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  2. 2. • Born 1642 • Went to University of Cambridge in England as a student and taught there as a professor after • Never married • Gave his attention mostly to physics and mathematics, but he also gave his attention to religion and alchemy • Newton was the first to solve three mysteries that intrigued the scientists • Laws of Motion • Laws of Planetary Orbits • Calculus
  4. 4. 4-2 Newton’s First Law of Motion Newton’s first law is often called the law of inertia. Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it.
  5. 5. 4-2 Newton’s First Law of Motion Inertial reference frames: An inertial reference frame is one in which Newton’s first law is valid. This excludes rotating and accelerating frames.
  6. 6. 4-3 Mass Mass is the measure of inertia of an object. In the SI system, mass is measured in kilograms. Inertia  The tendency of a body to maintain its state of rest or motion. Mass is not weight: Mass is a property of an object. Weight is the force exerted on that object by gravity. If you go to the moon, whose gravitational acceleration is about 1/6 g, you will weigh much less. Your mass, however, will be the same.
  7. 7. Newton’s first law The ball remains in its state of rest till an external force is acting on it.
  8. 8. Newton’s first law (contd..) Push the ball When an external force (push) is acting on a ball (object) at rest it changes its state (moves over the steps ).
  9. 9. Newton’s first law (contd..) Push the ball When external force (Push of the rabbit) acting on the object (ball) which is in motion the object changes its position.
  10. 10. • An Object at rest remains at rest, and an object in motion continues in motion with constant velocity (that is, constant speed in a straight line), unless it experiences a net external force. • The tendency to resist change in motion is called inertia • People believed that all moving objects would eventually stop before Newton came up with his laws
  11. 11. • When there is no force exerted on an object, the motion of the object remains the same like described in the diagram • Because the equation of Force is F=ma, the acceleration is 0m/s². So the equation is 0N=m*0m/s² • Therefore, force is not needed to keep the object in motion, when • The object is in equilibrium when it does not change its state of motion
  12. 12. • 1st Law: If no net force acts, object remains at rest or in uniform motion in straight line. • What if a net force acts? Do Experiments. • Find, if the net force F  0  The velocity v changes (in magnitude or direction or both). • A change in the velocity v (Δv)  There is an acceleration a = (Δv/Δt) OR A net force acting on a body produces an acceleration! Fa
  13. 13. Newton’s Second Law Force equals mass times accelerati on. F = ma
  14. 14. • The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass Fnet Acceleration
  15. 15. • From experiments: The net force F on a body and the acceleration a of that body are related. • HOW? Answer by EXPERIMENTS! • The outcome of thousands of experiments over hundreds of years: a  F/m (proportionality) • We choose the units of force so that this is not just a proportionality but an equation: a  F/m OR: (total!) F = ma
  16. 16. • Newton’s 2nd Law: F = ma F = the net (TOTAL!) force acting on mass m m = the mass (inertia) of the object. a = acceleration of object. Description of the effect of F F is the cause of a. • To emphasize that the F in Newton’s 2nd Law is the TOTAL (net) force on the mass m, text writes: ∑F = ma ∑ = a math symbol meaning sum (capital sigma)
  17. 17. • Newton’s 2nd Law: ∑F = ma Based on experiment! Not derivable mathematically!! Force is a vector, so ∑F = ma is true along each coordinate axis. ∑Fx = max ∑Fy = may ∑Fz = maz ONE OF THE MOST FUNDAMENTAL & IMPORTANT LAWS OF CLASSICAL PHYSICS!!!
  18. 18. 4-4 Newton’s Second Law of Motion Newton’s second law is the relation between acceleration and force. Acceleration is proportional to NET force and inversely proportional to mass. (4-1)
  19. 19. 4-4 Newton’s Second Law of Motion The unit of force in the SI system is the newton (N). Note that the pound is a unit of force, not of mass, and can therefore be equated to newtons but not to kilograms.
  20. 20. As the man jumps off the boat, he exerts the force on the boat and the boat exerts the reaction force on the man. The man leaps forward onto the pier, while the boat moves away from the pier.
  21. 21. Foil deflected up Engine pushed forward Flow backwardpushed backward Flow Foil deflected down deflected Foil down
  22. 22. • If two objects interact, the force exerted on object 1 by object 2 is equal in magnitude but opposite in direction to the force exerted on object 2 by object 1 • Forces always come in pair when two objects interact • The forces are equal, but opposite in direction Fn Fg
  23. 23. 4-5 Newton’s Third Law of Motion Any time a force is exerted on an object, that force is caused by another object. Newton’s third law: Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first. Law of action-reaction: “Every action has an equal & opposite reaction”. (Action-reaction forces act on DIFFERENT objects!)
  24. 24. 4-5 Newton’s Third Law of Motion A key to the correct application of the third law is that the forces are exerted on different objects. Make sure you don’t use them as if they were acting on the same object.
  25. 25. 4-5 Newton’s Third Law of Motion Rocket propulsion can also be explained using Newton’s third law: hot gases from combustion spew out of the tail of the rocket at high speeds. The reaction force is what propels the rocket. Note that the rocket does not need anything to “push” against.
  26. 26. 4-5 Newton’s Third Law of Motion Helpful notation: the first subscript is the object that the force is being exerted on; the second is the source. This need not be done indefinitely, but is a good idea until you get used to dealing with these forces. (4-2)
  27. 27. Example 4-11 Fpx  (40.0 N )(cos30.0 0 )  34.6 N Fpy  (40.0 N )(sin 30.00 )  20.0 N Fpx  max Fpx 34.6 N  ax   m 10.0 Kg  3.46 m / s 2 Fy  ma y  FN  Fpy  mg  ma y FN  20.0  98.0  0 ; a y  0 FN  78.0 N
  28. 28. Example 4-13 FT  mE g  mE aE   mE a (1) FT  mC g  mC aC   mC a ( 2) subtract (1) from (2) (mE  mC ) g  (mE  mC )a mE  mC a g  0.68 m / s 2 mE  mC FT  mE g  mE a  mE ( g  a)  10500 N
  29. 29. Example 4-14 The advantage of a pulley 2 FT  mg  ma const. speed  a  0 mg FT  2
  30. 30. Example 4-15 Fx  FRC cos  FRB cos  0 ; a  0  FRB  FRC Fy  Fp  2 FT sin   0 FT  Fp 2 sin   1700 N
  31. 31. Example 4-16
  32. 32. Example 4-18
  33. 33. 4-6 Weight – the Force of Gravity; and the Normal Force Weight is the force exerted on an object by gravity. Close to the surface of the Earth, where the gravitational force is nearly constant, the weight is:
  34. 34. 4-6 Weight – the Force of Gravity; and the Normal Force An object at rest must have no net force on it. If it is sitting on a table, the force of gravity is still there; what other force is there? The force exerted perpendicular to a surface is called the normal force. It is exactly as large as needed to balance the force from the object (if the required force gets too big, something breaks!)
  35. 35. m = 10 kg The normal force is NOT always equal to the weight!!
  36. 36. Example 4-7 m = 10 kg ∑F = ma FP – mg = ma
  37. 37. Example 4-19
  38. 38. 4-7 Solving Problems with Newton’s Laws – Free-Body Diagrams 1. Draw a sketch. 2. For one object, draw a free-body diagram (force diagram), showing all the forces acting on the object. Make the magnitudes and directions as accurate as you can. Label each force. If there are multiple objects, draw a separate diagram for each one. 3. Resolve vectors into components. 4. Apply Newton’s second law to each component. 5. Solve.
  39. 39. 4-7 Solving Problems with Newton’s Laws – Free-Body Diagrams When a cord or rope pulls on an object, it is said to be under tension, and the force it exerts is called a tension force.
  40. 40. 4-8 Applications Involving Friction, Inclines On a microscopic scale, most surfaces are rough. The exact details are not yet known, but the force can be modeled in a simple way. We must account for Friction to be realistic!
  41. 41. 4-8 Applications Involving Friction, Inclines • Exists between any 2 sliding surfaces. • Two types of friction: Static (no motion) friction Kinetic (motion) friction • The size of the friction force: Depends on the microscopic details of 2 sliding surfaces. • • • • The materials they are made of Are the surfaces smooth or rough? Are they wet or dry? Etc., etc., etc.
  42. 42. 4-8 Applications Involving Friction, Inclines For kinetic (sliding) friction, we write: k is the coefficient of kinetic friction, and is different for every pair of surfaces. k depends on the surfaces & their conditions k is dimensionless & < 1
  43. 43. 4-8 Applications Involving Friction, Inclines Static friction is the frictional force between two surfaces that are not moving along each other. Static friction keeps objects on inclines from sliding, and keeps objects from moving when a force is first applied.
  44. 44. 4-8 Applications Involving Friction, Inclines The static frictional force increases as the applied force increases, until it reaches its maximum. Then the object starts to move, and the kinetic frictional force takes over.
  45. 45. 4-8 Applications Involving Friction, Inclines
  46. 46. 4-8 Applications Involving Friction, Inclines An object sliding down an incline has three forces acting on it: the normal force, gravity, and the frictional force. • The normal force is always perpendicular to the surface. • The friction force is parallel to it. • The gravitational force points down. If the object is at rest, the forces are the same except that we use the static frictional force, and the sum of the forces is zero.
  47. 47. 4-9 Problem Solving – A General Approach 1. Read the problem carefully; then read it again. 2. Draw a sketch, and then a free-body diagram. 3. Choose a convenient coordinate system. 4. List the known and unknown quantities; find relationships between the knowns and the unknowns. 5. Estimate the answer. 6. Solve the problem without putting in any numbers (algebraically); once you are satisfied, put the numbers in. 7. Keep track of dimensions. 8. Make sure your answer is reasonable.
  48. 48.  FT1 m1g   FT2  FT2 Take up as positive! m1 = m2 = 3.2 kg m1g = m2g = 31.4 N a Bucket 1: FT1 - FT2 - m1g = m1a Bucket 2: FT2 -m2g = m2a a ∑F =  m2 g ma (y direction), for EACH bucket separately!!!
  49. 49. FT  Take up positive! m = 65 kg mg = 637 N  FT FT + FT - mg = ma 2FT -mg = ma FP = - FT (3rd Law!)  FP a ∑F = ma (y direction) on woman + bucket!  mg
  50. 50. • Kinetic Friction: Experiments determine the relation used to compute friction forces. • Friction force Ffr is proportional to the magnitude of the normal force FN between two sliding surfaces. DIRECTIONS of Ffr & FN are  each other!! Ffr  FN FN Ffr a Fa mg
  51. 51. • Static Friction: Experiments are used again. • The friction force Ffr exists || two surfaces, even if there is no motion. Consider the applied force Fa: FN Ffr Fa mg ∑F = ma = 0 & also v = 0  There must be a friction force Ffr to oppose Fa Fa – Ffr = 0 or Ffr = Fa The object is not moving
  52. 52. • Experiments find that the maximum static friction force Ffr (max) is proportional to the magnitude (size) of the normal force FN between the two surfaces. DIRECTIONS of Ffr & FN are  each other!! Ffr  FN • Write the relation as Ffr (max) = sFN s  Coefficient of static friction • Depends on the surfaces & their conditions • Dimensionless & < 1 • Always find s > k  Static friction force: Ffr  sFN
  53. 53. FN  mA g  5.0  9.8  49 N F fr   k FN  0.20  49  9.8 N FBy  mB g  FT  mB a FAx  FT  F fr  mA a Example 4-20  FT  F fr  mA a  mB g  F fr  mA a  mB a mB g  F fr 19.6  9.8 a   1.4 m / s 2 mA  mB 5.0  2.0 FT  F fr  mA a  17 N
  54. 54. Example 4-21 FGx  mg sin  FGy  mg cos Fy  FN  mg cos  0 Fx  mg sin    k FN  max  FN  mg cos  mg sin    k (mg cos )  max
  55. 55. Problem 48 FN Ffr Fp (m1+m2) g FN   m1  m2  g  Ffr   k FN   k  m1  m2  g  Fx  FP  Ffr   m1  m2  a  a a = 1.9 m/s2 For contact forces analyze forces on the two blocks separately.
  56. 56. 1. What acceleration will result when a 12 N net force applied to a 3 kg object? A 6 kg object? 2. A net force of 16 N causes a mass to accelerate at a rate of 5 m/s2. Determine the mass. 3. How much force is needed to accelerate a 66 kg skier 1 m/sec/sec? 4. What is the force on a 1000 kg elevator that is falling freely at 9.8 m/sec/sec?
  57. 57. Summary of Chapter 4 • Newton’s first law: If the net force on an object is zero, it will remain either at rest or moving in a straight line at constant speed. • Newton’s second law: • Newton’s third law: • Weight is the gravitational force on an object. • The frictional force can be written: (kinetic friction) or (static friction) • Free-body diagrams are essential for problemsolving
  58. 58. • Henderson, Tom. Physics. Course home page. 16 May 2008>. • Serway, Raymond A., and Jerry S. Faughn. "The Laws of Motion." College Physics . Fifth ed. 1999. Benson, Tom. Newton’s Third Law applied to Aerodynamics 21 May 2008 • • Introduction to Rocket Performance. Newton’s Third Law. 12 March 2004 • Stern, David P. (16) Newton’s Laws of Physics. 1. Force and Inertia. 9 October 2004