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Static Friction: no surface motion.

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Presentation on theme: "Static Friction: no surface motion."— Presentation transcript:

1 Static Friction: no surface motion.
ch.5

2 Friction is dependent on FN --- Not on contact area
Friction is dependent on FN --- Not on contact area ch.5

3 Kinetic & Rolling Friction: surface motion
Sliding Block: ch.5

4 ch.5

5 A pickup truck accelerates smoothly and a box in back does not slide
A pickup truck accelerates smoothly and a box in back does not slide. Is there surface motion as defined? Yes No 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 ch.5

6 As defined, coefficients of friction are
Dependent on normal force. Dependent on materials used. Dependent on both the above. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 ch.5

7 Block at rest. Draw a Force Diagram for the block.
ch.5

8 Three boxes are pushed by force F with v > 0 along a horizontal surface with mk = 0.291.
F=26N 3kg 5kg 2kg fk ch.5

9 Motion Along a Curved Path
Motion Along a Curved Path ch.5

10 ch.5

11 ch.5

12 Assume the car has m = 1200kg and ms = 0.92.
Assume the car has m = 1200kg and ms = 0.92. At what speed would the car begin to slide? ch.5

13 Which force diagram applies to the object at Point D?

14 5-4 Numerical Integration: Euler’s Method
Skip 5-4 Numerical Integration: Euler’s Method ch.5

15 End ch.5

16 Both have same centripetal acceleration
A disk rotates at a constant rate. Consider two points on the disk, one at radius R/2 and one at radius R. Which point has larger centripetal acceleration? Point at R/2 Point at R Both have same centripetal acceleration 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 ch.5

17 Is it possible for an object to have a non-zero acceleration when the velocity is constant?
Yes No Cannot be determined 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 ch.5

18 Maximum angle block remains at rest:
Derive the Angle of Repose relation: ch.5

19 Atwood with Friction: ch.5

20 Can you stop in time? Buggy rolls. You slide. ch.5

21 Insert values, determine ax.
Given m = 75kg, M = 20kg, D = 3.5m, vo =1.1m/s. What frictional coefficient is needed? Insert values, determine ax. ch.5

22 Diagramming Refresher:
ch.5

23 Stopping with 4W Disc-Brakes
Accelerating with F2WD. Stopping with 4W Disc-Brakes ch.5

24 (60° is close to maximum angle)
A 3kg box at rest on level surface with ms = What is the largest F acting 60° below horizontal for which the box remains at rest? fs (60° is close to maximum angle) ch.5

25 Relative vs. Absolute Velocity
vpc vpg P ch.5

26 Which force diagram applies to the object at Point B?

27 Which force diagram applies to the object at Point C?

28 Assume mass = 1. 2kg and radius = 45cm. If speed at Point D is 3
Assume mass = 1.2kg and radius = 45cm. If speed at Point D is 3.6m/s, what is the size of the normal force acting at Point D? -cen +cen ch.5

29 Q. Assume mass = 1. 2kg and radius = 45cm. If speed at Point B is 5
Q. Assume mass = 1.2kg and radius = 45cm. If speed at Point B is 5.1m/s, what is the size of the normal force acting at Point B? +cen -cen ch.5

30 Given: T = 50N, q = 30°, r = 1m Find: mg and v.
Net ch.5

31 The speed is now 6.5m/s and r = 1.0m. Angle, tension, mass?
Net For example, if m = 1.0kg, then T = 42.3N. ch.5

32 Q. The speed of a mass on a string of length L is 6. 5m/s
Q. The speed of a mass on a string of length L is 6.5m/s. The radius r = 2.0m. Find angle, tension, mass, and L. Net For example, if m = 1.0kg, then T = 23.9N. ch.5

33 Practice Q: What is F such that 0
Practice Q: What is F such that 0.5kg block stays at rest if all surfaces are frictionless? ch.5

34 Banked Turn: ch.5

35 Banked Turn: ch.5

36 Modified Atwood Machine with friction.
Objects are in CW motion. Let m1 = 2kg, m2 = 3kg, q = 30°, sliding friction coeff. 0.44 ch.5

37 Q. Recalculate last problem with m1 = 6kg m2 = 1kg
Q. Recalculate last problem with m1 = 6kg m2 = 1kg. (All else remaining the same) ch.5

38 Practice Q: Find the variable relationships.
ch.5

39 ch.5

40 Figures ch.5

41 ch.5

42 ch.5

43 ch.5

44 ch.5

45 ch.5

46 ch.5

47 ch.5

48 ch.5

49 ch.5

50 ch.5

51 ch.5

52 ch.5

53 ch.5

54 ch.5

55 Q. Assume the car has m = 1200kg and ms = 0.92.
Q. Assume the car has m = 1200kg and ms = 0.92. How large is the frictional force if v = 15m/s? ch.5

56 5-3 Drag Forces ch.5

57 Drag Forces Can be approximated as, Fdrag = bvn
Drag Forces Can be approximated as, Fdrag = bvn where b and n are constants ch.5

58 Example: Air drag, n = 2. If b = 25N/(m/s)2, at what speed would the object be resisted by 10N? At what speed would the same object be resisted by 30N? ch.5

59 Drag force grows quickly with v:
Terminal Velocity: Reached when drag force equals weight force ch.5


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