2. Physics 207: Lecture 8, Pg 1 Lecture 8 l Goals:  Differentiate between Newton’s 1 st, 2 nd and 3 rd Laws  Use Newton’s 3 rd Law in problem solving Assignment: HW4, (Chapters 6 & 7, due 2/18, Wednesday) Finish Chapter 7 1 st Exam Wed., Feb. 18 th from 7:15-8:45 PM Chapters 1-7 in room 2103 Chamberlin Hall

3. Physics 207: Lecture 8, Pg 2 Inclined plane with “Normal” and Frictional Forces 1.Static Equilibrium Case 2.Dynamic Equilibrium (see 1) 3.Dynamic case with non-zero acceleration Block weight is mg Normal Force Friction Force “Normal” means perpendicular   mg cos  f  x y mg sin   F = 0 F x = 0 = mg sin  – f F y = 0 = mg cos  – N with mg sin  = f ≤  S N if mg sin  >  S N, must slide Critical angle  s = tan  c

4. Physics 207: Lecture 8, Pg 3 Inclined plane with “Normal” and Frictional Forces 1.Static Equilibrium Case 2.Dynamic Equilibrium Friction opposite velocity (down the incline) mg Normal Force Friction Force “Normal” means perpendicular   mg cos  fKfK  x y mg sin   F = 0 F x = 0 = mg sin  – f k F y = 0 = mg cos  – N f k =  k N =  k mg cos  F x = 0 = mg sin  –  k mg cos    k = tan   (only one angle) v

5. Physics 207: Lecture 8, Pg 4 Inclined plane with “Normal” and Frictional Forces 3. Dynamic case with non-zero acceleration Result depends on direction of velocity Weight of block is mg Normal Force Friction Force Sliding Down   v mg sin  f k Sliding Up F x = ma x = mg sin  ± f k F y = 0 = mg cos  – N f k =  k N =  k mg cos  F x = ma x = mg sin  ±  k mg cos  a x = g sin  ±  k g cos 

6. Physics 207: Lecture 8, Pg 5 The inclined plane coming and going (not static): the component of mg along the surface > kinetic friction Putting it all together gives two different accelerations, a x = g sin  ± u k g cos . A tidy result but ultimately it is the process of applying Newton’s Laws that is key.  F x = ma x = mg sin  ± u k N  F y = ma y = 0 = -mg cos  + N

7. Physics 207: Lecture 8, Pg 6 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.

8. Physics 207: Lecture 8, Pg 7 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),  sea-level density of air, and velocity, v (m/s), the drag force is: D = ½ C  A v 2  c A v 2 in Newtons c = ¼ kg/m 3 In falling, when D = mg, then at terminal velocity l 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

9. Physics 207: Lecture 8, Pg 8 “Free” Fall l Terminal velocity reached when F drag = F grav (= mg) l 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

10. Physics 207: Lecture 8, Pg 9 Trajectories with Air Resistance l 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.

11. Physics 207: Lecture 8, Pg 10 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. FFa Law 2: For any object, F NET =   F = ma FF 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

12. Physics 207: Lecture 8, Pg 11 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 IMPORTANT: Newton’s 3 rd law concerns force pairs which act on two different objects (not on the same object) ! For every “action” there is an equal and opposite “reaction”

13. Physics 207: Lecture 8, Pg 12 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:

14. Physics 207: Lecture 8, Pg 13 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!!!

15. Physics 207: Lecture 8, Pg 14 Example (non-contact) Consider the forces on an object undergoing projectile motion F B,E = - m B g EARTH 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)

16. Physics 207: Lecture 8, Pg 15 Example Consider the following two cases (a falling ball and ball on table), Compare and contrast Free Body Diagram and Action-Reaction Force Pair sketch

17. Physics 207: Lecture 8, Pg 16 Example The Free Body Diagram mgmgmgmg F B,T = N Ball Falls For Static Situation N = mg

18. Physics 207: Lecture 8, Pg 17 Normal Forces Certain forces act to keep an object in place. These have what ever force needed to balance all others (until a breaking point). F T,B F B,T Main goal at this point : Identify force pairs and apply Newton’s third law

19. Physics 207: Lecture 8, Pg 18 Example First: Free-body diagram Second: Action/reaction pair forces F B,E = -mg F B,T = N F E,B = mg F B,E = -mg F E,B = mg F T,B = -N

20. Physics 207: Lecture 8, Pg 19 The flying bird in the cage l 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? l So, what is holding the airplane up in the sky?

21. Physics 207: Lecture 8, Pg 20 Exercise Newton’s Third Law A. greater than B. equal to C. less than A fly is deformed by hitting the windshield of a speeding bus.  v The force exerted by the bus on the fly is, that exerted by the fly on the bus.

22. Physics 207: Lecture 8, Pg 21 Exercise 2 Newton’s Third Law A. greater than B. equal to C. less than 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 that of the fly. Same scenario but now we examine the accelerations

23. Physics 207: Lecture 8, Pg 22 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).  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 | Thus the forces are the same Answer for acceleration is (C)

24. Physics 207: Lecture 8, Pg 23 Exercise 3 Newton’s 3rd Law A. 2 B. 4 C. 6 D. Something else a b l Two blocks are being pushed by a finger on a horizontal frictionless floor. l How many action-reaction force pairs are present in this exercise?

25. Physics 207: Lecture 8, Pg 24 6 Exercise 3 Solution: a b F F a,f F F f,a F F b,a F F a,b F g,a F a,g F g,b F b,g F E,a F a,E F E,b F b,E

26. Physics 207: Lecture 8, Pg 25 Example: Friction and Motion l A box of mass m 1 = 1 kg is being pulled by a horizontal string having tension T = 40 N. It slides with friction (  k = 0.5) on top of a second box having mass m 2 = 2 kg, which in turn slides on a smooth (frictionless) surface.  What is the acceleration of the second box ? Central question: What is force on mass 2? (A) a = 0 N (B) a = 5 N (C) a = 20 N (D) can’t tell m2m2m2m2 T m1m1m1m1  slides with friction (  k =0.5  ) slides without friction a = ? v

27. Physics 207: Lecture 8, Pg 26 Example Solution l First draw FBD of the top box: m1m1 N1N1 m1gm1g T f k =  K N 1 =  K m 1 g v

28. Physics 207: Lecture 8, Pg 27 l Newtons 3 rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2. m1m1 f f 1,2 =  K m 1 g = 5 N m2m2 f= f f 2,1 = -f 1,2 l As we just saw, this force is due to friction: Example Solution Action Reaction (A) a = 0 N (B) a = 5 N (C) a = 20 N (D) can’t tell

29. Physics 207: Lecture 8, Pg 28 l Now consider the FBD of box 2: m2m2 f f 2,1 =  k m 1 g m2gm2g N2N2 m1gm1g Example Solution

30. Physics 207: Lecture 8, Pg 29 l Finally, solve F x = ma in the horizontal direction: m2m2 f f 2,1 =  K m 1 g  K m 1 g = m 2 a Example Solution = 2.5 m/s 2

31. Physics 207: Lecture 8, Pg 30 Home Exercise Friction and Motion, Replay A box of mass m 1 = 1 kg, initially at rest, is now pulled by a horizontal string having tension T = 10 N. This box (1) is on top of a second box of mass m 2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are  s =1.5 and  k = 0.5. The second box can slide freely (frictionless) on an smooth surface. Compare the acceleration of box 1 to the acceleration of box 2 ? m2m2m2m2 T m1m1m1m1  friction coefficients  s =1.5 and  k =0.5 slides without friction a2a2 a1a1

32. Physics 207: Lecture 8, Pg 31 Home Exercise Friction and Motion, Replay in the static case A box of mass m 1 = 1 kg, initially at rest, is now pulled by a horizontal string having tension T = 10 N. This box (1) is on top of a second box of mass m 2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are  s =1.5 and  k = 0.5. The second box can slide freely on an smooth surface (frictionless). If no slippage then the maximum frictional force between 1 & 2 is (A) 20 N (B) 15 N (C) 5 N (D) depends on T m2m2m2m2 T m1m1m1m1  friction coefficients  s =1.5 and  k =0.5 slides without friction a2a2 a1a1

33. Physics 207: Lecture 8, Pg 32 Home Exercise Friction and Motion f s = 10 N and the acceleration of box 1 is Acceleration of box 2 equals that of box 1, with |a| = |T| / (m 1 +m 2 ) and the frictional force f is m 2 a (Notice that if T were in excess of 15 N then it would break free) m2m2m2m2 T m1m1m1m1  friction coefficients  s =1.5 and  k =0.5 slides without friction a2a2 a1a1 T g m 1 g N fSfS f S   S N =  S m 1 g = 1.5 x 1 kg x 10 m/s 2 which is 15 N (so m 2 can’t break free)

34. Physics 207: Lecture 8, Pg 33 Lecture 8 Recap Assignment: HW4, (Chapters 6 & 7, due 2/18, Wednesday) Finish Chapter 7 1 st Exam Wed., Feb. 18 th from 7:15-8:45 PM Chapters 1-7 in room 2103 Chamberlin Hall