Lecture 9

1. Lecture 9 • Goals • Describe Friction in Air (Ch. 6), (not on 1st Exam) • Differentiate between Newton’s 1st, 2nd and 3rd Laws • Use Newton’s 3rd Law in problem solving 1st Exam Thurs., Oct. 6th 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. Friction in a viscous mediumDrag Force Quantified • With a cross sectional area, A (in m2), 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  Av2 c A v2in Newtons c = ¼ kg/m3 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 m2 exerts a force of ~30 Newtons • At low speeds air drag is proportional to v but at high speeds it is v2 • Minimizing drag is often important

3. Newton’s Third Law: If object 1 exerts a force on object 2 (F2,1) then object 2 exerts an equal and opposite force on object 1 (F1,2) F1,2 = -F2,1 For every “action” there is an equal and opposite “reaction” IMPORTANT: Newton’s 3rd law concerns force pairs which act ontwo different objects(not on the same object) !

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. Forces just on a single body (1st and 2nd Laws only) mg FB,T= N mg Ball Falls For Static Situation N = mg

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

7. Example (non-contact) FB,E = - mB g FB,E = - mB g FE,B = mB g FE,B = mB g EARTH Consider the forces on an object undergoing projectile motion Question: By how much does g change at an altitude of 40 miles? (Radius of the Earth ~4000 mi)

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 m1and m2 are separated by distance r, each object “pulls” on the other with a force given by Newton’s law of gravity, as follows:

9. Example (non-contact) FB,E = - mB g FB,E = - mB g FE,B = mB g FE,B = mB g EARTH Consider the force on a satellite undergoing projectile motion 40 km above the surface of the earth: Compare: g= G m2 / 40002 g’ = G m2 / (4000+40)2 g’ / g = 40002 / (4000+40)2 = 0.98

10. A conceptual question: A flying bird in a 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? 

11. 3rd Law : Static Friction with a bicycle wheel • You are pedaling hard and the bicycle is speeding up. What is the direction of the frictional force? • You are breaking and the bicycle is slowing down What is the direction of the frictional force?

12. Static Friction with a bicycle wheel • 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 Ffriction, on B from E is to the right Ffriction, on E from,B is to the left

13. ExerciseNewton’s Third Law  • greater than • equal to • 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. ExerciseNewton’s 3rd Law  • greater than • equal to • less than 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 that of the fly.

15. Exercise Newton’s 3rd LawSolution 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 Fnet = ma or a = Fnet/m. So Fb, f = -Ff, b = F0 but |abus | = |F0 / mbus | << | afly | = | F0/mfly | Answer for acceleration is (C)

16. Exercise Newton’s 3rd Law a b • 2 • 4 • 6 • Something else • Two blocks are being pushed by a finger on a horizontal frictionless floor. • How many action-reaction force pairs are present in this exercise?

17. Force pairs on an Inclined plane Forces on the block (static case) Normal Force Friction Force f= mN Forces on the plane by block y q x

18. Force pairs on an Inclined plane Forces on the block (sliding case, no friction) Normal Force 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 cosq and N sin q along vertical and horizontal) y q x

19. Example: Friction and Motion v • A box of mass m1 = 1 kg is being pulled by a horizontal string having tension T = 40 N. It slides with friction (mk= 0.5) on top of a second box having mass m2 = 2 kg, which in turn slides on a smooth (frictionless) surface. (g is said to be 10 m/s2) • What is the acceleration of the secondbox ? 1st Question: What is the force on mass 2 from mass 1? slides with friction (mk=0.5) T m1 a = ? m2 slides without friction

20. ExampleSolution v • First draw FBD of the top box: N1 m1 fk = mKN1 = mKm1g T m1g

21. ExampleSolution • As we just saw, this force is due to friction: • Newtons 3rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2. Action Reaction f2,1= -f1,2 f1,2 = mKm1g = 5 N m1 m2 f = 5 N

22. ExampleSolution • Now consider the FBD of box 2: N2 f2,1 = mkm1g= 5 N m2 m1g m2g

23. ExampleSolution • Finally, solve Fx = ma in the horizontal direction: mK m1g = m2a = 2.5 m/s2 f2,1 = mKm1g m2

24. Recap • Wednesday: Review for exam