UNIT1Forces and Motion: Dynamics2OVERALL EXPECTATIONSANALYZE, predict, and explain the motion of selected objects in vertical, horizontal, and inclined planes.INVESTIGATE, represent, and analyze motion and forces in linear, projectile, and circular motion.RELATE your understanding of dynamics to the development and use of motion technologies.UNIT CONTENTSCHAPTER 1 Fundamentals of Dynamics CHAPTER 2 Dynamics in Two DimensionsCHAPTER 3 Planetary and Satellite DynamicsSpectators are mesmerized by trapeze artistsmaking perfectly timed releases, glidingthrough gracefu l arcs, and intersecting the paths oftheir partners. An error in timing and a graceful arccould become a trajectory of panic. Trapeze artistsknow that tiny differences in height, velocity, andtiming are critical. Swinging from a trapeze, the performer forces his body from its natural straight-line path. Gliding freely through the air, he is subject only to gravity. Then, the outstretched handsof his partner make contact, and the performer isacutely aware of the forces that change his speedand direction.In this unit, you will explore the relationshipbetween motion and the forces that cause it andinvestigate how different perspectives of the samemotion are related. You will learn how to analyzeforces and motion, not only in a straight line, butalso in circular paths, in parabolic trajectories, andon inclined surfaces. You will discover how themotion of planets and satellites is caused, described,and analyzed.Refer to pages 126127 before beginning this unit.In the unit project, you will design and build aworking catapult to launch small objects throughthe air.IWhat launching devices have you used, watched,or read about? How do they develop and controlthe force needed to propel an object?IWhat projectiles have you launched? How doyou direct their ight so that they reach a maximum height or stay in the air for the longest possible time?UNIT PROJECT PREP3C H A P T E RFundamentals of Dynamics1How many times have you heard the saying, It all dependson your perspective? The photographers who took the twopictures of the roller coaster shown here certainly had differentperspectives. When you are on a roller coaster, the world looksand feels very different than it does when you are observing themotion from a distance. Now imagine doing a physics experimentfrom these two perspectives, studying the motion of a pendulum,for example. Your results would denitely depend on your perspective or frame of reference. You can describe motion fromany frame of reference, but some frames of reference simplify theprocess of describing the motion and the laws that determine that motion. In previous courses, you learned techniques for measuring anddescribing motion, and you studied and applied the laws ofmotion. In this chapter, you will study in more detail how tochoose and dene frames of reference. Then, you will extend your knowledge of the dynamics of motion in a straight line.IUsing the kinematic equations foruniformly accelerated motion.PREREQUISITECONCEPTS AND SKILLSMulti-LabThinking Physics 51.1 Inertia and Frames of Reference 6Investigation 1-AMeasuring Inertial Mass 81.2 Analyzing Motion 151.3 Vertical Motion 27Investigation 1-BAtwoods Machine 341.4 Motion along an Incline 46CHAPTER CONTENTS4 MHR Unit 1 Forces and Motion: DynamicsM U L T IL A BThinking PhysicsTARGET SKILLSPredictingIdentifying variablesAnalyzing and interpretingSuspended SpringTape a plastic cup to one end of a short section of a large-diameter spring, such as a Slinky. Hold the other end of the springhigh enough so that the plastic cup is at least1 m above the oor. Before yourelease the spring, predict theexact motion of the cupfrom the instant that it isreleased until the momentthat it hits the oor. Whileyour partner watches thecup closely from a kneel-ing position, release thetop of the spring. Observethe motion of the cup.Analyze and Conclude1. Describe the motion of the cup and thelower end of the spring. Compare themotion to your prediction and describeany differences.2. Is it possible for any unsupported object to be suspended in midair for any lengthof time? Create a detailed explanation toaccount for the behaviour of the cup at themoment at which you released the top ofthe spring.3. Athletes and dancers sometimes seem tobe momentarily suspended in the air. How might the motion of these athletes be related to the springs movement in this lab?Thought ExperimentsWithout discussing the following questionswith anyone else, write down your answers. 1. Student A and Student B sit inidentical ofcechairs facingeach other, asillustrated.Student A, whois heavier than Student B, suddenly push-es with his feet, causing both chairs tomove. Which of the following occurs?(a) Neither student applies a force to theother.(b) A exerts a force that is applied to B, but A experiences no force.(c) Each student applies a force to theother, but A exerts the larger force.(d) The students exert the same amount of force on each other.2. A golf pro drives a ball through the air.What force(s) is/are acting on the golf ballfor the entirety of its ight?(a) force of gravity only(b) force of gravity and the force of the hit(c) force of gravity and the force of airresistance (d) force of gravity, the force of the hit,and the force of air resistance3. A photographer accidentally dropsa camera out of asmall airplane as it ies horizontally.As seen from theground, which path would the cameramost closely follow as it fell?Analyze and ConcludeTally the class results. As a class, discuss theanswers to the questions.A B C DA BChapter 1 Fundamentals of Dynamics MHR 5Imagine watching a bowling ball sitting still in the rack. Nothingmoves; the ball remains totally at rest until someone picks it upand hurls it down the alley. Galileo Galilei (15641642) and laterSir Isaac Newton (16421727) attributed this behaviour to the property of matter now called inertia, meaning resistance tochanges in motion. Stationary objects such as the bowling ballremain motionless due to their inertia. Now picture a bowling ball rumbling down the alley.Experience tells you that the ball might change direction and, ifthe alley was long enough, it would slow down and eventuallystop. Galileo realized that these changes in motion were due tofactors that interfere with the balls natural motion. Hundreds of years of experiments and observations clearly show that Galileowas correct. Moving objects continue moving in the same direc-tion, at the same speed, due to their inertia, unless some externalforce interferes with their motion.You assume that an inanimate object such as a bowling ballwill remain stationary until someone exerts a force on it. Galileo andNewton realized that this lack of motion is a very important property of matter.Analyzing ForcesNewton rened and extended Galileos ideas about inertia andstraight-line motion at constant speed now called uniformmotion.NEWTONS FIRST LAW: THE LAW OF INERTIAAn object at rest or in uniform motion will remain at rest or inuniform motion unless acted on by an external force.Figure 1.1Inertia and Frames of Reference 1. 16 MHR Unit 1 Forces and Motion: Dynamics Describe and distinguishbetween inertial and non-inertial frames of reference. Dene and describe the concept and units of mass. Investigate and analyze linear motion, using vectors,graphs, and free-body diagrams. inertia inertial mass gravitational mass coordinate system frame of reference inertial frame of reference non-inertial frame of reference ctitious force T E R M SK E YE X P E C T AT I O N SS E C T I O NNewtons rst law states that a force is required to change anobjects uniform motion or velocity. Newtons second law thenpermits you to determine how great a force is needed in order tochange an objects velocity by a given amount. Recalling thatacceleration is dened as the change in velocity, you can stateNewtons second law by saying, The net force (

F ) required toaccelerate an object of mass m by an amount (

a ) is the product of the mass and acceleration.Inertial MassWhen you compare the two laws of motion, you discover that therst law identies inertia as the property of matter that resists a change in its motion; that is, it resists acceleration. The secondlaw gives a quantitative method of nding acceleration, but it doesnot seem to mention inertia. Instead, the second law indicates that the property that relates force and acceleration is mass. Actually, the mass (m) used in the second law is correctlydescribed as the inertial mass of the object, the property thatresists a change in motion. As you know, matter has another prop-erty it experiences a gravitational attractive force. Physicistsrefer to this property of matter as its gravitational mass. Physicistsnever assume that two seemingly different properties are relatedwithout thoroughly studying them. In the next investigation, youwill examine the relationship between inertial mass and gravita-tional mass.QuantitySymbol SI unitforce F N (newtons)mass m kg (kilograms)acceleration ams2(metres per secondsquared)Unit analysis(mass)(acceleration) = (kilogram)

metressecond2

kgms2 =kg ms2 = NNote: The force (

F ) in Newtons second law refers to the vector sum of all of the forces acting on the object.

F = m

aNEWTONS SECOND LAWThe word equation for Newtons second law is: Net force isthe product of mass and acceleration.Chapter 1 Fundamentals of Dynamics MHR 7The Latin root of inertia means sluggish or inactive. An inertialguidance system relies on a gyro-scope, a sluggish mechanical devicethat resists a change in the directionof motion. What does this suggestabout the chemical properties of aninert gas?LANGUAGE LINKI N V E S T I G A T I O N 1-AMeasuring Inertial MassTARGET SKILLSHypothesizingPerforming and recordingAnalyzing and interpreting8 MHR Unit 1 Forces and Motion: DynamicsProblemIs there a direct relationship between an objectsinertial mass and its gravitational mass?HypothesisFormulate an hypothesis about the relationshipbetween inertial mass and its gravitational mass.EquipmentIdynamics cartIpulley and stringIlaboratory balanceIstandard mass (about 500 g)Imetre stick and stopwatch or motion sensor Iunit masses (six identical objects, such as small C-clamps) Iunknown mass (measuring between one and six unitmasses, such as a stone)Procedure1. Arrange the pulley, string, standard mass,and dynamics cart on a table, as illustrated.2. Set up your measuring instruments to deter-mine the acceleration of the cart when it ispulled by the falling standard mass. Find the acceleration directly by using computersoftware, or calculate it from measurementsof displacement and time.3. Measure the acceleration of the empty cart.4. Add unit masses one at a time and measurethe acceleration several times after eachaddition. Average your results.5. Graph the acceleration versus the number ofunit inertial masses on the cart.6. Remove the unit masses from the cart andreplace them with the unknown mass, thenmeasure the acceleration of the cart.7. Use the graph to nd the inertial mass of theunknown mass (in unit inertial masses).8. Find the gravitational mass of one unit ofinertial mass, using a laboratory balance.9. Add a second scale to the horizontal axis ofyour graph, using standard gravitational massunits (kilograms).10. Use the second scale on the graph to predictthe gravitational mass of the unknown mass.11. Verify your prediction: Find the unknownsgravitational mass on a laboratory balance.Analyze and Conclude1. Based on your data, are inertial and gravitational masses equal, proportional, or independent?2. Does your graph t a linear, inverse, expo-nential, or radical relationship? Write therelationship as a proportion (a ?).3. Write Newtons second law. Solve the expression for acceleration. Compare thisexpression to your answer to question 2.What inferences can you make?4. Extrapolate your graph back to the verticalaxis. What is the signicance of the point atwhich your graph now crosses the axis?5. Verify the relationship you identied inquestion 2 by using curve-straightening techniques (see Skill Set 4, MathematicalModelling and Curve Straightening). Write aspecic equation for the line in your graph.pulleystandardmassdynamicscartOver many years of observations and investigations, physicistsconcluded that inertial mass and gravitational mass were two different manifestations of the same property of matter. Therefore,when you write m for mass, you do not have to specify what typeof mass it is.Action-Reaction ForcesNewtons rst and second laws are sufcient for explaining andpredicting motion in many situations. However, you will discoverthat, in some cases, you will need Newtons third law. Unlike the rst two laws that focus on the forces acting on one object,Newtons third law considers two objects exerting forces on eachother. For example, when you push on a wall, you can feel thewall pushing back on you. Newtons third law states that this condition always exists when one object exerts a force onanother, the second force always exerts a force on the rst. Thethird law is sometimes called the law of action-reaction forces. To avoid confusion, be sure to note that the forces described inNewtons third law refer to two different objects. When you applyNewtons second law to an object, you consider only one of theseforces the force that acts on the object. You do not include any forces that the object itself exerts on something else. If thisconcept is clear to you, you will be able to solve the horse-cartparadox described below. The famous horse-cart paradox asks, If the cart is pulling onthe horse with a force that is equal in magnitude and opposite indirection to the force that the horse is exerting on the cart, howcan the horse make the cart move? Discuss the answer with aclassmate, then write a clear explanation of the paradox.Conceptual Problem

FAonB =

FBonANEWTONS THIRD LAWFor every action force on an object (B) due to another object(A), there is a reaction force, equal in magnitude but oppositein direction, on object A, due to object B.Chapter 1 Fundamentals of Dynamics MHR 910 MHR Unit 1 Forces and Motion: DynamicsBend a WallBend a Wall Q U I C KL A BTARGET SKILLSInitiating and planningPerforming and recordingAnalyzing and interpretingSometimes it might not seem as though anobject on which you are pushing is exhibitingany type of motion. However, the proper appa-ratus might detect some motion. Prove that youcan move or at least, bend a wall.Do not look into the laser.Glue a small mirror to a 5 cm T-head dissect-ing pin. Put a textbook on a stool beside thewall that you will attempt to bend. Place thepin-mirror assembly on the edge of the textbook.As shown in the diagram, attach a metre stick tothe wall with putty or modelling clay and restthe other end on the pin-mirror assembly. Thepin-mirror should act as a roller, so that anymovement of the metre stick turns the mirrorslightly. Place a laser pointer so that its beamreects off the mirror and onto the oppositewall. Prepare a linear scale on a sheet of paperand fasten it to the opposite wall, so that youcan make the required measurements.Push hard on the wall near the metre stick andobserve the deection of the laser spot. MeasureIthe radius of the pin (r)Ithe deection of the laser spot (S)Ithe distance from the mirror to the oppositewall (R)Analyze and Conclude1. Calculate the extent of the movement (s) or how much the wall bent using the formula s = rS2R.2. If other surfaces behave as the wall does, list other situations in which an apparentlyinexible surface or object is probably moving slightly to generate a resisting or supporting force.3. Do your observations prove that the wallbent? Suppose a literal-minded observerquestioned your results by claiming that youdid not actually see the wall bend, but thatyou actually observed movement of the laserspot. How would you counter this objection? 4. Is it scientically acceptable to use a mathe-matical formula, such as the one above, without having derived or proved it? Justifyyour response. 5. If you have studied the arc length formula inmathematics, try to derive the formula above.(Hint: Use the fact that the angular displace-ment of the laser beam is actually twice theangular displacement of the mirror.)Apply and Extend6. Imagine that you are explaining this experi-ment to a friend who has not yet taken aphysics course. You tell your friend thatWhen I pushed on the wall, the wallpushed back on me. Your friend says,Thats silly. Walls dont push on people.Use the laws of physics to justify your original statement.7. Why is it logical to expect that a wall willmove when you push on it?8. Dentists sometimes check the health of yourteeth and gums by measuring tooth mobility.Design an apparatus that could be used tomeasure tooth mobility.rod or metre stickscaleposter puttylaserdissectingpintextbookmirrorwall opposite wallRSCAUTIONFrames of ReferenceIn order to use Newtons laws to analyze and predict the motion ofan object, you need a reference point and denitions of distanceand direction. In other words, you need a coordinate system. Oneof the most commonly used systems is the Cartesian coordinatesystem, which has an origin and three mutually perpendicularaxes to dene direction. Once you have chosen a coordinate system, you must decidewhere to place it. For example, imagine that you were studyingthe motion of objects inside a car. You might begin by gluing metresticks to the inside of the vehicle so you could precisely expressthe positions of passengers and objects relative to an origin. Youmight choose the centre of the rearview mirror as the origin andthen you could locate any object by nding its height above orbelow the origin, its distance left or right of the origin, and itsposition in front of or behind the origin. The metre sticks woulddene a coordinate system for measurements within the car, asshown in Figure 1.2. The car itself could be called the frame ofreference for the measurements. Coordinate systems are alwaysattached to or located on a frame of reference.Establishing a coordinate system and dening a frame of reference are fundamental steps in motion experiments.An observer in the cars frame of reference might describe themotion of a person in the car by stating that The passenger didnot move during the entire trip. An observer who chose Earthssurface as a frame of reference, however, would describe the pas-sengers motion quite differently: During the trip, the passengermoved 12.86 km. Clearly, descriptions of motion depend verymuch on the chosen frame of reference. Is there a right or wrongway to choose a frame of reference?The answer to the above question is no, there is no right orwrong choice for a frame of reference. However, some frames ofreference make calculations and predictions much easier than do others. Think again about the coordinate system in the car.Imagine that you are riding along a straight, smooth road at a constant velocity. You are almost unaware of any motion. Then Figure 1.2Chapter 1 Fundamentals of Dynamics MHR 11Reference FramesA desire to know your location on Earth has made GPS receiversvery popular. Discussion aboutlocation requires the use offrames of reference concepts.Ideas about frames of referenceand your Course Challenge arecued on page 603 of this text.COURSE CHALLENGEthe driver suddenly slams on the brakes and your upper body fallsforward until the seat belt stops you. In the frame of reference ofthe car, you were initially at rest and then suddenly began toaccelerate. According to Newtons rst law, a force is necessary to cause amass your body to accelerate. However, in this situation youcannot attribute your acceleration to any observable force: Noobject has exerted a force on you. The seat belt stopped yourmotion relative to the car, but what started your motion? It wouldappear that your motion relative to the car did not conform toNewtons laws. The two stages of motion during the ride in a car movingwith a constant velocity or accelerating illustrate two classes offrames of reference. A frame of reference that is at rest or movingat a constant velocity is called an inertial frame of reference. When you are riding in a car that is moving at a constant velocity, motion inside the car seems similar to motion inside aparked car or even in a room in a building. In fact, imagine thatyou are in a laboratory inside a trucks semitrailer and you cannotsee what is happening outside. If the truck and trailer ran perfectlysmoothly, preventing you from feeling any bumps or vibrations,there are no experiments that you could conduct that would allowyou to determine whether the truck and trailer were at rest ormoving at a constant velocity. The law of inertia and Newtons second and third laws apply in exactly the same way in all inertialframes of reference. Now think about the point at which the driver of the car abrupt-ly applied the brakes and the car began to slow. The velocity waschanging, so the car was accelerating. An accelerating frame of reference is called a non-inertial frame of reference. Newtonslaws of motion do not apply to a non-inertial frame of reference.By observing the motion of the car and its occupant from outsidethe car (that is, from an inertial frame of reference, as shown inFigure 1.3), you can see why the law of inertia cannot apply. In the rst three frames, the passengers body and the car aremoving at the same velocity, as shown by the cross on the car seatand the dot on the passengers shoulder. When the car rst beginsto slow, no force has yet acted on the passenger. Therefore, his12 MHR Unit 1 Forces and Motion: DynamicsAlbert Einstein used the equiva-lence of inertial and gravitationalmass as a foundation of his general theory of relativity, published in 1916. According toEinsteins principle of equiva-lence, if you were in a laboratoryfrom which you could not seeoutside, you could not make any measurements that wouldindicate whether the laboratory(your frame of reference) wasstationary on Earths surface or in space and accelerating at avalue that was locally equal to g.PHYSICS FILEThe crosses on thecar seat and the dots on the passengers shoulder representthe changing locations of the carand the passenger at equal timeintervals. In the rst three frames,the distances are equal, indicatingthat the car and passenger aremoving at the same velocity. Inthe last two frames, the crossesare closer together, indicating thatthe car is slowing. The passenger,however, continues to move at the same velocity until stopped by a seat belt.Figure 1.3body continues to move with the same constant velocity until aforce, such as a seat belt, acts on him. When you are a passenger,you feel as though you are being thrown forward. In reality, the carhas slowed down but, due to its own inertia, your body tries tocontinue to move with a constant velocity. Since a change in direction is also an acceleration, the same situation occurs when a car turns. You feel as though you arebeing pushed to the side, but in reality, your body is attempting tocontinue in a straight line, while the car is changing its direction. Clearly, in most cases, it is easier to work in an inertial frame ofreference so that you can use Newtons laws of motion. However,if a physicist chooses to work in a non-inertial frame of referenceand still apply Newtons laws of motion, it is necessary to invokehypothetical quantities that are often called ctitious forces:inertial effects that are perceived as forces in non-inertial framesof reference, but do not exist in inertial frames of reference. Passengers in a high-speed elevator feel as though they are beingpressed heavily against the oor when the elevator starts movingup. After the elevator reaches its maximum speed, the feelingdisappears.(a) When do the elevator and passengers form an inertial frame of reference? A non-inertial frame of reference?(b) Before the elevator starts moving, what forces are acting onthe passengers? How large is the external (unbalanced) force?How do you know?(c) Is a person standing outside the elevator in an inertial ornon-inertial frame of reference?(d) Suggest the cause of the pressure the passengers feel whenthe elevator starts to move upward. Sketch a free-body diagram to illustrate your answer.(e) Is the pressure that the passengers feel in part (d) a ctitiousforce? Justify your answer.Conceptual ProblemINERTIAL AND NON-INERTIAL FRAMES OF REFERENCEAn inertial frame of reference is one in which Newtons rstand second laws are valid. Inertial frames of reference are atrest or in uniform motion, but they are not accelerating.A non-inertial frame of reference is one in which Newtonsrst and second laws are not valid. Accelerating frames of reference are always non-inertial.Chapter 1 Fundamentals of Dynamics MHR 13Earth and everything on it are incontinual circular motion. Earth is rotating on its axis, travellingaround the Sun and circling thecentre of the galaxy along withthe rest of the solar system. Thedirection of motion is constantlychanging, which means themotion is accelerated. Earth is anon-inertial frame of reference,and large-scale phenomena suchas atmospheric circulation aregreatly affected by Earths contin-ual acceleration. In laboratoryexperiments with moving objects,however, the effects of Earthsrotation are usually notdetectable. PHYSICS FILEYou can determine the nature of aframe of reference by analyzing itsacceleration.Figure 1.4Concept OrganizeryesnoNewtons lawsof motionframe ofreferenceat restconstantvelocitySome amusement park rides make you feel as though you are being thrown to the side, although no force is pushing you outward from the centre. Your frame of reference is moving rapidly along a curved path and therefore it is accelerating. You are in a non-inertial frame of reference, so it seems as though your motion is not following Newtons laws of motion.changingvelocityIs

a = 0?inertial frame of referenceNewtons lawsapplynon-inertial frame of referenceNewtons lawsdo not apply1.1 Section Review1. State Newtons rst law in two differentways.2. Identify the two basic situations thatNewtons rst law describes and explain howone statement can cover both situations.3. State Newtons second law in words andsymbols.4. A stage trick involves covering a tablewith a smooth cloth and then placing dinner-ware on the cloth. When the cloth is sudden-ly pulled horizontally, the dishes magicallystay in position and drop onto the table.(a) Identify all forces acting on the dishesduring the trick.(b) Explain how inertia and frictional forcesare involved in the trick.5. Give an example of an unusual frame of reference used in a movie or a televisionprogram. Suggest why this viewpoint waschosen.6. Identify the dening characteristic ofinertial and non-inertial frames of reference.Give an example of each type of frame of reference.7. In what circumstances is it necessary toinvoke cticious forces in order to explainmotion? Why is this term appropriate todescribe these forces?8. Compare inertial mass and gravitationalmass, giving similarities and differences.9. Why do physicists, who take pride in precise, unambiguous terminology, usuallyspeak just of mass, rather than distinguish-ing between inertial and gravitational mass?CCCK/UK/UMCK/UCK/UIWhat frame of reference would be the bestchoice for measuring and analyzing the performance of your catapult?IWhat forces will be acting on the payload ofyour catapult when it is being accelerated?When it is ying through the air?IHow will the inertia of the payload affect its behaviour? How will the mass of the payload affect its behaviour?Test your ideas using a simple elastic band orslingshot. Take appropriate safety precautionsbefore any tests. Use eye protection.CAUTIONUNIT PROJECT PREP14 MHR Unit 1 Forces and Motion: DynamicsThe deafening roar of the engine of a competitors tractor conveysthe magnitude of the force that is applied to the sled in a tractor-pull contest. As the sled begins to move, weights shift to increasefrictional forces. Despite the power of their engines, most tractorsare slowed to a standstill before reaching the end of the 91 mtrack. In contrast to the brute strength of the tractors, dragsterssprint to the nish line. Many elements of the two situations are identical, however, since forces applied to masses change thelinear (straight-line) motion of a vehicle.In the previous section, you focussed on basic dynamics the cause of changes in motion. In this section, you will analyzekinematics the motion itself in more detail. You will consider objects moving horizontally in straight lines. Kinematic EquationsTo analyze the motion of objects quantitatively, you will use thekinematic equations (or equations of motion) that you learned inprevious courses. The two types of motion that you will analyzeare uniform motion motion with a constant velocity and uniformly accelerated motion motion under constant accelera-tion. When you use these equations, you will apply them to onlyone dimension at a time. Therefore, vector notations will not benecessary, because positive and negative signs are all that you will need to indicate direction. The kinematic equations are summarized on the next page, and apply only to the type ofmotion indicated.In a tractor pull, vehicles develop up to 9000 horsepower to accelerate a sled, until they can no longer overcome the constantlyincreasing frictional forces. Dragsters, on the other hand, accelerate right up to the nish line.Figure 1.5Analyzing Motion 1. 2Chapter 1 Fundamentals of Dynamics MHR 15 Analyze, predict, and explainlinear motion of objects in horizontal planes. Analyze experimental data todetermine the net force actingon an object and its resultingmotion. dynamics kinematics uniform motion uniformly accelerated motion free-body diagram frictional forces coefcient of static friction coefcient of kinetic friction T E R M SK E YE X P E C T AT I O N SS E C T I O N The equations above are the most fundamental kinematic equations. You can derive many more equations by making combinations of the above equations. For example, it is some-times useful to use the relationship d = v2t 12at2. Derive this equation by manipulating two or more of the equationsabove. (Hint: Notice that the equation you need to derive is verysimilar to one of the equations in the list, with the exceptionthat it has the nal velocity instead of the initial velocity. Whatother equation can you use to eliminate the initial velocity fromthe equation that is similar to the desired equation?)Combining Dynamics and KinematicsWhen analyzing motion, you often need to solve a problem in twosteps. You might have information about the forces acting on anobject, which you would use to nd the acceleration. In the nextstep, you would use the acceleration that you determined in orderto calculate some other property of the motion. In other cases, youmight analyze the motion to nd the acceleration and then use theacceleration to calculate the force applied to a mass. The followingsample problem will illustrate this process.Conceptual Problema = vtora =v2 v1tv2 = v1 + atd =(v1 + v2)2 td = v1t +12at2v22 = v21 + 2adUniformly accelerated motionIdenition of accelerationISolve for nal velocity in terms of initial velocity, acceleration, and time interval.Idisplacement in terms of initial velocity,nal velocity, and time intervalIdisplacement in terms of initial velocity,acceleration, and time intervalI nal velocity in terms of initial velocity,acceleration, and displacementv = dtd = vtUniform motionIdenition of velocityISolve for displacement in terms of velocity and time.16 MHR Unit 1 Forces and Motion: DynamicsRefer to your Electronic LearningPartner to enhance your under-standing of acceleration and velocity.ELECTRONICLEARNING PARTNERFinding Velocity from Dynamics DataIn television picture tubes and computer monitors (cathode ray tubes),light is produced when fast-moving electrons collide with phosphor molecules on the surface of the screen. The electrons (mass 9.1 1031kg)are accelerated from rest in the electron gun at the back of the vacuumtube. Find the velocity of an electron when it exits the gun after experi-encing an electric force of 5.8 1015N over a distance of 3.5 mm.Conceptualize the ProblemIThe electrons are moving horizontally, from the back to the front of thetube, under an electric force.IThe force of gravity on an electron is exceedingly small, due to the electrons small mass. Since the electrons move so quickly, the timeinterval of the entire ight is very short. Therefore, the effect of the forceof gravity is too small to be detected and you can consider the electricforce to be the only force affecting the electrons.IInformation about dynamics data allows you to nd the electrons acceleration.IEach electron is initially at rest, meaning that the initial velocity is zero.IGiven the acceleration, the equations of motion lead to other variables of motion.ILet the direction of the force, and therefore the direction of the accelera-tion, be positive.Identify the GoalThe nal velocity, v2, of an electron when exiting the electron gunIdentify the Variables and Constants Known Implied Unknownme = 9.1 1031kgF = 5.8 1015Nd = 3.5 103mv1 = 0msav2Develop a Strategy

a = 6.374 1015ms2[toward the front of tube]Nkgis equivalent to ms2.

a = +5.8 1015N9.1 1031kgSubstitute and solve.

a =

FmWrite Newtons second law in terms of acceleration.

F = m

aApply Newtons second law to ndthe net force.SAMPLE PROBLEM Chapter 1 Fundamentals of Dynamics MHR 17continuedThe nal velocity of the electrons is about 6.7 106m/s in the directionof the applied force.Validate the SolutionElectrons, with their very small inertial mass, could be expected to reach high speeds. You can also solve the problem using the concepts of work andenergy that you learned in previous courses. The work done on the electrons was converted into kinetic energy, so W = Fd =12mv2. Therefore, v =

2Fdm =

2(5.8 1015N)(3.5 103m)9.1 1031kg = 6.679 106ms 6.7 106ms. Obtaining the same answer by two different methods is a strong validation of the results.1. A linear accelerator accelerated a germaniumion (m = 7.2 1025kg) from rest to a velocity of 7.3 106m/s over a time intervalof 5.5 106s. What was the magnitude of the force that was required to acceleratethe ion? 2. A hockey stick exerts an average force of 39 N on a 0.20 kg hockey puck over a displacement of 0.22 m. If the hockey puckstarted from rest, what is the nal velocity ofthe puck? Assume that the friction betweenthe puck and the ice is negligible.PRACTICE PROBLEMSv22 = v21 + 2adv22 = 0 + 2

6.374 1015ms2

(3.5 103m)v2 = 6.67 967 106msv2 6.7 106msApply the kinematic equation thatrelates initial velocity, acceleration,and displacement to nal velocity.18 MHR Unit 1 Forces and Motion: Dynamicscontinued from previous pageDetermining the Net ForceIn almost every instance of motion, more than one force is actingon the object of interest. To apply Newtons second law, you needto nd the resultant force. A free-body diagram is an excellent toolthat will help to ensure that you have correctly identied andcombined the forces.To draw a free-body diagram, start with a dot that representsthe object of interest. Then draw one vector to represent each forceacting on the object. The tails of the vector arrows should all startat the dot and indicate the direction of the force, with the arrow-head pointing away from the dot. Study Figure 1.6 to see how afree-body diagram is constructed. Figure 1.6 (A) illustrates a cratebeing pulled across a oor by a rope attached to the edge of thecrate. Figure 1.6 (B) is a free-body diagram representing the forcesacting on the crate.Two of the most common types of forces that inuence themotion of familiar objects are frictional forces and the force ofgravity. You will probably recall from previous studies that themagnitude of the force of gravity acting on objects on or nearEarths surface can be expressed as F = mg, where g (which isoften called the acceleration due to gravity) has a value 9.81 m/s2.Near Earths surface, the force of gravity always points toward thecentre of Earth.Whenever two surfaces are in contact, frictional forces opposeany motion between them. Therefore, the direction of the friction-al force is always opposite to the direction of the motion. Youmight recall from previous studies that the magnitudes of friction-al forces can be calculated by using the equation Ff = FN. Thenormal force in this relationship (FN) is the force perpendicular to the surfaces in contact. You might think of the normal force asthe force that is pressing the two surfaces together. The nature ofthe surfaces and their relative motion determines the value of the coefcient of friction (). These values must be determinedexperimentally. Some typical values are listed in Table 1.1.Table 1.1 Coefcients of Friction for Some Common SurfacesIf the objects are not moving relative to each other, you woulduse the coefcient of static friction (s). If the objects are moving,the somewhat smaller coefcient of kinetic friction (k) applies tothe motion. As you begin to solve problems involving several forces, youwill be working in one dimension at a time. You will select a coordinate system and resolve the forces into their components in each dimension. Note that the components of a force are notvectors themselves. Positive and negative signs completelydescribe the motion in one dimension. Thus, when you applyNewtons laws to the components of the forces in one dimension,you will not use vector notations. Surfacerubber on dry, solid surfacesrubber on dry concreterubber on wet concreteglass on glasssteel on steel (unlubricated)steel on steel (lubricated)wood on woodice on iceTeflon on steel in airball bearings (lubricated)joint in humansCoefficient ofstatic friction(s)141.000.700.940.740.150.400.100.04 FAonB. The calculated forcesagree with this relationship. You would also expect that the forceexerted by the tractor on trailer A would be the force necessary toaccelerate the sum of the masses of trailers A and B at 0.30 m/s2.FTonA = (31 500 kg + 19 600 kg)

0.30ms2

= 15 330 N 1.5 104NThis value agrees with the value above.8. A 1700 kg car is towing a larger vehicle withmass 2400 kg. The two vehicles accelerateuniformly from a stoplight, reaching a speedof 15 km/h in 11 s. Find the force needed toaccelerate the connected vehicles, as well asthe minimum strength of the rope betweenthem. 9. An ice skater pulls three small children, onebehind the other, with masses 25 kg, 31 kg,and 35 kg. Assume that the ice is smoothenough to be considered frictionless.(a) Find the total force applied to the trainof children if they reach a speed of 3.5 m/s in 15 s.(b) If the skater is holding onto the 25 kgchild, nd the tension in the arms of thenext child in line.PRACTICE PROBLEMSFTonA = Ftotal onA FBonAFTonA = 9.45 103N (5.88 103N)FTonA = 1.533 104NFTonA 1.5 104NThe force that the rst hitch must withstand isthe force that the truck exerts on trailer A.Solve the force equation above for FTonAandcalculate the value. According to Newtonsthird law, FBonA = FAonB.Ftotal = FTonA + FBonAUse the free-body diagram to help write theexpression for total (horizontal) force on trailer A.Ftotal onA = mAaFtotal onA = (31 500 kg)

0.30ms2

Ftotal onA = 9.45 103kg ms2Ftotal onA 9.5 103NUse Newtons second law to nd the total forcenecessary to accelerate trailer A at 0.30 m/s2.FAonB = mBaFAonB = (19 600 kg)

0.30ms2

FAonB = 5.88 103kg ms2FAonB 5.9 103NUse Newtons second law to nd the force nec-essary to accelerate trailer B at 0.30 m/s2. Thisis the force that the second trailer hitch mustwithstand.Chapter 1 Fundamentals of Dynamics MHR 252400 kg 1700 kgv1 = 0 km/h v2 = 15 km/ht = 11 scontinued10. A solo Arctic adventurer pulls a string of twotoboggans of supplies across level, snowyground. The toboggans have masses of 95 kgand 55 kg. Applying a force of 165 N causesthe toboggans to accelerate at 0.61 m/s2. (a) Calculate the frictional force acting on thetoboggans.(b) Find the tension in the rope attached tothe second (55 kg) toboggan. 26 MHR Unit 1 Forces and Motion: Dynamics1. 2 Section Review1. How is direction represented when ana-lyzing linear motion?2. When you pull on a rope, the rope pullsback on you. Describe how the rope createsthis reaction force.3. Explain how to calculate(a) the horizontal component (Fx) of a force F(b) the vertical component (Fy) of a force F(c) the coefcient of friction () between two surfaces(d) the gravitational force (Fg) acting on anobject4. Dene (a) a normal force and (b) theweight of an object.5. An object is being propelled horizontal-ly by a force F. If the force doubles, useNewtons second law and kinematic equations to determine the change in(a) the acceleration of the object(b) the velocity of the object after 10 s6. A 0.30 kg lab cart is observed to acceler-ate twice as fast as a 0.60 kg cart. Does thatmean that the net force on the more massivecart is twice as large as the force on thesmaller cart? Explain.7. A force F produces an acceleration awhen applied to a certain body. If the massof the body is doubled and the force isincreased vefold, what will be the effect on the acceleration of the body?8. An object is being acted on byforces pictured inthe diagram. (a) Could the objectbe acceleratinghorizontally? Explain.(b) Could the object be moving horizontally?Explain.9. Three identical blocks, fastened togetherby a string, are pulled across a frictionlesssurface by a constant force, F.(a) Compare the tension in string A to themagnitude of the applied force, F.(b) Draw a free-body diagram of the forcesacting on block 2.10. A tall person and a short person pull on a load at different angles but with equalforce, as shown. (a) Which person applies the greater horizon-tal force to the load? What effect does thishave on the motion of the load?(b) Which person applies the greater verticalforce to the load? What effect does thishave on frictional forces? On the motionof the load?K/U3B A F2 1C

F4 F1

F3

F2K/UK/UK/UK/UK/UK/UK/UK/Ucontinued from previous pageCatapulting a diver high into the air requires a force. How large aforce? How hard must the board push up on the diver to overcomeher weight and accelerate her upward? After the diver leaves theboard, how long will it take before her ascent stops and she turnsand plunges toward the water? In this section, you will investigatethe dynamics of diving and other motions involving rising andfalling or straight-line motion in a vertical plane.After the diver leaves the diving board and before she hits thewater, the most important force acting on her is the gravitational forcedirected downward. Gravity affects all forms of vertical motion.Weight versus Apparent WeightOne of the most common examples of linear vertical motion is riding in an elevator. You experience some strange sensationswhen the elevator begins to rise or descend or when it slows andcomes to a stop. For example, if you get on at the rst oor andstart to go up, you feel heavier for a moment. In fact, if you arecarrying a book bag or a suitcase, it feels heavier, too. When theelevator slows and eventually stops, you and anything you are carrying feels lighter. When the elevator is moving at a constantvelocity, however, you feel normal. Are these just sensations thatliving organisms feel or, if you were standing on a scale in the elevator, would the scale indicate that you were heavier? You cananswer that question by applying Newtons laws of motion to aperson riding in an elevator. Figure 1.8Vertical Motion 1. 3Chapter 1 Fundamentals of Dynamics MHR 27 Analyze the motion of objects invertical planes. Explain linear vertical motion interms of forces. Solve problems and predict themotion of objects in verticalplanes. apparent weight tension counterweight free fall air resistance terminal velocity T E R M SK E YE X P E C T AT I O N SS E C T I O NImagine that you are standing on a scale in an eleva-tor, as shown in Figure 1.9. When the elevator is standingstill, the reading on the scale is your weight. Recall thatyour weight is the force of gravity acting on your mass.Your weight can be calculated by using the equationFg = mg, where g is the acceleration due to gravity. Vector notations are sometimes omitted because the force due to gravity is always directed toward the centre of Earth.Find out what happens to the reading on the scale bystudying the following sample problem.When you are standing on a scale, you exert a forceon the scale. According to Newtons third law, the scale mustexert an equal and opposite force on you. Therefore, the readingon the scale is equal to the force that you exert on it.Figure 1.928 MHR Unit 1 Forces and Motion: DynamicsApparent WeightA 55 kg person is standing on a scale in an elevator. If the scale iscalibrated in newtons, what is the reading on the scale when theelevator is not moving? If the elevator begins to accelerate upwardat 0.75 m/s2, what will be the reading on the scale?Conceptualize the ProblemIStart framing the problem by drawing a free-body diagram of the person on the scale. A free-body diagram includes all of the forcesacting on the person.IThe forces acting on the person are gravity (

Fg) and the normalforce of the scale.IAccording to Newtons third law, when the person exerts a force(

FPS) on the scale, it exerts an equal and opposite force (

FSP) on theperson. Therefore, the reading on the scale is the same as the forcethat the person exerts on the scale.IWhen the elevator is standing still, the persons acceleration is zero. IWhen the elevator begins to rise, the person is accelerating at the samerate as the elevator.ISince the motion is in one dimension, use only positive and negativesigns to indicate direction. Let up be positive and down be negative.IApply Newtons second law to nd the magnitude of

FSP.IBy Newtons third law, the magnitudes of

FPSand

FSPare equal toeach other, and therefore to the reading on the scale.

Fg

FSP

aSAMPLE PROBLEM weight ofperson onscalenormal forceof scaleon personIdentify the GoalThe reading on the scale,

FSP

, when the elevator is standing still andwhen it is accelerating upwardIdentify the VariablesKnown Implied Unknownm = 55 kg

a = +0.75ms2g = 9.81ms2

FPS

Fg

FSPDevelop a StrategyWhen the elevator is not moving, the reading on the scale is 5.4 102N,which is the persons weight.When the elevator is accelerating upward, the reading on the scale is 5.8 102N.Validate the SolutionWhen an elevator rst starts moving upward, it must exert a forcethat is greater than the persons weight so that, as well as supportingthe person, an additional force causes the person to accelerate. The reading on the scale should reect this larger force. It does. The acceleration of the elevator was small, so you would expect that the increase in the reading on the scale would not increase by a large amount. It increased by only about 7%.

F = m

a

Fg +

FSP = m

a

FSP =

Fg + m

a

FSP = (mg) + m

a

FSP = (55 kg)

9.81ms2

+ (55 kg)

+0.75ms2

FSP = 580.8 N

FSP 5.8 102N[up]Apply Newtons second law to the case inwhich the elevator is accelerating upward. The acceleration is positive.

F = m

a

Fg +

FSP = m

a

FSP =

Fg + m

a

FSP = (mg) + m

a

FSP = (55 kg)

9.81ms2

+ 0

FSP = 539.55kg ms2

FSP 5.4 102NApply Newtons second law and solve for theforce that the scale exerts on the person. The force in Newtons second law is the vector sum of all of the forces acting on the person. In the rst part of the problem, the accelerationis zero.Chapter 1 Fundamentals of Dynamics MHR 29continued30 MHR Unit 1 Forces and Motion: Dynamics11. A 64 kg person is standing on a scale in anelevator. The elevator is rising at a constantvelocity but then begins to slow, with anacceleration of 0.59 m/s2. What is the sign ofthe acceleration? What is the reading on thescale while the elevator is accelerating?12. A 75 kg man is standing on a scale in an elevator when the elevator begins to descendwith an acceleration of 0.66 m/s2. What isthe direction of the acceleration? What is thereading on the scale while the elevator isaccelerating?13. A 549 N woman is standing on a scale in an elevator that is going down at a constantvelocity. Then, the elevator begins to slowand eventually comes to a stop. The magni-tude of the acceleration is 0.73 m/s2. What is the direction of the acceleration? What isthe reading on the scale while the elevator is accelerating?PRACTICE PROBLEMSAs you saw in the problems, when you are standing on a scalein an elevator that is accelerating, the reading on the scale is not the same as your true weight. This reading is called yourapparent weight. When the direction of the acceleration of the elevator is positive it starts to ascend or stops while descending yourapparent weight is greater than your true weight. You feel heavierbecause the oor of the elevator is pushing on you with a greaterforce than it is when the elevator is stationary ormoving with a constant velocity. When the direction of the acceleration is negative when the elevator is rising and slows to a stop or begins to descend your apparent weightis smaller than your true weight. The oor of the elevator is exerting a force on you that is smaller than your weight, so you feel lighter.Tension in Ropes and CablesWhile an elevator is supporting or lifting you, what is supporting the elevator? The simple answer is cables exceedingly strong steel cables. Constructioncranes such the one in Figure 1.10 also use steel cablesto lift building materials to the top of skyscrapersunder construction. When a crane exerts a force on one end of a cable, each particle in the cable exerts anequal force on the next particle in the cable, creatingtension throughout the cable. The cable then exerts aforce on its load. Tension is the magnitude of the forceexerted on and by a cable, rope, or string. How doengineers determine the amount of tension that thesecables must be able to withstand? They applyNewtons laws of motion.Mobile construction cranes canwithstand the tension necessary to lift loads ofup to 1000 t.Figure 1.10continued from previous pageTo avoid using complex mathematical analyses, you can makeseveral assumptions about cables and ropes that support loads.Your results will be quite close to the values calculated by computers that are programmed to take into account all of thenon-ideal conditions. The simplifying assumptions are as follows.IThe mass of the rope or cable is so much smaller than the massof the load that it does not signicantly affect the motion orforces involved.IThe tension is the same at every point in the rope or cable.IIf a rope or cable passes over a pulley, the direction of the tension forces changes, but the magnitude remains the same.This statement is the same as saying that the pulley is friction-less and its mass is negligible.Chapter 1 Fundamentals of Dynamics MHR 31Tension in a CableAn elevator lled with people has a total mass of 2245 kg. As the elevatorbegins to rise, the acceleration is 0.55 m/s2. What is the tension in thecable that is lifting the elevator?Conceptualize the ProblemITo begin framing the problem, draw a free-body diagram.IThe tension in the cable has the same magnitude as the forceit exerts on the elevator.ITwo forces are acting on the elevator: the cable (

FT) and gravity (

Fg).IThe elevator is rising and speeding up, so the acceleration isupward.INewtons second law applies to the problem.IThe motion is in one dimension, so let positive and negative signsindicate direction. Let up be positive and down be negative.Identify the GoalThe tension, FT, in the ropeIdentify the VariablesKnown Implied Unknownm = 2245 kg

a = 0.55ms2[up]g = 9.81ms2

FT

Fg

Fg

FT

aSAMPLE PROBLEM continuedDevelop a StrategyThe magnitude of the tension in the cable is 2.3 104N[up].Validate the SolutionThe weight of the elevator is (2245 kg)(9.81 m/s2) 2.2 104N. The tension in the cable must support the weight of the elevator andexert an additional force to accelerate the elevator. Therefore, youwould expect the tension to be a little larger than the weight of theelevator, which it is.14. A 32 kg child is practising climbing skills on a climbing wall, while being belayed(secured at the end of a rope) by a parent.The child loses her grip and dangles from thebelay rope. When the parent starts loweringthe child, the tension in the rope is 253 N.Find the acceleration of the child when sheis rst being lowered.15. A 92 kg mountain climber rappels down arope, applying friction with a gure eight (apiece of climbing equipment) to reduce hisdownward acceleration. The rope, which isdamaged, can withstand a tension of only675 N. Can the climber limit his descent to aconstant speed without breaking the rope? Ifnot, to what value can he limit his down-ward acceleration?16. A 10.0 kg mass is hooked on a spring scalefastened to a hoist rope. As the hoist startsmoving the mass, the scale momentarilyreads 87 N. Find(a) the direction of motion(b) the acceleration of the mass(c) the tension in the hoist rope 17. Pulling on the strap of a 15 kg backpack, astudent accelerates it upward at 1.3 m/s2.How hard is the student pulling on the strap?18. A 485 kg elevator is rated to hold 15 peopleof average mass (75 kg). The elevator cablecan withstand a maximum tension of3.74 104N, which is twice the maximumforce that the load will create (a 200% safetyfactor). What is the greatest acceleration thatthe elevator can have with the maximumload?PRACTICE PROBLEMS

F = (2245 kg)

9.81ms2

+ (2245 kg)

0.55ms2

FT = 23 258.2kg ms2

FT 2.3 104N[up]Substitute values and solve.

F = m

a

FT +

Fg = m

a

FT =

Fg + m

a

FT = (mg) + m

aApply Newtons second law and insert all ofthe forces acting on the elevator. Then solve forthe tension.32 MHR Unit 1 Forces and Motion: Dynamicscontinued from previous pageConnected ObjectsImagine how much energy it would require to lift an elevatorcarrying 20 people to the main deck of the CN Tower inToronto, 346 m high. A rough calculation using the equationfor gravitational potential energy (Eg = mgh), which youlearned in previous science courses, would yield a value ofabout 10 million joules of energy. Is there a way to avoidusing so much energy?Elevators are not usually simply suspended from cables.Instead, the supporting cable passes up over a pulley andthen back down to a heavy, movable counterweight, asshown in Figure 1.11. Gravitational forces acting downwardon the counterweight create tension in the cable. The cablethen exerts an upward force on the elevator cage. Most of the weight of the elevator and passengers is balanced by the counterweight. Only relatively small additional forcesfrom the elevator motors are needed to raise and lower the elevator and its counterweight. Although the elevator and counterweight move in different directions, they are connected by a cable, so they accelerate at the same rate. Elevators are only one of many examples of machines that have large masses connected by a cable that runs over a pulley. In fact, in 1784, mathematician George Atwood(17451807) built a machine similar to the simplied illustration in Figure 1.12. He used his machine to test and demonstrate the laws of uniformly accelerated motionand to determine the value of g, the acceleration due to gravity. The acceleration of Atwoods machine depended on g, but was small enough to measure accurately. In the following investigation, you will use an Atwood machine to measure g.An Atwood machine uses a counterweight to reduce acceleration due to gravity.Figure 1.12FtFgm1FtFgm2Chapter 1 Fundamentals of Dynamics MHR 33Most elevators are connected by a cable to a counter-weight that moves in the oppositedirection to the elevator. A typicalcounterweight has a mass that is the same as the mass of the emptyelevator plus about half the mass of a full load of passengers.Figure 1.11I N V E S T I G A T I O N 1-BAtwoods MachineTARGET SKILLSPredictingPerforming and recordingAnalyzing and interpretingGeorge Atwood designed his machine to demon-strate the laws of motion. In this investigation,you will demonstrate those laws and determinethe value of g.ProblemHow can you determine the value of g, theacceleration due to gravity, by using an Atwood machine?PredictionIPredict how changes in the difference betweenthe two masses will affect the acceleration ofthe Atwood machine if the sum of the massesis held constant.IWhen the difference between the two massesin an Atwood machine is held constant, predict how increasing the total mass (sum ofthe two masses) will affect their acceleration. EquipmentIretort stand IclampsImasses: 100 g (2), 20 g (1), 10 g (10), or similar identi-cal masses, such as 1 inch plate washersI2 plastic cups to hold masses Ilight stringTraditional instrumentationlab pulleylab timermetre stickProbewareSmart Pulley or photogates or ultrasonic range ndermotion analysis softwarecomputerProcedureConstant Mass Difference1. Set up a data table to record m1, m2, totalmass, d and t (if you use traditional equipment), and a.2. Set up an Atwood machine at the edge of atable, so that m1 = 120 g and m2 = 100 g. 3. Lift the heavier mass as close as possible tothe pulley. Release the mass and make themeasurements necessary for nding itsdownward acceleration. Catch the massbefore it hits the oor. IUsing traditional equipment, nd displace-ment (d) and the time interval (t) whilethe mass descends smoothly.IUsing probeware, measure velocity (v) and graph velocity versus time. Find acceleration from the slope of the line during an interval when velocity wasincreasing steadily.4. Increase each mass by 10 g and repeat theobservations. Continue increasing mass andnding acceleration until you have ve totalmass-acceleration data pairs.5. Graph acceleration versus total mass. Draw a best-t line through your data points.34 MHR Unit 1 Forces and Motion: DynamicsChapter 1 Fundamentals of Dynamics MHR 35Constant Total Mass6. Set up a data table to record m1, m2, massdifference (m), d and t (if you use traditional equipment), and a.7. Make m1 = 150 g and m2 = 160 g. Makeobservations to nd the downward accelera-tion, using the same method as in step 3.8. Transfer one 10 g mass from m1to m2. Themass difference will now be 30 g, but thetotal mass will not have changed. Repeatyour measurements.9. Repeat step 8 until you have data for vemass difference-acceleration pairs.10. Graph acceleration versus mass difference.Draw a best-t line or curve through yourdata points.Analyze and Conclude1. Based on your graphs for step 5, what type ofrelationship exists between total mass andacceleration in an Atwood machine? Useappropriate curve-straightening techniques to support your answer (see Skill Set 4,Mathematical Modelling and CurveStraightening). Write the relationship symbolically.2. Based on your graphs for step 10, what type of relationship exists between mass difference and acceleration in anAtwood machine? Write the relationshipsymbolically.3. How well do your results support your prediction? 4. String that is equal in length to the stringconnecting the masses over the pulley issometimes tied to the bottoms of the twomasses, where it hangs suspended betweenthem. Explain why this would reduce experimental errors. Hint: Consider the massof the string as the apparatus moves and howthat affects m1and m2.5. Mathematical analysis shows that the accel-eration of an ideal (frictionless) Atwood machine is given by a = g m1 m2m1 + m2. Use this relationship and your experimental results to nd an experimental result for g.6. Calculate experimental error in your value of g. Suggest the most likely causes of experimental error in your apparatus and procedure.Apply and Extend7. Start with Newtons second law in the form

a =

Fmand derive the equation for a in question 5 above. Hint: Write

F and min terms of the forces and masses in theAtwood machine.8. Using the formula a = g m1 m2m1 + m2for anAtwood machine, nd the acceleration when m1 = 2m2.9. Under what circumstances would the accel-eration of the Atwood machine be zero?10. What combination of masses would make the acceleration of an Atwood machine equal to 12 g?www.mcgrawhill.ca/links/physics12For some interactive activities involving the Atwoodmachine, go to the above Internet site and click on Web Links.WEB LINKAssigning Direction to the Motion of Connected ObjectsWhen two objects are connected by a exible cable or rope thatruns over a pulley, such as the masses in an Atwood machine,they are moving in different directions. However, as you learnedwhen working with trains of objects, connected objects move as a unit. For some calculations, you need to work with the forcesacting on the combined objects and the acceleration of the combined objects. How can you treat the pair of objects as a unitwhen two objects are moving in different directions?Since the connecting cable or rope changes only the direction ofthe forces acting on the objects and has no effect on the magnitudeof the forces, you can assign the directionof the motion as being from one end of thecable or rope to the other. You can call one end negative and the other end positive, as shown in Figure 1.13.When you have assigned the direc-tions to a pair of connected objects, youcan apply Newtons laws to the objects as a unit or to each object independently.When you treat the objects as one unit, you must ignore the tension in the ropebecause it does not affect the movement ofthe combined objects. Notice that the forceexerted by the rope on one object is equalin magnitude and opposite in direction to the force exerted on the other object.However, when you apply the laws ofmotion to one object at a time, you mustinclude the tension in the rope, as shownin the following sample problem.36 MHR Unit 1 Forces and Motion: DynamicsYou can assign the bottom of the left-hand sideof the machine to be negative and the bottom of the right-hand side to be positive. You can then imagine the connectedobjects as forming a straight line, with left as negative andright as positive. When you picture the objects as a lineartrain, make sure that you keep the force arrows in the samerelative directions in relation to the individual objects.Figure 1.13

Fg1

Fg1

Fg2

Fg2

FT

FT

FT

FT++m1m2 m1m2m1m2Motion of Connected ObjectsAn Atwood machine is made of two objects connected by a rope that runs over a pulley.The object on the left (m1) has a mass of 8.5 kg and the object on the right (m2) has a mass of 17 kg. (a) What is the acceleration of the masses?(b) What is the tension in the rope?8.5 kg17 kgm1m1m2m2 Fg1 Fg2

FT

FTSAMPLE PROBLEM Conceptualize the ProblemITo start framing the problem, draw free-body diagrams. Draw one diagram of the system moving as a unit anddiagrams of each of the two individual objects.ILet the negative direction point from the centre to the8.5 kg mass and the positive direction point from thecentre to the 17 kg mass.IBoth objects move with the same acceleration. IThe force of gravity acts on both objects.IThe tension is constant throughout the rope. IThe rope exerts a force of equal magnitude and opposite direction on each object.IWhen you isolate the individual objects, the tension in the rope is one of the forces acting on the object.INewtons second law applies to the combination of the twoobjects and to each individual object.Identify the Goal(a) The acceleration,

a , of the two objects(b) The tension,

FT

, in the ropeIdentify the VariablesKnown Implied Unknownm1 = 8.5 kgm2 = 17 kgg = 9.81ms2

Fg1

Fg2

FTDevelop a Strategy(a) The acceleration of the combination of objects is 3.3 m/s2to the right.

F = m

a

Fg1 +

Fg2 = (m1 + m2)

am1g + m2g = (m1 + m2)

a

a =(m2 m1)gm1 + m2

a =(17 kg 8.5 kg)9.8ms28.5 kg + 17 kg

a = 3.27ms2

a 3.3ms2[to the right]Apply Newtons second law to the combina-tion of masses to nd the acceleration. The mass of the combination is the sum ofthe individual masses.

FT

Fg1

Fg2

Fg2

FTm1 + m2

Fg1m1 m2+ + + Chapter 1 Fundamentals of Dynamics MHR 37continued(b) The tension in the rope is 1.1 102N.Validate the SolutionYou can test your solution by applying Newtons second law to the second mass.

Fg2 +

FT = m2

am2g +

FT = m2

a

FT = m2

a m2g

FT = (17 kg)

3.27ms2

(17 kg)

9.81ms2

FT = 111.18 N

FT = 1.1 102NThe magnitudes of the tensions calculated from the two masses independentlyagree. Also, notice that the application of Newtons second law correctly gavethe direction of the force on the second mass.19. An Atwood machine consists of masses of3.8 kg and 4.2 kg. What is the acceleration ofthe masses? What is the tension in the rope?20. The smaller mass on an Atwood machine is5.2 kg. If the masses accelerate at 4.6 m/s2,what is the mass of the second object? Whatis the tension in the rope?21. The smaller mass on an Atwood machine is45 kg. If the tension in the rope is 512 N,what is the mass of the second object? Whatis the acceleration of the objects?22. A 3.0 kg counterweight is connected to a 4.5 kg window that freely slides vertically inits frame. How much force must you exert tostart the window opening with an accelera-tion of 0.25 m/s2?23. Two gymnasts of identical 37 kg mass danglefrom opposite sides of a rope that passes overa frictionless, weightless pulley. If one of thegymnasts starts to pull herself up the ropewith an acceleration of 1.0 m/s2, what happens to her? What happens to the othergymnast?PRACTICE PROBLEMS

F = m

a

Fg1 +

FT = m1

am1g +

FT = m1

a

FT = m1g + m1

a

FT = (8.5 kg)

9.81ms2

+ (8.5 kg)

3.27ms2

FT = 111.18kg ms2

FT 1.1 102NApply Newtons second law to m1andsolve for tension.38 MHR Unit 1 Forces and Motion: Dynamicscontinued from previous pageObjects Connected at Right AnglesIn the lab, a falling weight is often used to provide a constant forceto accelerate dynamics carts. Gravitational forces acting downwardon the weight create tension in the connecting string. The pulleychanges the direction of the forces, so the string exerts a horizontalforce on the cart. Both masses experience the same accelerationbecause they are connected, but the cart and weight move at rightangles to each other. You can approach problems with connected objects such as thelab cart and weight in the same way that you solved problemsinvolving the Atwood machine. Even if a block is sliding, with friction, over a surface, the mathematical treatment is much thesame. Study Figure 1.14 and follow the directions below to learnhow to treat connected objects that are moving both horizontallyand vertically.IAnalyze the forces on each individual object, then label the diagram with the forces.IAssign a direction to the motion.IDraw the connecting string or rope as though it was a straightline. Be sure that the force vectors are in the same direction relative to each mass.IDraw a free-body diagram of the combination and of each individual mass.IApply Newtons second law to each free-body diagram.When you visualize the string straightened, the force ofgravity appears to pull down on mass 1, but to the side on mass 2. Althoughit might look strange, be assured that these directions are correct regardingthe way in which the forces affect the motion of the objects.Figure 1.14

Fg1

Fg1

Fg2

Fg2

FT

FT

FT

FT

FN

FN

Ff

Ff m2m2m1+ +m1Chapter 1 Fundamentals of Dynamics MHR 39Connected ObjectsA 0.700 kg mass is connected to a 1.50 kg lab cart by a lightweight cable passing over a low-frictionpulley. How fast does the cart accelerate and whatis the tension in the cable? (Assume that the cartrolls without friction.)Conceptualize the ProblemIMake a simplied diagram of the connectedmasses and assign forces.IVisualize the cable in a straight conguration.ISketch free-body diagrams of the forces acting on each object and of the forces acting on thecombined objects.IThe force causing the acceleration of both masses is the force ofgravity acting on mass 2.INewtons second law applies to the combined masses and to eachindividual mass.ILet left be the negative direction and right be the positive direction.Identify the GoalThe acceleration of the cart,

a , and the magnitude of the tension forcein the cable, FTIdentify the Variables and ConstantsKnown Implied Unknownm1 = 1.50 kgm2 = 0.700 kgg = 9.81ms2

a

FT

Fg1

Fg2

Fg1

Fg2

FNm2m2m2m1m1 m1 + m2 +

FT

FT

Fg2m21.50 kg0.700 kgSAMPLE PROBLEM 40 MHR Unit 1 Forces and Motion: DynamicsDevelop a StrategyThe cart accelerates at about 3.1 m/s2. Since the sign is positive, it accelerates to the right.The tension in the cable is about 4.7 N.Validate the SolutionThe acceleration of the combined masses is less than 9.81 m/s2, which is reasonable since only part of the mass is subject to unbalanced gravitationalforces. Also, the tension calculated at m2is also about 4.7 N.

F = m

a

Fg +

FT = m2

a

FT = m2

a

Fg

FT = (0.700 kg)

3.121 36ms2

(0.700 kg)

9.81ms2

FT 4.7 N24. A Fletchers trolley apparatus consists of a1.90 kg cart on a level track attached to alight string passing over a pulley and holdinga 0.500 kg mass suspended in the air.Neglecting friction, calculate (a) the tension in the string when the suspended mass is released(b) the acceleration of the trolley25. A 40.0 g glider on an air track is connectedto a suspended 25.0 g mass by a string pass-ing over a frictionless pulley. When the massis released, how long will it take the glider totravel the 0.85 m to the other end of thetrack? (Assume the mass does not hit theoor, so there is constant acceleration duringthe experiment.)PRACTICE PROBLEMS

F = m

a

FT = m1

a

FT = (1.5 kg)

3.121 36ms2

FT = 4.682 04 N

FT 4.7 NApply Newtons second law to mass 1to nd the tension in the rope.

a =(0.700 kg)

9.81ms2

0.700 kg + 1.5 kg

a = 3.121 36ms2

a 3.1ms2Substitute values and solve.

F = m

a

Fg2 = (m1 + m2)

am2g = (m1 + m2)

a

a = m2gm1 + m2Apply Newtons second law to thecombined masses and solve for acceleration. Chapter 1 Fundamentals of Dynamics MHR 41Free Fall Have you ever dared totake an amusement parkride that lets you fallwith almost no supportfor a short time? A rollercoaster as it drops froma high point in its trackcan bring you close tothe same feeling of freefall, a condition inwhich gravity is theonly force acting on you.To investigate free fallquantitatively, imagine,once again, that you arestanding on a scale in anelevator. If the cable wasto break, there were nosafety devices, and friction was negligible,what would be yourapparent weight?If gravity is the onlyforce acting on the elevator, it will acceler-ate downward at the acceleration due to gravity, or g. Substitutethis value into Newtons second law and solve for your apparent weight. The reading on the scale is zero. Your apparent weight is zero.This condition is often called weightlessness. Your mass has not changed, but you feel weightless because nothing is pushingup on you, preventing you from accelerating at the accelerationdue to gravity.FN = mg mgFN = 0ISolve for the normal force.FN mg = mg IThe force of gravity is mg.FN + Fg = mg ILet up be positive and down benegative. The total force acting on youis the downward force of gravity andthe upward normal force of the scale.Your acceleration is g downward.

F = m

aIWrite Newtons second law.42 MHR Unit 1 Forces and Motion: DynamicsWhen you are on a free-fallamusement park ride, you feel weightless.Figure 1.15 How would a person on a scale in a freely falling elevator analyze the forces that were acting? Make a free-body analysissimilar to the one in the sample problem (Apparent Weight) onpage 28, using the elevator as your frame of reference. Considerthese points.(a) To an observer in the elevator, the person on the scale wouldnot appear to be moving.(b) The reading on the scale (the normal force) would be zero.Close to Earths surface, weightlessness is rarely experienced,due to the resistance of the atmosphere. As an object collides withmolecules of the gases and particles in the air, the collisions act asa force opposing the force of gravity. Air resistance or air frictionis quite different from the surface friction that you have studied.When an object moves through a uid such as air, the force of friction increases as the velocity of the object increases. A falling objecteventually reachesa velocity at whichthe force of frictionis equal to the forceof gravity. At thatpoint, the net forceacting on the objectis zero and it nolonger acceleratesbut maintains a constant velocitycalled terminalvelocity. The shapeand orientation ofan object affects itsterminal velocity. For example, skydivers control their velocity bytheir position, as illustrated in Figure 1.16. Table 1.2 lists theapproximate terminal velocities for some common objects.Table 1.2 Approximate Terminal VelocitiesObjectlarge featherfluffy snowflakeparachutistpennyskydiver (spread-eagled)Terminal velocity (m/s downward)0.417958Conceptual ProblemChapter 1 Fundamentals of Dynamics MHR 43In 1942, Soviet air force pilot I. M. Chisov was forced to parachute from a height ofalmost 6700 m. To escape beingshot by enemy ghters, Chisovstarted to free fall, but soon lostconsciousness and never openedhis parachute. Air resistanceslowed his descent, so he probably hit the ground at about193 km/h, plowing through ametre of snow as he skiddeddown the side of a steep ravine.Amazingly, Chisov survived withrelatively minor injuries andreturned to work in less than four months.PHYSICS FILEAir resistance is of great concern tovehicle designers, who can increasefuel efciency by using body shapesthat reduce the amount of air frictionor drag that is slowing the vehicle.Athletes such as racing cyclists andspeed skaters use body position and specially designed clothing to minimize drag and gain a competitiveadvantage. Advanced computer hardware and modelling software aremaking computerized simulations of air resistance a practical alternative totraditional experimental studies usingscale models in wind tunnels.TECHNOLOGY LINKGravity is not theonly force affecting these sky-divers, who have become expertsat manipulating air friction andcontrolling their descent.Figure 1.1644 MHR Unit 1 Forces and Motion: DynamicsCANADIANS IN PHYSICSFather of the Canadian Space ProgramCan you imagine sending one of the very rstsatellites into space? How about writing a reportthat changed the entire direction of Canadas spaceefforts, or being involved in a telecommunicationsprogram that won an Emmy award? These are justa few of the accomplishments that earned John H.Chapman the nickname Father of the CanadianSpace Program.Chapman, who was born in 1921, wasa science graduate ofMcGill University inMontral. In 1951, the London, Ontario,native became sectionleader of the DefenceResearch Boards unitat Shirleys Bay,Ontario. While there,he played a key rolein several ground-breaking projects.Lift Off!Early in the history of space exploration, Canadianspace scientists focussed on the study of Earthsupper atmosphere and ionosphere. They wantedto understand the behaviour of radio waves inthese lofty regions, especially above CanadasNorth. As head of the government team research-ing this area, Chapman was a moving force in theAlouette/International Satellites for IonosphericStudies (ISIS) program. With the successful launch of Alouette I in 1962,Canada became the third nation to reach space,following the Soviet Union and the United States.Designed to last for one year, Alouette I functionedfor ten. It has been hailed as one of the greatestachievements in Canadian engineering in the pastcentury. The ISIS satellites lasted for 20 years,earning Canada an international reputation forexcellence in satellite design and engineering.During this time, Chapman brought Canadianindustry into the space age. He argued that privatecompanies, not just government laboratories, hadthe right stuff to design and build space hard-ware. As a result, Canadian industry was given asteadily increasing role in the manufacture ofAlouette II and the ISIS satellites.Connecting Canada and the WorldChapman also inuenced the very purpose forwhich Canadas satellites were built. The ChapmanReport, issued in 1967, helped turn Canadas spaceprogram away from space science and towardtelecommunications. Chapman believed that satellites could deliver signals to rural and remoteregions of the country. This was achieved in 1972,when Canada placed the Anik A1 satellite into stationary orbit above the equator and became the rst country to have its own communicationssatellite system of this type.Today, live news reports can be delivered fromremote locations, due to technology that Chapmanand his team helped pioneer in co-operation withNASA and the European Space Agency. Before theHermes satellite was launched in 1976, videotapesof news events were own to a production centreand distributed. This was a time-consumingprocess. With Hermes in place, a telecommunica-tions dish on location could beam news up to thesatellite and, from there, to anywhere in the world.Hermes was also revolutionary because it sent andreceived television signals on high frequencies thatdid not interfere with frequencies already in use.For this innovation, the Hermes satellite programwon an Emmy in 1987.At the time of his death in 1979, John Chapmanwas the Assistant Deputy Minister for Space in theCanadian Department of Communications. OnOctober 2, 1996, in recognition of his distinguishedcareer, the headquarters of the Canadian SpaceAgency was dedicated as the John H. ChapmanSpace Centre.www.mcgrawhill.ca/links/physics12For more information about the Canadian Space Agencyand the Alouette, Hermes, and ISIS space programs, visit the above Internet site and click on Web Links.WEB LINKJohn Herbert ChapmanBend a WallDescending DropsQ U I C KL A BTARGET SKILLSPerforming and recordingModelling conceptsAnalyzing and interpretingYou can observedrops of waterfalling at terminalvelocity throughcooking oil in a testtube. Use an eye-dropper to carefullyinject drops ofcold water below the surface of thecooking oil. Measurethe diameter of the drops and the speed of their descent. Analyze and Conclude1. Assume that the drops are spherical and arepure water with density 1.0 g/cm3. Using the formulas for volume of a sphere

V =43r3

and density

D = mV

, calculate the mass of each drop.2. Calculate the gravitational force and theretarding force on each drop.3. What force(s) are retarding the downwardforce of gravity acting on the drops? Comparethese forces to those acting on an objectfalling through air.4. The curved sides of the test tube act like alens, producing some optical magnicationof objects inside. Describe in detail how thismight be affecting your results.5. How well does this activity model the movement of an object through air and thephenomenon of terminal velocity? Justifyyour answer.1. 3 Section Review1. Explain why your apparent weight issometimes not the same as your true weight.2. Explain how Newtons third law appliesto connected objects that are all pulled byone end.3. How does an Atwood machine make iteasier to determine g (the acceleration due to gravity), rather than by just measuring theacceleration of a free-falling object?4. Suppose you are standing on a scale in amoving elevator and notice that the scalereading is less than your true weight.(a) Draw a free-body diagram to represent theforces acting on you.(b) Describe the elevator motion that wouldproduce the effect.5. List the simplifying assumptions usual-ly made about supporting cables and ropes.Why are simplifying assumptions needed?6. Two objects are moving in differentdirections. Under what circumstances canyou treat this as a one-dimensional problem?7. By the mid-1800s, steam-driven elevatorswith counterweights had been developed.However, they were not in common use until1852, when Elisha Otis invented an elevatorwith a safety device that prevented the eleva-tor from falling if the cable broke. How doyou think that the invention of a safe eleva-tor changed modern society?8. Describe a situation in which you couldbe standing on a scale and the reading on thescale would be zero. (Note: The scale is func-tioning properly and is accurate.) What is thename of this condition?CMCK/UK/UCCK/UK/UChapter 1 Fundamentals of Dynamics MHR 45

Fg+y+x = 30Fg||Fg||FgWhen you watch speed skiers, it appears as though there is nolimit to the rate at which they can accelerate. In reality, theiracceleration is always less that that of a free-falling object, becausethe skier is being accelerated by only a component of the force of gravity and not by the total force. Using the principles ofdynamics and the forces affecting the motion, you can predictdetails of motion along an inclined plane.Gravitational forces acting on downhill skiers have producedspeeds greater than 241 km/h, even though only part of the total gravitation-al force accelerates a skier.Choosing a Coordinate System for an InclineThe key to analyzing the dynamics and motion of objects on aninclined plane is choosing a coordinate system that simplies theprocedure. Since all of the motion is along the plane, it is conven-ient to place the x-axis of the coordinate system parallel to theplane, making the y-axis perpendicular to the plane, as shown in Figure 1.18. Figure 1.17Motion along an Incline 1. 446 MHR Unit 1 Forces and Motion: Dynamics Analyze the motion of objectsalong inclined planes. Use vector and free-body diagrams to analyze forces. Predict and explain motionalong inclined planes.E X P E C T AT I O N SS E C T I O NTo nd the compo-nents of the gravitational forcevector, use the shaded triangle.Note that

Fgis perpendicular tothe horizontal line at the bottomand Fgis perpendicular to theplane of the ramp. Since theangles between two sets of perpendicular lines must be equal, the angle () in the triangleis equal to the angle that theinclined plane makes with the horizontal.Figure 1.18The force of gravity affects motion on inclined planes, but theforce vector is at an angle to the plane. Therefore, you mustresolve the gravitational force vector into components parallel toand perpendicular to the plane, as shown in Figure 1.18. The component of force parallel to the plane inuences the accelera-tion of the object and the perpendicular component affects themagnitude of the friction. Since several forces in addition to thegravitational force can affect the motion on an inclined plane, free-body diagrams are essential in solving problems, as shown inthe sample problem below.Chapter 1 Fundamentals of Dynamics MHR 47Sliding Down an Inclined PlaneYou are holding an 85 kg trunk at the top of a ramp that slopes froma moving van to the ground, making an angle of 35 with the ground.You lose your grip and the trunk begins to slide. (a) If the coefcient of friction between the trunk and the ramp is 0.42,what is the acceleration of the trunk?(b) If the trunk slides 1.3 m before reaching the bottom of the ramp,for what time interval did it slide?Conceptualize the ProblemIdentify the Goal(a) The acceleration, a||, of the trunk along the ramp(b) The time interval, t, for the trunk to reach the end of the ramp ITo start framing the problem, draw a free-body diagram. IBeside the free-body diagram, draw a coordi-nate system with the x-axis parallel to theramp. On the coordinate system, draw theforces and components of forces acting on the trunk. ILet the direction pointing down the slope bethe positive direction.ITo nd the normal force that is needed todetermine the magnitude of the frictionalforce, apply Newtons second law to the forcesor components of forces that are perpendicu-lar to the ramp.IThe acceleration perpendicular to the ramp iszero.IThe component of gravity parallel to the trunkcauses the trunk to accelerate down the ramp.I Friction between the trunk and the rampopposes the motion.IIf the net force along the ramp is positive, thetrunk will accelerate down the ramp.ITo nd the acceleration of the trunk down theramp, apply Newtons second law to theforces or components of forces parallel to theramp.IGiven the acceleration of the trunk, you canuse the kinematic equations to nd otherquantities of motion.SAMPLE PROBLEM

FN

FN

Ff

Ff

Fg

Fg+y+x Fg||FgcontinuedIdentify the VariablesKnown Implied Unknownm = 85 kg = 0.42 = 35d = 1.3 mg = 9.81ms2vi = 0a = 0

FgFg||Fg

Ff

FNvfa||Develop a Strategy(a) The acceleration of the trunk down the ramp is 2.3 m/s2.(b) The trunk slid for 1.1 s before reaching the end of the ramp.Validate the Solution(a) Since the ramp is not at an extremely steep slope and since there is a signicantamount of friction, you would expect that the acceleration would be muchsmaller than 9.81 m/s2, which it is.(b) The ramp is very short, so you would expect that it would not take long for thetrunk to reach the bottom of the ramp. A time of 1.1 s is quite reasonable.t =

2(1.3 m)2.251 71ms2t = 1.075 st 1.1 sInsert values and solve.d = vit +12at2t2=2dat =

2daApply the kinematic equation that relates dis-placement, acceleration, initial velocity, andtime interval. Given that the initial velocitywas zero, solve the equation for the time interval.a|| =(85 kg)

9.81ms2

sin35 (0.42)(683.05 N)85 kga|| = 2.251 71ms2a|| 2.3ms2Insert values and solve.

F = m

aFg|| + Ff = ma||Ff = FNin negative directionmg sin FN = ma||a|| = mg sin FNmApply Newtons second law to the forces par-allel to the ramp. Refer to the diagram to ndall of the forces that are parallel to the ramp.Solve for the acceleration parallel to the ramp.FN = (85 kg)

9.81ms2

cos 35 + 0FN = 683.05 NInsert values and solve. Note that the accelera-tion perpendicular to the ramp (a) is zero.

F = m

aFN + Fg = maFN mg cos = maFN = mg cos + maApply Newtons second law to the forces per-pendicular to the ramp. Refer to the diagram tond all of the forces that are perpendicular tothe ramp. Solve for the normal force.48 MHR Unit 1 Forces and Motion: Dynamicscontinued from previous pageChapter 1 Fundamentals of Dynamics MHR 49Bend a WallThe Slippery SlopeQ U I C KL A BTARGET SKILLSPerforming and recordingAnalyzing and interpretingYou can determine the coefcients of static andkinetic friction experimentally. Use a coin orsmall block of wood as the object and a textbookas a ramp. Find the mass of the object.Experiment to nd the maximum angle of incli-nation possible before the object begins to slidedown the ramp (l). Then, use a slightly greaterangle (2), so that the object slides down theramp. Make appropriate measurements of dis-placement and time, so that you can calculatethe average acceleration. If the distance is tooshort to make accurate timings, use a longerramp, such as a length of smooth wood ormetal.Analyze and Conclude1. Calculate the gravitational force on the object(weight). Resolve the gravitational force intoparallel and perpendicular components.2. Draw a free-body diagram of the forces actingon the object and use it to nd the magnitudeof all forces acting on the object just before it started to slide (at angle l). Note: If theobject is not accelerating, no net force is acting on it, so every force must be balancedby an equal and opposite force.3. Calculate the coefcient of static friction, s,between the object and the ramp, using youranswer to question 2.4. Use the data you collected when the rampwas inclined at 2to calculate the accelera-tion of the object. Find the net force necessary to cause this acceleration.5. Use the net force and the parallel componentof the objects weight to nd the force of friction between the object and the ramp.6. Calculate the coefcient of kinetic friction,k, between the object and the ramp.7. Compare sand k. Are they in the expectedrelationship to each other? How well do yourexperimental values agree with standard values for the materials that you used foryour object and ramp? (Obtain coefcients of friction from reference materials.) 26. A 1975 kg car is parked at the top of a steep42 m long hill inclined at an angle of 15. Ifthe car starts rolling down the hill, how fastwill it be going when it reaches the bottom ofthe hill? (Neglect friction.)27. Starting from rest, a cyclist coasts down thestarting ramp at a professional biking track. Ifthe ramp has the minimum legal dimensions(1.5 m high and 12 m long), nd (a) the acceleration of the cyclist, ignoringfriction(b) the acceleration of the cyclist if allsources of friction yield an effective coefcient of friction of = 0.11(c) the time taken