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Lecture 9

Lecture 9. Goals Describe Friction in Air (Ch. 6) Differentiate between Newton’s 1 st , 2 nd and 3 rd Laws Use Newton’s 3 rd Law in problem solving. Assignment: HW4, (Chap. 6 & 7, due 10/5) 1 st Exam Thurs., Oct. 7 th from 7:15-8:45 PM Chapters 1-7

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Lecture 9

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  1. Lecture 9 • Goals • Describe Friction in Air (Ch. 6) • Differentiate between Newton’s 1st, 2nd and 3rd Laws • Use Newton’s 3rd Law in problem solving Assignment: HW4, (Chap. 6 & 7, due 10/5) 1st Exam Thurs., Oct. 7th from 7:15-8:45 PM Chapters 1-7 in rooms 2103, 2141, & 2223 Chamberlin Hall

  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. Free body diagrams mg FB,T= N mg Ball Falls For Static Situation N = mg

  6. Force Pairs 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. Cavendish’s Experiment F = m1 g = G m1 m2 / r2 g = G m2 / r2 If we know big G, little g and r then will can find m2 the mass of the Earth!!!

  10. 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’ = / (4000+40)2 / 40002 = 0.98

  11. 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? 

  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? • You are breaking and the bicycle is slowing down What is the direction of the frictional force?

  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. Exercise 2Newton’s Third 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 3Newton’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?

  16. Force pairs on an Inclined plane Forces on the block “Normal” means perpendicular Normal Force Friction Force f mg sin q q y mg cos q q q x Block weight is mg

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

  18. Force pairs on an Inclined plane Forces on the block (non equilibrium case) Normal Force Friction Force f y Forces on the plane 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. • What is the acceleration of the secondbox ? 1st Question: What is the force on mass 2 from mass 1? (A) a = 0 N (B) a = 5 N (C) a = 20 N(D) can’t tell 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 (A) a = 0 N(B) a = 5 N (C) a = 20 N(D) can’t tell

  22. ExampleSolution • Now consider the FBD of box 2: N2 f2,1 = mkm1g 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. Home Exercise Friction and Motion, Replay • A box of mass m1 = 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 m2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are s=1.5 and mk= 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 ? a1 friction coefficients ms=1.5andmk=0.5 T m1 a2 slides without friction m2

  25. Home Exercise Friction and Motion, Replay in the static case a1 friction coefficients ms=1.5andmk=0.5 T m1 a2 slides without friction m2 • A box of mass m1 = 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 m2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are s=1.5 and mk= 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

  26. Home ExerciseFriction and Motion fs = 10 N and the acceleration of box 1 is Acceleration of box 2 equals that of box 1, with |a| = |T| / (m1+m2) and the frictional force f ism2a (Notice that if T were in excess of 15 N then it would break free) N fSS N = S m1 g = 1.5 x 1 kg x 10 m/s2 which is 15 N (so m2 can’t break free) fS T m1 g a1 friction coefficients ms=1.5andmk=0.5 T m1 a2 slides without friction m2

  27. Recap • Wednesday: Review for exam

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