1. Copyright © 2010 Pearson Education, Inc. Lecture Outline Chapter 5 Physics, 4 th Edition James S. Walker

2. Copyright © 2010 Pearson Education, Inc. Chapter 5 Newton’s Laws of Motion

3. Copyright © 2010 Pearson Education, Inc. Units of Chapter 5 Force and Mass Newton’s First Law of Motion Newton’s Second Law of Motion Newton’s Third Law of Motion The Vector Nature of Forces: Forces in Two Dimensions Weight Normal Forces

4. Copyright © 2010 Pearson Education, Inc. 5-1 Force and Mass Force: push or pull Force is a vector – it has magnitude and direction

5. Copyright © 2010 Pearson Education, Inc. 5-1 Force and Mass After the net force acting on an object, the second key ingredient in Newton’s laws is the mass of an object. Mass is the measure of how hard it is to change an object’s velocity. Mass can also be thought of as a measure of the quantity of matter in an object.

6. Copyright © 2010 Pearson Education, Inc. 5-2 Newton’s First Law of Motion If you stop pushing an object, does it stop moving? When you stop pushing the bags, it is not true that they stop moving because they no longer have a acting on them. What is missing in this analysis is the force of friction between the bags and the floor. Only if there is friction! In the absence of any net external force, an object will keep moving at a constant speed in a straight line, or remain at rest. This is also known as the law of inertia.

7. Copyright © 2010 Pearson Education, Inc. In order to change the velocity of an object – magnitude or direction – a net force is required. Newton’s First Law : -An object at rest remains at rest as long as no net force acts on it. -An object moving with constant velocity continues to move with the same speed and in the same direction as long as no net force acts on it. An inertial reference frame is one in which the first law is true that’s a frame of reference in which the law of inertia holds. Accelerating reference frames are not inertial. 5-2 Newton’s First Law of Motion

8. Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion Two equal weights exert twice the force of one; this can be used for calibration of a spring: With two weights, the force exerted by the scale is twice the force when only a single weight is attached.

9. Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion Now that we have a calibrated spring, we can do more experiments. Acceleration is proportional to force: If the force is doubled, the acceleration is also doubled.

10. Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion Acceleration is inversely proportional to mass: If the mass of an object is doubled but the force remains the same, the acceleration is halved.

11. Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion If we pull with a force F1, we find an acceleration equal to 1/2a 1. Thus the acceleration is inversely proportional to mass- the greater the mass, the less the acceleration. Combining these two observations gives Rearranging the equation yields the form of Newton’s law

12. Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion An object may have several forces acting on it; the acceleration is due to the net force: (5-1)

13. Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion

14. Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion The net force on a Jaguar XK8 has a magnitude of 6800 N. If the car’s acceleration is 3.8 m/s 2, what is its mass? m = ∑ F = 6800 N = 1800 kg a 3.8 m/s 2

15. Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion Free-body diagrams: A free-body diagram shows each and every external force acting on a given object. Sketch the forces - identify and sketch all of the external forces acting on an object. Isolate the object of interest - replace the object with a point particle of the same mass. Apply each of the forces acting on the object to that point. Choose a convenient coordinate system - Any coordinate system will work; however, if the object moves in a known direction, it is often convenient to pick that direction for one of the coordinate axes. Resolve the forces into components - determine the components of each force in the free-body diagram Apply Newton’s second law to each coordinate direction - analyze motion in each coordinate direction using the component form of Newton’s second law.

16. Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion Example of a free-body diagram: (a)The physical situation. (b) the free-body diagram for the book, showing the two external forces acting on it. Our choice for the a coordinate system is also indicated.

17. Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion Example of a free-body diagram: (a)Note that two forces act on the book: (i) the downward force of gravity, W, and (ii) the upward force, F, exerted by your hand. (b). Now that the free body diagram is drawn, we indicate a coordinate system so that the forces can be resolved into components. In this case all the forces are vertical. Note that we have chosen the upward to be the positive direction. ( c) The y components of the forces are F y = F and W y = -W. (d) It follows that ∑F y = F – W. using the Newton’s 2 nd law ( ∑F y = ma y ) (e) Since the book remains at rest, its acceleration is zero. Thus a y = 0 F – W = ma y =0 or F = W.

18. Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion Example of a free-body diagram: Moe, Larry and Curly push on a 752-kg boat that floats next to a dock. They each exert an 80.5-N force parallel to the dock. (a) What is the acceleration of the boat if they all push in the same direction? Give both direction and magnitude. (b) What are the magnitude and direction of the Boat’s acceleration if Larry and Curly push in the opposite direction to Moe’s push? FLFL FMFM FCFC X-component for each of the three forces F M,x = F L,x = F c,x = 80.5 N Sum the x components of force and set equal to ma x : ∑Fx = F M,x + F L,x + F C,x = 241.5 N = ma x Divide by the mass to find a x = a x = ∑F x /m = 241.5 N / 752 kg = 0.321 m/s 2 (b) The x component for each force : F M,x = 80.5 N F L,x = F C,x = -80.5 N Sum the x components of force and set equal to ma x : ∑Fx = F M,x + F L,x + F C,x = 80.5 N – 80.5 N – 80.5 N = -80.5 N = ma x Divide by the mass to find a x = a x = ∑F x /m = -80.5 N / 752 kg = - 0.107 m/s 2 Three forces in the same direction cause more acceleration than three forces in opposing directions. FCFC FMFM FLFL

19. Copyright © 2010 Pearson Education, Inc. 5-4 Newton’s Second Law of Motion A pitcher throws a 0.15-kg baseball, accelerating it from rest to a speed of about 90 mi/h. Estimate the force exerted by the pitcher on the ball. 1 mi = 5280 ft1 m = 3.281 ft1 h = 3600 s V = 90 mi/h = (90 mi/h)(5280 ft / 1mi)(1 m/3.281 ft)(1 h/3600 s) = 40.23 m/s a x = (V f 2 – V 0 2 ) / 2∆x = (40.23 m/s) 2 – 0 / 2(2.0 m) = 405 m/s 2 F ind the corresponding force with F x = ma x = (0.15 kg) (405 m/s 2 ) = 60.7 N

20. Copyright © 2010 Pearson Education, Inc. 5-4 Newton’s Third Law of Motion Forces always come in pairs. In addition, the forces in a pair, which always act on different objects, are equal in magnitude and opposite in direction. This is Newton’s Third law of motion. Newton’s Third Law states that for every force that acts on an object, there is a reaction force acting on a different object that is equal in magnitude and opposite in direction. In a somewhat more specific form: If object 1 exerts a force on object 2, then object 2 exerts a force – on object 1. These forces are called action-reaction pairs.

21. Copyright © 2010 Pearson Education, Inc. 5-4 Newton’s Third Law of Motion Some action-reaction pairs: Car on ground Ground on car Object on person Person on object Earth on shuttle Shuttle on earth

22. Copyright © 2010 Pearson Education, Inc. 5-4 Newton’s Third Law of Motion When objects are touching one another, the action-rection forces are often referred to as contact forces. Contact forces: When the external force is applied to the small box, the only force acting on the large box (mass m 1 ) is the contact force; hence the contact force must have a magnitude equal to m 1 a. In the second case, the only force acting on the small box (mass m 2 ) is the contact force, and so the magnitude of the contact force is m 2 a. Since m 1 is greater than m 2, it follows that the force of contact is larger when you push on the small box, m 2 a, than when you push on the large box, m 2 a To summarize, the contact force is larger when it must push the larger box.

23. Copyright © 2010 Pearson Education, Inc. 5-4 Newton’s Third Law of Motion A box of m1 = 10.0 kg rests on a smooth horizontal floor next to a box of mass m2 = 5.0 kg. If you push on box 1 with a horizontal force of magnitude F = 20.0 N, what is the acceleration of the boxes? (b) what is the force of contact between the boxes? 1 F =20 N a x = ∑F x / (m 1 + m 2 ) = 20.0 N/ (10 kg + 5 kg) = 20 N/15 kg = 1.33 m/s 2 F ind the net horizontal force acting on box 2 ∑F x = F 2,x = f = m 2 a x (b). f = m 2 a x = (5.0 kg) (1.33 m/s 2 ) = 6.67 N 10.0 kg 5.0 kg

24. Copyright © 2010 Pearson Education, Inc. 5-5 The Vector Nature of Forces: Forces in Two Dimensions The easiest way to handle forces in two dimensions is to treat each dimension separately, as we did for kinematics.

25. Copyright © 2010 Pearson Education, Inc. 5-6 Weight The weight of an object on the Earth’s surface is the gravitational force exerted on it by the Earth.

26. Copyright © 2010 Pearson Education, Inc. 5-6 Weight Apparent weight: Your perception of your weight is based on the contact forces between your body and your surroundings. If your surroundings are accelerating, your apparent weight may be more or less than your actual weight. ∑F y = W a - W

27. Copyright © 2010 Pearson Education, Inc. 5-7 Normal Forces The normal force is the force exerted by a surface on an object.

28. Copyright © 2010 Pearson Education, Inc. 5-7 Normal Forces The normal force may be equal to, greater than, or less than the weight.

29. Copyright © 2010 Pearson Education, Inc. 5-7 Normal Forces The normal force is always perpendicular to the surface.

30. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 5 Force: a push or pull Mass: measures the difficulty in accelerating an object Newton’s first law: if the net force on an object is zero, its velocity is constant Inertial frame of reference: one in which the first law holds Newton’s second law: Free-body diagram: a sketch showing all the forces on an object

31. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 5 Newton’s third law: If object 1 exerts a force on object 2, then object 2 exerts a force – on object 1. Contact forces: an action-reaction pair of forces produced by two objects in physical contact Forces are vectors Newton’s laws can be applied to each component of the forces independently Weight: gravitational force exerted by the Earth on an object

32. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 5 On the surface of the Earth, W = mg Apparent weight: force felt from contact with a floor or scale Normal force: force exerted perpendicular to a surface by that surface Normal force may be equal to, lesser than, or greater than the object’s weight