1. 1 7/1/04 Midterm 1: July 9 Will cover material from Chapters 1-6 Go to the room where you usually have recitation July 6 Recitation will be a review session Practice exam available Friday Note: HW 3 is still due on Wed, July 7

2. 2 7/1/04 Review Newton’s 1 st Law: Inertia Newton’s 2 nd Law: F=ma Newton’s 3 rd Law: “Action and Reaction” Forces are vectors Units: Newtons [kg m/s 2 ]

3. 3 7/1/04 Review: Free Body Diagrams The first step in solving any force problem: 1. Sketch the object in question 2. Draw an arrow for each force acting on the object 3. Label each force 4. Indicate the direction of acceleration off to the side (acceleration is NOT a force) a F1F1 F3F3 F2F2

4. 4 7/1/04 Review: How to Solve a Force Problem 1. Draw a free body diagram Guess at forces of unknown magnitude or direction 2. Break forces into components 3. Sum of all the forces in one direction = mass * acceleration in that direction i.e. F net,x =m a x 4. Repeat step 3 for each direction 5. Solve for unknown quantities

5. 5 7/1/04 Mass on a string If we pull steadily on the bottom string, which will break first? A) Top B) Bottom C) It’s a matter of luck

6. 6 7/1/04 Mass on a string F Tension in a string transmits a force along the direction of the string.

7. 7 7/1/04 Mass on a string Free body diagram for bottom string (at rest): F T bottom F T bottom = F

8. 8 7/1/04 Mass on a string Free body diagram for mass m (at rest): F T top mg F T top = F+mg

9. 9 7/1/04 What happens if we pull fast? Which breaks first? A) Top B) Bottom C) It’s a matter of luck

10. 10 7/1/04 Why did the bottom string break? If we attempt to accelerate too fast tension on bottom string becomes too large and string snaps. Newton’s First Law: mass is at rest and cannot accelerate instantaneously to speed of the hand yanking down…

11. 7/1/04 11 Chapter 6 Forces (Part the Second)

12. 12 7/1/04 Dissipative Forces So far - Forces don’t depend on history Direction of motion Velocity of particle Dissipative Forces DO Friction Viscosity (Air Resistance)

13. 13 7/1/04 Frictional Forces Frictional forces are from object's surface interacting with another material Frictional forces always oppose (intended) motion F push F friction

14. 14 7/1/04 Kinetic Friction For an object that is moving, the magnitude of the frictional force is proportional to the magnitude of the normal force on an object:  k is the “coefficient of kinetic friction”

15. 15 7/1/04 Static Friction The maximum magnitude of the static frictional force is proportional to the magnitude of the normal force on an object: Once an object begins to move, kinetic friction takes over.  k is the “coefficient of kinetic friction”

16. 16 7/1/04 Kinetic vs. Static Friction MaterialKineticStatic Glass on Glass0.400.94 Copper on Glass0.530.68 Rubber on Concrete (dry)0.81.0 Rubber on Concrete (wet)0.250.30 Teflon on Teflon0.04 Objects are harder to start moving than to keep moving  s >  k Some typical values: No need to memorize, this will be given when needed

17. 17 7/1/04 Static and Kinetic Friction If you push with steadily increasing force: F friction F push μsNμsN μkNμkN F friction

18. 18 7/1/04 A Classic Demo… Pull the “table cloth” slowly, then quickly

19. 19 7/1/04 Pig on a Frictional Plane What angle should we tilt the plane so that the pig slides?  μ s =0.5 N FgFg F fr y x

20. 20 7/1/04 Pig on a Frictional Plane F g,x = mg sin θ  N FgFg FfFf y x Breaking F g into components: F g,y = mg cos θ x direction: Summing forces in the y direction:

21. 21 7/1/04 Pig on a Frictional Plane  N FgFg FfFf y x We have: So for the pig to slip:

22. 22 7/1/04 Pig on a Frictional Plane  N FgFg FfFf y x x direction: Summing forces in the y direction: Say we tilt the ramp just pass  = 26.6° and a = 2m/s 2. What is μ k ?

23. 23 7/1/04 Pig on a Frictional Plane  N FgFg FfFf y x

24. 24 7/1/04 Heavy Box Say you can bench 500 N (about 110 lbs), can you push a 1000 N box across the floor? Max static friction = μ s N = (0.75)(1000 N) = 750 N F friction μ s = 0.75 and μ k = 0.45 Kinetic friction = μ k N = (0.45)(1000 N) = 450 N Not a chance… No problem!

25. 25 7/1/04 Pulling a Box μ k =0.25 F rope θ x: F rope cosθ - F fr = ma  F rope cosθ - μ k (mg - F rope sinθ) = ma y: F rope sinθ + N - F g = 0  N = mg - F rope sinθ F rope FgFg N F fr a Summing forces:

26. 26 7/1/04 Pulling a Box F rope μ k =0.25 θ To find the maximum:

27. 27 7/1/04 Accelerating a Car The coefficients of friction for rubber on concrete are: Dry:  s =1.0  k =0.80 Wet:  s =0.30  k =0.25 A car of mass 1000 kg tries to accelerate from a stop sign. What is the minimum time to accelerate to 30 m/s on dry pavement? On wet pavement?

28. 28 7/1/04 Accelerating a Car Dry:  s =1.0  t = (30 m/s)/(1.0 × 9.8 m/s 2 ) = 3.06 s Wet:  s =0.30  t = (30 m/s)/(0.30 × 9.8 m/s 2 ) = 10.2 s If a tire does not slip, friction is static Ff=μsNFf=μsN

29. 29 7/1/04 Stopping a Car What are the minimum stopping distances for a 1000 kg car at 30 m/s in the following circumstances? Dry pavement, wheels rolling? (ABS) Dry pavement, wheels locked? Wet pavement, wheels rolling? (ABS) Wet pavement, wheels locked?

30. 30 7/1/04 Stopping a Car Wheels locked: Ff=μkNFf=μkNv 0 =30 m/s Dry: μ k = 0.80  x f = 57.4 m Wet: μ k = 0.25  x f = 183.7 m so…

31. 31 7/1/04 Stopping a car Wheels rolling: Ff=μsNFf=μsN v i =30 m/s Dry: μ s = 1.0  x f = 45.9 m Wet: μ s = 0.30  x f = 153.1 m Same as before, but we replace  k with  s. Distance is shorter with ABS

32. 32 7/1/04 Drag Force Opposes Motion Depends on velocity A correct treatment is complicated, but an approximation of the “drag force” in the case of high velocity is given by:  is the fluid density, C is the “drag coefficient”

33. 33 7/1/04 Terminal Velocity An object stops accelerating when F g =F D FgFg FDFD This is the “terminal velocity”

34. 34 7/1/04 Force and Uniform Circular Motion Object travels around a circle at constant speed a v R Recall: centripetal acceleration Centripetal Force

35. 35 7/1/04 Demo: Water in a bucket

36. 36 7/1/04 Why Didn’t I Get Wet (hopefully...)? a v Accelerates faster than g downward, water cannot fall out and is pushed in a circle.

37. 37 7/1/04 Ferris Wheel Radius: R = 9m Period: T = 20 sec (fast) What is the force on an 80 kg rider from the seat when he is at the top and at the bottom?

38. 38 7/1/04 Ferris Wheel The centripetal force is given by: Recall: Then,

39. 39 7/1/04 Ferris Wheel At the top: At the bottom: N FgFg acac N FgFg acac

40. 40 7/1/04 Rotation with Friction How fast can a car round an unbanked curve (dry pavement) with a radius of 50 m without slipping sideways? R=50 m v FfFf  v max = μ s gR Dry: μ s = 1.0  v max = 1.0×9.8×50 = 22.1 m/s fsfs acac

41. 41 7/1/04 Rotation with Friction How much does the curve need to be banked to provide the centripetal acceleration without using sideways frictional force from the tires? θ N mg y x y: N cosθ = mg  N = mg / cosθ x: N sinθ = ma c but a c = v 2 /R  N sinθ = mv 2 /R Thus:

42. 42 7/1/04 Example: Problem 6.41 A puck of mass m slides on a frictionless table while attached to a hanging cylinder of mass M by a cord through a hole in the table. What speed must the puck travel at to keep the cylinder at rest?