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Forces and the Laws of Motion

Forces and the Laws of Motion. Ch 4. FORCES. The cause of an acceleration or the change in an object’s velocity. Forces are measured in Newtons. SI system 1N = 1 kg m/s 2 1N =(approx) ¼ lb or 1 stick of butter. Types of Forces. 1. Contact

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Forces and the Laws of Motion

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  1. Forces and the Laws of Motion Ch 4

  2. FORCES • The cause of an acceleration or the change in an object’s velocity

  3. Forces are measured in Newtons • SI system • 1N = 1 kg m/s2 1N =(approx) ¼ lb or 1 stick of butter

  4. Types of Forces 1. Contact • Force that arises from physical contact of two objects • Field Force that can exist between objects, even in the absence of physical contact between objects

  5. Contact Forces EX: pull on a spring- it stretches pull on a wagon- the wagon moves

  6. Field Force Field- region of influence EX: gravity, attraction and repulsion between electrical charges, particle physics

  7. Force Diagrams The arrows represent forces, which act in pairs on an object.

  8. Force Diagram • Vectors- use to identify forces • Free body diagrams • Help analyze a situation used to analyze only the forces affecting the motion of a single object • Isolate an object and the forces acting on it

  9. Newton’s First Law (1687) Anobject at rest remains at rest; An object in motion continues to remain in motion; Unless the object is acted upon by an unbalanced force

  10. Newton’s First Law • Also called the “ Law of Inertia”

  11. Weight Weight - mass X gravity W = m g • Elevation: further away from the center of the earth gravity decreases • -9.81m/s2 is an average (round to 10 m/s2) • Weight is downward vector in force diagrams.

  12. Mass Mass- the amount of matter in a substance -same no matter where you go in the universe W = m g or m = W / g HW: Weight vs mass WS

  13. Newton’s Second Law • Force = mass X acceleration F = m a EX: F = measured in Newtons m = measured in kg a = measured in m/s2

  14. Predicting • If mass remains the same and the force needs to be increased then … F = m a • If the force remains the same and the mass increases then.. F = m a

  15. Vertical Problems • F Net = F App - W F Net = ma F App W

  16. Applied Force Fapp is the applied force. It is the force that you apply to the object

  17. Net Force • FNET is the net force • F = ma is FNET = ma • Net force is the total force. • If the variable is just listed as F it is the net force.

  18. Vertical Problems • Example 1: A 5 kg ball is struck upward with a force of 70 N ( Use g = 10 m/s2). What is the acceleration? • W = m g = 5 kg (10 m/s2 ) = 50 N 70 N 20 N FNET 50N

  19. Vertical Problems – EX:1 FNet = Fapp - W FNet = 70 N - 50 N FNet = 20 N FNet = ma a = FNet / m a = 20 N / 5 kg = 4m/s2

  20. Vertical Problems-EX: 2 EX 2: A 5 kg ball is struck downward with a force of 70 N ( Use g = 10 m/s2). What is the acceleration? • W = m g = 5 kg (10 m/s2 )= 50 N -50 N -70 N -120 N FNET

  21. Vertical #2 FNet = ma - a = FNet / m a = -120 N / 5 kg = -24m/s2

  22. Vertical Problems-EX: 3 EX 3: A 5 kg ball is struck upward with a force of 40 N ( Use g = 10 m/s2). What is the acceleration? W = m g = 5 kg (10 m/s2 )= 50 N 40N -10 N FNet -50 N

  23. Vertical Problems – EX: 3 FNet = Fapp - W FNet = 40 N - 50 N FNet = -10 N FNet = ma - a = FNet / m a = -10 N / 5 kg = -2m/s2 Vertical WS

  24. Elevator Problems What would a bathroom scale read on an elevator????? Fapp = ma + W up a = + down a = -

  25. Elevator Problems - EX 1 A 100 kg person is accelerating upward at 3 m / s2. What would the scale read? W = mg = 100kg (10 m / s2) = 1000 N Fapp = ma + W Fapp = (100kg)(+ 3 m / s2 ) + 1000 N Fapp = 1300 N

  26. Elevator Problems - EX 2 A 150 kg person is accelerating downward at 2m/s2. What would the scale read? W = mg = 150kg (10 m / s2) = 1500 N Fapp = ma + W Fapp = (150kg)(- 2 m / s2 ) + 1500 N Fapp = 1200 N

  27. Elevator Problems - EX 3 The elevator cable breaks and a 120 kg person in the elevator is free falling down the elevator shaft . What would the scale read? W = mg = 120kg (10 m / s2) = 1200 N Fapp = ma + W Fapp = (120kg)(- 10 m / s2 ) + 1200 N Fapp = 0 N “weightless” Elevator WS

  28. Forces of Friction When an object is moved how much resistance you feel depends on • how heavy the object is 2. The surfaces

  29. Forces of Friction FN FF Fapp W

  30. Forces of friction Surfaces- look it up in table 2 pg 138 coefficient of static friction ( ms) - ratio of friction when an object is just starting to move Coefficient of kinetic friction ( mk) - ratio of friction when an object is already moving and you want to keep it moving

  31. Forces are all measured in Newtons. Normal is a mathematical term for a line that is perpendicular to a surface FN – Normal Force FF– the Force of Friction or resistance

  32. Forces of Friction Equations W = mg FF = FNm FNet = Fapp - FF FNet = ma (a = FNet / m )

  33. Force of Friction- Example A 10 kg wooden box is pushed from rest with a force of 120 N on a wooden surface. What is the acceleration of the wooden box? 1.Determine the mass and weight. Mass = 10 kg Weight = mg = 10 kg ( 10 m/s2) = 100 N

  34. Force of Friction- Example • If the surface is perpendicular to the the object then the normal force is the same as the weight. The weight and the normal force need to be the same so that the box does not move vertically. 100 N 100 N

  35. Force of Friction- Example • Find the force of friction by using FF = m FN = 0.4 (100 N) = 40 N FN =100 N FF = 40 N Fapp W = 100 N

  36. Force of Friction- Example • Find the net force. FNet = Fapp - FF = 120 N - 40 N = 80 N 100 N 40 N 120 N 100 N

  37. Force of Friction- Example • Calculate the acceleration. FNet = ma (a = FNet / m ) a = 80 N / 10 kg = 8 m /s2 8 m /s2 Horizontal WS

  38. Horizontal with an Angle An object is being moved but the force is applied at an angle. FN Fapp y FF x W

  39. Horizontal with an Angle A 10 kg wooden box is pushed from rest with a force of 120 N at an angle of 30o on a wooden surface. What is the acceleration of the wooden box? 1.Determine the mass and weight. Mass = 10 kg Weight = mg = 10 kg ( 10 m/s2) = 100 N

  40. Horizontal with an Angle • Find the x and y. 120 N (sin 30o ) = y = 60 N 120 N (cos 30o ) = x = 104 N 3. Determine FN W = FN + y FN = W – y = 100N – 60 N = 40 N

  41. Horizontal with an Angle 4. Find the force of friction by using FF = m FN = 0.4 (40 N) = 16 N 5. Find the FNET. FNET = x - FF = 104 N - 16 N = 88 N 6. Find the acceleration. FNET = ma a = FNET / m = 88 N / 10 kg = 8.8 m/s2 Horizontal with an angle WS

  42. Incline Fll W F l

  43. Incline Sin q = Fll / W Cos q = F / W q q

  44. Incline A 30o incline has a 5 kg box on it. Find the parallel and perpendicular forces. W = mg W = 5 kg (10 m/s2) = 50 N 50 N (sin 30o) = Fll = 25 N 50 N (cos 30o) = F = 43 N Incline WS

  45. Newton’s Third Law For every action there is an equal and opposite reaction Forces always act in pairs Action-Reaction pair a pair of simultaneous equal but opposite forces resulting from the interaction of two objects

  46. Action- Reaction Pairs Bullet being fired is action and the recoil of the gun is the reaction Pulling out from a stop sign you go forward and your head goes backward

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