1. 1 Chapter 4: Forces and Newton’s Laws of Motion Forces Newton’s Three Laws of Motion The Gravitational Force Contact Forces (normal, friction, tension) Application of Newton’s Second Law Apparent Weight Air Resistance Fundamental Forces CQ: 16, 18. P: 3, 5, 21, 37, 41, 63, 73, 87, 97, 147.

2. 22 Force Concept Contact Forces Ex: car on road, ball bounce Non-Contact Ex: magnetism, gravity /

3. 33 units Force units (SI): newton, N 1N ≈ ¼ lb. 1N = (1kg)(1m/s/s) N/kg = m/s/s

4. 44 Inertia is ‘resistance’ to change in velocity Measurement: Mass SI Unit: Kilogram (Kg) /

5. 55 Universal Law of Gravity all matter is weakly attracted attraction is inverse-square with distance G = 6.67x10 -11 N·m 2 /kg 2 Example: Two 100kg persons stand 1.0m apart

6. 66 g vs G G is universal g ~ Mass and Radius /

7. 7 Contact Forces Surfaces in contact are often under compression: each surface pushes against the other. The outward push of each object is called the Normal Force. If the objects move (even slightly) parallel to their surface the resistance force experienced is called the frictional force.

8. 88 Normal forces are? 1.Always vertically upward. 2.Always vertically downward. 3.Can point in any direction.

9. 9 Tension & Compression Compressed objects push outward away from their center (aka Normal Force). Stretched objects pull toward their center. This is called the Tension Force.

10. 10 Force Label Notation F = general force F N = normal force f = frictional force w = mg = F g = weight T = tension force /

11. 11 Net Force = change of motion vector sum of all forces acting on an object

12. 12 constant velocity Force Diagram F net = 0 a = 0 Example: Net Force = 0, Ball rolls along a smooth level surface table force weight force

13. 13 Example: Net-force on 0.5kg Net-force = 4N: Acceleration = 4N/0.5kg = 8m/s/s 5N, Right; 3N Left; Net-force = 2N Acceleration = 2N/0.5kg = 4m/s/s Falling; Net-force = mg Acceleration = mg/m = g = 9.8m/s/s /

14. 14 1. An object maintains constant velocity when the Net-Force on it is zero. 3. Forces always occur in pairs equal in size and opposite in direction. 2. An object’s acceleration equals the Net-Force on it divided by its mass. Newton’s Laws of Motion

15. 15 Force Diagrams Object is drawn as a “point” Each force is drawn as a “pulling” vector Each force is labeled Relevant Angles are shown x, y axes are written offset from diagram Only forces which act ON the object are shown

16. 16 Example of a Force Diagram for a Sled net force equals the mass times its acceleration.

17. 17 g’s one “g” of acceleration = 9.8m/s/s “two g’s” = 19.6m/s/s, etc. Example: What is the net force on a 2100kg SUV that is accelerating at 0.75g? Net-force = ma = m(0.75g) = 0.75mg = ¾ weight of car. /

18. 18 Block on Frictionless Incline a = w x /m =mgsin  /m = gsin . F n = w y.

19. 19 Newton’s 3 rd Law of Motion equal-sized oppositely-directed forces Independent of mass Pair-notation x x

20. 20 Newton’s 3 rd Law Pair Notation use “x” marks on forces that are 3 rd Law pairs. Use “xx” for a different interaction, etc.

21. 21 Force Diagram each object. Which has greater acceleration when released? Spring Force Spring Force xx Acceleration = F/m Acceleration = F/(2m)

22. 22 Friction Static Friction “sticking force” Kinetic Friction “sliding force” Coefficients: 0 = min, 1 ~ max e.g. teflon around 0.05 Rubber on concrete around 1.0

23. 23 Using Coefficients of Friction Ex. 10kg block. F N = weight = mg = 98N. Static coef. = 0.50; Kinetic coef. = 0.30.

24. 24 Applications

25. 25 A 3kg object sits on a frictionless table. Two horizontal forces act, one is 2N in the y-direction, the other 4N in the x- direction. A top-view diagram will be shown. F net What is the magnitude of the net-force acting? 4 2 2

26. 26 What direction does the 3kg mass accelerate? parallel to Fnet by Newton ’ s 2 nd Law. We are in Quadrant I since x and y are both + F net 4 2 2

27. 27 The magnitude of the acceleration is:

28. 28 Two 1kg Blocks; a = 1m/s/s Fnet = F = (2m)a = (2kg)(1m/s/s) = 2N Fnet = T = ma = (1kg)(1m/s/s) = 1N / F

29. 29 Two 1kg Blocks; F = 10N a = F/(2m) = 10N/2kg = 5 m/s/s T = ma = (1kg)(5m/s/s) = 5N / F

30. 30 4 Summary F net = ma (F net = 0, v = constant) forces always occur in pairs of equal size and opposite direction various force types (& symbols) equilibrium problems (a = 0) dynamic problems (a ≠ 0)

31. 31 30 90 60 30 Mg, 300 deg.

32. 32 Inclined Plane Forces Fxnet = FNcos90 + mgcos300 = (0.02)(a) = 0 + (0.02)(9.8)(0.5) = (0.02)a accel = 4.9 m/s/s Fynet = FNsin90 + mgsin300 = (0.02)(0) FN + (0.02)(9.8)(-.866) = 0 FN = 0.17N

33. 33 F net acceleration Ex: Newton’s 2 nd Law

34. 34 Coefficients of Friction Ex: Block&Load = 580grams If it takes 2.4N to get it moving and 2.0N to keep it moving

35. 35 1. 3kg box on level frictionless surface. F=86N acts 60° below horizontal. Example:

36. 36 1.(cont)

37. 37 Q1. What are a x and F N if angle is 30?

38. 38 Interaction Notation Since all forces are ‘pairs’, label as interactions, e.g. 1 on 2, 2 on 1, etc. F12 = “force of object 1 on object 2” F21 = “force of object 2 on object 1” F34 = “force of object 3 on object 4” Etc.

39. 39 Interaction Notation Symbols F12 – general force, 1 on 2 N12 – normal contact force, 1 on 2 f12 – frictional force, 1 on 2 W12 – gravitational force, 1 on 2 T12 – tension force, 1 on 2 m12 – magnetic force, 1 on 2 e12 – electrical force, 1 on 2

40. 40 Gravitational Force All masses attract via gravitational force Attraction is weak for two small objects Ex: Attraction between two bowling balls is so small it is hard to measure. Force is proportional to mass product Force is inversely proportional to the square of the distance between objects

41. 41 Example: Net Force = 0. Block on a surface inclined 30° from horizontal. Applied force F acts 40° below horizontal. Net Force = 0 velocity = constant

42. 42 Diagrams with Interaction Notation If f21 exists, then f12 also exists, and is opposite in direction to f21. f21 and f12 act on different objects.

43. 43 A 10kg box is being pushed along a horizontal surface by a force of 15N. A frictional force of 5N acts against the motion. We will want to (a) Calculate the net-force acting and (b) calculate the acceleration of the box. The net-horizontal force determines its x-acceleration The y-acceleration is known to be zero because it remains in horizontal motion, thus The net-force is 10N horizontal (0 vertical) The x-acceleration is: Example: