1. Physics 207: Lecture 9, Pg 1 Lecture 9 l Goals  Describe Friction in Air (Ch. 6), (not on 1 st Exam)  Differentiate between Newton’s 1 st, 2 nd and 3 rd Laws  Use Newton’s 3 rd Law in problem solving 1 st Exam Thurs., Oct. 6 th from 7:15-8:45 PM Chapters 1-6 & 7 (“light”, direct applications of the third law) Rooms: 2103 (302, 303, 306, 309, 310, 313) 2141 (304, 307, 308, 312), 2223 (311) Chamberlin Hall (plus quiet room)

2. Physics 207: Lecture 9, Pg 2 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 a force of ~30 Newtons  At low speeds air drag is proportional to v but at high speeds it is v 2  Minimizing drag is often important

3. Physics 207: Lecture 9, Pg 3 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”

4. Physics 207: Lecture 9, Pg 4 Force Pairs vs. Free Body Diagrams Consider the following two cases (a falling ball and ball on table), Compare and contrast Free Body Diagram and Action-Reaction Force Pair sketch

5. Physics 207: Lecture 9, Pg 5 Forces just on a single body (1 st and 2 nd Laws only) mgmgmgmg F B,T = N Ball Falls For Static Situation N = mg

6. Physics 207: Lecture 9, Pg 6 Force Pairs (3 rd Law) 1 st and 2 nd Laws  Free-body diagram Relates force to acceleration 3 rd Law  Action/reaction pairs Shows how forces act between objects F B,E = -mg F B,T = N F E,B = mg F B,E = -mg F E,B = mg F T,B = -N

7. Physics 207: Lecture 9, Pg 7 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)

8. Physics 207: Lecture 9, Pg 8 Note on Gravitational Forces 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:

9. Physics 207: Lecture 9, Pg 9 Example (non-contact) Consider the force on a satellite undergoing projectile motion 40 km above the surface of the earth: 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 Compare: g = G m 2 / 4000 2 g ’ = G m 2 / (4000+40) 2 g ’ / g = 4000 2 / (4000+40) 2 = 0.98

10. Physics 207: Lecture 9, Pg 10 A conceptual question: A flying bird in a 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 Follow up question: So, what is holding the airplane up in the sky?

11. Physics 207: Lecture 9, Pg 11 3 rd Law : Static Friction with a bicycle wheel l You are pedaling hard and the bicycle is speeding up. What is the direction of the frictional force? l You are breaking and the bicycle is slowing down What is the direction of the frictional force?

12. Physics 207: Lecture 9, Pg 12 Static Friction with a bicycle wheel l You are pedaling hard and the bicycle is speeding up. What is the direction of the frictional force? Hint…you are accelerating to the right a = F / m F friction, on B from E is to the right F friction, on E from,B is to the left

13. Physics 207: Lecture 9, Pg 13 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.

14. Physics 207: Lecture 9, Pg 14 Exercise Newton’s 3 rd 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

15. Physics 207: Lecture 9, Pg 15 Exercise Newton’s 3 rd 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)

16. Physics 207: Lecture 9, Pg 16 Exercise 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?

17. Physics 207: Lecture 9, Pg 17 Force pairs on an Inclined plane Forces on the block (static case) Normal Force Friction Force  f=  N x y Forces on the plane by block

18. Physics 207: Lecture 9, Pg 18 Force pairs on an Inclined plane Forces on the block (sliding case, no friction) Normal Force  x y Just one force on the plane by block so if plane is to remain stationary these two components must be offset by other force pairs (N cos  and N sin  along vertical and horizontal)

19. Physics 207: Lecture 9, Pg 19 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. (g is said to be 10 m/s 2 )  What is the acceleration of the second box ? 1 st Question: What is the force on mass 2 from mass 1? m2m2m2m2 T m1m1m1m1  slides with friction (  k =0.5  ) slides without friction a = ? v

20. Physics 207: Lecture 9, Pg 20 Example Solution l First draw FBD of the top box: m1m1 N1N1 m1gm1g T f k =  K N 1 =  K m 1 g v

21. Physics 207: Lecture 9, Pg 21 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 f = 5 N

22. Physics 207: Lecture 9, Pg 22 l Now consider the FBD of box 2: m2m2 f f 2,1 =  k m 1 g= 5 N m2gm2g N2N2 m1gm1g Example Solution

23. Physics 207: Lecture 9, Pg 23 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

24. Physics 207: Lecture 9, Pg 24 Recap l Wednesday: Review for exam