Midterm 1: July 9

1. 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. Review • Newton’s 1st Law: Inertia • Newton’s 2nd Law: F=ma • Newton’s 3rd Law: “Action and Reaction” • Forces are vectors • Units: Newtons [kg m/s2]

3. a Review: Free Body Diagrams The first step in solving any force problem: • Sketch the object in question • Draw an arrow for each force acting on the object • Label each force • Indicate the direction of acceleration off to the side (acceleration is NOT a force) F2 F1 F3

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

5. 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. F Mass on a string Tension in a string transmits a force along the direction of the string.

7. F Mass on a string Free body diagram for bottom string (at rest): Tbottom F Tbottom = F

8. F Mass on a string Free body diagram for mass m (at rest): Ttop F mg Ttop = F+mg

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

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

11. Chapter 6 Forces (Part the Second)

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

13. Fpush Ffriction Frictional Forces Frictional forces are from object's surface interacting with another material Frictional forces always oppose (intended) motion

14. 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: mk is the “coefficient of kinetic friction”

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

16. Kinetic vs. Static Friction Objects are harder to start moving than to keep moving ms > mk Some typical values: No need to memorize, this will be given when needed

17. Ffriction Fpush Ffriction μsN μkN Fpush Static and Kinetic Friction If you push with steadily increasing force:

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

19. y x N Ffr μs=0.5 q Fg Pig on a Frictional Plane What angle should we tilt the plane so that the pig slides?

20. y x N Ff q Fg Pig on a Frictional Plane Summing forces in the y direction: x direction: Breaking Fg into components: Fg,x = mg sin θ Fg,y = mg cos θ

21. y x N Ff q Fg Pig on a Frictional Plane So for the pig to slip: We have:

22. y x N Ff q Fg Pig on a Frictional Plane Summing forces in the y direction: x direction: Say we tilt the ramp just pass q = 26.6° and a = 2m/s2. What is μk?

23. y x N Ff q Fg Pig on a Frictional Plane

24. Ffriction Heavy Box μs = 0.75 and μk= 0.45 Say you can bench 500 N (about 110 lbs), can you push a 1000 N box across the floor? Max static friction = μsN = (0.75)(1000 N) = 750 N Not a chance… Kinetic friction = μkN = (0.45)(1000 N) = 450 N No problem!

25. Frope Frope N θ Ffr μk=0.25 Fg a Pulling a Box Summing forces: y: Fropesinθ + N - Fg = 0  N = mg - Fropesinθ x: Fropecosθ - Ffr = ma Fropecosθ - μk(mg - Fropesinθ) = ma

26. Pulling a Box Frope θ μk=0.25 To find the maximum:

27. 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. Ff=μsN Accelerating a Car If a tire does not slip, friction is static Dry: s=1.0  t = (30 m/s)/(1.0 × 9.8 m/s2) = 3.06 s Wet: s=0.30  t = (30 m/s)/(0.30 × 9.8 m/s2) = 10.2 s

29. 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. v0=30 m/s Ff=μkN Stopping a Car so… Wheels locked: Dry: μk = 0.80  xf = 57.4 m Wet: μk = 0.25  xf = 183.7 m

31. vi=30 m/s Ff=μsN Stopping a car Same as before, but we replace mk with ms. Wheels rolling: Dry: μs = 1.0  xf = 45.9 m Wet: μs = 0.30  xf = 153.1 m Distance is shorter with ABS

32. 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: r is the fluid density, C is the “drag coefficient”

33. Terminal Velocity FD An object stops accelerating when Fg=FD Fg This is the “terminal velocity”

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

35. Demo: Water in a bucket

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

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

38. Ferris Wheel The centripetal force is given by: Recall: Then,

39. N ac Fg N ac Fg Ferris Wheel At the top: At the bottom:

40. v Ff  vmax = μsgR Dry: μs = 1.0  vmax = 1.0×9.8×50 = 22.1 m/s Rotation with Friction How fast can a car round an unbanked curve (dry pavement) with a radius of 50 m without slipping sideways? fs ac R=50 m

41. N θ mg y x 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? y: N cosθ = mg  N = mg / cosθ x: N sinθ = mac but ac = v2/R  N sinθ = mv2/R Thus:

42. 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?