Lecture 8. Static and Kinetic Friction

Transcription

1 Lecture 8 Goals: Solve 1D motion with friction Differentiate between Newton s 1 st, 2 nd and 3 rd Laws Begin to use Newton s 3 rd Law in problem solving Assignment: HW4, (Chapter 6, due 2/17, Wednesday) Finish Chapter 7 1 st Exam Wed., Feb. 17 th from 7:15-8:45 PM Chapters 1-6 in room 2103 Chamberlin Hall Physics 207: Lecture 8, Pg 1 Static and Kinetic Friction Friction exists between objects and its behavior has been modeled. At Static Equilibrium: A block, mass m, with a horizontal force F applied, Direction: A force vector to the normal force vector N and the vector is opposite to the direction of acceleration if µ were 0. Magnitude: f is proportional to the applied forces such that f s µ s N µ s called the coefficient of static friction Physics 207: Lecture 8, Pg 2 Page 1

2 Dynamics: Case study... big F x-axis i : ma x = F µ K N y-axis j : ma y = 0 = N mg or N = mg so F µ K mg = m a x f k v N j F f k µ K mg mg ma x i Physics 207: Lecture 8, Pg 3 Dynamics: Case study... little F x-axis i : ma x = F µ K N y-axis j : ma y = 0 = N mg or N = mg so F µ K mg = m a x f k v N j F i f k µ K mg mg ma x Physics 207: Lecture 8, Pg 4 Page 2

3 Friction: Static friction Static equilibrium: A block with a horizontal force F applied, Σ F x = 0 = -F + f s f s = F FBD Σ F y = 0 = - N + mg N = mg As F increases so does f s F mg N m 1 f s Physics 207: Lecture 8, Pg 5 Static friction, at maximum (just before slipping) Equilibrium: A block, mass m, with a horizontal force F applied, Direction: A force vector to the normal force vector N and the vector is opposite to the direction of acceleration if µ were 0. Magnitude: f S is proportional to the magnitude of N f s = µ s N F N m f s mg Physics 207: Lecture 8, Pg 6 Page 3

4 Kinetic or Sliding friction (f k < f s ) Dynamic equilibrium, moving but acceleration is still zero FBD Σ F x = 0 = -F + f k f k = F Σ F y = 0 = - N + mg N = mg As F increases f k remains nearly constant (but now there acceleration is acceleration) v F mg N m 1 f k f k = µ k N Physics 207: Lecture 8, Pg 7 Sliding Friction: Modeling Direction: A force vector to the normal force vector N and the vector is opposite to the velocity. Magnitude: f k is proportional to the magnitude of N f k = µ k N ( = µ K mg in the previous example) The constant µ k is called the coefficient of kinetic friction Logic dictates that µ S > µ K for any system Physics 207: Lecture 8, Pg 8 Page 4

5 Coefficients of Friction Material on Material steel / steel add grease to steel metal / ice brake lining / iron tire / dry pavement tire / wet pavement µ s = static friction µ k = kinetic friction Physics 207: Lecture 8, Pg 9 An experiment Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find µ S Static equilibrium: Set m 2 and add mass to m 1 to reach the breaking point. Requires two FBDs Mass 1 Σ F y = 0 = T m 1 g T m 1 m 1 g Mass 2 Σ F x = 0 = -T + f s Σ F y = 0 = N m 2 g f S N T = -T + µ S N m 2 m 2 g T = m 1 g = µ S m 2 g µ S = m 1 /m 2 Physics 207: Lecture 8, Pg 10 Page 5

6 A 2 nd experiment Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find µ K. Dynamic equilibrium: Set m 2 and adjust m 1 to find place when a = 0 and v 0 Requires two FBDs Mass 1 Σ F y = 0 = T m 1 g T m 1 m 1 g Mass 2 Σ F x = 0 = -T + f f Σ F y = 0 = N m 2 g f k N T = -T + µ k N m 2 m 2 g T = m 1 g = µ k m 2 g µ k = m 1 /m 2 Physics 207: Lecture 8, Pg 11 An experiment (with a 0) Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find µ K. N T Non-equilibrium: Set m 2 and adjust m 1 to find regime where a 0 Requires two FBDs T m 1 m 1 g f k m 2 m 2 g Mass 1 Mass 2 Σ F y = m 1 a = T m 1 g Σ F x = m 2 a = -T + f k = -T + µ k N Σ F y = 0 = N m 2 g T = m 1 g + m 1 a = µ k m 2 g m 2 a µ k = (m 1 (g+a)+m 2 a)/m 2 g Physics 207: Lecture 8, Pg 12 Page 6

7 Sample Problem You have been hired to measure the coefficients of friction for the newly discovered substance jelloium. Today you will measure the coefficient of kinetic friction for jelloium sliding on steel. To do so, you pull a 200 g chunk of jelloium across a horizontal steel table with a constant string tension of 1.00 N. A motion detector records the motion and displays the graph shown. What is the value of µ k for jelloium on steel? Physics 207: Lecture 8, Pg 13 Sample Problem Σ F x =ma = F - f f = F - µ k N = F - µ k mg Σ F y = 0 = N mg µ k = (F - ma) / mg & x = ½ a t m = ½ a 4 s 2 a = 0.40 m/s 2 µ k = ( ) / ( ) = 0.46 Physics 207: Lecture 8, Pg 14 Page 7

8 Inclined plane with Normal and Frictional Forces 1. Static Equilibrium Case 2. Dynamic Equilibrium (see 1) 3. Dynamic case with non-zero acceleration Normal means perpendicular Normal Force Σ F = 0 F x = 0 = mg sin θ f F y = 0 = mg cos θ + N with mg sin θ = f µ S N if mg sin θ > µ S N, must slide Critical angle µ s = tan θ c f Friction Force θ mg sin θ mg cos θ θ Block weight is mg θ Physics 207: Lecture 8, Pg 15 y x Inclined plane with Normal and Frictional Forces 1. Static Equilibrium Case 2. Dynamic Equilibrium Friction opposite velocity (down the incline) v Normal means perpendicular Normal Force Σ F = 0 f k F x = 0 = mg sin θ f k F y = 0 = mg cos θ + N = µ k N = µ k mg cos θ f K F x = 0 = mg sin θ µ k mg cos θ Friction Force µ k = tan θ (only one angle) mg sin θ mg cos θ θ θ mg θ Physics 207: Lecture 8, Pg 16 y x Page 8

9 Inclined plane with Normal and Frictional Forces 3. Dynamic case with non-zero acceleration Result depends on direction of velocity v Friction Force Sliding Down Normal Force θ mg sin θ f k Sliding Up θ F x = ma x = mg sin θ ± f k F y = 0 = mg cos θ + N f k = µ k N = µ k mg cos θ F x = ma x = mg sin θ ± µ k mg cos θ a x = g sin θ ± µ k g cos θ Weight of block is mg Physics 207: Lecture 8, Pg 17 The inclined plane coming and going (not static): the component of mg along the surface > kinetic friction Σ F x = ma x = mg sin θ ± u k N Σ F y = ma y = 0 = -mg cos θ + N Putting it all together gives two different accelerations, a x = g sin θ ± u k g cos θ. A tidy result but ultimately it is the process of applying Newton s Laws that is key. Physics 207: Lecture 8, Pg 18 Page 9

10 Velocity and acceleration plots: Notice that the acceleration is always down the slide and that, even at the turnaround point, the block is always motion although there is an infinitesimal point at which the velocity of the block passes through zero. At this moment, depending on the static friction the block may become stuck. Physics 207: Lecture 8, Pg 19 Friction in a viscous medium Drag Force Quantified With a cross sectional area, A (in m 2 ), coefficient of drag of 1.0 (most objects), ρ sea-level density of air, and velocity, v (m/s), the drag force is: D = ½ C ρ A v 2 c A v 2 in Newtons c = ¼ kg/m 3 In falling, when D = mg, then at terminal velocity Example: Bicycling at 10 m/s (22 m.p.h.), with projected area of 0.5 m 2 exerts ~30 Newtons Minimizing drag is often important Physics 207: Lecture 8, Pg 20 Page 10

11 Fish Schools Physics 207: Lecture 8, Pg 21 By swimming in synchrony in the correct formation, each fish can take advantage of moving water created by the fish in front to reduce drag. Fish swimming in schools can swim 2 to 6 times as long as individual fish. Physics 207: Lecture 8, Pg 22 Page 11

12 Free Fall Terminal velocity reached when F drag = F grav (= mg) For 75 kg person with a frontal area of 0.5 m 2, v term 50 m/s, or 110 mph which is reached in about 5 seconds, over 125 m of fall Physics 207: Lecture 8, Pg 23 Trajectories with Air Resistance Baseball launched at 45 with v = 50 m/s: Without air resistance, reaches about 63 m high, 254 m range With air resistance, about 31 m high, 122 m range Vacuum trajectory vs. air trajectory for 45 launch angle. Physics 207: Lecture 8, Pg 24 Page 12

13 Newton s Laws Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. Law 2: For any object, F NET = Σ F = ma Law 3: Forces occur in pairs: F A, B = - F B, A (For every action there is an equal and opposite reaction.) Read: Force of B on A Physics 207: Lecture 8, Pg 25 Newton s Third Law: If object 1 exerts a force on object 2 (F 2,1 ) then object 2 exerts an equal and opposite force on object 1 (F 1,2 ) F 1,2 = -F 2,1 For every action there is an equal and opposite reaction IMPORTANT: Newton s 3 rd law concerns force pairs which act on two different objects (not on the same object)! Physics 207: Lecture 8, Pg 26 Page 13

14 Gravity Newton also recognized that gravity is an attractive, long-range force between any two objects. When two objects with masses m 1 and m 2 are separated by distance r, each object pulls on the other with a force given by Newton s law of gravity, as follows: Physics 207: Lecture 8, Pg 27 Cavendish s Experiment F = m 1 g = G m 1 m 2 / r 2 g = G m 2 / r 2 If we know big G, little g and r then will can find m 2 the mass of the Earth!!! Physics 207: Lecture 8, Pg 28 Page 14

15 Example (non-contact) Consider the forces on an object undergoing projectile motion EARTH F B,E = - m B g F E,B = m B g F B,E = - m B g F E,B = m B g Question: By how much does g change at an altitude of 40 miles? (Radius of the Earth ~4000 mi) Physics 207: Lecture 8, Pg 29 Example (non-contact) Consider the forces on an object undergoing projectile motion EARTH F B,E = - m B g F E,B = m B g F B,E = - m B g F E,B = m B g Compare: g = G m 2 / g = G m 2 / ( ) 2 g / g = / ( ) 2 / = 0.98 Physics 207: Lecture 8, Pg 30 Page 15

16 The flying bird in the cage You have a bird in a cage that is resting on your upward turned palm. The cage is completely sealed to the outside (at least while we run the experiment!). The bird is initially sitting at rest on the perch. It decides it needs a bit of exercise and starts to fly. Question: How does the weight of the cage plus bird vary when the bird is flying up, when the bird is flying sideways, when the bird is flying down? Follow up question: So, what is holding the airplane up in the sky? Physics 207: Lecture 8, Pg 31 Exercise Newton s Third Law A fly is deformed by hitting the windshield of a speeding bus. v The force exerted by the bus on the fly is, A. greater than B. equal to C. less than that exerted by the fly on the bus. Physics 207: Lecture 8, Pg 32 Page 16

17 Exercise Newton s Third Law A fly is deformed by hitting the windshield of a speeding bus. v The force exerted by the bus on the fly is, B. equal to that exerted by the fly on the bus. Physics 207: Lecture 8, Pg 33 Exercise 2 Newton s Third Law Same scenario but now we examine the accelerations A fly is deformed by hitting the windshield of a speeding bus. v The magnitude of the acceleration, due to this collision, of the bus is A. greater than B. equal to C. less than that of the fly. Physics 207: Lecture 8, Pg 34 Page 17

18 Exercise 2 Newton s Third Law Solution By Newton s third law these two forces form an interaction pair which are equal (but in opposing directions). Thus the forces are the same However, by Newton s second law F net = ma or a = F net /m. So F b, f = -F f, b = F 0 but a bus = F 0 / m bus << a fly = F 0 /m fly Answer for acceleration is (C) Physics 207: Lecture 8, Pg 35 Exercise 3 Newton s 3rd Law Two blocks are being pushed by a finger on a horizontal frictionless floor. How many action-reaction force pairs are present in this exercise? a b A. 2 B. 4 C. 6 D. Something else Physics 207: Lecture 8, Pg 36 Page 18

19 Exercise 3 Solution: F a,f F f,a F b,a F a,b F g,a a F E,a F g,b b F E,b F a,g F b,g 6 F a,e F b,e Physics 207: Lecture 8, Pg 37 Lecture 8 Recap Assignment: HW4, (Chapter 6, due 2/17, Wednesday) Finish Chapter 7 1 st Exam Wed., Feb. 17 th from 7:15-8:45 PM Chapters 1-6 in room 2103 Chamberlin Hall Physics 207: Lecture 8, Pg 38 Page 19