Newton’s 1st Law of Inertia

1. Newton’s 1st Law of Inertia Any object continues in its state of rest or in its uniform velocity unless it is made to change that state by an unbalanced force is acting upon it. • An object does not accelerate itself and it wants to retain a state of zero acceleration. Every object possesses inertia and the amount depends on the amount of matter or mass.

2. The greater the mass, the greater the inertia or resistance to acceleration. • Mass is a measure of the inertia. Mass is the amount of matter contained in an object. • Mass is a scalar quantity meaning it has magnitude only. • Mass is measured in g, kg, or slugs. • For nonrelativistic speeds, the mass of an object remains constant.

3. Do not use mass and weight interchangeably! • Weight is a measure of the gravitational attraction between an object and the earth. • Weight is a vector quantity because it has both magnitude and direction (the direction is always assumed to be towards the center of the earth). • The weight of an object varies with location as it is dependent on the distance from the center of the earth.

4. Newton’s 2nd Law of Acceleration The acceleration of an object is directly proportional to the net force acting on the object and is inversely proportional to its mass. Inertia is the tendency to resist changes in motion and Newton’s 2nd law expresses this mathematically. a is directly proportional to the Fnet. • By whatever factor Fnet changes, a changes by the same factor.

5. a α Fnet. If you double the force, you double the acceleration. If you decrease the Fnet by 1/3, you decrease the acceleration by 1/3. A graph of Acceleration vs Force would be a straight line passing through the origin.

6. a is inversely proportional to m. • a α 1/m • If you double the mass, the acceleration is reduced by ½. • If you decrease the mass by a factor of 1/3, you would triple the acceleration. • A graph of Acceleration vs Mass would be a hyperbola.

7. An object always accelerates in the direction of the net force. • If the net force is applied in the direction of the object’s motion (velocity), the object accelerates positively (speeds up). • If the net force is applied in the opposite direction of the object motion (velocity), the object decelerates.

8. Mathematically, Newton’s 2nd Law is • Fnet = ma where m is the mass of the object in kg, a is the acceleration in m/s2, and Fnet is the net force in N. • We now can formally define 1 N of force. If you have a mass of 1.0 kg and the net force causes it to accelerate at 1.0 m/s/s, then it is by definition 1.0 N of force.

9. Which Is It? Which is true? • a = Δv/Δt or a = Fnet/m? • Acceleration was previously defined to be the rate at which the velocity changes. • Now we are defining acceleration to be Fnet/m. • Why the difference?

10. Both are true! Previously, we looked at kinematics or how do we describe motion? • Do objects move at a constant velocity or a constant acceleration? Newton’s Laws describe the dynamics or why do objects move as they do? • Is the net force equal to greater than zero?

11. Newton’s 2nd Law Problems A car traveling at 32 m/s slows down to a stop and travels a distance of 52 m. If the mass of the car is 1375 kg, what net force acted on the car? vi = 32 m/s vf = 0 m = 1375 kg Δx = 52 m vf2 = vi2 + 2aΔx 0 = (32 m/s)2 + 2 × a × 52 m

12. a = -9.8 m/s2 Fnet = ma = 1375 kg × -9.8 m/s2 = -1.4 × 104 N The negative values for a and F make sense because the car decelerated in coming to a stop requiring a force in the opposite direction to its motion.

13. A stone weighs 7.4 N. What force must be applied to make it accelerate upward at 4.2 m/s2? Fw = 7.4 N g = 9.80 m/s2 a = 4.2 m/s2 FT FT = Fnet + Fw • Fw

14. Fnet = ma Fw = mg m = 7.4 N/9.80 m/s2 = 0.76 kg Fnet = 0.76 kg × 4.2 m/s2 = 3.2 N FT = 3.2 N + 7.4 N = 11 N

15. Some notes from the previous problem: • If the acceleration of the stone is upward, then the Fnet must also act upwards. • This implies that FT > Fw because the rope must provide the total force to support the weight of the object and also provide the net force.

16. Free Fall Free fall exists when an object’s weight is the only force acting on it (straight down, towards the center of the earth). In the absence air resistance, all objects accelerate at the same rate. • a = Fnet/m = Fw/m • a = g = 9.80 m/s2 = 980 cm/s2 = 32 ft/s2

17. Air resistance can usually be ignored for small dense objects that travel short distances but there can be exceptions: • Every baseball fan has heard the expression that the ball was headed out but the wind knocked it down. • A ping pong ball will never be confused with a small dense object.

18. Sometimes air resistance is not what you want: • When throwing a football for distance, a tight spiral minimizes air resistance. • Long range rifles have grooves in the barrel so the bullets come out spiraling. • If either the football or the bullet started to topple end over end, well … • The reason will be explained in another set of slides discussing angular momentum.

19. Sometimes you want air resistance. • Just ask any parachutist. • Air resistance depends on both velocity and surface area. a = g = Fnet/m = Fw – R/m At t = 0, R = 0. As an object accelerates downward, R increases.

20. a = g = Fnet/m = Fw – R/m At t = 0, R = 0. As an object accelerates downward, R increases. When Fw = R, Fnet = Fw – FR = 0, and a = 0. When the acceleration equals zero, the object is said to be moving with a terminal velocity.

21. Two brothers, Pete and Repeat, jump from the same helicopter and their parachutes are initially opened. The parachutes are the same size and Pete weighs 500 N and Repeat weighs 450 N. Who hits the ground first?

22. True Weight vs Apparent Weight A man stands on a bathroom scale in an elevator. The scale reads 917 N. • What is the man’s weight? Fup is the force that the bathroom scale pushes up on the man. Fw = Fup = 917 N and the man appears to weigh 917 N. Fup • Fw

23. (b) What is the man’s mass? Fw = 917 N g = 9.80 m/s2 Fw = mg m = 917 N/9.80 m/s2 = 93.6 kg

24. (c) As the elevator moves up, the scale reading increases to 1017 N. Determine the upward acceleration of the elevator. Fnet = ma = Fup - Fw Fup • Fnet = 1017 N – 917 N = 100. N Fw a= Fnet/m = 100. N/93.6 kg a= 1.1 m/s2 straightup

25. (d) As the elevator approaches the 13th floor, the scale reading decreases to 798 N. What is the acceleration of the elevator? Fnet = ma = Fw - Fup Fup • Fnet = 917 N – 798 N = 119 N Fw a= Fnet/m = 119 N/93.6 kg a= 1.3 m/s2 straightdown

26. (e) When the elevator reaches the 13th floor it stops. After about 5 sec the man looks down and the scale is reading 0. What is going on? • Fnet = Fw = mg Fw The guy is in a heap of trouble as he is in a state of free fall!

27. Thoughts To Ponder If the elevator was sound proof and there was no visible connection to the outside world, there is nothing the man could do to detect uniform motion. • When the acceleration of a system is zero, there is no experiment that distinguishes between an object at rest (∑F = 0) or an object moving in a straight line at constant speed (∑F = 0).

28. There are no relativistic speeds involved, so that the mass of the man remains constant. Whenever there is an acceleration involved, the net force will always be in the same direction as the acceleration.

29. Friction Friction is a force that resists the relative motion of solid objects that are in contact with each other. • If the solid is in a fluid (a liquid or a gas), then it is called viscosity. Friction is caused by uneven surfaces of touching objects.

30. Six Principles of Friction Friction acts parallel to the surfaces that are in contact and always opposes motion. Friction depends on the nature or composition of the solid surfaces in contact. Rolling Friction < Sliding Friction < Starting Friction

31. Sliding friction is practically independent of surface area for a given object. Sliding friction is practically independent of medium speeds. Sliding friction is directly proportional to the force pressing the two surfaces together.

32. Coefficient of Friction The formula for friction is given by Ff = µFN where Ff is the frictional force in N (newtons) and FN is the normal (perpendicular) force pressing the two surfaces together. • The normal force, FN, will not always equal the weight of the object!

33. µ (mu) is the coefficient of friction which is determined by what the two solid surfaces consist of (glass on glass, wood on wood, etc.). µ = Ff/FN is the ratio of the frictional force to the normal force. • µs > µk where µs is the coefficient of starting friction and µk is the coefficient of sliding friction.

34. Friction Problems A crate weighing 475 N is pulled along a level floor at a uniform speed by a rope which makes an angle of 30.0° with the floor. The applied force on the rope is 232 N. (a) Draw a free-body diagram of the box. FN F Fv θ Ff FH Fw

35. Ff FH F × cosθ µ = = = (b) Determine the coefficient of friction. (c) How much force is needed to pull the box? Fw - F × sinθ FN FW - FV 232 N × 0.866 µ = = 0.560 475 N – 232 N × 0.500 201 N FH = F × cosθ 232 N × 0.866 = =

36. (d) Compare the force in (c) to the weight of the box. It is easier to pull the crate, 201 N, than it is to lift the crate, 475 N.

37. An Inclined Plane Problem A roller coaster reaches the top of a steep hill with a speed of 7.0 km/h. It then descends the hill, which is at an angle of 45° and is 35.0 m long. If µk is 0.12, what is the speed when it reaches the bottom? Vi = 7.0 km/h Δx = 35.0 m θ = 45° = 0.12 µk

38. µk mg cos θ Fnet = mg sinθ - FN Ff • Fp θ FN . θ Fw µk FN Fnet = Fp - Ff =Fw sin θ - Fnet = ma

39. a = km 103 m 1 h vi × × 1.9 m/s = 7.0 = 3600 s 1 km h mgsinθ - µmgcosθ m 9.80 m/s2 × 0.707 – 0.12 × 9.80 m/s2 × 0.707 a = . a = 6.10 m/s2

40. vf2 = vi2 + 2a(x-xi) vf2 = (1.9 m/s)2 + 2 × 6.10 m/s2 × 35.0 m vf = 21 m/s

41. Another Inclined Plane Problem A block is given an initial speed of 4.2 m/s up a 24.0° inclined plane. Ignoring frictional effects, calculate: • How far up the inclined plane will the block travel? FN • Fp FN θ θ Fw

42. Fnet m vi = 4.2 m/s θ = 24.0° µ = 0 vf = 0 Fnet = ma -Fp -Fw sinθ -mgsinθ = = = = a m m m -9.80 m/s2 × 0.407 = - 3.99 m/s2 = a

43. vf2 = vi2 + 2a(x-xi) 0 = (4.2 m/s)2 + 2 × (- 3.99m/s2)× Δx Δx = 2.2 m (b) How long does it take before the block returns to its starting point? vf = vi + aΔt 0 = 4.2 m/s/-3.99 m/s2 Δt = 1.1 s ΔtT = 2 × 1.1 s = 2.2 s

44. Newton’s 3rd Law When one object exerts a force on a second object, the second object exerts a force on the first that is equal in magnitude but opposite in direction. • These two forces are called an action-reaction pair of forces. • F1 = - F2

45. To apply Newton’s 3rd Law, you must distinguish between forces acting on an object and forces exerted by the object. • When using Newton’s 3rd Law, you must have two different objects!

46. Examples of Newton’s 3rd Law Consider a 10. N ball falling freely in a vacuum where there is no air resistance. • What is the action force? • What is the reaction force?

47. • • The action force could be the earth pulling down on the ball with a force of 10. N. The reaction force would be the 10. N ball pulling up on the earth. Fa = Fw = 10. N Fr = Fw = 10. N

48. Fb Fw mbg ab = g = 9.80 m/s2 = = = mb mb mb It is easy to see why the ball falls toward the center of the earth. From Newton’s 2nd Law: straight down

49. 10. N 5.96 x 1024 kg Fb Fw me me It is easy to see why the earth remains stationary. ae = = = ae = 1.68 x 10-24 m/s2 straight up

50. At The Firing Range What happens when you fire a long range rifle? • The action force can be considered to be the force the gun exerts on the bullet. • The reaction force would be the force the bullet exerts on the gun.