Forces - solutions.pdf

2.1.2 Mass, Weight and Force of GravityMassThe mass of an object refers to how much (the quantity) matter there is in an object.Mass is a scalar quantity and is measured in kilograms (kg). The mass of an objectdoes not change depending on where the object is – the amount of matter presentremains the same. For example, if a person has a mass of 65 kg on the Earth, theywill have the same mass of 65 kg on the moon. Mass is a property of the object itself.Mass can also be considered to be a measure of the inertia of an object – the moremass an object has, the more force that is required to change its motion. Forexample, it takes more force to stop a truck that is moving at the same speed as atoy car. This is because the truck has more mass than the toy car.Force of Gravity and WeightThe force of gravity is a common force that everyone is familiar with. Although mostpeople are familiar with this force when it is exerted by an object such as a planet,this force is a non-contact force of attraction that is present between any two objects.The magnitude of this force is directly proportional to the mass of each object (andinversely proportional to the distance squared); 2 the bigger the masses of theobjects, the bigger the force of gravity. The force of gravity between two tennis balls,for example, is extremely weak and is not noticeable because the mass of theseobjects is fairly small; the force of gravity between the Earth and one of these tennisballs is much larger because of the large mass of the Earth. The result of this forcecan be seen if you drop a tennis ball – it is pulled toward the Earth. The Earth is alsopulled toward the tennis ball, but this will be explained later.Because the mass of the Earth is so much larger than any other object near us, theforce of gravity between an object on (or near) the Earth and the Earth itself is muchmore significant than any other force of gravity that is acting on the object; similarly,an object on the surface of the moon will experience a force of attraction with themoon that is larger than other forces of gravity that may be acting on it. We refer tothe force of gravitational attraction between an object and the planet (or othercelestial body) as the weight of the object. Where mass was a property of an object,weight is the force of gravity exerted on an object.We learned in Unit 2 that the acceleration due to gravity ( g ) on the surface of theEarth is 9.80 m/s 2 toward the center of the Earth, and does not depend on the massof the object. All objects at the same location on the Earth’s surface fall at the same2 This will be studied in more detail in a future unit.UNIT 2 Dynamics RRHS PHYSICS Page 6 of 100

Since all forces are measured in newtons ( N ), it can be seen from equation (3.1)that a newton can be defined as21 N 1 kg m / sRemember, weight is defined as the force of gravity acting on an object. Since gvaries over the Earth as well as between planets, it is obvious that the weight of anobject does change depending on location although mass does not. In themetric system, mass is measured in kilograms and weight is measured in newtons.Most people have seen video footage of the astronauts walking on the moon. Eventhough they are wearing heavy spacesuits that increase their mass to significantlymore than they are used to carrying, they appear to bounce or float somewhat whenthey walk on the moon. This is because their weight is significantly reduced on themoon as compared to the Earth, since the acceleration due to gravity is only onesixthwhat it is on Earth.UNIT 2 Dynamics RRHS PHYSICS Page 8 of 100

Check Your LearningA person has a weight of 639 N at the North Pole.a. What is the person’s mass?Remember that weight is the same as the force of gravity. Since we are atthe North Pole, the acceleration due to gravity can be obtained from Table1. Since the directions here are all toward the center of the Earth, we willuse the scalar form of the equation and ignore directions.Fg 639Ng 9.83 m / sm ?2Fg mg639 m(9.83)m 65.0kgb. What is the person’s mass at the equator?Since mass is an indication of the amount of matter present in an objectand is not dependent on location, their mass is still 65.0 kg.c. What is the person’s weight at the equator?Since we are looking for weight, we actually need to find the force of gravityon the person. Again, we can obtain the value of g from Table 1.m 65.0kgF mgg 9.78 m / sFg ?2g (65.0)(9.78) 636Nd. What is the person’s mass on the moon?Since mass is an indication of the amount of matter present in an objectand is not dependent on location, the mass is still 65.0 kg.e. What is the person’s weight on the moon?Since we are looking for weight, we actually need to find the force of gravityon the person. This time we can obtain the value of g from Table 2.m 65.0kgF mgg 1.64 m / sFg ?2g (65.0)(1.64) 107NUNIT 2 Dynamics RRHS PHYSICS Page 9 of 100

2.1.2 In Class or Homework Exercise1. You are a person in car traveling at 80 km/h. Explain in terms of inertia whathappens to you if the driver brakes suddenly.Since the person in the car is also traveling at 80 km/h, they will continuetraveling forward at 80 km/h until some force changes their motion. If the driverbrakes suddenly, the person will continue traveling forward until they hitsomething (or is restrained by their seatbelt).2. While you are in a car traveling at a constant velocity (with the windowsclosed), you throw a ball in the air. When the ball comes back down, does itland in front of your hand, in your hand, or behind your hand? Explain.Since the ball is in the car with the person, they are both traveling at the samevelocity as the car. When the ball is tossed in the air, no horizontal force actson the ball to change its motion, so it continues traveling forward at the samevelocity (inertia). It will stay above the person’s hand, which is also moving atthe same velocity. When it comes back down, it will land in the person’s hand.3. When a person diets, is their goal to lose mass or to lose weight? Explain.The person generally want to become smaller and lose body mass; if theywanted to lose weight, they could simply go to a different location (such as thetop of a mountain!)4. A person states that “he has a weight of 82 kg”. What is wrong with thisstatement? Provide values for his weight and mass.The problem is that he is using units of mass and referring it to as weight. Hismass is probably 82 kg. To find his weight, we must calculate the force ofgravity: (82)(9.80) 800NThe person’s weight would therefore be 800 N.Fg mg5. If you weigh 784 N on Earth, how much would you weigh on the moon?Since it is mass that is the same on both Earth and moon, we must find thisfirst:EarthFg 784Ng 9.80 m / sm ?2Fg mg784 m(9.80)m 80.0kgMoonm 80.0kgg 1.64 m / sFg ?2(from Table 2)Fg mg (80.0)(1.64) 131NUNIT 2 Dynamics RRHS PHYSICS Page 10 of 100

6. How much would a 3.2 kg rock weigh on Jupiter?Using Table 2 to obtain the acceleration due to gravity on Jupiter,m 3.2kgF mgg 25.9 m / sFg2 (3.2)(25.9) ? 83NgUNIT 2 Dynamics RRHS PHYSICS Page 11 of 100

2.1.3 Forces and Free Body DiagramsCommon ForcesWhen analyzing a problem involving forces, it is important to be able to identify theforces acting on the object in question. Forces are identified using the notation Fwith a subscript that is appropriate for the particular force. 3 You have already seenan example of this notation where F was used to represent the force of gravity.gIn addition to the force of gravity, there are some common forces that you should beaware of when analyzing a given situation. Some of these forces are identified in3 Both the scalar notation F and the vector notation F will be used in this unit. The scalar notationwill be used when referring to only the magnitude of the force, while the vector notation will be usedwhen referring to both the magnitude and direction of the force.UNIT 2 Dynamics RRHS PHYSICS Page 12 of 100

Table 3. All of the forces inUNIT 2 Dynamics RRHS PHYSICS Page 13 of 100

Table 3 are contact forces; the only non-contact force that we will deal with now isgravity. Other non-contact forces will be introduced in Unit 6.UNIT 2 Dynamics RRHS PHYSICS Page 14 of 100

Table 3 - Common Forces and their SymbolsName of Force Common Symbol DescriptionForce of frictionNormal ForceAir ResistanceTensionApplied ForceFfFNFaFTFpThis is the force of resistancethat a surface exerts on anobject. It acts in a directionopposite that of the motion ofthe object. This will beexamined more closely in thenext section.This is the force exerted by asurface on an object. It isalways perpendicular to thesurface and in a directionaway from the surface.This is similar to the force offriction, but refers to the forceof resistance exerted by airmolecules on an object. It actsin a direction opposite that ofthe motion of the object.This is the force exerted by arope. The tension force isdirected along the length ofthe rope and pulls equally onthe objects on the oppositeends of the rope.A force applied to an object bya person or another object.Before trying to analyze a complex situation, it is first necessary to identify all of theforces acting on the object in question.UNIT 2 Dynamics RRHS PHYSICS Page 15 of 100

Check Your LearningIdentify the forces (with their direction) acting on the book in each of the followingsituations:a. A book is at rest on a table.F - force of gravity acting downwardgFN- the normal force of the table pushing upb. A book is being pushed by a person horizontally to the right at a constantspeed.F - force of gravity acting downwardgFN- the normal force of the table pushing upFp- the force of the person pushing to the rightFf- the force of friction acting on the box to the leftc. The book in the previous part is let go, allowing it to slow down and come torest.Since there is no longer any interaction between the person and the book,there is no longer a F acting on the book. The other forces remain the same.Fg- force of gravity acting downwardpFN- the normal force of the table pushing upFf- the force of friction acting on the box to the leftd. A book is falling through the air, accelerating downward.F - force of gravity acting downwardgFa- force of air resistance acting upwardUNIT 2 Dynamics RRHS PHYSICS Page 16 of 100

Free Body DiagramsThere are often so many forces in a particular situation that it can be confusing andintimidating to try and analyze the motion. A free body diagram is essential whensetting up a problem involving forces. This diagram identifies all of the forces(including their directions) acting on a single object, and only the forces acting onthat object.For simplicity, we will be using a dot to represent the object in most of our free bodydiagrams. When drawing a free body diagram, each force should be represented byan arrow that is directed outward from the body in the direction of the force. Thearrows should be drawn to approximate scale; in other words, the larger the force (ifknown) the longer the arrow should be. Each force must be labeled with anappropriate symbol identifying the force..Consider the following example:Example 1:Draw a free body diagram for a box sitting on a table.Solution:First we can draw a picture representing this situation:Next, we must ask ourselves what forces are acting on the box in this situation?Obviously, gravity ( F ) is acting on the box in a downward direction. This will be onegof the forces acting on the box. It is obviously not the only force acting on the box,though, since the box is not falling. The table is pushing up on the box. Although wecould refer to this force as Ftor Ftable, the table is a surface and the force exerted bya surface is referred to as a normal force ( FN).We now know that there is an upward force FNand a downward forceFgacting onthe box. These are the only 2 vertical forces present. Are there any horizontal forcespresent? It is not moving horizontally and there is no obvious object applying aUNIT 2 Dynamics RRHS PHYSICS Page 17 of 100

horizontal force on the object, so there are no horizontal forces present in thissituation.The last thing that we must consider is the relative size of the two forces. Since theobject is not moving (and there is therefore no acceleration), we know that the totalforce upward must be the same as the total force downward so that the net force iszero; therefore, when we draw our force vectors we should make sure that they aredrawn the same length.By looking at the length of the force vectors in a free body diagram, it can bedetermined in what direction (if any) the net force is pointing. Remember, net force isthe total overall force that is obtained by adding all of the forces (with theirappropriate signs).Example 2:Consider the following free body diagram:In what direction is the net force?Solution:If we look at the vertical forces, we see that the force upward is equal in magnitudeto the force downward (since the force vectors are the same length); therefore, thevertical net force is zero. If we look at the horizontal forces, it is obvious that the forceto the right is bigger than the force to the left; therefore, there is a net force to theright and the object will have an acceleration to the right.UNIT 2 Dynamics RRHS PHYSICS Page 18 of 100

Check Your LearningDraw a free body diagram for the book in each of the following situations:a. A book is at rest on a table.Fg- force of gravity acting downwardFN- the normal force of the table pushing upb. A book is being pushed by a person horizontally to the right at a constantspeed.Fg- force of gravity acting downwardFN- the normal force of the table pushing upFp- the force of the person pushing to the rightFf- the force of friction acting on the box to the leftc. The book in the previous part is let go, allowing it to slow down and come torest.Since there is no longer any interaction betweenthe person and the book, there is no longer a Facting on the book. The other forces remain thesame.F - force of gravity acting downwardgFN- the normal force of the table pushing upFf- the force of friction acting on the box to theleftd. A book is falling through the air, accelerating downward.pFg- force of gravity acting downwardFa- force of air resistance acting upwardUNIT 2 Dynamics RRHS PHYSICS Page 19 of 100

2.1.3 In Class or Homework Exercise1. Draw free body diagrams for the following situations:a. A car sitting at rest on the road.b. A car moving at a constant velocity along the highway.c. A car accelerating to the right along the highway.d. A car coasting to a stop (to the right) along the highway.UNIT 2 Dynamics RRHS PHYSICS Page 20 of 100

2. Draw free-body diagrams for the following situations:a. A pail hanging from a rope is held motionless in a well.b. A pail hanging from a rope is lifted at a constant speed.c. A pail hanging from a rope is lifted with an upward acceleration.3. Predict the motion that each object would undergo based on the free-bodydiagram provided:a.It will accelerate downward.b.UNIT 2 Dynamics RRHS PHYSICS Page 21 of 100

It will accelerate to the right.c.It will accelerate upward.UNIT 2 Dynamics RRHS PHYSICS Page 22 of 100

2.1.4 FrictionIn many idealized situations, friction is sometimes ignored in order to simplify theproblem; however; friction must be taken into account for most practical situations.We will be concerned here with the friction between two solid surfaces (there is alsoa rolling friction, air friction, and friction in fluids but we will not be dealing with thesetypes in depth). When dealing with solids, even smooth surfaces are rough on amicroscopic scale. Friction between two surfaces exists because of an interactionbetween these surfaces - the high points of one surface temporarily bond to the highpoints of the other surface. There can also be a lot of friction between extremelysmooth surfaces, since the molecules of each surface are extremely close to oneanother and can form bonds easily.Before we talk about friction, it is necessary to again mention the normal force, whichwas discussed previously. Remember that the normal force is the force exerted byone surface on the other (or in other words, the force pushing the surfaces together).By definition, the normal force is always perpendicular to the surface. Since thenormal force tells us with how much force the two objects are being pushed together,it is logical to assume that an increased normal force will increase the force offriction.Consider a situation where a box is sitting on a horizontal table, as shown below.There is no general equation to calculate a normal force; in order to find the normalforce, it is necessary to draw a free body diagram for the box:In this diagram, notice that the only two forces acting on the box are the downwardforce of gravity and the upward force that the table is exerting (which is the normalforce). Since there is no vertical acceleration, then we know that FN Fgin thissituation. Although we will often see this relationship ( FN Fg), it is not alwaystrue!!! It is necessary in every problem to draw a free body diagram anddetermine how to calculate the normal force from the free body diagram. (seeExample 2 in this section)UNIT 2 Dynamics RRHS PHYSICS Page 23 of 100

Kinetic FrictionWhen a body is already in motion, a force of kinetic friction (also sometimes calledsliding friction) always acts to oppose the sliding of the two surfaces past each other.The magnitude of this kinetic friction depends on two things:1. The nature of the two sliding surfaces – different surfaces will have differentamounts of friction.2. The normal force – the bigger the normal force, the more force there ispushing the two surfaces together.For a given pair of surfaces, it can be found that the force of friction (proportional to the normal force ( FN)Ff) isF FfNTo write the proportionality as an equation, a constant is needed. This constant canbe found by plotting a graph of F vs FN:fThe slope of this graph will be different for different surfaces; the value obtained forthis slope is referred to as the coefficient of kinetic friction kand the equation forthe force of friction can then be written asF F(3.2)f k NThis coefficient of friction depends on the two surfaces in question. For example, iceon ice may have a of 0.03, while for wood on wood it may be 0.2. See Table 4 forksome more sample values of k. These are just approximate values, since thecoefficient of kinetic friction depends on whether surfaces are wet or dry, if they havebeen sanded or not, etc. The coefficient of kinetic friction is independent of speed orsurface area.Notice that if you rearrange equation (3.2) to solve for k,UNIT 2 Dynamics RRHS PHYSICS Page 24 of 100

kFFfNthat kis just a ratio of two forces. The units (newtons in the numerator and in thedenominator) cancel out so that there are no units for the coefficient of friction.Static FrictionWhen a body is at rest (relative to the surface that it is in contact with), a force ofstatic friction always acts to resist any attempt to start a body moving. This type offriction is modeled very similarly to kinetic friction:F F(3.3)f s NFor Ff (max) s Nwhere sis the coefficient of static friction. This coefficient of static frictiongenerally has a larger value than the coefficient of kinetic friction. As with thecoefficient of kinetic friction, the coefficient of static friction does not have any units.The equation above represents the maximum value of the force of static friction. If abox is sitting on a table with no horizontal force being applied, there is no frictionforce. If you apply a small force and the box still doesn’t move, then the force ofstatic friction is exactly balancing your force. As your force increases, so does theforce of static friction until you apply enough force to overcome the maximum valueof the static friction force.Think of the force of static friction as an elastic band connecting molecules from eachsurface. You have to exert enough force to stretch the band past its breaking point toovercome the friction; if you exert less force than the maximum force of the band, theband will stretch just enough to counteract your force.Note that the force of friction and the normal force are both forces exerted by onesurface on the other, but they are never in the same direction. The normal force isalways perpendicular to the surfaces, while the force of friction is always parallel tothe surfaces.Friction can be a hindrance, since it often opposes what we are trying to do. Weoften try to minimize friction using lubricants or a layer of air between the twosurfaces. Can friction ever be helpful? Think about it and suggest some examples.UNIT 2 Dynamics RRHS PHYSICS Page 25 of 100

Table 4 - Sample values of kinetic and static coefficients of frictionSurfacesksWood on wood 0.2 0.4Ice on ice 0.03 0.1Rubber on dry concrete 0.8 1.0Rubber on wet concrete 0.5 0.7Rubber on other surfaces 1 1-4Metal on metal (lubricated) 0.07 0.15Note that most of the time the coefficient of friction is less than one, but this is notalways true.Example 1A 7.0 kg box is being pushed horizontally at a constant speed. If the coefficient offriction is 0.30, how much force is being used to push the box?SolutionFirst, draw a free body diagram.Looking at our vertical forces, we see that Fg FN(since these are the only verticalforces and must balance one another); thereforeFN Fg mg (7.0)(9.80) 68.6NKnowing the normal force FN, we can now find the force of frictionFf FN (0.30)(68.6) 21NFf:UNIT 2 Dynamics RRHS PHYSICS Page 26 of 100

Since the speed is constant, the net force must be zero. The force that is pushing thebox must balance the force of friction (the net force must be zero since there is nochange in motion).Fp Ff 21NExample 2:You are pushing horizontally on a book against a wall so that it does not slide downthe wall.a. Draw a free body diagram for this situation. What condition is necessary for thebox to not slide down the wall?b. If you lessen the horizontal push that you are exerting, the box will start to slidedown the wall. Explain why this happens.Solution:a. Assuming that you are pushing to the rightIn order for the box to not slide down the wall, the force of gravity must bebalanced by the force of friction ( Fg Ff)b. Notice in the free body diagram that FNandFpmust be equal, since the box isnot accelerating horizontally (the net horizontal force must be zero). If youpush with less force, then the normal force that the wall is exerting on the boxwill be less. Since Ff FN, this means that the Ffwill be less. Since Ff Fg,the box will now accelerate down the wall.UNIT 2 Dynamics RRHS PHYSICS Page 27 of 100

Check Your LearningA friend pushes a 625 g textbook horizontally along a table at a constant velocity with3.50 N of force.a. Determine the normal force supporting the textbook.As can be seen in the free body diagram, the normal force must equal the forceof gravity (since there is no vertical acceleration).m 625g 0.625kgg 9.80 m / sFFpN 3.50N ?2FN Fg mg (0.625)(9.80) 6.13Nb. Calculate the force of friction between the book and the bench.Since the textbook has a constant velocity, the net force must be zero;therefore, the force pushing to the right must equal the force of friction to theleft:F Ffp 3.50Nc. Calculate the coefficient of friction between the book and the bench.Since we already know the force of friction and the normal force, we can findthe coefficient of friction using equation (3.2)F 3.50NF FFfN ? 6.13NfN3.50 (6.13) 0.571d. Which coefficient of friction have you found, static or kinetic?Since the textbook was moving at a constant velocity, this is the coefficient ofkinetic friction.UNIT 2 Dynamics RRHS PHYSICS Page 28 of 100

2.1.4 In Class or Homework Exercise1. A force of 280 N is required to start a 24.0 kg box moving across a concretefloor. What is the coefficient of static friction between the box and the floor?In order to start the box moving, the force of the push must be greater than theforce of friction.FFp 280NN FFp Ffg mgFp FNm 24.0kg (24.0)(9.80)280 (235)s ? 235N 1.22. The coefficient of static friction between a 5.4 kg box and the floor is 0.43, whilethe coefficient of kinetic friction is 0.26.a. How much force is required to start the box moving?Since the box is not yet moving, we must use static friction (and Fp Ffs 0.43FN FFp Ffgm 5.4kg mgFp FN (5.4)(9.80)Fp 0.43(52.9) 52.9NF 23NpUNIT 2 Dynamics RRHS PHYSICS Page 29 of 100

. How much force is needed to keep the box moving at a constantspeed?Since the box is moving at a constant speed, we must use kineticfriction (and Fp Ff) 0.26km 5.4kgFp Ff FN (0.26)(52.9) 14N3. Suppose that you are trying to push a book against a vertical wall and you findthat it takes a horizontal force of 63 N in order to prevent it from falling. If themass of the book is 2.2 kg, what is the coefficient of static friction between thebook and the wall?Fp 63Nm 2.2kgg 9.80 m / s2FN Fp 63 N (since there is no horizontal acceleration)If the box is not sliding down the wall, then Ff FgsfN (63) 2.2(9.80)sF Fg 0.34s F mgUNIT 2 Dynamics RRHS PHYSICS Page 30 of 100

4. You are pushing horizontally on a 3.0 kg block of wood, pressing it against awall. If the coefficient of static friction between the block and the wall is 0.60,how much force must you exert on the block to prevent it from sliding down? 0.60m 3.0kgg 9.80 m / s2If the box is not sliding down the wall, then Ff FgsF FfN(0.60) F 3.0(9.80)FNN Fg mg 49NAnd since Fp FN(since there is no horizontal acceleration)Fp FN 49 NUNIT 2 Dynamics RRHS PHYSICS Page 31 of 100

2.1.5 Module SummaryIn this module you have learned thato Inertia is the tendency of an object to maintain its motion and that a net forceis requires to change the motion of an object.o The force of gravity can be found using the equationweight of an object is the force of gravity acting on that object.Fg mg and that theo Free body diagrams can be used to identify and analyze the forces in aproblem.o The force of friction can be found using the equation Ff FN.UNIT 2 Dynamics RRHS PHYSICS Page 32 of 100

MODULE 2.2 NEWTON’S LAWS OF MOTION ANDAPPLICATIONS2.2.1 Newton’s LawsNewton’s First LawNewton’s First Law is sometimes called the Law of Inertia, as it very closelyresembles Galileo’s concept of inertia. Newton developed Galileo’s idea more fully toinclude the effect of a net force. Remember that net force is the overall externalforce that is applied to an object.Consider two people pushing on a box, one to the left and one to the right. Sinceforce is a vector, we can assign signs to each of the forces. We will consider theforce to the right to be positive, and the force to the left to be negative. If the personpulling to the right exerts a bigger force, then the net force is positive; if the personpulling to the left exerts a bigger force, than the net force is negative. If the forces areequal in magnitude, then the net force is zero.Remember that we said inertia is the tendency of an object to maintain its motion. If itis at rest, it will tend to stay at rest; if it is moving at a constant velocity it will continuemoving at a constant velocity. The inertia of an object can only be changed byapplying a net force.Newton’s First Law: An object in uniform motion (or at rest) will remain in uniformmotion (or at rest) unless acted on by an outside net force.Consider a rugby ball at rest in someone’s hand. It stays at rest unless acted on bysome force; It does not spontaneously start to move. Once the ball is thrown, itcontinues moving at a constant horizontal speed through the air until it hitssomething. As you can see, a force must be applied to stop the ball. This outsideforce changes the ball’s motion.Inertial Frames of ReferenceAn inertial frame of reference can be defined as one in which Newton’s First Lawapplies. For example, consider a car that is moving at a constant speed. If you placeUNIT 2 Dynamics RRHS PHYSICS Page 33 of 100

a cup on the dash, it stays at rest relative to the car and does not appear to bemoving from your point of view (which makes sense since there are no horizontalforces acting on it). The car is an inertial frame of reference.Consider now what happens if somebody puts the brakes on the car so that the carbegins slowing down quickly. The cup begins accelerating forward (from your point ofview), even though no force is acting on it. The car is no longer an inertial frame ofreference since Newton’s First Law does not seem to hold.Somebody outside the car on the ground, however, sees the cup continuing itsuniform motion (no force acting on it) and the car slowing down because of friction.This is because the person outside the car is on the ground, which is an inertialframe of reference.Any frame of reference that moves at a constant velocity relative to an inertial frameof reference is also an inertial frame of reference; any frame of reference thataccelerates relative to an inertial frame of reference is not an inertial frame ofreference (it can be referred to as a noninertial frame of reference). In this course,we will be dealing with inertial frames of reference.UNIT 2 Dynamics RRHS PHYSICS Page 34 of 100

Check Your Learning1. A physics book is motionless on the top of a table. If you give it a hard push, itslides across the table and slowly comes to a stop. Use Newton’s first law ofmotion to answer the following questions.(a) Why does the book remain motionless before the force is applied?The book is rest; it remains at rest since there is no outside force actingon it.(b) Why does the book move when the hand pushes on it?The hand provides an outside force, so the book no longer remains atrest.(c) Why does the book eventually come to a stop?When you remove your hand, the only horizontal force acting on the bookis friction. Because of this outside force, the book does not maintain itsuniform motion and begins slowing down.(d) Under what conditions would the book remain in motion at a constantspeed?If there were no friction or if you push the book with a force exactly equalto friction (so that the net force were zero) then Newton’s First Law appliesand the book remains in uniform motion.2. What determines if a frame of reference is inertial or not? Give an example ofeach type of frame of reference.A frame of reference is inertial if Newton’s First Law is true. This will occur if theframe of reference is at rest or moving at a constant velocity. A frame ofreference is not inertial if Newton’s First Law does not hold; this will occur if theframe of reference is accelerating.An example of an inertial frame of reference would be an elevator moving at aconstant speed. If the elevator starts accelerating upward, it is no longer aninertial frame of reference.UNIT 2 Dynamics RRHS PHYSICS Page 35 of 100

Newton’s Second LawNewton’s First Law addresses what happens if there is no net force acting on theobject – it maintains uniform motion; but what about the situation where there is a netforce? A net force must provide an acceleration if uniform motion is no longer beingmaintained. How much will an object accelerate if there is a net force?Consider pushing a child on a wagon. The harder that you push, the more quickly thewagon’s velocity will change; it will therefore have a larger acceleration. So the largerthe net force, the larger the acceleration. Acceleration can be said to be directlyproportional to the net force ( a F ).netConsider now that the child is in a car instead of on the wagon. The car has muchmore mass than the wagon, and it is obvious that the same force that was applied tothe wagon will not change the motion of the car as quickly; therefore, theacceleration is smaller. The larger the mass, the smaller the acceleration.Acceleration can be said to be inversely proportional to the mass ( a 1 ). Becausema larger object has a smaller acceleration (for the same force applied), we say that abody with a larger mass has more inertia.Putting these proportionalities together, we geta 2But remember from equation (3.1) we said that a newton was defined as a kg m / s .The constant in this equation is therefore 1 and the proportionality can then bewritten as an equationF netma F netmor, more commonly asFnet ma(3.4)UNIT 2 Dynamics RRHS PHYSICS Page 36 of 100

Newton’s Second Law: The net force needed to accelerate an object is aproduct of the object’s mass and acceleration.Fnet maNotice in this equation that both net force and acceleration are vectors. They musthave the same direction – the acceleration is always in the same direction as the netforce.It is important to note that the net force here is an overall force requirement – otherforces must act to provide this net force. Net force should never be a force vectorthat appears in a free body diagram.ExampleA 1300 kg car is moving at a constant speed when the brakes are applied, providinga frictional force of 6500N. What is the acceleration?Solution:Drawing a free body diagram,m 1300kgFfa ? 6500NThe acceleration that we are looking for here is a horizontal acceleration (the car isslowing down horizontally). Since the net force and the acceleration must be in thesame direction, the net force can only be provided by horizontal forces. In this case,the only horizontal force is friction. (The vertical forces of gravity and the normalforce balance one another, resulting in no vertical acceleration)Using Newton’s Second Law (and using the right as the positive direction)F manet6500 (1300)aa 5.0 m / s2UNIT 2 Dynamics RRHS PHYSICS Page 37 of 100

Notice that the acceleration is negative, meaning that it is to the left. This is thesame direction that the force of friction was pointing.UNIT 2 Dynamics RRHS PHYSICS Page 38 of 100

Check Your LearningA race car has a mass of 710 kg. It starts from rest and travels 40.0 m in 3.0 s. Whatnet force is applied to it?m 710kgt 3.0sv 0id40.0mWe must first find the acceleration using our kinematics equations:d v t ati121240.0 0 a(3.0)a 8.9 m / s222We can then apply Newton’s Second law,Fnet ma (710)(8.9) 6300NNote that the force is positive, meaning that it is in the same direction as theacceleration.UNIT 2 Dynamics RRHS PHYSICS Page 39 of 100

Newton’s Third LawSuppose that you punch a brick wall. You are obviously applying a force to the wall.Does your hand hurt afterwards? Of course it does. The reason that your hand hurtswhen you punch the wall is that the wall exerts a force on your hand equal to (but inthe opposite direction) that exerted by your hand on the wall. The harder you hit thewall, the bigger the force that the wall exerts on your hand and the more that it hurts.Suppose that you and a friend are on skateboard (or roller skates) facing oneanother. Your friend pushes you. Which one of you moves? The answer is that youboth move. While it is true that he exerted a force on you, causing you to move, youalso exert a force on him that is the same magnitude; therefore, he will move too.Newton’s Third Law addresses this reaction force. Anytime that an object applies aforce to something, it receives the same force back.Newton’s Third Law: For every action force, there is an equal and oppositereaction force.These forces always act on different objects!Newton’s Third Law and Net ForceSince Newton’s Third Law says that for every force there is an equal and oppositereactive force, how does anything ever move? A common misconception is thatthese forces will cancel out, giving a net force of zero; however, it is important toremember that these forces act on different objects.Consider the situation where we are picking up a ball. If we exert an upward force onthe ball, the ball exerts an equal and opposite force downward on our hand. Butremember, when analyzing the forces acting on the object, we must draw a free bodydiagram that includes only the forces acting on that object. In this case, the object isthe ball so the free body diagram would look like this:where Fpis the force that your hand is exerting on the ball. As long as this force isgreater than the force of gravity, the ball will accelerate upward. Notice that theforce that the ball exerts on your hand is not included since this force is actingon a different object and cannot contribute to the net force on the ball.UNIT 2 Dynamics RRHS PHYSICS Page 40 of 100

The net force acting on an object should never include the two forces that aredescribed by Newton’s Third Law since these forces never act on the same object.ExampleSuppose you are floating around in space (many km from any planet so that you feelno gravity) outside of your spaceship. You get frustrated and decide to kick yourspaceship. Does your foot hurt?SolutionYes, your foot will hurt. Even though there is no gravity, Newton’s Third law stillapplies. If you kick the spaceship, it applies an equal and opposite force on your foot.UNIT 2 Dynamics RRHS PHYSICS Page 41 of 100

Check Your LearningA 60.0 kg boy and a 40.0 kg girl use an elastic rope while engaged in a tug of war ona frictionless icy surface. If the acceleration of the girl toward the boy is 3.0 m/s 2 ,what is the acceleration of the boy toward the girl?Looking at the girl first,m 40.0kga 3.0 m / sFT ?2We can calculate the force on the girl using Newton’s Second Law.FnetFT ma ma (40.0)(3.0) 120NAccording to Newton’s Third Law, an equal and opposite force will be applied tothe boy:UNIT 2 Dynamics RRHS PHYSICS Page 42 of 100

m 60.0kga ?FT120NNewton’s Second Law can be used again here.FnetFT ma ma120 (60.0)aa 2.0 m / s2Notice that the negative sign indicates that the boy’s acceleration is in theopposite direction.UNIT 2 Dynamics RRHS PHYSICS Page 43 of 100

2.2.1 In Class or Homework Exercise1. If you are in a car that is struck from behind, you can receive whiplash.(a) Using Newton’s laws of motion, explain what happens.Your body is initially at rest; it therefore needs a force to accelerate itforward (1 st Law). The back of the seat applies this force to yourbody, but your head is above the seat. In order for your head toaccelerate forward at the same rate as your body, a force must beapplied to it. This force must come from your neck, as nothing elseis in contact with your head. This strain that your neck experiencesapplying this force is what results in the whiplash.(b) How does a headrest reduce whiplash?The headrest is in contact with your head and can therefore exertthe force needed to accelerate your head with the rest of your body.Your neck does not have to apply the required force in this situation.2. If you push forward on the back of an automobile, does that mean that itsvelocity has to be in the forward direction? Explain and give an example.No. If the force that you are exerting is the only horizontal force, this meansthat the net force is in the direction that you are pushing and theacceleration would have to be in that direction, not the velocity. The carcould be moving toward you and slowing down.There could also be other forces acting on the car. For example, the carcould be on a hill with you pushing up the hill. In this case, the net forcecould be down the hill even though you are pushing up the hill. In this case,the acceleration would not even be in the same direction that you arepushing.3. According to legend, a horse learned Newton’s laws. When told to pull acart, he refused, saying that if he pulled the cart forward, according toNewton’s Third Law there would be an equal force backwards. Thus, therewould be balanced forces and, according to Newton’s Second Law, the cartwould not accelerate. How would you reason with this rather weird horse?Although there is an equal force backwards, the two forces are acting ondifferent objects. The horse is exerting a force in the positive direction onthe cart; the cart will exert a negative force on the horse. To determine ifthe cart accelerates, it is necessary to look at only the forces acting on thecart:UNIT 2 Dynamics RRHS PHYSICS Page 44 of 100

Since the only other horizontal force is friction, as long as the force that thehorse is exerting is larger than the force of friction the cart will accelerate inthe positive direction.4. Suppose that you exert a force on a black block and measure itsacceleration. You then exert the same force on a brown block. Itsacceleration is three times that of the black block. What can you concludeabout the masses of the blocks?We will assume no friction in this explanation, so that the force that you areexerting is the net force in this situation. Since according to Newton’sFSecond Law a , if the acceleration is 3 times as large (and the forcemstays the same) the mass must be only 1/3 as large.5. A common trick involves covering a table with a cloth and placingdinnerware on the cloth. When the cloth is suddenly pulled horizontally, thedishes stay in position and drop onto the table.(a) Identify all forces acting on the dishes during the trick.Assuming that the cloth is being pulled to the left, the free bodydiagram on one of the dishes would look like this:Notice that the only horizontal force acting on the dishes is the forceof friction between the dishes and the cloth. It is this force of frictionthat attempts to accelerate the dishes.(b) Explain how the trick works.When you pull the cloth, you are giving it an acceleration. The forceof friction attempts to accelerate the dishes at the same rate, butbecause of inertia they attempt to stay at rest. The force of friction islimited by the coefficient of friction and the normal force, so there isUNIT 2 Dynamics RRHS PHYSICS Page 45 of 100

a maximum force of friction that the cloth can apply to the dishes;since there is a maximum force of friction, there is also a maximumacceleration that can be given to the dishes.If the acceleration of the cloth is greater than the acceleration thatcan be provided to the dishes, the cloth will slide out from under thedishes and they will be left behind on the table. That is why the keyto the trick is to pull the cloth quickly; pull it too slow and the disheswill keep up with the cloth and fall on the floor. This trick is alsoeasier with a smaller coefficient of friction between the cloth and thedishes.Demonstration: This trick can easily be demonstrated in class. If adish and cloth are not available, a piece of paper and a mass can beused to show the same effect.6. A stone hangs by a fine thread from the ceiling and a section of the samethread dangles from the bottom of the stone. If a person gives a sharp pullon the dangling thread, where is the thread likely to break, below the stoneor above it? What if the person gives a slow and steady pull? Explain youranswers.A demonstration is easy to do to answer what happens in this situation.Simply tie a two threads to a mass (500 g or so). Use one thread to holdthe mass in the air, and pull on the other thread hanging below.Quick Pull: A quick pull should result in the thread breaking below themass. The mass has a tendency to stay at rest (Newton’s First Law), sodoes not move much (or attempt to move) initially. The pull on the bottomthread attempts to accelerate the mass (Newton’s Second Law). If the pullis quick enough and large enough, the tension in the bottom thread is largeenough to break it before the force is transferred to the top thread.Slow Pull: A slow pull should result in the thread breaking above the mass.As most trainees may have predicted, the top thread is subjected to boththe pull on the bottom thread and the weight of the mass. The bottomthread does not have to support the weight of the mass and is onlysubjected to the pull on the thread. Since there is more tension in the topthread, it is more likely to break.7. A force of 9000. N is used to stop a 1500 kg car traveling at 20.0 m/s. Whatbraking distance is needed to bring the car to a halt?UNIT 2 Dynamics RRHS PHYSICS Page 46 of 100

Using the forward direction as the positive direction,Fnet9000.Nm 1500kgv 20.0 m / svif 0Fnet ma9000. 1500aa 6.0 m / s2d2 2vf vi2a0 20.02( 6.0) 33m2 28. When you drop a 0.40 kg apple, Earth exerts a force on it that causes it toaccelerate at 9.8 m/s 2 . According to Newton’s third law, the apple mustexert an equal and opposite force on Earth. If the mass of the Earth is5.9810 24 kg, what is the Earth’s acceleration?Looking at the apple first, we can see that the only force acting on it is theforce of gravitym 0.40kga 9.8 m / sFg ?2FnetFg ma ma (0.40)(9.8) 3.92NAccording to Newton’s Third Law, there will be an equal and opposite forceupward on the Earth:m 5.9810Fga ?3.92N24kgFnet ma 243.92 (5.98 10 )aa 6.610m / s25 2UNIT 2 Dynamics RRHS PHYSICS Page 47 of 100

2.2.2 Dynamics ProblemsMultiple Force ProblemsIn the previous section, you used Newton’s Second Law ( Fnet ma ) to solveproblems where a single force provided the net force that resulted in an acceleration.In most real world situations, there are multiple forces acting on an object thatcontribute to the net force. In this section, you will be learning to put together theprevious concepts of free body diagrams, friction, and Newton’s Laws of Motion tosolve complex problems involving a number of vertical and horizontal forces.It is extremely important to draw a free body diagram when there are multiple forcesinvolved. Remember that this type of diagram identifies all forces that are acting onthe object in question; only forces that are acting on the object in question should beincluded in this diagram. Each force should be represented by an arrow indicating itsdirection and must be labeled, indicating what force it is. Do not go any further withthe problem until you have drawn a free body diagram!This free body diagram must be referred to in order to help you set up Newton’sSecond Law. The free body diagram helps you to do a number of things: Identify the forces involved, with their directions Separate the vertical from horizontal forces Determine what forces must be equal to one another (in order to determinethe normal force, for example) Determine what forces contribute to the acceleration (if there is one) Assign a positive or negative sign to the forces to indicate directionsBefore going any further, it is important to be reminded of the difference between Fand F . The vector notation F refers to the force vector (including the direction); thescalar notation F refers to the magnitude of the force only – the direction must bespecified otherwise (using words, a positive or negative sign in front of the symbol, ora vector in a diagram). Both notations will be used in this and future units.Example 1:A 43.0 kg chair is being pushed across a floor with an acceleration of 3.1 m/s 2 to theright. If the coefficient of friction between the chair and the floor is 0.44, with howmuch force is the chair being pushed?Solution:First, draw a free body diagram:UNIT 2 Dynamics RRHS PHYSICS Page 48 of 100

Since the chair is accelerating in the direction that it is being pushed, the forcepushing the chair must be greater than the force of friction. In this problem, thereare both vertical forces and horizontal forces – these must be looked atseparately.Writing down our information we getm 43.0kga 3.1 m / s 0.442g 9.80 m / s2Fg mgand (43.0)(9.80) 421NLooking at our vertical forces, we only have the normal force and the force of gravity.Since there is no vertical acceleration, these forces must have the same magnitude:FN Fg 421NLooking at our horizontal forces, we can calculate the force of friction:Ff FN (0.44)(421)185NOur acceleration in this problem is in the horizontal direction; therefore, the net forceis in a horizontal direction and when we set up Newton’s Second Law, we can onlyinclude horizontal forces.Next, we set up the net force equation:Fnet maor as it is often written,ma Fwhere F refers to the net force and is just all the forces added up (with theirproper signs).UNIT 2 Dynamics RRHS PHYSICS Page 49 of 100

Using right as the positive direction:ma (43.0)(3.1) F 185ppppFma F Fma F FF 320NffThe person would have to push with 320 N of force to the right (since we chose theright to be our positive direction).UNIT 2 Dynamics RRHS PHYSICS Page 50 of 100

Check Your LearningA force of 40. N accelerates a 5.0 kg block at 6.0 m/s 2 to the right along a horizontalsurface. What is the coefficient of friction?a 6.0 m / sm 5.0kgFp 40. N2g 9.80 m / s ?2Fg mg (5.0)(9.80) 49NLooking at the vertical forces, the normal force must equal the force ofgravity since there is no vertical acceleration.F FNg 49NLooking at the horizontal forces (and using the right as positive)ma 5.0(6.0) 40.FfpFma F FF10.NffFand 10. (49) 0.20f FNUNIT 2 Dynamics RRHS PHYSICS Page 51 of 100

Apparent WeightIn section 3.1.2 you learned that weight was equal to the force of gravity, and can becalculated using Fg mg . Have you ever been in an elevator that was acceleratingup or down and felt heavier or lighter? Your mass m hasn’t changed in this situationand neither has g , so your weight is still the same; however, we say that yourapparent weight has changed – you feel heavier or lighter.Consider a person in an elevator standing on a bathroom scale (for our purposes allscales will measure force using metric units, so they will be calibrated in newtons).When the elevator is standing still, the reading on the scale is equal to your weight.Since there is no acceleration, the downward force of gravity on you must equal theupward force that the scale is exerting on you.According to Newton’s Third Law, the upward force that the scale is exerting on youis the same magnitude as the downward force that you are exerting on the scale;therefore, the reading on the scale is equal to the force that you exert on it.Consider now the case where the elevator is accelerating upward. 4 Since theacceleration is upward, there must be a net force upward. This means that theupward force that the scale is exerting on you must be greater than the downwardforce of gravity on you.Since the force that the scale exerts on you is equal in magnitude to the force thatyou exert on the scale, this means that you are exerting more force on the scale than4 This does not mean that the elevator is moving upward. To have an upward acceleration, theelevator could be moving upward with an increasing speed or it could be moving downward with adecreasing speed.UNIT 2 Dynamics RRHS PHYSICS Page 52 of 100

you would if you were at rest; you appear to be heavier than you are. In this situation,we would say that your apparent weight is greater than your true weight. You feelheavier because the floor (or scale) is pushing up on you with a greater force thanwhen it is at rest or moving at a constant velocity.What if the elevator is accelerating downward? In this case the net force must bedownward, so the force that the scale is exerting upward must be less than the forceof gravity.In this case, you are exerting a force on the scale that is less than your true weight.You apparent weight is less than your actual weight because the floor is not pushingup on you with as much force as when it is at rest.Example 2:A person has a mass of 82 kg. He is standing on a scale (which measures force innewtons) in an elevator. If the elevator is accelerating upward at 2.3 m/s 2 , what is theperson’s apparent weight?Solution:First, draw a free body diagram for the person in the elevator:The only forces acting on the person are gravity Fg(downward) and the scale Fs(pushing upward). Since the acceleration is upward, the upward force ( Fs) must begreater than the downward force ( Fg), as can be seen in the free body diagram.UNIT 2 Dynamics RRHS PHYSICS Page 53 of 100

What the scale actually measures is how much force it is pushing upward with ( Fs),so this is what we are looking for.Writing down all of our information (and using up as positive),m 82kga 2.3 m / s2g 9.80 m / sF ?s2If we take the upward direction as positive (it often makes things simpler if youchoose the direction of the acceleration as being positive), then Fsis positive and Fgis negative. We can include these signs in the equation.ma ma FssFFgma F FNotice that the vector notation on the forces has been dropped, since we havereplaced it with a positive or negative sign indicating the direction.Substituting our numbers in, we get(82)(23) F (82)(98)Fssg 990NSince the force of gravity acting on the person is F mg (82)(98) 800N, thismeans that the scale is pushing upward with more force than it would if the elevatorwere at rest (as we noted in the beginning of the problem). In other words, theperson will feel heavier than usual because the scale (or floor) is pushing upwardwith more force than usual and their apparent weight is 990 N.gUNIT 2 Dynamics RRHS PHYSICS Page 54 of 100

Check Your LearningYou are in an elevator at rest and you are standing on a scale. The scale reads 715N. When the elevator begins accelerating, the scale reads 820 N.a. What is the acceleration of the elevator?When the elevator is at rest, the upward force that the scale is exerting willequal the downward force of gravity.Using up as positive,F 715Nsm ?It is first necessary to calculate the mass:F FFss715 m(9.80)g mgm 73.0kgWhen the elevator is accelerating upward, the scale must exert a forcegreater than the force of gravity:UNIT 2 Dynamics RRHS PHYSICS Page 55 of 100

F 820NFsg 715Nm 73.0kga ?Using up as positive,ma (73.0) a 820 715ssFma F Fma F Fgg2a 1.4 m / sThe positive sign on the acceleration indicates that it is upward.b. If the scale were calibrated in kg (as many typical bathroom scales are),what would the two readings be?At rest, the scale would read 73.0 kg (which corresponds to a weight of 715N on Earth). While accelerating upward, the scale would read 84 kg (whichcorresponds to the apparent weight of 820 N).c. Has your mass actually changed?No, your mass has not actually changed. The scale measures force, notmass. It is simply calibrated to provide the mass that corresponds to themeasured weight on Earth. Since the scale is measuring your apparentweight (not your real weight which is simply the force of gravity that isacting on you), the scale gives your “apparent mass” which is what you feellike. Your mass is still 73.0 kg, although you feel as heavy as you would ifyou had a mass of 84 kg.UNIT 2 Dynamics RRHS PHYSICS Page 56 of 100

2.2.2 In Class or Homework Exercise1. A sled of mass 50.0 kg is pulled along snow-covered, flat ground. Thestatic friction coefficient is 0.30, and the sliding coefficient is 0.10.(a) What force is needed to start the sled moving?In order to start the sled moving, the force of the push must begreater than the force of friction ( Fp Ff). Since the sled is not yetmoving, this is a static situation. 0.30sm 50.0kgg 9.80 m / sFp ?2Fg mg (50.0)(9.80) 490. NF Ff s N Fsg (0.30)(490.)so F 147N147N(b) What force is needed to keep the sled moving at constant velocity?In order to keep the sled moving at constant velocity, the force of thepush must be equal to the force of friction ( Fp Ff). Since the sledis moving, this is a kinetic situation.k 0.10Fg mgm 50.0kg (50.0)(9.80)2g 9.80 m / s 490. NF ?ppUNIT 2 Dynamics RRHS PHYSICS Page 57 of 100

F Ff k N Fkg (0.10)(490.) 49Nso F 49Np(c) Once moving, what force must be applied to the sled to accelerate itat 3.00 m/s 2 ?In order to keep the accelerate the sled, the force of the push mustbe greater than the force of friction ( Fp Ff); however, since wehave an acceleration, we are able to calculate the specific forcerequired. Since the sled is moving, this is a kinetic situation.F 49NFfg 490. Na 3.00 m / sm 50.0kgFp ?2Applying Newton’s Second Law,ma (50.0)(3.00) F 49pppFma F FF 199Nf2. A 65 kg swimmer jumps off a 10.0 m tower. Assume the swimmer’s initialvelocity is zero.(a) Find the swimmer’s velocity when hitting the water.Since the swimmer is in free fall, their downward acceleration is 9.80m/s 2 . Using down as positive,UNIT 2 Dynamics RRHS PHYSICS Page 58 of 100

a 9.80 m / sv 0id10.0mvf ?22 2vf vid2a2vf 010.0 2(9.80)vf 14 m / s(b) The swimmer comes to a stop 2.0 m below the surface of the water.Find the average net force exerted on the swimmer over this 2.0 m.We must first find the acceleration of the swimmer in the water, Wewill continue to use down as positive.m 65kg2 2vf vivi 14 m / sd2avf 020 142.0 d2.0m2a2a ?a 49 m / sWe can nor find the net force:Fnet ma(65)( 49)3200NNotice that the negative sign tells us that the net force must beupward (in the same direction as the acceleration)(c) Find the force exerted by the water on the swimmer.Since we are now looking for an individual force, we must draw afree body diagram:m 65kgg 9.80 m / sa 49 m / sFW ?22Fg mg (65)(9.80) 637NUNIT 2 Dynamics RRHS PHYSICS Page 59 of 100

ma W(65)( 49) F 637WFma F FFWg 3800NNotice that the sign of the force is positive, even though it is upward and wechose down as positive. This is because we already put the negative signin the equation ( F ) so WFWis a scalar quantity. The force that the watermust exert is 3800 N upward.3. A person fishing hooks a 2.0 kg fish on a line that can only sustain amaximum of 38 N of force before breaking. At one point while reeling in thebass, it fights back with a force of 40 N (including its own weight). What isthe minimum acceleration with which you must play out the line during thistime in order to keep the line from breaking?Using up as positive,m 2.0kgFFlinefish 38N40Nma 2.0a38 40a ?2a 1.0 m / slineFma F FThe person would have to let the line out at a rate of 1.0 m/s 2 .4. The maximum force a grocery bag can withstand and not rip is 250 N. If20.0 kg of groceries are lifted from the floor to the table with anacceleration of 5.0 m/s 2 , will the bag hold?In order for the bag not to rip, the force that the bag exerts ( Fb) must beless than or equal to 250N. Looking at our other information, and using upas positive,a 5.0 m / sm 20.0kg2g 9.80 m / sF ?b2Fg mgfish (20.0)(9.80)196NUNIT 2 Dynamics RRHS PHYSICS Page 60 of 100

ma (20.0)(5.00) F 196bbbFma F FFg 296NSince F 250N, the bag will break.b5. A student stands on a bathroom scale in an elevator at rest on the 64 th floorof a building. The scale reads 836 N.(a) As the elevator moves up, the scale reading increases to 935 N,then decreases back to 836 N. Find the acceleration of the elevator.At rest, the force of gravity will be equal to the force that the scalereads. We can use this information to find the mass of the student.Fs 836Ng 9.80 m / sm ?2FFss F836 m(9.80)g mgm 85.3kgWhen the scale reading increases, the force that the scale exerts isno longer equal to the force of gravity.Using up as positiveUNIT 2 Dynamics RRHS PHYSICS Page 61 of 100

Fs 935Nm 85.3kga ?ma (85.3) a 935 836ssFma F Fma F Fgga 1.16 m / sThe elevator is accelerating upward at 1.16 m/s 2 .2(b) As the elevator approaches the 74 th floor, the scale reading drops aslow as 782 N. What is the acceleration of the elevator?Continuing to use up as positive,Fs 782Nma Fm 85.3kgma FsFgFg 836Nma FsFga ?(85.3) a 782 836a 0.633 m / sThe elevator is accelerating downward at 0.633 m/s 2 .6. A person weighing 490 N stands on a scale (calibrated in newtons) in anelevator.(a) What does the scale read when the elevator is at rest?a 0Fg 490NAt rest, the scale pushing up and the force of gravity pushing downmust be equal:2UNIT 2 Dynamics RRHS PHYSICS Page 62 of 100

This can be shown with Newton’s Second Law as well (using up aspositive):ma m(0) F FsssgFma F FF Fgg 490N(b) What is the reading on the scale when the elevator ascends at aconstant acceleration of 2.7 m/s 2 ?Using up as positive again,a 2.7 m / sFg 490N2g 9.80 m / sF ?s2Fg mg490 m(9.80)m 50. kgma (50.)(2.7) F 490sssFma F FFg 630N(c) What is the reading when the elevator ascends at a constant speedof 7.5 m/s?Since the speed is constant, the acceleration is zero; therefore, theanswer is the same as (a) and the scale will read 490 N.(d) What is the reading when the elevator slows down to a stop at thetop floor, assuming he starts decelerating from his constant speed of7.5 m/s when he is 3.5 m away from the top floor?Using up as positive, we must first find the acceleration:UNIT 2 Dynamics RRHS PHYSICS Page 63 of 100

a ?v 7.5 m / svif 0d3.5m2 2vf vid2a20 7.53.5 2aa 8.0 m / s2ma (50.)( 8.0) F 490sssFma F FF 90Ng(e) Suppose the cable snapped and the elevator fell freely. What wouldthe scale read?Continuing to use up as positive,a 9.80 m / sFg 490Nm 50. kgF ?s2ma (50.)( 9.80) F 490sssFma F FF 0NAs expected, the scale will not read anything – the person will feelweightless. This is apparent weightlessness, since gravity is stillacting on the person.7. A 5.5 kg object is being pushed to the right with a force of 65 N. If theacceleration of the object is 3.0 m/s 2 to the right, what is the coefficient offriction?gUNIT 2 Dynamics RRHS PHYSICS Page 64 of 100

a 3.0 m / sm 5.5kgFp 65N2g 9.80 m / s ?2Fg mg (5.5)(9.80) 53.9NLooking at the vertical forces, the normal force must equal the force ofgravity since there is no vertical acceleration.F FNg 53.9NLooking at the horizontal forces (and using the right as positive)ma 5.5(3.0) 65FfpFma F FF 48.5Nffand 48.5 (53.9)Ff FN0.908. What is the acceleration of a falling 65.0 kg skydiver if air resistance exertsa force of 250. N?m 65.0kgFg mg2g 9.80 m / s (65.0)(9.80)Fa 250. N 637Na ?Using up as positive,ma aaFma F Fma F Fgg65.0a250. 6372a 5.95 m / sThe acceleration is 5.95 m/s 2negative sign).down (the direction is indicated by theUNIT 2 Dynamics RRHS PHYSICS Page 65 of 100

9. The cable supporting a 1650 kg elevator has a maximum strength of 22500N. What maximum upward acceleration can the cable give the elevatorwithout breaking?m 1650kgFT 22500Ng 9.80 m / sa ?2Fg mg (1650)(9.80)16170NUsing up as positive,ma TTFma F Fma F F1650a22500 16170a gg3.84 m / s210. You are driving a car at a constant speed of 14.0 m/s along an icy road. Allof a sudden, your physics teacher jumps in front of your car! You slam onthe brakes. Your wheels lock, the tires begin skidding, and the car slides toa halt in a distance of 25.0 m. What is the coefficient of sliding frictionbetween your tires and the icy roadbed?v 14.0 m / svif 0d25.0m ?2 2vf vid2a20 14.025.0 2aa 3.92 m / sc2UNIT 2 Dynamics RRHS PHYSICS Page 66 of 100

ma Fma Ffma Ffma FNma mg (since the normal force must equal the force of gravity)3.92 (9.80) 0.4011. A train locomotive pulling three identical 8850 kg cars accelerates at 0.225m/s 2 . With what force does the first car pull the second?Since the first car is actually pulling cars 2 and 3, we can treat them as oneobject. Since there is no obvious friction in this problem, this force is theonly horizontal force.m 17700kg(the mass of cars 2 and 3)a 0.225 m / sFp ?2ma ma F17700(0.225) FFpppF 3980N12. Suppose that the locomotive in the last problem has a mass of 23500 kg,and that the friction force on each of the three cars is 1100 N. If theUNIT 2 Dynamics RRHS PHYSICS Page 67 of 100

acceleration is the same as in the last problem, what must be the horizontalforce that the locomotive exerts against the rails?This time, we can treat the locomotive and all 3 cars as one object. Whenthe locomotive exerts a horizontal force backward against the rails, the railsexert an equal forward force against the locomotive (Newton’s Third Law);this is the pushing force that accelerates the train and what we are trying tofind.m 50050kg(the total mass of the 3 cars and the locomotive)Ff 3300Na 0.225 m / sFp ?2Since the acceleration is horizontal, we can only include horizontal forcesto calculate the acceleration.ma (50050)(0.225) F 3300pppFma F FF 14600NfSince there is 14600 N of force pushing the locomotive forward, this is alsothe size of the force that the locomotive must push against the rails with.13. A horizontal force of 85 N is applied to a box on a table, resulting in anacceleration of 3.2 m/s 2 . If the coefficient of friction between the box andthe table is 0.41, find the mass of the box.UNIT 2 Dynamics RRHS PHYSICS Page 68 of 100

Fp 85Na 3.2 m / s 0.412g 9.80 m / sm ?verticalFN Fg mg 9.80m2ma horizontalpppFma F Fma F Fma F Fm(3.2) 85 0.41(9.80 m)3.2m85 4.0m7.2m 85m 12kgffN14. One paint bucket weighing 20.0 N is hanging by a massless rope fromanother paint bucket also weighing 20.0 N, and the two are being pulledupward with an acceleration of 1.65 m/s 2 by another massless ropeattached to the upper bucket. Calculate the tension in each rope.To find1 TF , we will treat both paint cans as one big object (since the toprope is lifting both cans)UNIT 2 Dynamics RRHS PHYSICS Page 69 of 100

FgT1 40.0Ng 9.80 m / sa 1.65 m / sF ?22Fg mg40.0 m(9.80)m 4.08kgUsing up as positive,ma T1T1T14.08(1.65) F 40.0T1Fma F Fma F FFgg 46.7NTo find FT 2, we only need to look at the bottom can (since the top rope islifting just the bottom can)FgT 2 20.0Ng 9.80 m / sa 1.65 m / sF ?22Fg mg20.0 m(9.80)m 2.04kgUsing up as positive,UNIT 2 Dynamics RRHS PHYSICS Page 70 of 100

ma T 2T2T22.04(1.65) F 20.0T 2Fma F Fma F FF 23.4NggNotice that the top rope is exerting twice as much force as the bottom rope,which makes sense since it is lifting twice the mass.15. A 5000 kg helicopter accelerates upward at 0.550 m/s 2 while lifting a 1500kg car.(a) What is the lift force exerted by the air on the propellers?The lift force exerted on the propellers is actually lifting thehelicopter and the car, so we will treat the two of them as one object.Using up as positive,m 6500kga 0.550 m / sg 9.80 m / sFL ?22Fg mg (6500)(9.80) 63700Nma 6500(0.550) F 63700LLLLFma F Fma F FFgg 67300N(b) What is the tension in the cable that connects car to helicopter?This time, the cable is lifting only the car so we will use the car aloneas our object.UNIT 2 Dynamics RRHS PHYSICS Page 71 of 100

m 1500kga 0.550 m / sg 9.80 m / sF ?c22Fg mg (1500)(9.80)14700Nma 1500(0.550) F 14700ccccFma F Fma F FggF 15500N16. A motorcyclist is coasting with the engine off at a steady speed of 12 m/s,but enters a sandy stretch where the coefficient of friction is 0.80. Will themotorcyclist emerge from the sandy stretch without having to start theengine if the sand lasts for 15 m? If so, what will be the speed uponemerging?Since we do not know if the motorcyclist makes it or not, we will assumethat he does not make it and find out how far he travels before stopping. Ifthis is less than 15 m, then our assumption was right.v 12 m / svif 0 0.80d?Looking at the vertical forces, we know thatUNIT 2 Dynamics RRHS PHYSICS Page 72 of 100

FN FgSince there is no vertical acceleration.Looking at the horizontal forces (and using forward as positive),ma ma Fma FfFma FfNma mga (0.80)(9.80)2 2vf vid2a20 122( 7.84) 9.2m27.84 m/sSince 9.2m < 15m, the motorcyclist did not make it out of the sandy stretch.17. You are pushing a 3.2 kg box horizontally with your hand against a verticalwall. The coefficient of friction between the box and the wall is 0.35.(a) If the box slides down the wall with an acceleration of 2.3 m/s 2 , withhow much force were you pushing with your hand?m 3.2kgg 9.80 m / s 0.35a 2.3 m / sFp ?22(down)Fg mg (3.2)(9.80) 31.4NLooking at the vertical forces (and using down as positive),ma 3.2(2.3) F 31.4ffFma F Fff 24Ngma F FFgFfand 24 (0.35) FSince Fp FN(from the free body diagram)FN FN 69NNUNIT 2 Dynamics RRHS PHYSICS Page 73 of 100

Fp FN 69N(b) How hard would you have to push to stop the box from moving?For no motion, the acceleration is zero soF Ffg 31.4NAndF FfN31.4 (0.35) FFN 90NNsoFp FN 90N18. A person falls from a tower 5.0 m high. When he strikes the ground below,he bends his knees so that his torso decelerates over an approximatedistance of 0.70 m. If the mass of his torso (excluding legs) is 50.0 kg, findthe average force exerted on his torso by his legs during the deceleration.It is necessary to break this problem up into two parts – the falling part andthe landing part. We will use down as positive in both parts.Part 1 – Fallinga 9.80 m / sv 0id5.0mvf ?22 2vf vid2a2vf 05.0 2(9.80)vf 9.9 m / sPart 2 – Landing2 2vf vivi 9.9 m / sd2avf 020 9.90.70 d0.70m2a2a ?a 70.0 m / sKnowing the acceleration, we can know look at the forces:UNIT 2 Dynamics RRHS PHYSICS Page 74 of 100

a 70.0 m / sm 50.0kgg 9.80 m / sFL ?ma 50.0( 70.) F 490L22LFma F FLLgma F FF 4000NgFg mg (50.0)(9.80) 490N19. An exceptional standing jump would raise a person 0.80 m off the ground.To do this, what force must a 70.0 kg person exert against the ground?Assume the person lowers himself 0.20 m prior to jumping.It is necessary to break this problem up into two parts, on the ground and inthe air. We will use up as the positive direction.On the groundd0.20mv 0iIn the aird0.80mm 70.0kg2a 9.80 m / s 0 (at the top of the jump)m 70.0kgIf we can find the speed at which the person leaves the ground (the initialvelocity from part 2) we can use this as the final velocity in part 1.2 2vf vid2a20 vi0.80 2( 9.80)v 3.96 m / siWe can now find the acceleration while pushing against the ground that theperson needs to get this speed:UNIT 2 Dynamics RRHS PHYSICS Page 75 of 100vf

vd v2 2f i2a23.96 00.20 2a2a 39.2 m / sThe forces acting on the person are the force of the ground (the normalforce) and the force of gravity:ma (70.0)(39.2) F (70.0)(9.80)NNNNFma F Fma F mgFg 3400NSince the ground is exerting an upward force of 3400 N on the person, theperson must be exerting an equal downward force on the ground(Newton’s Third Law).F FpN 3400N20. A box is given an initial push so that it slides across the floor. How far will itgo, given that the coefficient of friction is 0.30 and the push imparts aninitial speed of 3.0 m/s?Since the box was only given an initial push, there is no longer any forcepushing the box forward.UNIT 2 Dynamics RRHS PHYSICS Page 76 of 100

v 3.0 m / svif 0 0.30d?To find the displacement, we first need the accelerationma ma Fma FfFma FfNma mga (0.30)(9.80)and then2 2vf vid2a20 3.02( 2.94) 1.5m2a 2.94 m / s21. Two crates, of mass 80. kg and 120. kg, are in contact and rest on ahorizontal surface. A 700. N horizontal force is exerted on the 80. kg cratetoward the second crate. If the coefficient of kinetic friction is 0.250,calculate(a) the acceleration of the systemTo find the acceleration of the system, we can treat both boxes asone object:Using right as positive in this problem,m 200. kgFg mg 0.2502 (200.)(9.80)g 9.80 m / s1960NF 700. NpFf FN (0.250)(1960) 490. Nma (200.) a 700. 490.ppFma F Fma F Fa 1.05 m / sff2UNIT 2 Dynamics RRHS PHYSICS Page 77 of 100

(b) the force that each crate exerts on the otherFor this part, it is necessary to look at the individual boxes:Looking at the 80. kg box (box A),m 80. kga 1.05 m / s2Fg mgFf FN 0.25 (80.)(9.80) (0.250)(784)Fp 700. N 784 N196Ng 9.80 m / s2ma p f Bp f B(80.)(1.05) 700. 196 FBFma F F Fma F F FF 420NBLooking at the 120. Kg box (box B),Notice that the 700. N force is not included since it is not actingdirectly on the 120. kg box.UNIT 2 Dynamics RRHS PHYSICS Page 78 of 100

m 120.kg 0.25g 9.80 m / s2Fg mg (120.)(9.80)1176NFf FN (0.250)(1176) 294Nma (120.)(1.05) F 294AAAAFma F Fma F FF 420NffNotice that FAand FBare the same size! Of course this should betrue according to Newton’s Second Law. Because of this, it was onlynecessary to calculate one or the other, since we know that theymust be equal.(c) Repeat part (b) assuming the 700 N force is exerted on the 120 kgcrate (in the opposite direction)For this part, it is necessary to look at the individual boxes:Looking at the 80. kg box (box A),m 80. kgFf196Na 1.05 m / sF ?B2(since the boxes will now be moving in the opposite direction)UNIT 2 Dynamics RRHS PHYSICS Page 79 of 100

ma (80.)( 1.05) F196BBFma F Fma F FFBB 280NAnd because of Newton’s Third Law.F FFAAB 280Nff22. A flatbed truck is carrying a 2800 kg crate of heavy machinery. If thecoefficient of static friction between the crate and the bed of the truck is0.55, what is the maximum rate at which the driver can slow down whencoming to a stop in order to avoid crushing the cab with the crate?In order for the crate to no crush the cab, it will have to have the sameacceleration as the truck. The only horizontal force acting on the crate toslow it down is the force of static friction; therefore, it is necessary to findthe maximum acceleration that this force of static friction can provide.m 2800kg 0.55g 9.80 m / sa ?2Fg mg (2800)(9.80) 27400NFf FN (0.55)(27400)15000Nma Fma Ffma F2800a15000fa 5.4 m / s2Alternate Solution:UNIT 2 Dynamics RRHS PHYSICS Page 80 of 100

ma ma Fma FfFma Fa (0.55)(9.80)a 5.4 m / sfNma mga g223. A 40.0 kg child wants to escape from a third story window to avoidpunishment. Unfortunately, a makeshift rope made of sheets can onlysupport a mass of 30.0 kg.(a) How can the child use this “rope” to escape? Give a quantitativeanswer.Force that rope can exert :Weight of child:Fr mg (30.0)(9.80)Fg mg (40.0)(9.80) 294N 392NSince the rope cannot support the child’s weight, it will be necessaryfor the child to accelerate downward so that the rope does not haveto exert as much upward force.A quantitative answer requires a value for this acceleration. Usingup as positive:ma 40.0a294 392rrFma F Fma F Fa 2.45 m / sgg2UNIT 2 Dynamics RRHS PHYSICS Page 81 of 100

Basically, the child is using the “rope” to slow himself down so thathe only accelerates downward at 2.45 m/s 2 , rather than 9.80 m/s 2 ifhe jumped from the window.(b) Give one example of a pair of forces that demonstrates Newton’sThird Law in this problem.Some acceptable answers include: The force of the Earth pulling down on the child and the forceof the child pulling up on the Earth The force of the child pulling down on the bedsheets, theforce of the bedsheets pulling up on the child The force of the bedsheets pulling down on the bed (orwhatever they are tied to) and the force of the bed pulling upon the bedsheetsUNIT 2 Dynamics RRHS PHYSICS Page 82 of 100

2.2.3 Systems of MassesIn industry, most forces are exerted by machines that have many moving parts, andoften include cables (or chains) and pulleys to lift large loads. By using pulleys andcables, we can change the direction of forces and use forces such as gravity to ourbenefit even if they are not in the direction that we need. This involves using variousmasses that are connected by ropes or cables. It is therefore important to be able toanalyze these situations to predict motion, to know the limits of a machine, and to beable to design required equipment.When dealing with cables or ropes, the tension in a cable can be considered to bethe same everywhere in the cable (or rope); as a result, the cable exerts a force onwhatever it is connected to that is equal to its tension. For our purposes, it can beassumed that the mass of a rope or cable is negligible compared to the objects thatthey are connected to and can be ignored. We will also assume that all pulleys arefrictionless and their masses can be ignored, so if a rope or cable passes over apulley the direction of the tension changes but the magnitude of the tension remainsthe same.A system of masses consists of masses that interact and are connected with oneanother. The systems that we are dealing with here consist of masses that areconnected to one another by a rope and involve pulleys that change the direction,but not the magnitude, of forces. An example of this is an elevator (see below) that isconnected to a counterweight suspended over a pulley. The counterweight usesgravity to exert an upward force on the elevator; this upward force assists the motorthat is moving the elevator so that the motor does not have to exert as much force.Up until now, you have been dealing with forces on single objects, or objects movingtogether in a linear direction. Systems such as the one above involve multiplemasses moving in different directions; however, sometimes a system that does notappear to be linear can in fact be treated as a linear system. Since the masses areconnected, they will have the same size acceleration, even though the accelerationsUNIT 2 Dynamics RRHS PHYSICS Page 83 of 100

may be in different directions. In these situations, it is easier if you assign thedirection of motion to be from one end of the rope to the other (instead of using up,down, to the right etc.) You can then call one end of the rope positive and the otherend negative.Consider the situation below where m2 m1We will assign the lighter mass m1to be the negative end and the heavier mass m2tobe the positive end.We can then linearize the system; in other words, redraw the system to be a linearsystem that would have similar forces and would result in the same acceleration:Figure 1: Linearized system treated as two separate objectsOr, simplifying to only one object (since tension is an internal force and notnecessary when looking at the whole system as one)Figure 2: Linearized system treated as one objectUNIT 2 Dynamics RRHS PHYSICS Page 84 of 100

After you have assigned the direction to be positive, you can work with the system aseither two separate objects (see Figure 1) or as one big object (see Figure 2). Bothof these methods will be used in the example below.When treating the system as individual objects:1. Use the individual mass of the object that you are studying when applyingNewton’s Second Law.2. Draw a free body diagram for each object individually. Tension forcesexerted along a cable or rope do contribute to the motion of an individualobject and should be included as part of the net force.3. Set up Newton’s Second Law for each object. It may then be necessary tosolve a system of equations using the equation for each object.When treating the system as one big object:1. The mass of the system is the total mass of all of the objects. This is themass that should be used when applying Newton’s Second Law.2. Tension forces exerted along a rope or a cable between any two objects inthe system are called internal forces and should not be included as part ofthe net force. These internal forces do not affect the motion of the systemas a whole; only external forces such as friction and gravity affect themotion of the system and should be included in Newton’s Second Law.Example:An Atwood machine is a simple machine consisting of two objects connected by arope hanging over a pulley as shown in the picture below.What is the acceleration of this system?Solution:There are two approaches to solve this problem. For each method, we will use the3.0 kg mass as the negative end and the 4.0 kg mass as the positive end; in otherwords, the direction of acceleration (which is clockwise) will be positive.Method 1 (Individual objects):This method involves looking at each object individually. Our free body diagramswould look like these:UNIT 2 Dynamics RRHS PHYSICS Page 85 of 100

Or, if we were to linearize it,We will use the end of the rope with the 4.0 kg mass as the positive end and the endwith the 3.0 kg mass as the negative end, as shown above.Looking at the 3.0 kg mass first,m 3.0kg1g 9.80 m / sa ?2F m gg1 1 (3.0)(9.80) 29.4NNewton’s Second Law for this mass would give us the following:m a F1m a F F1 T g1m a F F1 T g13.0a FT29.4Notice that there are 2 unknown variables in this equation; this requires anotherequation in order to solve. This other equation can be obtained by examining theother mass.Looking at the 4.0 kg mass now,UNIT 2 Dynamics RRHS PHYSICS Page 86 of 100

m2 4.0kgg 9.80 m / sa ?2Fg m g2 2 (4.0)(9.80) 39.2NUsing the 4.0 kg mass end of the rope as positive, Newton’s Second Law for thismass would give us the following:m a F2m a F F2 T g2m a F F2 T g24.0a F 39.2TNotice that a is the same for each object (since they are connected) and FTis alsothe same on each object, since the tension is the same everywhere in the rope and itmust exert the same force at each end. We can now solve our system of equations:3.0a F 29.4T4.0a F 39.27.0a39.2 29.2T7.0a 9.8a 1.4 m / s2Note that this acceleration is significantly smaller than the acceleration due to gravity,even though gravity is being used to accelerate the system.Method 2 (One big object):This method involves looking at the whole system and “linearizing” it. Looking at theoriginal picture, we can see that the acceleration is going to be clockwise (the 3.0 kgmass will accelerate upward and the 4.0 kg mass will accelerate downward). We willuse clockwise as our positive direction 5 , instead of up or down.It is also necessary to look at all of the forces acting to accelerate this system, eitherto speed it up or slow it down. Note that the tension in the rope will not be includedhere, since the only purpose of the rope is to connect the two objects. It can beconsidered to be an internal force, similar to the force of attraction between themolecules of a box to hold it together.5 In other words, the positive direction will be directed from the 3.0 kg mass toward the 4.0 kg mass.UNIT 2 Dynamics RRHS PHYSICS Page 87 of 100

Since the only other two forces are the two forces of gravity, which are actingopposite one another, we can model this situation as a single object with twoopposing forces acting on it:Since we are looking at the system as a whole, the mass that we must use is thetotal mass that is being accelerated.m a tFm a F Ft g1 g 2m a F Ft g1 g 2(7.0)( a) 29.4 39.2a 1.4 m / s2Systems can also involve forces that are at angles to one another, such as a fallingweight that exerts a horizontal force on a lab cart, as shown below:UNIT 2 Dynamics RRHS PHYSICS Page 88 of 100

This situation can be approached in the same way as the Atwood machine in theexample, keeping the following things in mind:Analyze the forces on each individual object.Assign a direction to the motion along the rope.Pretend the motion is linear, including only forces that are parallel to themotion in the calculation of net force.Draw both individual free body diagrams and a combination free bodydiagram of the system.UNIT 2 Dynamics RRHS PHYSICS Page 89 of 100

Check Your LearningIgnoring friction, calculate the acceleration of the system below and the tension in therope.Drawing individual free body diagrams we getandm 4.0kgm12 7.0kgg 9.80 m / sa ?FT ?2F m gg1 1 (4.0)(9.80) 39.2NFg m g2 2 (7.0)(9.80) 68.6NFor the 4.0 kg mass, the only force parallel to the acceleration (which ishorizontal) is the tension force FTin the rope; the forces FNand Fg1areperpendicular to the acceleration and should not be included. For the 7.0kg mass, both of the forces FTand Fg 2are parallel to the acceleration(which is vertical).When looking at the forces causing the system to accelerate, tension is notincluded since this is an internal force. The only remaining force istherefore Fg 2. This would be the only horizontal force (or the only forceUNIT 2 Dynamics RRHS PHYSICS Page 90 of 100

parallel to the motion) in our linearized free body diagram for the wholesystem:Using clockwise (or toward the 7.0 kg mass) as positive,ma m a Ftm a Ftg2g2F11.0a 68.6a 6.2 m / s2To find the tension FTit is necessary to set up an equation for one of themasses individually. If we do this for the 4.0 kg mass, we getma Fma F(4.0)(6.2) FFTTT 25NUNIT 2 Dynamics RRHS PHYSICS Page 91 of 100

2.2.3 In Class or Homework Exercise1. A 2.00 kg mass and a 3.00 kg mass are attached to a lightweight cord thatpasses over a pulley. The hanging masses are free to move.(a) What is its acceleration?m 2.00kgm12 3.00kgg 9.80 m / sa ?2F m gg1 1 (2.00)(9.80)19.6NFg m g2 2 (3.00)(9.80) 29.4NTreating the system as one object and using the 3.00 kg mass asthe positive end (clockwise)ma Fm a F Ft g 2 g1m a F Ft g 2 g15.00a29.4 19.6a 1.96 m / s2(b) What is the tension force acting in the cord?To find the tension it is necessary to look at just one of the objects.Looking at the 3.00 kg object,ma 2 g22 g2Fm a F Fm a F F3.00(1.96) 29.4FTTTFT 23.5N2. Do the Check Your Learning question again if the coefficient of frictionbetween the top box and the table is 0.31.UNIT 2 Dynamics RRHS PHYSICS Page 92 of 100

m 4.0kgmF12T 7.0kg ?g1 1 0.31 (4.0)(9.80)a ? 39.2NF m gWe also need the force of friction on m1:F Ff1 N1 Fg1 (0.31)(39.2)12.2NFg m g2 2 (7.0)(9.80) 68.6NUsing the 7.0 kg mass end of the rope as positive,ma Fm a F Ft g 2 f 1m a F Ft g 2 f 111.0a68.6 12.22a 5.1 m / sTo find the tension, we need to look at just one object. Looking at the 4.0kg object,UNIT 2 Dynamics RRHS PHYSICS Page 93 of 100

ma 1 T f11 T f14.0(5.1) F 12.2TTFm a F Fm a F FF 33N3. What coefficient of static friction is necessary to prevent motion in thediagram below?m1 3.0kgFg1 m1gFg2 m2gm2 5.0kg (3.0)(9.80) (5.0)(9.80)s ? 29.4N 49Na 0Since there is no motion, the acceleration must be zero.ma Fm a F FFFFt g 2 f 10 Fg2 f1g2 f1g 2 s N1g 2 s g149 (29.4)s 1.7s F F FFUNIT 2 Dynamics RRHS PHYSICS Page 94 of 100

4. What coefficient of kinetic friction is necessary in the previous question fora constant speed?m 3.0kgm12 5.0kg ?ka 0Since the acceleration is again zero, the same forces have to balance as inthe previous question, so the math is the exact same as the previousquestion. The only difference is that this time we are finding the coefficientof kinetic friction since the system is in motion. 1.7k5. Find the acceleration of the following system(a) if there is no friction.m 2.0kgmm123a ? 5.0kg 6.0kgF m gg1 1 (2.0)(9.80)19.6NFg m g2 2 (5.0)(9.80) 49NFg m g3 3 (6.0)(9.80) 58.8NSince all of the tension forces are internal forces, the only forces thatcontribute to the motion are Fg1and Fg 2. We will use the 6.0 kg endof the rope as the positive direction:UNIT 2 Dynamics RRHS PHYSICS Page 95 of 100

ma Fm a F Ft g1 g3m a F Ft g1 g313.0a 19.6 58.8a 3.0 m / s2(b) if the coefficient of friction between the box and the table is 0.12.m 2.0kgmm123 5.0kg 6.0kg 0.12a ?F Ff 2 N2 Fg 2 (0.12)(49) 5.9NF m gg1 1 (2.0)(9.80)19.6NFg m g2 2 (5.0)(9.80) 49NFg m g3 3 (6.0)(9.80) 58.8NSince all of the tension forces are internal forces, the only forces thatcontribute to the motion are Fg1, Ff 2and Fg 2. We will use the 6.0 kgend of the rope as the positive direction:UNIT 2 Dynamics RRHS PHYSICS Page 96 of 100

ma Fm a F F Ft g1 f 2 g3m a F F Ft g1 f 2 g313.0a 19.6 5.9 58.8a 2.6 m / s26. A 2.5 kg counterweight is connected (over a pulley) to a 4.5 kg window thatslides vertically in its frame. How much force must you exert to start thewindow opening with an upward acceleration of 0.18 m/s 2 ?m 4.5kgm12 2.5kgg 9.80 m / sa 0.18 m / sFp ?22Fg m g1 1 (4.5)(9.80) 44.1NFg m g2 2 (2.5)(9.80) 24.5NLooking at the system as one object (so we will ignore the internal tensionforces) and using the counterweight end of the rope as the positive end,ma Fm a F F Ft g1 g 2 pm a F F Ft g1 g 2 p7.0(0.18) 44.1 24.5 FFp 21Np7. A 1.2 kg mass and a 3.2 kg mass are attached by a massless rope hangingover a pulley. The two masses are initially 1.60 m above the ground andUNIT 2 Dynamics RRHS PHYSICS Page 97 of 100

the pulley is 4.0 m above the ground. What maximum height does the 1.2kg object reach after the system is released from rest?The 3.2 kg mass will fall 1.6 m to the floor; during this time, the 1.2 kg masswill be accelerating upward. We must find this acceleration so that we candetermine the speed of the 1.2 kg mass when it stops accelerating upward(at a height of 3.2 m from the floor).m 1.2kgm12 3.2kgg 9.80 m / sa ?2F m gg1 1 (1.2)(9.80)11.8NFg m g2 2 (3.2)(9.80) 31.4NTreating the system as one object and using the 3.2 kg mass as thepositive end (clockwise)ma Fm a F Ft g 2 g1m a F Ft g 2 g14.4a31.4 11.8a 4.45 m / s2We can now find the upward velocity of the 1.2 kg mass when it is at aheight of 3.2m:v 0vif ?a 4.45 m / sd1.6m22 2vf vid2a2vf 01.6 2(4.45)vf 3.77 m / sUNIT 2 Dynamics RRHS PHYSICS Page 98 of 100

We can now calculate how much higher it goes before gravity brings it to astop:v 3.77 m / svif 0a 9.80 m / sd?22 2vf vid2a20 3.772( 9.80) 0.73mSo the 1.2 kg mass will be1.6 1.6 0.7 3.9 mabove the floor (and still below the 4.0 m high pulley)UNIT 2 Dynamics RRHS PHYSICS Page 99 of 100

2.2.4 Module SummaryIn this module, you learnedo Newton’s First Law – An object in uniform motion (or at rest) will remain inuniform motion (or at rest) unless acted on by an outside net force.o Newton’s Second Law - The net force needed to accelerate an object is aproduct of the object’s mass and acceleration.Fnet mao Newton’s Third Law – For every action force, there is an equal and oppositereaction force.o How to solve complex single body problems using free body diagrams andNewton’s Lawso How to calculate apparent weighto How to solve systems of masses involving ropes and pulleysUNIT 2 Dynamics RRHS PHYSICS Page 100 of 100