1. Apparent Weight

2. Apparent Weight of an object is the reading on a ___________ scale when that object is placed on it.

3. Apparent Weight Apparent Weight of an object is the reading on a spring scale when that object is placed on it.

4. Apparent Weight Apparent Weight of an object is the reading on a spring scale when that object is placed on it. We can have a stand-on scale. Spring scale

5. Apparent Weight Apparent Weight of an object is the reading on a spring scale when that object is placed on it. We can have a stand-on scale. Spring scale Person exerts a force down on the spring scale, producing a “reading” in Newtons on the scale. This reading is the apparent weight.

6. Apparent Weight Apparent Weight of an object is the reading on a spring scale when that object is placed on it. We can have a stand-on scale. Spring scale Person exerts a force down on the spring scale, producing a “reading” in Newtons on the scale. This reading is the apparent weight. What is the reaction to this force?

7. Apparent Weight Apparent Weight of an object is the reading on a spring scale when that object is placed on it. We can have a stand-on scale. Spring scale Person exerts a force down on the spring scale, producing a “reading” in Newtons on the scale. This reading is the apparent weight. The spring scale exerts an equal but opposite Normal force up on the person. The size of the normal force is also equal to the apparent weight. FNFN

8. Apparent Weight Apparent Weight of an object is the reading on a spring scale when that object is placed on it. We can have a stand-on scale. Spring scale Person exerts a force down on the spring scale, producing a “reading” in Newtons on the scale. This reading is the apparent weight. The spring scale exerts an equal but opposite Normal force up on the person. The size of the Normal force is also equal to the apparent weight. FNFN For a stand-on scale, the apparent weight is equal to the magnitude of the normal force ǀF N l on the object.

9. Apparent Weight Apparent Weight of an object is the reading on a spring scale when that object is placed on it. We can have a hanging scale. Hanging spring scale

10. Apparent Weight Apparent Weight of an object is the reading on a spring scale when that object is placed on it. We can have a hanging scale. Hanging spring scale The person pulls down on the hanging scale producing a “reading” equal in magnitude to the “apparent weight”

11. Apparent Weight Apparent Weight of an object is the reading on a spring scale when that object is placed on it. We can have a hanging scale. Hanging spring scale The person pulls down on the hanging scale producing a “reading” equal in magnitude to the “apparent weight” What is the reaction to this force?

12. Apparent Weight Apparent Weight of an object is the reading on a spring scale when that object is placed on it. We can have a hanging scale. Hanging spring scale The person pulls down on the hanging scale producing a “reading” equal in magnitude to the “apparent weight” The hanging spring scale pulls up on the person with an equal tension force. The size of the tension force of the hanging scale on the person equals the apparent weight. FTFT

13. Apparent Weight Apparent Weight of an object is the reading on a spring scale when that object is placed on it. We can have a hanging scale. Hanging spring scale The person pulls down on the hanging scale producing a “reading” equal in magnitude to the “apparent weight” The hanging spring scale pulls up on the person with an equal tension force. The size of the tension force of the hanging scale on the person equals the apparent weight. For a hanging scale, the apparent weight is equal to the magnitude of the tension force ǀF T l on the object. FTFT

14. Deriving an Apparent Weight Formula

15. A person is in an elevator and is standing on a “stand-on Newton spring scale.” Spring scale

16. Deriving an Apparent Weight Formula A person is in an elevator and is standing on a “stand-on Newton spring scale.” The elevator could be accelerating up or down. Spring scale a

17. Deriving an Apparent Weight Formula A person is in an elevator and is standing on a “stand-on Newton spring scale.” The elevator could be accelerating up or down. Let’s draw an FBD of the person Spring scale a

18. Deriving an Apparent Weight Formula A person is in an elevator and is standing on a “stand-on Newton spring scale.” The elevator could be accelerating up or down. Let’s draw an FBD of the person Spring scale a Up + Down - a

19. Deriving an Apparent Weight Formula A person is in an elevator and is standing on a “stand-on Newton spring scale.” The elevator could be accelerating up or down. Let’s draw an FBD of the person Spring scale a Up + Down - a FNFN FgFg

20. Deriving an Apparent Weight Formula A person is in an elevator and is standing on a “stand-on Newton spring scale.” The elevator could be accelerating up or down. Let’s draw an FBD of the person Spring scale a Up + Down - a FNFN Short-cut formula for F g ?

21. Deriving an Apparent Weight Formula A person is in an elevator and is standing on a “stand-on Newton spring scale.” The elevator could be accelerating up or down. Let’s draw an FBD of the person Spring scale a Up + Down - a FNFN F g =mg

22. Deriving an Apparent Weight Formula A person is in an elevator and is standing on a “stand-on Newton spring scale.” The elevator could be accelerating up or down. Let’s draw an FBD of the person Spring scale a Up + Down - a FNFN F g = mg Let’s start with the equation of motion. What is it?

23. Deriving an Apparent Weight Formula A person is in an elevator and is standing on a “stand-on Newton spring scale.” The elevator could be accelerating up or down. Let’s draw an FBD of the person Spring scale a Up + Down - a FNFN F g = mg F net Y = ma y

24. Deriving an Apparent Weight Formula A person is in an elevator and is standing on a “stand-on Newton spring scale.” The elevator could be accelerating up or down. Let’s draw an FBD of the person Spring scale a Up + Down - a FNFN F g = mg F net Y = ma y ? Vector Statement

25. Deriving an Apparent Weight Formula A person is in an elevator and is standing on a “stand-on Newton spring scale.” The elevator could be accelerating up or down. Let’s draw an FBD of the person Spring scale a Up + Down - a FNFN F g = mg F net Y = ma y F N + F g = ma y Vector Statement

26. Deriving an Apparent Weight Formula A person is in an elevator and is standing on a “stand-on Newton spring scale.” The elevator could be accelerating up or down. Let’s draw an FBD of the person Spring scale a Up + Down - a FNFN F g = mg F net Y = ma y F N + F g = ma y Vector Statement ? Scalar Statement

27. Deriving an Apparent Weight Formula A person is in an elevator and is standing on a “stand-on Newton spring scale.” The elevator could be accelerating up or down. Let’s draw an FBD of the person Spring scale a Up + Down - a FNFN F g = mg F net Y = ma y F N + F g = ma y Vector Statement F N - mg = ma y Scalar Statement

28. Deriving an Apparent Weight Formula A person is in an elevator and is standing on a “stand-on Newton spring scale.” The elevator could be accelerating up or down. Let’s draw an FBD of the person Spring scale a Up + Down - a FNFN F g = mg F net Y = ma y F N + F g = ma y Vector Statement F N - mg = ma y Scalar Statement F N = ?

29. Deriving an Apparent Weight Formula A person is in an elevator and is standing on a “stand-on Newton spring scale.” The elevator could be accelerating up or down. Let’s draw an FBD of the person Spring scale a Up + Down - a FNFN F g = mg F net Y = ma y F N + F g = ma y Vector Statement F N - mg = ma y Scalar Statement F N = ma y + mg

30. Deriving an Apparent Weight Formula A person is in an elevator and is standing on a “stand-on Newton spring scale.” The elevator could be accelerating up or down. Let’s draw an FBD of the person Spring scale a Up + Down - a FNFN F g = mg F net Y = ma y F N + F g = ma y Vector Statement F N - mg = ma y Scalar Statement F N = ma y + mg ǀ F N ǀ is apparent weight

31. Deriving an Apparent Weight Formula A person is in an elevator and is standing on a “stand-on Newton spring scale.” The elevator could be accelerating up or down. Let’s draw an FBD of the person Spring scale a Up + Down - a FNFN F g = mg F net Y = ma y F N + F g = ma y Vector Statement F N - mg = ma y Scalar Statement F N = ma y + mg ǀ F N ǀ is apparent weight = m (a y + g)

32. General Apparent Weight Formula

33. For a stand-on Newton weigh scale ǀ F N ǀ ( apparent weight ) = m (a y + g)

34. General Apparent Weight Formula For a stand-on Newton weigh scale ǀ F N ǀ ( apparent weight ) = m (a y + g) For a hanging Newton weigh scale ǀ F T ǀ ( apparent weight ) = m (a y + g)

35. General Apparent Weight Formula For a stand-on Newton weigh scale ǀ F N ǀ ( apparent weight ) = m (a y + g) For a hanging Newton weigh scale ǀ F T ǀ ( apparent weight ) = m (a y + g) Note that these equations a combination of vectors and scalars. The mass m and gravitational field strength g are positive quantities, but a y is a vector and could be up (+) or down (-)

36. Apparent Weight ǂ Actual Weight Apparent Weight Actual Weight

37. Apparent Weight ǂ Actual Weight Apparent Weight reading on a spring scale when that object is placed on it. Actual Weight

38. Apparent Weight ǂ Actual Weight Apparent Weight reading on a spring scale when that object is placed on it. Actual Weight Force of earth’s gravity on an object

39. Apparent Weight ǂ Actual Weight Apparent Weight reading on a spring scale when that object is placed on it. Formula is… ǀ F N ǀ or ǀ F T ǀ = m (a y + g) Actual Weight Force of earth’s gravity on an object

40. Apparent Weight ǂ Actual Weight Apparent Weight reading on a spring scale when that object is placed on it. Formula is… ǀ F N ǀ or ǀ F T ǀ = m (a y + g) Actual Weight Force of earth’s gravity on an object Formula is… F g = m g

41. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut)

42. a)a y = ?

43. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) a)a y = 0

44. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) a)a y = 0 ǀ F N ǀ = m (a y + g) = ?

45. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) a)a y = 0 ǀ F N ǀ = m (a y + g) = 65(0 +10) = ?

46. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) a)a y = 0 ǀ F N ǀ = m (a y + g) = 65(0 +10) =650 N

47. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) a)a y = 0 ǀ F N ǀ = m (a y + g) = 65(0 +10) =650 N At rest, the apparent weight equal to the actual weight.

48. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) a)a y = 0 ǀ F N ǀ = m (a y + g) = 65(0 +10) =650 N At rest, the apparent weight equal to the actual weight. b)a y = ?

49. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) a)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 +10) =650 N At rest, the apparent weight equal to the actual weight. b)a y = 0

50. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) a)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 +10) =650 N At rest, the apparent weight equal to the actual weight. b)a y = 0 ǀ F ǀ N = m (a y + g) = ?

51. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) a)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 +10) =650 N At rest, the apparent weight equal to the actual weight. b)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 + 10) = ?

52. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) a)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 +10) =650 N At rest, the apparent weight equal to the actual weight. b)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 + 10) = 650 N

53. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) a)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 +10) =650 N At rest, the apparent weight equal to the actual weight. b)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 + 10) = 650 N At constant velocity, the apparent weight equal to the actual weight.

54. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) a)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 +10) =650 N At rest, the apparent weight equal to the actual weight. b)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 + 10) = 650 N At constant velocity, the apparent weight equal to the actual weight. c)a y = ?

55. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) a)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 +10) =650 N At rest, the apparent weight equal to the actual weight. b)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 + 10) = 650 N At constant velocity, the apparent weight equal to the actual weight. c)a y = + 4.0 m/s 2

56. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) a)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 +10) =650 N At rest, the apparent weight equal to the actual weight. b)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 + 10) = 650 N At constant velocity, the apparent weight equal to the actual weight. c)a y = + 4.0 m/s 2 ǀ F ǀ N = m (a y + g) = ?

57. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) a)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 +10) =650 N At rest, the apparent weight equal to the actual weight. b)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 + 10) = 650 N At constant velocity, the apparent weight equal to the actual weight. c)a y = + 4.0 m/s 2 ǀ F ǀ N = m (a y + g) = 65( +4+10) = ?

58. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) a)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 +10) =650 N At rest, the apparent weight equal to the actual weight. b)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 + 10) = 650 N At constant velocity, the apparent weight equal to the actual weight. c)a y = + 4.0 m/s 2 ǀ F ǀ N = m (a y + g) = 65( +4+10) = 910 N

59. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) a)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 +10) =650 N At rest, the apparent weight equal to the actual weight. b)a y = 0 ǀ F ǀ N = m (a y + g) = 65(0 + 10) = 650 N At constant velocity, the apparent weight equal to the actual weight. c)a y = + 4.0 m/s 2 ǀ F ǀ N = m (a y + g) = 65( +4+10) = 910 N With an upward acceleration, the apparent weight is bigger than the actual weight.

60. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) d)a y = ?

61. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) d)a y = - 4.0 m/s 2

62. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) d)a y = - 4.0 m/s 2 ǀ F ǀ N = m (a y + g) = ?

63. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) d)a y = - 4.0 m/s 2 ǀ F ǀ N = m (a y + g) = 65 (-4 + 10) = ?

64. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) d)a y = - 4.0 m/s 2 ǀ F ǀ N = m (a y + g) = 65 (-4 + 10) = 390 N

65. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) d)a y = - 4.0 m/s 2 ǀ F ǀ N = m (a y + g) = 65 (-4 + 10) = 390 N With a downward acceleration, the apparent weight is smaller than the actual weight.

66. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) d)a y = - 4.0 m/s 2 ǀ F ǀ N = m (a y + g) = 65 (-4 + 10) = 390 N With a downward acceleration, the apparent weight is smaller than the actual weight. e)a y = ?

67. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) d)a y = - 4.0 m/s 2 ǀ F ǀ N = m (a y + g) = 65 (-4 + 10) = 390 N With a downward acceleration, the apparent weight is smaller than the actual weight. e)a y = - 10.0 m/s 2

68. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) d)a y = - 4.0 m/s 2 ǀ F ǀ N = m (a y + g) = 65 (-4 + 10) = 390 N With a downward acceleration, the apparent weight is smaller than the actual weight. e)a y = - 10.0 m/s 2 ǀ F ǀ N = m (a y + g) = ?

69. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) d)a y = - 4.0 m/s 2 ǀ F ǀ N = m (a y + g) = 65 (-4 + 10) = 390 N With a downward acceleration, the apparent weight is smaller than the actual weight. e)a y = - 10.0 m/s 2 ǀ F ǀ N = m (a y + g) = 65 (-10 + 10) = ?

70. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) d)a y = - 4.0 m/s 2 ǀ F ǀ N = m (a y + g) = 65 (-4 + 10) = 390 N With a downward acceleration, the apparent weight is smaller than the actual weight. e)a y = - 10.0 m/s 2 ǀ F ǀ N = m (a y + g) = 65 (-10 + 10) = 0 N

71. Example : A 65.0 kg lady is standing on a Newton weigh scale in an elevator. What is the apparent weight (or normal force) on the lady if the elevator is … a) at rest b) moving down at a constant velocity of 4.0 m/s c) accelerating up at 4.0 m/s 2 d) accelerating down at 4.0 m/s 2 e) in free fall (cable has been cut) d)a y = - 4.0 m/s 2 ǀ F ǀ N = m (a y + g) = 65 (-4 + 10) = 390 N With a downward acceleration, the apparent weight is smaller than the actual weight. e)a y = - 10.0 m/s 2 ǀ F ǀ N = m (a y + g) = 65 (-10 + 10) = 0 N In free fall, the apparent weight is zero. There is apparent weightlessness.