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2. Resources Chapter Presentation Visual Concepts Transparencies Sample Problems Standardized Test Prep

3. Chapter 4 Forces and the Laws of Motion Table of Contents Section 1 Changes in Motion Section 2 Newton's First Law Section 3 Newton's Second and Third Laws Section 4 Everyday Forces

4. Chapter 4 Section 1 Changes in Motion Objectives • Describehow force affects the motion of an object. • Interpretand constructfree body diagrams.

5. Chapter 4 Section 1 Changes in Motion Force • A forceis an actionexerted on an object which may change the object’s state of rest or motion. • Forcescan causeaccelerations. • The SI unit of force is thenewton, N. • Forces can act throughcontactor at adistance.

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

7. Chapter 4 Section 1 Changes in Motion Force Diagrams • The effect of a force depends on both magnitude anddirection.Thus, force is avectorquantity. • Diagrams that show force vectors as arrows are calledforce diagrams. • Force diagrams that show only the forces acting on a single object are calledfree-body diagrams.

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

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

10. Chapter 4 Section 1 Changes in Motion Many orangutans spend their entire lives among the trees and are well adapted to move in this arboreal habitat. They have long arms (about two-thirds of their body height) and powerful chest muscles. Suppose an adult orangutan is hanging by its arms from a tree branch. The angle between each of the animal’s arms and the vertical is 15° with each arm exerting a force 430 N. The gravitational force acting on it is 830 N. Draw a free-body diagram of the animal.

11. Chapter 4 Section 1 Changes in Motion Solution

12. Chapter 4 Section 1 Changes in Motion Rock wall climbing is a challenging activity. Suppose that on one ascent, a climbing piton is wedged into a crevice on the rock wall. A climber negotiating the wall exerts a force of 350 N at an angle of 65° below the horizontal on a ropesecured to the piton. The rock surrounding the piton exerts an upward force of 320 N and a horizontal force of 150 N to the left on the pin. Draw a free-body diagram of the pin.

13. Chapter 4 Section 1 Changes in Motion Solution

14. Chapter 4 Section 1 Changes in Motion An experienced pilot can keep an engineless glider aircraft aloft for hours by using only the lift provided by rising air currents. Suppose during a flight an upward force of 5.00 × 103 N is exerted on the glider. Simultaneously, the glider experiences a forward force of 0.50 × 103N. The gravitational force acting on the glider and its pilot is 4.80 × 103N. Draw a free-body diagram of the glider

15. Chapter 4 Section 1 Changes in Motion Solution

16. Chapter 4 Section 2 Newton’s First Law Objectives • Explainthe relationship between the motion of an object and the net external force acting on the object. • Determinethe net external force on an object. • Calculatethe force required to bring an object into equilibrium.

17. 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 with constant velocity (that is, constant speed in a straight line) unless the object experiences a net external force. • In other words, when the net external force on an object is zero, the object’s acceleration (or the change in the object’s velocity) is zero.

18. Chapter 4 Section 2 Newton’s First Law Net Force • Newton's first law refers to thenet forceon an object.Thenet forceis thevector sumof all forces acting on an object. • The net force on an object can be found by using the methods for finding resultant vectors. Although several forces are acting on this car, the vector sum of the forces is zero. Thus, the net force is zero, and the car moves at a constant velocity.

19. Chapter 4 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.

20. 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 = ?

21. 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.

22. 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).

23. 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

24. 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.

25. Chapter 4 Section 2 Newton’s First Law Sample Problem Determining Net Force Two tugboats pull a barge across the harbor. One boat exerts a force of 7.5 x 104 N north, while the second boat exerts a force of 9.5 x 104 N at 15.0° north of west. Precisely, in what direction does the barge move?

26. Chapter 4 Section 2 Newton’s First Law Sample Problem Determining Net Force 47° north of west

27. Chapter 4 Section 2 Newton’s First Law Sample Problem Determining Net Force A four-way tug-of-war has four ropes attached to a metal ring. The forces on the ring are as follows: F1 = 4.00 x 103 N east, F2 = 5.00 x 103 N north, F3 = 7.00 x 103 N west, and F4 = 9.00 x 103 N south. What is the net force on the ring? What would be that force’s precise direction?

28. Chapter 4 Section 2 Newton’s First Law Sample Problem Determining Net Force 5.00 × 103N @ 53.1° south of west

29. Chapter 4 Section 2 Newton’s First Law Inertia • Inertiais the tendency of an object to resist being moved or, if the object is moving, to resist a change in speed or direction. • Newton’s first lawis often referred to as thelaw of inertiabecause it states that in the absence of a net force, a body will preserve its state of motion. • Massis a measure of inertia.

30. 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.

31. 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. Tip: To determine whether a body is in equilibrium, find the net force. If the net force is zero, the body is in equilibrium. If there is a net force, a second force equal and opposite to this net force will put the body in equilibrium.

32. Section 3 Newton’s Second and Third Laws Chapter 4 Objectives • Describean object’s acceleration in terms of its mass and the net force acting on it. • Predictthe direction and magnitude of the acceleration caused by a known net force. • Identifyaction-reaction pairs.

33. Section 3 Newton’s Second and Third Laws Chapter 4 Newton’s Second Law The acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to the object’s mass. SF = ma net force = mass  acceleration SFrepresents the vector sum of all external forces acting on the object, or the net force.

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

35. Section 3 Newton’s Second and Third Laws Chapter 4 Newton’s Third Law • If two objects interact, the magnitude of the force exerted on object 1 by object 2 is equal to the magnitude of the force simultaneously exerted on object 2 by object 1, and these two forces are opposite in direction. • In other words, for every action, there is an equal and opposite reaction. • Because the forces coexist, either force can be called the action or the reaction.

36. 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. • The action-reaction forces are equal and opposite, but either object may still have a net force on it. Consider driving a nail into wood with a hammer. The force that the nail exerts on the hammer is equal and opposite to the force that the hammer exerts on the nail. But there is a net force acting on the nail, which drives the nail into the wood.

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

38. 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.

39. 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.

40. 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)

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

42. Chapter 4 Section 4 Everyday Forces Normal Force • Thenormal forceacts on a surface in a directionperpendicular to the surface. • The normal force is not always opposite in direction to the force due to gravity. • In the absence of other forces, the normal force is equal and opposite to the component of gravitational force that is perpendicular to the contact surface. • In this example, Fn = mg cos q.

43. Chapter 4 Section 4 Everyday Forces Normal Force

44. Chapter 4 Section 4 Everyday Forces Friction • Static frictionis a force that resists the initiation of sliding motion between two surfaces that are in contact and at rest. • Kinetic frictionis the force that opposes the movement of two surfaces that are in contact and are sliding over each other. • Kinetic frictionis alwayslessthan themaximum static friction.

45. Chapter 4 Section 4 Everyday Forces Friction

46. 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. • The force ofkinetic frictionis always opposite the direction of motion. • To determine the direction of the force ofstatic friction,use the principle of equilibrium. For an object in equilibrium, the frictional force must point in the direction that results in a net force of zero.

47. 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:

48. Chapter 4 Section 4 Everyday Forces

49. 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.

50. 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: