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Inertia Newton's First Law

Chapter 4. Section 2 Newton’s First Law. Inertia Newton's First Law. Inertia tendancy to resist a change in speed or direction. Newton’s first law is often referred to as the law of inertia because it states that in the absence of a net force, a body will preserve its state of motion.

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Inertia Newton's First Law

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  1. Chapter 4 Section 2 Newton’s First Law Inertia Newton's First Law • Inertia tendancy to resist a change in speed or direction. • Newton’s first law is often referred to as the law of inertia because it states that in the absence of a net force, a body will preserve its state of motion. • Mass is a measure of inertia.

  2. Chapter 4 Section 1 Changes in Motion Force

  3. Section 1 Changes in Motion Force • A force is an actionexerted on an object which may change the object’s state of rest or motion. • Forces can cause accelerations. • The SI unit of force is the newton, N. • Forces contact like friction or field forces

  4. Field Forces • Gravity- • Electromagnetic- • Strong Nuclear- • Weak Nuclear- • All other contact forces result from gravity or electromagnetic force.

  5. Chapter 4 Section 1 Changes in Motion Comparing Contact and Field Forces

  6. Chapter 4 Section 1 Changes in Motion Force Diagrams • force is avector quantity. • Force diagrams showing only the forces acting on a single object are calledfree-body diagrams. • Only forces that cause a possible change in motion state of object in question.

  7. Chapter 4 Section 1 Changes in Motion Force Diagrams, continued Force Diagram (rarely used) Free-Body Diagram In a force diagram,vector arrows represent all the forces acting in a situation. • A free-body diagramshows only the forces acting on the object of interest—in this case, the car.

  8. Chapter 4 Section 2 Newton’s First Law Newton’s First Law • Anobject at rest remains at rest, and an object in motion continues in motion at constant velocity unless acted on by a net external force. • Equilibrium- net external force =0 • the object’s acceleration = 0.

  9. Chapter 4 Section 2 Newton’s First Law Net Force • Use vector analysis to determine net force!!! Oh Yeah! How many forces are acting on this car? Consider the magnitude and direction of the forces, What can you infer?

  10. P 129 of holt in class 1-3 ,5,6

  11. Chapter 4 Section 1 Changes in Motion Drawing a Free-Body Diagram

  12. Section 2 Newton’s First Law Sample Problem Determining Net Force Derek leaves his physics book on top of a drafting table that is inclined at a 35° angle. The free-body diagram below shows the forces acting on the book. Find the net force acting on the book.

  13. Chapter 4 Section 2 Newton’s First Law Sample Problem, continued 1. Define the problem, and identify the variables. Given: Fgravity-on-book = Fg = 22 N Ffriction = Ff = 11 N Ftable-on-book = Ft = 18 N • Unknown: • Fnet = ?

  14. Chapter 4 Section 2 Newton’s First Law Sample Problem, continued 2. Select a coordinate system, and apply it to the free-body diagram. Choose the x-axis parallel to and the y-axis perpendicular to the incline of the table, as shown in (a). This coordinate system is the most convenient because only one force needs to be resolved into x and y components. Tip:To simplify the problem, always choose the coordinate system in which as many forces as possible lie on the x- and y-axes.

  15. Chapter 4 Section 2 Newton’s First Law Sample Problem, continued 3. Find the x and y components of all vectors. Draw a sketch, as shown in (b), to help find the components of the vector Fg. The angle q is equal to 180– 90 – 35 = 55. Add both components to the free-body diagram, as shown in (c).

  16. Chapter 4 Section 2 Newton’s First Law Sample Problem, continued For the y direction: SFy = Ft – Fg,y SFy = 18 N – 18 N SFy = 0 N 4. Find the net force in both the x and y directions. Diagram (d) shows another free-body diagram of the book, now with forces acting only along the x- and y-axes. For the x direction: SFx = Fg,x – Ff SFx = 13 N – 11 N SFx = 2 N

  17. Chapter 4 Section 2 Newton’s First Law Sample Problem, continued 5. Find the net force. Add the net forces in the x and y directions together as vectors to find the total net force. In this case, Fnet = 2 N in the +x direction, as shown in (e). Thus, the book accelerates down the incline.

  18. Chapter 4 Section 2 Newton’s First Law Mass and Inertia

  19. Chapter 4 Section 2 Newton’s First Law Inertia and the Operation of a Seat Belt • While inertia causes passengers in a car to continue moving forward as the car slows down, inertia also causes seat belts to lock into place. • The illustration shows how one type of shoulder harness operates. • When the car suddenly slows down, inertia causes the large mass under the seat to continue moving, which activates the lock on the safety belt.

  20. Chapter 4 Section 2 Newton’s First Law Equilibrium • Equilibriumis the state in which the net force on an object is zero. • Objects that are eitherat restor moving withconstant velocityare said to be in equilibrium. • Newton’s first lawdescribes objects in equilibrium.

  21. Section 3 Newton’s Second and Third Laws Chapter 4 Newton’s Second Law SF = ma net force = mass  acceleration SFrepresents the vector sum of all external forces acting on the object, or the net force.

  22. Section 3 Newton’s Second and Third Laws Chapter 4 Newton’s Second Law

  23. Section 3 Newton’s Second and Third Laws Chapter 4 Newton’s Third Law • action, there is an equal and opposite reaction.

  24. Section 3 Newton’s Second and Third Laws Chapter 4 Action and Reaction Forces • Action-reaction pairs do not imply that the net force on either object is zero. .

  25. Section 3 Newton’s Second and Third Laws Chapter 4 Newton’s Third Law

  26. Chapter 4 Section 4 Everyday Forces Objectives • Explainthe difference between mass and weight. • Find the direction and magnitude of normal forces. • Describeair resistance as a form of friction. • Usecoefficients of friction to calculate frictional force.

  27. Chapter 4 Section 4 Everyday Forces Weight • The gravitational force(Fg) exerted on an object by Earth is avector quantity, directed toward the center of Earth. • The magnitudeof this force (Fg) is a scalar quantity calledweight. • Weight changes with the location of an object in the universe.

  28. Chapter 4 Section 4 Everyday Forces Weight, continued • Calculating weight at any location: Fg = mag ag = free-fall acceleration at that location • Calculating weight on Earth's surface: ag = g = 9.81 m/s2 Fg = mg = m(9.81 m/s2)

  29. Chapter 4 Section 4 Everyday Forces Comparing Mass and Weight

  30. Chapter 4 Section 4 Everyday Forces Normal Force • Thenormal force is perpendicular to the surface. • not always opposite in direction g. • Fn = mg cos q.

  31. Chapter 4 Section 4 Everyday Forces Normal Force

  32. Chapter 4 Section 4 Everyday Forces Friction • Static frictionforce resisting initiation of sliding motion between two surfaces at rest. • Kinetic friction=force that opposes the movement of two surfaces that are sliding over each other. • Kinetic frictionis alwayslessthan themaximum static friction.

  33. Chapter 4 Section 4 Everyday Forces Friction

  34. Chapter 4 Section 4 Everyday Forces Friction Forces in Free-Body Diagrams • In free-body diagrams, the force of friction is alwaysparallelto the surface of contact. • kinetic frictionalways opposite the direction of motion. • for static friction,always opposite the direction of likey motion.

  35. Chapter 4 Section 4 Everyday Forces The Coefficient of Friction • The quantity that expresses the dependence of frictional forces on the particular surfaces in contact is called thecoefficient of friction, m. • Coefficient of kinetic friction: • Coefficient of static friction:

  36. Chapter 4 Section 4 Everyday Forces

  37. Chapter 4 Section 4 Everyday Forces Sample Problem Overcoming Friction A student attaches a rope to a 20.0 kg box of books.He pulls with a force of 90.0 N at an angle of 30.0° with the horizontal. The coefficient of kinetic friction between the box and the sidewalk is 0.500. Find the acceleration of the box.

  38. Chapter 4 Section 4 Everyday Forces Sample Problem, continued 1. Define Given: m = 20.0 kg mk = 0.500 Fapplied = 90.0 N at q = 30.0° Unknown: a= ? Diagram:

  39. Chapter 4 Section 4 Everyday Forces Sample Problem, continued The diagram on the right shows the most convenient coordinate system, because the only force to resolve into components is Fapplied. 2. Plan Choose a convenient coordinate system, and find the x and y components of all forces. • Fapplied,y = (90.0 N)(sin 30.0º) = 45.0 N (upward) • Fapplied,x = (90.0 N)(cos 30.0º) = 77.9 N (to the right)

  40. Chapter 4 Section 4 Everyday Forces Sample Problem, continued Choose an equation or situation: A. Find the normal force, Fn, by applying the condition of equilibrium in the vertical direction: SFy= 0 B. Calculate the force of kinetic friction on the box: Fk = mkFn C. Apply Newton’s second law along the horizontal direction to find the acceleration of the box: SFx= max

  41. Chapter 4 Section 4 Everyday Forces Sample Problem, continued 3. Calculate A. To apply the condition of equilibrium in the vertical direction, you need to account for all of the forces in the y direction: Fg, Fn, and Fapplied,y. You know Fapplied,yand can use the box’s mass to find Fg. Fapplied,y= 45.0 N Fg= (20.0 kg)(9.81 m/s2) = 196 N Next, apply the equilibrium condition, SFy= 0, and solve for Fn. SFy= Fn + Fapplied,y– Fg = 0 Fn + 45.0 N – 196 N = 0 Fn = –45.0 N + 196 N = 151 N Tip: Remember to pay attention to the direction of forces. In this step, Fg is subtracted from Fn and Fapplied,y because Fg is directed downward.

  42. Chapter 4 Section 4 Everyday Forces Sample Problem, continued B. Use the normal force to find the force of kinetic friction. Fk = mkFn = (0.500)(151 N) = 75.5 N C. Use Newton’s second law to determine the horizontal acceleration. a = 0.12 m/s2 to the right

  43. Chapter 4 Section 4 Everyday Forces Sample Problem, continued 4. Evaluate The box accelerates in the direction of the net force, in accordance with Newton’s second law. The normal force is not equal in magnitude to the weight because the y component of the student’s pull on the rope helps support the box.

  44. Chapter 4 Section 4 Everyday Forces Air Resistance • Air resistanceis a form of friction. Whenever an object moves through a fluid medium, such as air or water, the fluid provides a resistance to the object’s motion. • For a falling object, when the upward force of air resistance balances the downward gravitational force, the net force on the object is zero. The object continues to move downward with a constant maximum speed, called theterminal speed.

  45. Chapter 4 Section 4 Everyday Forces Fundamental Forces • There are fourfundamental forces: • Electromagnetic force • Gravitational force • Strong nuclear force • Weak nuclear force • The four fundamental forces are all field forces.

  46. Chapter 4 Standardized Test Prep Multiple Choice • Use the passage below to answer questions 1–2. • Two blocks of masses m1 and m2 are placed in contact with each other on a smooth, horizontal surface. Block m1 is on the left of block m2. A constant horizontal force F to the right is applied to m1. • 1. What is the acceleration of the two blocks? • A. C. • B. D.

  47. Chapter 4 Standardized Test Prep Multiple Choice • Use the passage below to answer questions 1–2. • Two blocks of masses m1 and m2 are placed in contact with each other on a smooth, horizontal surface. Block m1 is on the left of block m2. A constant horizontal force F to the right is applied to m1. • 1. What is the acceleration of the two blocks? • A. C. • B. D.

  48. Chapter 4 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 1–2. Two blocks of masses m1 and m2 are placed in contact with each other on a smooth, horizontal surface. Block m1 is on the left of block m2. A constant horizontal force F to the right is applied to m1. 2. What is the horizontal force acting on m2? F.m1a G.m2a H. (m1 + m2)a J.m1m2a

  49. Chapter 4 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 1–2. Two blocks of masses m1 and m2 are placed in contact with each other on a smooth, horizontal surface. Block m1 is on the left of block m2. A constant horizontal force F to the right is applied to m1. 2. What is the horizontal force acting on m2? F.m1a G.m2a H. (m1 + m2)a J.m1m2a

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