Newton's Laws in Motion: Applications and Examples

1. Chapter 3 Forces and Motion in One Dimension

2. Applications of Newton’s Laws Introduction • We will learn how Newton’s Laws apply in various situations • We will begin with motion in one dimension • Idealized case – most real motion is three-dimensional • Mathematics is easier • Basic ideas and approaches can be applied to motion in higher dimensions

3. Motion of a Spacecraft • Assume the spacecraft travels along a straight-line path from one galaxy to another • Look at the motion in terms of position, velocity and acceleration • No forces acting on the spacecraft so no acceleration Section 3.1

4. Insights: Problem Set Up Section 3.1 • Start with a picture • Include the coordinate axes you have chosen for the problem • Determine a positive direction • Be careful with units • Be sure they are all in the same system of measurement • Include the units with your calculations • Check the units that they correctly match the quantity being calculated

5. Notation Simplification Section 3.1 For cases involving one-dimensional motion the usual vector notation can be simplified The direction is specified by the sign (+ or - ) of the quantity This notation corresponds to specifying the components of the quantity along the coordinate axis

6. Motion Example – Constant Velocity = 0 • The velocity is zero • On the graph, the line is at v = 0 • The position is constant, but generally not zero • Not moving, so no change in position • On the graph, a horizontal line Section 3.1

7. Motion Example – Constant Velocity ≠ 0 • The velocity is not zero • On the graph, the line is at v = constant value • The position is changing steadily • The same Δ x occurs each second • On the graph, an upward sloping straight line • The slope of the position-time curve is equal to the value of the velocity Section 3.1

8. Motion Example – Constant Acceleration • A constant force is applied • The acceleration is a constant • Value depends on the total force exerted on (and mass of) the object • The value is equal to the slope of the line on the v-t graph • The velocity is changing • On the graph, this is an upward sloping straight line • The position is changing • Not the same change each second • On the graph, this is a curved line Section 3.1

9. Equations to Describe Motion with Constant Acceleration Section 3.1 • v = vo + a t • vo is the velocity at some initial time t = 0 • It depends on what happened prior to t = 0 • x = xo + vo t + ½ a t2 • xo is the position at some initial time t = 0 • v2 = vo² + 2 a (xf - xo) • Eliminates t from the equation • Which equation to use depends on what information you are given in the problem and what you are asked to find

10. Constant Acceleration Equations, Summary Section 3.1

11. Note About Units • Units can still be cancelled and combined as shown in Chapter 1 • When dealing with units of force, it is simplest to express Newtons in terms of fundamental units before combining and/or canceling • Remember: 1 N = 1 kg m / s2 • Always include units in your calculations and check that they correctly match the quantity being calculated

12. Weight and Normal Forces Section 3.2 There are many cases where gravity plays an important role Weight is associated with gravity The normal force is another force often encountered in problems

13. Weight Section 3.2 • Weight is the force of gravity exerted by the Earth on an object • Denoted by • If an object has a mass m, then • The force of gravity is a consequence of Newton’s Law of Universal Gravitation • The value of g is approximately the same for all locations near the surface of the Earth • g ≈ 9.8 m/s² • The weight will be different on another planet • Since it is due to the gravitational attraction of that planet

14. Weight, cont. Section 3.2 • The value of g is independent of the mass of the object • The weight of an object is proportional to the mass of the object • g is commonly referred to as the “acceleration due to gravity” • Weight will be measured in Newtons • It is a force • Since weight acts vertically, it will be along the y-axis • Since the weight acts downward, Fgrav = - m g • It acts toward the center of the Earth

15. Normal Force Section 3.2 • A normal force acts perpendicularly to the plane of contact • Normal forces are common in nature • They occur whenever the surfaces of two objects come in contact

16. Weight and Normal Force, Example • The weight is directed downward • The normal force is directed upward • The person is at rest • a = 0 • ΣF = -m g + N = m a = 0 • N = m g • In this case, the normal is equal in magnitude to the weight and opposite in direction to the person’s weight Section 3.2

17. Free Body Diagram Section 3.2 A free body diagram should be used for analysis using Newton’s Second Law It is a simplified diagram showing all the forces acting on each object involved in the problem

18. Problem Solving with Newton’s Laws Section 3.2 • Recognize the objects of interest • List all the forces acting on each object • Sketch the problem • Start with a drawing that shows all the objects of interest • Include all the forces acting on the objects • Make a separate sketch showing all the forces acting on each using a dot for the object • This is the free body diagram • Forces in the free body diagram should be represented by arrows • The direction of the arrow will show the direction of the force • Include only the forces acting on that particular object

19. Problem Solving with Newton’s Laws, cont. Section 3.2 • Identify the relationships • Represent unknown quantities by an appropriate letter • Many unknowns have a particular symbol associated with them • Generally, you will use Newton’s Laws to solve for the unknowns • Solve • Use the information in the free body diagram to write Newton’s Second Law • You may need to include some algebraic steps to actually solve for the unknown • Check • Consider what your answer means and check that it makes sense

20. Apparent Weight • The normal force is not always equal to the weight • One example is moving upward in an accelerating elevator • Letting upward be positive: • ΣF = m a = N – mg • So N = m a + m g • If the elevator moved downward, N = m g – m a • In this case, the normal force is called the object’s apparent weight Section 3.2

21. Weight and Mass Section 3.2 • Weight and mass are closely related • They are not the same • Mass is an intrinsic property of the object • Weight depends on the location of the object • Example: • Assume mass on Earth is 10.0 kg • The weight on the surface of the Earth would be 98.0 N • Fgrav = m g • The mass on the Moon would be 10.0 kg • The weight on the Moon would be ~ 16.3 N • gMoon ~ 1/6 gEarth

22. Forces and Interactions Section 3.2 • Newton recognized that two objects could exert forces on each other even when they are not in direct contact • This is known as action-at-a-distance • Gravitational force is an example • Applies to many other forces • Contact on a microscopic scale • Contact forces are a result of electric forces between atoms that are in very close proximity • The atom-atom interaction is described by quantum mechanics, not Newtonian Mechanics

23. More About Mass Section 3.2 • The force of gravity depends on the mass of the object • This is called gravitational mass • The mass of an object also determines how the object will move in response to forces • This is called inertial mass • Physicists believe that the inertial mass is precisely equal to the gravitational mass • Why will be addressed later

24. Adding Friction Section 3.3 • Friction is another force commonly encountered in problems • Friction can be • Kinetic • Related to moving • Static • When objects are at rest

25. An Object Coming to Rest Section 3.3 • Aristotle’s idea was that rest was the natural state of terrestrial objects • Newton’s view is that an object comes to rest because a force acts on it • This force is often due to a phenomena called friction

26. Friction • The force of friction opposes the motion • The magnitude of the frictional force is related to the magnitude of the normal force • Force of kinetic friction • Ffriction = μk N • μk is called the coefficient of kinetic friction • A pure number with no units Section 3.3

27. Motion with Friction Section 3.3 • Set up the problem as usual, including the force of friction • For example, as seen in fig. 3.11: • ΣF = Ffriction = -μk N = m a • From the y-direction, Fgrav = N = m g • Therefore, -μk m g = m a and a = - μk g • Once you have found the acceleration, other quantities involved with motion can also be found • See Table 3.2 for values for coefficients of friction for various materials • The values depend on the smoothness of the surfaces

28. Static Friction • In many situations, the relevant surfaces are not slipping (moving) with respect to each other • This situation involves static friction • The amount of the push can vary without the object moving Section 3.3

29. Static Friction, cont. Section 3.3 • |Ffriction | ≤ μs N • μs is called the coefficient of static friction • Static indicates that the two surfaces are not moving relative to each other • If the push is increased, the force of static friction also increases and again cancels the force of the push • The magnitude of the static friction has an upper limit of μs N

30. Kinetic Friction Compared to Static Friction Section 3.3 • Consider both the kinetic and static friction cases • Use the different coefficients of friction • The force of kinetic friction is just Ffriction = μk N • The force of static friction varies by |Ffriction | ≤ μs N • For a given combination of surfaces, generally μs > μk • It is more difficult to start something moving than it is to keep it moving once started

31. Friction and Walking • The person “pushes” off during each step • The bottoms of his shoes exert a force on the ground • This is • If the shoes do not slip, the force is due to static friction • The shoes do not move relative to the ground Section 3.3

32. Friction and Walking, cont. • Newton’s Third Law tells us there is a reaction force • This force propels the person as he moves • If the surface was so slippery that there was no frictional force, the person would slip Section 3.3

33. Friction and Rolling • The car’s tire does not slip • There is a frictional force between the tire and road • There is a reaction force on the tire • This is the force that propels the car forward Section 3.3

34. Friction and Motion Section 3.3 The force of friction makes walking and running possible Friction plays a key role in rolling motion

35. Free Fall • A specific type of motion • Only gravity acts on the object when it is in free fall • There is actually some air drag, but it is generally considered negligible • We will analyze the motion in terms of acceleration, velocity, and position Section 3.4

36. Free Fall – Acceleration Section 3.4 • Take t = 0 to be the instant after the ball leaves the thrower’s hand • Before then, there is an additional force from her hand acting on the ball • This would not be an example of free fall since a force other than gravity would be acting on the object • Choose a coordinate system that measures position as the height y above the ground • Using Newton’s Second Law: • The negative sign means gravity is directed downward

37. Free Fall – Velocity and Position – Equations Section 3.4 Find the velocity and position as functions of time: Note the acceleration is constant (and equal to g) The velocity and acceleration are not always in the same direction

38. Free Fall – Velocity and Position – Graphs • The motion can be expressed graphically as well • The velocity can be positive, negative, or zero • The slope of the v-t graph is constant and negative for the entire motion • The constant negative acceleration is shown in graph C Section 3.4

39. Free Fall – Final Notes Section 3.4 • When the ball starts and stops at the same location (assume the ground), the ball’s speed just before it hits the ground is equal to its initial speed • The velocities are in opposite directions • The time spent on the way up is equal to the time spent falling back down • Shows the symmetry in the problem

40. Transmitting Forces Section 3.5 • Strings exert a force on the objects they are connected to • Also applies to cables, ropes, etc. • The mass of the cable may have to be taken into account • Pulleys can redirect forces • Forces can be amplified

41. Tension • Strings exert a force on the objects they are connected to • Cables and ropes act the same way • The strings exert force due to their tension • The ends of the string both exert a force of magnitude T on the supports where they are connected • T is the tension in the string Section 3.5

42. Tension Example – Elevator Cable • Two forces are acting on the compartment • Gravity acting downward • Tension in cable acting upward, T • Assume an acceleration upward • Applying Newton’s Second Law gives T = mg + ma Section 3.5

43. Tension Example – Elevator, cont. Section 3.5 • Now consider the cable • Assume the cable is massless • Applying Newton’s Second Law gives: TC = T • The tension is the same for all points along the cable • This is true for all massless cables • Tension has force units

44. Cables with Mass • Apply Newton’s Second Law to the cable • To support the cable, the upper tension, T1 must be larger than the tension from the box, T2 • T1 = T2 + mcable g • If no acceleration • Can assume a massless cable if the mass of the cable is small compared to the other tensions present Section 3.5

45. Single Pulleys • We often need to change the direction of the force • A simple pulley changes the direction of the force, but not the magnitude • See diagram • Assume the rope and pulley are both massless • Assume the cable does not slip on the pulley Section 3.5

46. Pulleys To Amplify Forces • The person exerts a force of T on the rope • The rope exerts a force of 2T on the pulley • This force can be used to lift an object • More complex sets of pulleys can amplify an applied force by greater factors • The distance decreases to compensate for the increase in force Section 3.5

47. Reasoning and Relationships – Problem Notes Section 3.6 • We may need to identify important information that is “missing” from the initial description of the problem • We need to recognize that additional information is needed • Then make reasonable estimates of the “missing” quantities • An approximate mathematical solution and an approximate numerical answer are generally sufficient • The estimates of the “missing” values will vary from case to case

48. Reasoning and Relationships – Problem Solving Strategy Section 3.6 • Recognize the principle • Determine the key physics ideas central to the problem • What principles connect the quantity you want to calculate with the quantities you know • Sketch the problem • Show all the given information • Draw a free body diagram, if needed • Include all the forces, velocities, etc. • Identify the relationships • Motion equations are an example of a set of relationships • If some values are unknown, make estimates for these values

49. Reasoning and Relationships – Problem Solving Strategy, cont. Section 3.6 • Solve • An exact mathematical solution typically is not needed • Cast the problem into one that is easy to solve mathematically • Check • Consider what your answer means • Check to be sure the answer makes sense

50. Parachutes, Air Drag, and Terminal Velocity Section 3.7 • Many constant acceleration problems are good approximations to real-life examples • Some interesting and important cases occur when acceleration is not constant • These cases include an object moving through a fluid • The object is subject to a resistive drag force arising from contact with the fluid molecules