therefore, I would immediately get up and go to bed feeling disgusted that I had ever liked the show in the first place. Vampires have never been proven to exist, nor is it known whether they have space travel; this is yet another mistake (despite its irrelevance to physics). UB, Phy101: Chapter 4, Pg 12

Example 1 F 1 M M=10 kg F 1 =200 N Find a a = F net /M = 200N/10kg = 20 m/s 2 F 1 M F 2 M=10 kg F 1 =200 N F 2 = 100 N Find a a = F net /M = (200N-100N)/10kg = 10 m/s 2 UB, Phy101: Chapter 4, Pg 13

Example 2 A force F acting on a mass m 1 results in an acceleration a 1. The same force acting on a different mass m 2 results in an acceleration a 2 = 2a 1. What is the mass m 2? m 1 m 2 F a 1 F a 2 = 2a 1 (a) 2m 1 (b) m 1 (c) 1/2 m 1 F=ma F= m 1 a 1 = m 2 a 2 = m 2 (2a 1 ) Therefore, m 2 = m 1 /2 Or in words twice the acceleration means half the mass UB, Phy101: Chapter 4, Pg 14

Newton s s Third Law For every action, there is an equal and opposite reaction. F finger box Finger pushes on box F finger box = force exerted on box by finger F box finger Box pushes on finger F box finger = force exerted on finger by box Third Law: F box finger = - F finger box UB, Phy101: Chapter 4, Pg 15

Newton's Third Law... F A,B = - F B,A. is true for all types of forces F w,m F m,w F m,f F f,m UB, Phy101: Chapter 4, Pg 17

Example of Bad Thinking Since F m,b = -F b,m why isn t F net = 0, and a = 0? a?? F b,m F m,b ice UB, Phy101: Chapter 4, Pg 18

Example of Good Thinking Consider only the box! F on box = ma box = F m,b Free Body Diagram (more on this later) What about forces on man? a box F b,m F m,b ice UB, Phy101: Chapter 4, Pg 19

Chapter 4, Pre-Flight Questions Suppose you are an astronaut in outer space giving a brief push to a spacecraft whose mass is bigger than your own (see Figure 4.8 in text). 1) Compare the magnitude of the force you exert on the spacecraft, F S, to the magnitude of the force exerted by the spacecraft on you, F A, while you are pushing: 1. F A = F S 2. F A > F S 3. F A < F S correct Third Law! 2) Compare the magnitudes of the acceleration you experience, a A, to the magnitude of the acceleration of the spacecraft, a S, while you are pushing: 1. a A = a S 2. a A > a S 3. a A < a S correct a=f/m F same lower mass give larger a UB, Phy101: Chapter 4, Pg 20

Summary: Newton s First Law: The motion of an object does not change unless it is acted on by a net force Newton s Second Law: F net = ma Newton s Third Law: F a,b = -F b,a UB, Phy101: Chapter 4, Pg 21

Lecture 6, Pre-Flight Questions 1-61 A B What are forces on A? What are forces on B? How does net force on A compare with net force on B? The answers: F ha >F BA F BA =F AB F B >F A UB, Phy101: Chapter 4, Pg 22

Newton sat in an orchard, and an apple, plumping down on his head, started a train of thought which opened the heavens to us. Had it been in California, the size of the apples there would have saved him the trouble of much thinking thereafter, perhaps, opening the heavens to him, and not to us. [clipped from "TheCourier-Journal," Louisville, KY] UB, Phy101: Chapter 4, Pg 24

Forces: 1. Gravity F 2,1 F 1,2 m 2 m 1 r 12 F 1,2 = force on m 1 due to m 2 = G m1m2 = F 2,1 = force on m 2 due to m 1 2 r12 Direction: on line connecting the masses; attractive G = universal gravitation constant = 6.67 x 10-11 N-m 2 /kg 2 Example: two 1-kg masses separated by 1 m Force = 6.67 x 10-11 N (very weak, but this holds the universe together!) UB, Phy101: Chapter 4, Pg 25

Gravity and Weight m mass on surface of Earth M e R e Force on mass: GM e g = m = gm = R e F 2 mg using M e = 5.98 x 10 GM g = R 24 e 2 e kg and R g = 9.81 m/s 2 e = 6.38 x 10 6 m } g F g W = mg UB, Phy101: Chapter 4, Pg 27

Forces: 2. Normal Force F N W book at rest on table: What are forces on book? Weight is downward System is in equilibrium (acceleration = 0 net force = 0) Therefore, weight balanced by another force F N = normal force = force exerted by surface on object F N is always perpendicular to surface and outward For this example F N = W UB, Phy101: Chapter 4, Pg 29

Forces: 2. Normal Force F N = normal force = force exerted by surface on object F N is always perpendicular to surface and outward (away from surface) Another Example: F hand F N Book at rest on table: Push down with F hand What is F N? W F N -W -F hand = 0 F N = W+F hand UB, Phy101: Chapter 4, Pg 30

Normal Force-- --more examples What is direction of F N? Always perpendicular to surface normal UB, Phy101: Chapter 4, Pg 31

Chapter 4, Pre-Flight Questions Suppose a box sits next to you on the floor of an elevator. During which of the following situations is the normal force exerted on the box by the floor of the elevator smallest? [see text 4.8] 1. When the elevator is accelerating upward. 2. When the elevator is moving upward with constant speed. 3. When the elevator is stationary. 4. When the elevator is moving downward with constant speed. 5. When the elevator is accelerating downward. correct Elevator Demo Fn = mg + ma. If the acceleration in downward, then a is negative. This will make F-n the smallest. Objects experiencing a net force accelerate. Therefore, for an object to accelerate downward, its net force must also be in that direction, meaning the normal force is less than the force of gravity. UB, Phy101: Chapter 4, Pg 33

y ΣF y = ma y F N -W = ma y a F N F N = W + ma y If a < 0 Then F N < W W Note: if free fall, a = -g Then F N = W- mg = 0! UB, Phy101: Chapter 4, Pg 34

Forces: 3. Sliding Friction f k F N direction of motion F Friction demo W Sliding Friction (aka Kinetic Friction): A force, f k, between two surfaces that opposes relative motion. f k = µ k F N µ k = coefficient of kinetic friction a property of the two surfaces UB, Phy101: Chapter 4, Pg 36

Forces: 3. Static Friction F N f s F W Static Friction: A force, f s, between two surfaces that prevents relative motion. f s µ s F N µ s = coefficient of static friction a property of the two surfaces UB, Phy101: Chapter 4, Pg 39

Chapter 4, Pre-Flight Questions Now suppose the box next to you on the elevator floor is annoying you, and you want to push it out of the way. Because of static friction, you need to push on the box with a minimum force F before it will start to move. During which of the following situations is the force F required to move the box smallest? [see text 4.9] 1. When the elevator is accelerating upward. 2. When the elevator is moving upward with constant speed. 3. When the elevator is stationary. 4. When the elevator is moving downward with constant speed. 5. When the elevator is accelerating downward. correct The normal force is minimized when the elevator accelerates downward, minimizing the friction on the box. UB, Phy101: Chapter 4, Pg 40

Forces: 4. Tension T Tension: force exerted by a rope (or string) Magnitude: same everywhere in rope Not changed by pulleys Direction: same as direction of rope. UB, Phy101: Chapter 4, Pg 41

Forces: 4. Tension example: box hangs from a rope attached to ceiling y T ΣF y = ma y T - W = ma y Elevator Demo W T = W + ma y In this case a y = 0 So T = W UB, Phy101: Chapter 4, Pg 42

Chapter 4, Pre-Flight Questions How often do you go to office hours? 1. At least once per week 2. From time to time but less than once per week 3. Never Office hours are a great opportunity for one-on-one teaching. Take advantage of it whenever you think it is necessary. UB, Phy101: Chapter 4, Pg 43

Solving Problems Identify forces using Free Body Diagram This is the most important step! Set up axes x and y Write F net =ma for each axis Solve! Strong suggestion:» work problem algebraically» plug in numbers only at the end UB, Phy101: Chapter 4, Pg 44

Chapter 4, Pre-Flight Questions You are driving a car up a hill with constant velocity. On a piece of paper, draw a Free Body Diagram (FBD) for the car. How many forces are acting on the car? 1 2 3 correct 4 5 F N f V W UB, Phy101: Chapter 4, Pg 45

Chapter 4, Pre-Flight Question The net force on the car is 1. Zero correct 2. Pointing up the hill 3. Pointing down the hill 4. Pointing vertically downward 5. Pointing vertically upward F N f V W f F N W ΣF = ma = 0 UB, Phy101: Chapter 4, Pg 46

Chapter 4, Pre-Flight Questions You are driving a car up a hill with constant acceleration. How many forces are acting on the car? 1 2 3 correct 4 5 F N f a W UB, Phy101: Chapter 4, Pg 47

Chapter 4, Pre-Flight Question The net force on the car is now: 1. Zero 2. Pointing up the hill correct 3. Pointing down the hill 4. Pointing vertically downward 5. Pointing vertically upward F N f a W f F N W ΣF = ma = up the hill UB, Phy101: Chapter 4, Pg 48

Lets do some examples Inclined Plane with Friction Example And more examples (bed time reading) UB, Phy101: Chapter 4, Pg 49