Chapter 4
`
`The Laws of Motion
`
`tions that a given force produces ondifferent bodies. Suppose a force acting on a
`body of mass m, produces an acceleration a,, and the sameforce acting on a body
`of mass my produces an acceleration ay. The
`ratio of the two masses is defined as
`«© accelerations produced by the sameforce: the ineverse
`
`=m
`
`4.1]
`
`44°
`
`Newton's Second Law
`
`87
`
`These observations are summarized in Newton's second law:
`
`
`The acceleration of an object is directly proportional to the net force acting
`on it and inversely proportional to its mass.
`
`
`Thus, we can relate mz
`2
`sand force through the following mathematical stat
`of Newton's second law:*
`
`
`
`[4.2]
`iF = ma
`Youshould note that Equation 4.2 is a vecorexpression and henceis equivalent
`
`to the following three component equ
`iF, = ma,
`=F, = ma,
`
`EF, = ma,
`
`[4.3]
`
`Newton's second law
`
`Newton's second law—
`componentform
`
`1 ke—the mass of an unknown object
`If one mass is standard and known—say,
`can be obtained from acceleration measurements. For example, if the standard
`
`
`|-kg mass undergoes an acceleration of 3 m/s" under the influence of some force,
`2-kg mass will undergo anacceleration of 1.5 m/s? under the action of the same
`force.
`Units of Force and Mass
`sur-
`Mass is an inherent property of a body, independent of the body's
`roundings and of the method used to measure it. [1 is an experimentalfact that
`The SI unit of force is the newton, which is defined as the force that, when acting
`mass is a scalar quantity. Finally, mass is a quantity that obeys the rules of
`
`on a l-kg mass, produces an aceeleration of | m/s*, Fromthis definition and New-
`
`ordinary arithmetic. That is, several masses can be combined in simple numerical
`ton's second law, we see that the newton can be expressed in terms of the funda-
`
`
`fashion. For example, ifyou combine a S-kg mass witha5-kg mass, their total mass
`mental units of mass, length, and time:
`is 8 ke. This can be verified experimentally by comparing the acceleration of each
`
`Definition of a newton
`IN = 1 kg-m/s?
`4]
`object produced by a known force with the acceleration of the combined system
`The units of force and mass are summarized in Table 4.1. Most of the
`using the same foree,
`
`calculations weshall make in our study of mechanicswill bein ST units. Conversion
`Mass should not be confused with weight. Mass and weight are two different
`quantities. The weight of a body is equal to the magnitude ofthe force exerted by
`factors between the three systems are given in Appendix A.
`
`the Earth on the bedy and varies with location, For example, a person who weighs
`
`
`180 IbonEarth weighs only about 50 tb on the Moon. However, the mass of a body
`Thinking Physics 2
`is the same everywhere. An object having a mass of 2 kg on Earth also has a mass
`of 2
`
`kg on the Moon.
`
`nce ofstanding in an clevator chat accelerates
`You have most likely had the experi
` da higherfloor. In
`
` is cast, youfret heavier,
`If you
`upward as it leaves to move towa
`are standing on a bathroom scaleat th
`it will measureaforce larger than your
`
`
`
`ed evidence that lead you to believe that you
`are heavier in this situation. Are you heavier?
`In order to provide the ac-
`Reasoning No, you are not—yourweight is unchanged.
`celeration upward, the floor or
`
`
`than your weight.
`It
`is this larger
`force that
`feel, which you interpret
`heavier. A bathroom scale reads this upward force, not your weight, soits reading also
`INCTCASES,
`
`
`
`
`
`‘AnEnglish physicist and
`
`4.4 +» NEWTON’S SECOND LAW
`
`Newton's first law expla
`what happens to an object when the net external force
`18 Al rest or moves ina straight line with c
`on itis zero: It citherr
`
`unit speed.
`
`Newton's second law answers the question of what happens to.an object that has a
`honzeroe net force acting onit.
`
`
`Imagine you-are pushing a block of ice across a frictionless horizontal surface.
`When you exert some horizontal force F, the block moves with some acceleration
`
`
`
`TABLE 4.1 Units of Force, Mass, and Acceleration”
`
`
`a.
`[If you apply a force twiceas large,
`the acceleration doubles. If you increase the
`
`Systems of Units
`Mass
`Acceleration
`Force
`applied force to SF, the original acceleration is tripled,
`and so on. From such ob-
`
`servations, we concludethat the acceleration ofan object is directly proportional
`oble to explain the motions o
`SI
`ke
`m/s?
`N= ke- m/s?
`planets, the ebb and
`to the net force acting onit.
`
`cys
`K
`cm/s
`dyne = g-cm/s
`of an object also depends
`the accelerat
`As stated in the preceding section,
`and monyspecialfeatures of 1
` e Moon and the Earth
`slug
`Ib = stug- ft,'s*
`British engineering
`
`on its mass. This can be understood by considering thefollowing set of experiments.
`
`reted many fyndament
`“1 N=1dyne=0.225Ib.
`ou apply a
`force F to a block of
`ice on a frictionless surface, the block will
`cerning the m
`lergo some acceleration a.
`If
`
`sof the block is doubled, the same applied
`ufions to phys
`
`rocuce an acceler
`
`scientific thou
`the mass is tripled, the same applied force
`2.
`If
` remai po
`¢
`an acceleration a/
`n. We concludethat the acceleration of
`
`nf An Resource,
`
`the object
`
`is much le
`
`in
`
`the
`
`speedof fight. We will
`
` nly when the speed of
` ion in- Chapter 9
`
`t
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`The Laws of Motion
`Chapter 4
`
`
`EXERCISE 3) An 1800-kg car is travelingina straight line with a speed of 25.0 m/s. What is
`
`the magnitude of the constant horizontal force that is weeded to bring the car to rest ina
`Fiaukine Perea
`distance of 80.0 m?
`Anisw
`7030 N
`
`45
`
`The Gravitational Force and Weight
`
`In a train, the ears are connectedby couplers,The couplers between the cars are under
`tension as the train is pulled by the locomotive in the front, As you move from the
`locomotive to the caboose, does the tension force in the couplers imerease, decrease, or
`stay the same as the wain speeds up? What if the engineer applies the brakes, so that
`
`the couplers are under compression? Howdoes the compression force vary from lo-
`comotive to caboose in this
`
`Reasoning The tension force decreases from the front ofthe train to the back. The
`coupler between the locomotive andthefirst car must apply enough force to accelerate
`all of the remaining cars. As we move back along the train, each coupleris accelerating
`less mass behind it. The last coupler onlyhas to accelerate the caboose, so it is under
`the least tension. Ifthe brakes are applied, the couplers are under compression. The
`force of compression decreases from the
`front to the back ofthe train in this case also.
`
`The first coupler, at the back of the locor
`just apply a kirge force to slow down
`
`
`all of
`the
`ning cars. The final coupler must onlyapply a force large enough to
`slow down the mass of the caboose.
`
`4.5 +» THE GRAVITATIONAL FORCE AND WEIGHT
`
`
`
`Weare well aware of the fact that all objects are attracted to the Earth. The force
`exerted by the Earth on an object is the gravitational force F. (Fig. 4,5). This force
`is directed toward the center of the Earth. The magnitude of the gravitational
`force is called the weight of the object, w.
`Wehave seen thata freelyfalling object experiences an acceleration g directed
`
`toward the center of the Earth. Applying Newton's second law to a freely falling
`
`object of mass m, we have F = ma. Because F. = mg,it follows that a = g
`and
`w= mg
`[4.5]
`Because it depends on g, weight varies with geographiclocation. (You should
`
`not confuse the italicized symbol g that we use for gravitational accel
`om with
`Astronaut Edwin E, Aldrin, fr.,
`the symbol g thatis used for grams.) Bodies weigh less at higher altitudes than at
`walking on the Moon after the
`sea level because g decreases with increasing distance fromthe center of the Earth.
`Example 4.1 An Accelerating Hockey Puck
`Apollo 11 lunar landing, The
`Hence, weight, unlike mass,
`is not an inherent property of a body. For example,
`if
`
`The resultant foree in the 9 directionis
`weight of the astronaut on the
`A. hockey puck with a mass of 0.30 kg slides on the horizontal
`a body has a mass of 70 kg, thenits weight in a location where g = 9.80 m,/'s* is
`frictionless sur
`Moonis bess than it
`is on Earth but
`
`ce of an ice rink, Two forces act on the puck
`mg = 686 N (about 154 Ib). At the top of a mountain where g=9.76 m/s*, the
`his mass remains the same.
`(Cow
`
`as shown in Figure 44. The force F, has a magnitude of
`SR=F,+#
`—F, sin 20° + Fsin 60°
`rxy of NASAJ
`tile of 8.0 N. Devermine the ac-
`=
`= (5.0 N) (0.342) + (8.0 N) (0.866) = 45.2 N
`
`5.0 N, and F, hasa er
`celeration of the puck
`
`Figure 4.5 Theonly force acting on an
`abject
`in free fall is the gravitationalforee,
`F,. The magnimdeof this force is the
`
`Now we can use Newton's second lawin component form to
`find the «and y components of acceleration;
`
`t= = = m/s
`EE ee cg
`ie
`m
`O.30-kp
`‘
`:
`
`ow. O28
`130 ke
`m
`:
`Theacceleration has a magnitude of
`a=
`+ (17)? m/s* =
`=
`(29)?
`= 34 m/s?
`and its direction is
`
`7
`é= tan! (*) =tan! (=) = 31°
`99
`a,
`relative tothe positive x axis.
`
`ge
`5ON
`B,
`Ber
`
`
`
`:
`
`© 4.1) An object moving onafrictionless
`direction of the
`revwitant force,
`
`
`
`weight of the object, mg. iatement ia simplification in that it ignores the fact
`
`ree inthe x direction is
`EXERCISE 1 Determine the components of a third force
`
`cos 20° + Fy cos 60°
`that, when applied to the puck, will cause it to be in equilib-
`N
`
`(8.0 Nj(0500)=8.7N rium. F,=—-8.7N,F =
`
`
`
`(a) What is the mag-
`A 6.0+kg object undergoes an acceleration of 2.0 m/s*.
`MERCISE 2
`> of
`the resultant force actin
`4
`he object? (b) If this same force is applied to a
`fa)
`1L2N
`(hb) 3.0 m/s*
`
`
`abject, what acceleration does it f
`
`
`that
`
`tt
`
`disoabotion of tee Earth is no
`
`
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`nitudeof this force is mg. The react
`
`force of theprojectile
`to this forceis the
`must accelerate the Earth toward
`
`
`on the Earth,
`F) = —F,. The reactionforce, F /
`the projectile just as the action force, F,, accelerates the projectile toward the Earth,
`
`However, because the Earth has such a large mass,
`its acceleration due to th
`reaction forceis negligibly small.
`
`
`4.6b, The
`Another
`example of Newton's third lawin action is shown in Figu
`CONCEPTUAL PROBLEM 3a
`force exerted by the hammer onthenail (the action) is equal to and opposite the
`force exerted by the nail on the hammer (the reaction). You directly experience
`
`Suppose you are
`g by interplanetary telephone to yourfriend, wholives on the Moon.
`the law if you slam your fist against a wall or kick a football with your bare foot. You
`
`
`
`Hetells you that he has just won
`a newton of gold ina contest. Youtell him that you entered
`should be able to identify the action and reaction forces in these cases.
`
`the Barthversion of the same contest and also won a newton of gold! Who is richer?
`
`
`h exerts
`5 we mentioned earl
`E
`force Fon any object. If the object
`
`CONCEPTUAL PROBLEM 4a
`isa TVat rest on atable, asinFigure 4.7a, the reaction force to F. is theforce the
`TV exerts on the Earth,
`F/. The TV does not accelerate, because it
`is held up by
`s neglected,
`
`hi
`
`the table. The table, therefore, exerts on the TV an upward action force, m, called
`rball of mass m is thrown upward with someinitial speed. If air resistan
`
`
`d (b) whenit
`fis the force on the ball (a) when it reaches half its maximumheight
`
`reaches its p
`is
`TV set
`(a) When a
`Figure 4.7
`
`
`
`ting onatable, the forcesacting
`are the nonmal force, a,
` force
`4.6 + NEWTON'S THIRD LAW
`wity,
`F_.
`The
`reaction tom is the force of the TV
`
`
`setou the
`ble, n'.
`Th
`c reaction
`»
`FB,
`is
`th
`TV set os
`
`
`Fe,
`he Earth,
`free-body
`(b) The
`diagram for the
`TV set
`
`4.6
`
`Newton's ThereLaw
`
`91
`
`
`
`Figure 4.6 Newton's third law
`1
`(a) Theforce exerted by object
`
`on abject 2 is equal in magnitude
`
`and opposite the force exerted by
`
`(b} The force
`object-2 on object
`1.
`exerted by the hi
`er on the nail
`is equal in magnitude and opposite
`
`the force exerted by the nail on
`foln Gillmowre,
`the hammer
`The
`Stock Market)
`
`Fu
`
`
`
`Chapter 4
`
`The Laws of Motion
`
`body's weight would be 683 N. Thus, if you want to lose weight without going on a
`diet, climb a mountain or weigh yourself at 30 000 ft during an airplane flight.
`
`Because w = mg, we can compare the masses of two bodies by measuring their
`
`weights with a spring seale or balance. Ata given locationthe ratio of the weights
`of two bodies equals the ratio of their masses.
`
`Physics 4
`Thinking
`absence ofair friction, itis claimed thatall objects fall with the sameacceleration.
`
`
`wier object is pulled to the Earth with more force than a light object. Why does
`the heavier object not fall faster?
`Reasoning [tis indeed true that the heavier object is pulled with a larger force. The
`
`strength of the force is determined by the gravitational mass of the object. But the
`refore, W
`resistance to the force and, th
`¢ change in motion of the object, is repre:
`
`
`sented by the object
`1 mass andgravitational mass are chosen to
`be equal in Newtonian mechanics. Thus,
`if an object has twiee as much massas another,
`
`
`itis pulled to the Earth with twice the force, but itulso exhibits
`twice the resistance to
`
`having its motion changed. These factors cancel, so that the change in motion, the
`acceleration, is the samefor all objects, regardless of mass,
`
`Aw
`
`A statement of Newton's third *
`law
`
`Newton's third law states that if two bodies interact, the force exerted on
`
`
`body 1 by body 2 is equal in magnitude but opposite
`ection to the
`force exerted on body 2 by body |
`
`[4.6]
`F.,= —Fy,
`
` This law, which is illustrated in Figure 4.6a, fk
`
`8 equiva
`always occur in pairs, or that a single isolated force cannot exist. The force that
`body 1] exerts on body 2 is sometimes called the action force, and the force of body
`2 on body 1
`is called the reaction force
`In reality, either force can be labeled the
`ction or reaction force, The action force is equal in magnitude to the reaction
`force and opposite in direction. In all cases, the action and reaction forces act
`ondifferent objects and mustbe ofthe same type. For example, the force acting
`on afreely falling projectile is the force of the Earth on theprojectile, F_, and the
`
`
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`Chapter 4
`
`The Laws of Motion
`
`Normalforce *
`
`the normal force.? This is the force that prevents the TV from falling through the
`table; it can have any value needed, up to the point of breaking the table. The
`normal force balances the weight and provides equilibrium. The reaction to m is
`the force of the TV on the table, n'. Therefore, we conclude that
`
`> —2,
`
`cand
`
`n=-n
`
`The forces m and n’ have the same magnitude, which is the same as F, unless the
`table has broken. Note that the forces acting on the TVare F, and n, as shown in
`Figure 4.7b. The two reaction forces, F/ and n’, are exerted on objects other than
`the TV. Remember, the two forces in an action=reaction pair always act on two
`different objects.
`From Newton's second law, we see that, because the TV is in equilibrium
`(a = 0), it follows that F = n = mg.
`
`in Figure 4.8,
`on the horse,
`ves the system
`
`Thinking Physics 5
`
`A horse pulls a sled with a horizontal force, causing it to accelera
`
`
`Newton's third law says that the sled exerts an equal and opposite fi
`
`
`te? Under what conditior
`In view of this, how can thesled accele
`(horse plus sled) move with constant velocity?
`is important to remember that the
`it
`Reasoning Whenapplying Newton's third law,
`
`forces involved act on different objects. When you are determining the motion ofan
`object, you must add the forces on tha
`ect alone. The force that accelerates the
`system (horseane sh
`1) is the force exe
`Ih on the horse's feet. Newton's
`
`third law tells us wh
`
`tthe horse exer
`equal and opposite force on the Earth, The
`
`horizontal forces exerted:on the sled are the forward force exerted by the horse and
`the backward force of friction betweensled and surface (Fig. 4.8b), When the forward
`
`force exer
`dd bythe horse on the sled exceeds the backward force, thesled accelerates
`
`to the right. The horivental forces exerted on the horse are the forward force of the
`Earth and the backward force of the sled (Fig. 4.8c¢). The resultant of these two forces
`causes the horse to aceelerate. When the
`forward force of the Earth on the horse
`
`balances the force of friction between sled and surface, the system moves with constant
`velocity
`
`
`
`
`
`
`
`
`
`Figure 4.8 (Thinking Physics 5)
`
`
`
`
` mal is used because the dir cular to the surface
`
`4.7
`
`Same Applications of Newton's Laws
`
`93
`
`CONCEPTUAL PROBLEM 5
`
`If a small sports car collides head-on with a massive truck, which vehicle experiences the
`
`greater impa
`ce? Which vehicle experiences the greater acceleration?
`
`CONCEPTUAL PROBLEM 6
`
`is directed toward the right,
`Twoforces, F, and F,, are exerted on an object of mass m. F,
`whereas F, is directed toward the left. Under what condition does the object accelerate
`toward the right? Toward the left? Can its acceleration ever be zere
`
`‘CONCEPTUAL PROBLEM 7
`
`Whatforce causes an automobile to move? A propeller-<dnven airplane? A rocket? A rowboat?
`
`CONCEPTUAL PROBLEM 8
`
`Alarge manand a small boy stand facing each other on frictionless ice. They put their hands
`together and push toward each other, so that they move away from each other. Whoexerts
`the larger force? Whoexperiences the larger acceleration? Who moves away with the higher
`velocity? Who moves over a longer distance while their hands are in contact?
`
`4.7 - SOME APPLICATIONS OF NEWTON’S LAWS
`
`In this section we present some simple applications of Newton's laws to bodies that
`
`are either in equilibrium (a = 0) or moving linearly
`under the action of constant
`external forces. For our model, we shall assume that the bodies behave as particles
`so that we need not worry about rotational motions. In this section we also neglect
`the effects of friction for those problems involving motion. This is equivalent to
`stating that the surfaces are frefioniess. Finally, we usually neglect the masses of any
`ropes involved.
`In this approximation, the magnitude of the force exerted at any
`point along theropeis the sameat all points along therope. In problemstatements,
`
`the terms fight and of negligible mass are used to
`indicate that a mass is to be ignored
`
`when you work the problem. These two terr
`re synonymous in this context.
`Whenwe apply Newton's laws to an object, we shall be interested only in those
`
`external forces that act on the object. For example,
`in Figure 4.7 the only external
`forces acting on the TVare n andF_.
`The
`
`actions to these forces, m’ and F_, act
`on the table and on the Earth, respectively, and do not appear in Newton's second
`law as applied to the TV.
`When an object such asa block is being pulled by a rope attached to it, the
`rope exerts a force on the object.
`In general, tension is a scalar and is definedas
`
`the magnitude of the force that the rope exerts on whatever is attachedto it.
`
`Consider a cratebeing pulled to theright on the frictionless. horizontal surface,
`as in Figure 4.9a. Suppose you are asked to find the acceleration of the crate
`and
`the force the floor exerts onit. First, note that the
`
`
`al force being applied
`
`
`
`son the cra’
`s denoted
`to the crate acts through the rope, The force the rope exe:
`
`In Figure
`in the rope.
`by the symbol T. The magnitude of T is equal to the tens
`ou to isolate it from its
`49a, a dotted circle is drawn around the block to ren
`surroundings.
`
`Tension
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`
`IPR2023-00783
`
`
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`
`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`
`IPR2023-00783
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`IPR2023-00783
`
`
`
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`Chapter 5
`
`More Applications of Newton's Laws
`
`(Focus on Sports)
`
`An athlete in the process of throwing the ham-
`mer, The force exerted by the chain on the
`hammer provides the centripetal acceleration.
`Only when the athlete releases the hammer
`will it move along a straightline path tangent
`to the cirche.
`
`A force causing a centripetal acceleration acts toward the center of the circular
`path and causes a change in the direction of the velocity vector. In the case of a
`ball rotating at the end ofa string, the force exerted by the string on the ball causes
`the centripetal acceleration of the ball. Fora satellite in a circular orbit around the
`Earth, the force of gravity causes the centripetal acceleration of the satellite, The
`
`foree acting on a car rounding a curve on a flat road to cause the centripetal
`acceleration ofthe car is theforceof static friction betweenthe tires and pavement,
`and so forth. In general, a body can move in a circular path under the influence
`of various types of forces or a combination offorces.
`if the
`Regardless of the type of force causing the centripetal acceleration,
`foree acting on an object should vanish, the object would no longer move in its
`
`circular path;
`instead it would move along a straight-line path tangent to the
`circle. This idea is illustrated in Figure 5.6 for the case of the ball whirling in a
`circle at the end of a string.If the string breaks at some instant, the ball will move
`
`along the straight-line path tangent to thecircle at the point at which the string
`broke.
`
`a ee eee
`
`Figure 5.6 When thestring breaks, the ball
`moves in the direction tangent to the cireu-
`lar path,
`
`4
`
`
`
`5.2
`
`
`Newton's Second Law Applied to Uniform Circular Motion
`
`117
`
`Thinking Physics 3
`The force causing centripetal acceleration is often called a centripetal foree, We are
`familiar with « variety of forces im nature—friction, gravity, nonmal forces, electrical
`
`forces, tension, and so. on. Should we
`add centripetal force to this list?
`
`Reasoning Centripetal force
`uld not be added to this list. This is « pitfall for many
`
`
`1g the force causing circular motion a name—cenwipetal force—leacls
`studerits
`
`many students to consider th
`asa newkindof force rather than a new role for force:
`
`A common mistake in force diagrams is to draw in all of the usual forces and then add
`
`
`another vector for the centripetal foree, But it
`is not a separate forcee—it is simply one
`of our familiar forces acting in the role of causing a arowar motion. Consider some ex-
` i the Sun, the centripetal force is grewity
`amples. For the motion of the Earth
`For an object sitting on a rotating turntable
`centripetal forceis friction, For a rock
`
`whirled on the end of a string, the centripe
`¢ is the tension in the string. For an
`
`amusement park patr
`pressed against the ir
`1 wall of a rapidly rotating circular
`room,
`the centripetal forceis the normalforce trom the wall. In addition, the centripetal
`
`force could be a combination of
`two or more forces. For example, as a Ferriewheel
`rider passes through the lowest point of the ride, the centripetal force on therider is
`the difference between the normal force fromthe seat and the force of gravity
`
`is, the roadway istilted toward theinside
`
`Thinking Physics 4
`
`
`High-speed,
`curved roadways are banked—that
`of tight curves. Why is this?
`a force providing the
`Reosoning If an automobileis rounding a curve, there must be
`
`tally,
`this is the static friction force between thetires and
`centripetal acceleration, N
`t
` roadway, parallel to the roadway surface and perpendicular to the centerlineof
`
`
`the car. This is the
`W horizontal force—the weight and normal force are both ver
`
` ton
`force may be larger
`tical. If the curve is: tight or the speed is high, the required fri
`
`
`n's first law takes ove
`than the maximum possible friction force. In this case.
`of
`and the car skids along a straight line, which is off the side
`the curved road.
`T
`
`
`situation is even morelikely if the road is icy, which reduces the
`friction coefficient
`
`
`from the vertical, so that
`it has a
`Banking the roadway tilts the normal force away
`roadway. Thus, the normal
`horizontal component toward the
`center of
`the
`circular
`
`
`force and friction work together to provide the necde
`al acceleration. The
`larger the acceleration, the stronger the banking, in oF
`ovide a tanger oom-
`
`ponent of the normal force in the horizontal direction
`
`As these cyclists
`gotiate
`
`tripet
`of static frictior
`track surface
`Spo
`Cimon
`
`Tour de France me-
`he
`a Mato
` k, the cen
`nis provided by the
`force
`en the
`tires and the
`he!
`Goureruarr, Phat N
`
`
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`sale cial ra
`
`
`
`! a
`IPR2023-00783
`
`rs
`oad
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`Chapter 5
`
`MoreApplications ofNewton's Laws
`
`5.5
`
`Numerical Modeling in Particle Dynamics
`
`127
`
`Example 5.9 Am Object Falling in a Vacuum
`We can illustrate the procedurejust described by considering_particle times its acceleration (taking upward to be the posi-
`a particle falling in a vacuum under the influence of the force
`tive direction):
`of gravity, as in Figure 5.15,
`~
`
`F= ma = —mg
`
`
`
`a
`
`me
`
`Figure 5.15 (Example 5.9) An object falling in vacuum
`under the influence of gravity.
`
`Thus, a = —g, a constant. Because dv/dt = a, the resulting
`differential equation for velocity is dv/dt = — g, which may be
`integrated to give
`
`u(t) = wy ~ gt
`Then, because dx/di = v, the position of the particle is ob-
`tained from anotherintegration, which yields the well-known
`result
`
`x(f) = xo + tet — det?
`
`Figure 5.14 (Conceptual Problem5) By spreading their arms and legs and by keeping their
`bodies parallel to the ground, skydivers experience maximum air resistance resulting in a mini-
`mum terminal speed.
`(Guy Suvage, Photo Researchers, Inc.)
`In this expression, x) and uy, represent the position and speed
`Reasoning and Solution Applying Newton's second law, we
`set the sum ofthe external forces equal to the mass of the—of the particle at (= 0.
`
`CONCEPTUAL PROBLEM 5SSTT
`The procedure just described is straightforward for some physical situations,
`Consider a skydiverfalling through air before reaching her terminal speed, asin Figure 5.14.
`such as the one described in the previous example. In the real world, however,
`Asthe speed of the skydiver increases, what happens to her acceleration? What is her accel-
`eration once she reaches terminal speed?
`complications often arise that make analytical solutions for manypractical situations
`difficult and perhaps beyond the mathematical abilities of most students taking
`introductory physics. For example, the force may depend on the position of the
`particle, as in cases in which the variation in gravitational acceleration with height
`must be taken into account. The force mayalso vary with velocity, as we have seen
`in cases of resistive forces caused by motion througha liquid or gas. The force may
`depend on both position and velocity, as in the case of an object falling through
`air where the resistive force depends on velocity and on height (air density). In
`rocket motion, the mass changes with time, so even if the force is constant, the
`acceleration is not.
`Another complication arises because the equations relating acceleration, ve-
`locity, position, and time are differential equations, not algebraic equations. Dif-
`ferential equationsare usually solved using integral calculus and other special tech-
`niques that introductorystudents may not have mastered.
`So how does one proceedto solve real-world problems without advanced math-
`
`ematics? One answer is to solve such problems on personal computers, using ele-
`mentary numerical methods, The simplest of these is the Euler method, named
`after the Swiss mathematician Leonhard Euler (1707-1783).
`
`5.5 + NUMERICAL MODELING IN PARTICLE DYNAMICS?
`
`As we have seen in this and the preceding chapter, the study of dynamics of a
`particle focuses on describing the position, velocity, and acceleration as functions
`of time. Cause-ancd-effect relationships exist between these quantities: A velocity
`causes position to change, an acceleration causes velocity to change, and an accel
`eration is the direct result of applied forces. Therefore, the study of motion usually
`begins with an evaluation of the net force on the particle,
`In this section, we confine our discussion to one-dimensional motion, so bold-
`face notation will not be used for vector quantities. If a particle of mass m moves
`under the influence of a net force F, Newton's second law tells us that the accel-
`eration of the particle is given by a = F/m. In general, we can then obtain the
`solution to a dynamics problem by using the following procedure:
`
`1. Sum all the forces on the particle to get the net force F.
`2. Use this force to determine the acceleration, using a = f/m.
`3. Use this acceleration to determine the velocity from du/dt = a
`4. Use this velocity to determine the position from dx/dt = wv.
`
`“The author is most grateful to Colonel James Hew
`section
`
`
`© US. Air Force Academy for preparing this
`
`The Euler Method
`
`In the Euler method ofsolving differential equations, derivatives are approximated
`as finite differences. Considering a small increment of time, At, the relationship
`between speed and acceleration may be approximated as
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`Chapter 5
`
`More Applications of Newton's Laws
`
`breaking. What range of speeds can the mass havebefore
`the string breaks?
`14. A 55.0-kg ice skater is moving at 4,00 m/s when she grabs
`the loose end of a rope the opposite end ofwhichis tied to
`a pole. She then moves in a circle of radius0.800 maround
`the pole. (a) Determine the force exerted by the rope on
`her arms. (b) Compare this force with her weight.
`15. In the Bohr model of the hydrogen atom, the speed of the
`electron is approximately 2.20 * 10° m/s. Find (a) the cen-
`tripetal acceleration of the electron and (b) the force acting
`on the electron as it revolves in a circular orbit of radius
`0.530 * 10°m.
`16. Accrate of eggs is located in the middle of the flat bed of a
`pickup truck as the truck negotiates an unbanked curve in
`the road. The curve may be regarded as an arc of a circle
`ofradius $5.0 m, If the coefficient ofstatic friction between
`crate and truck is 0.600, what must be the maximum speed
`of the truck if the crate is not to slide?
`[17)A coin placed 30.0 cm from the center ofa rotating, hori-
`zontal turntable slips when its speed is 50.0 cm/s. (a) What
`force provides the cenwipetal acceleration when the coin is
`stationary relative to the turntable? (b) What is the coeffi-
`cient of static friction between coin and turntable?
`is Asmall turtle, appropriately named Dizzy, is placed on a
`horizontal, rotating curable 20.0cm from its center.
`Dizzy's mass is 50.0, and the coefficient of static friction
`between his feet and the turntable is 0.900, Find (a) the
`maximum number of revolutions per second the turntable
`can have if Dizzy is to remain stationary relative to the turn-
`table and (b) Dizzy's speed and centripetal acceleration
`when he is on the verge of slipping.
`Consider a conical pendulum with an 80.0-kg bob on a
`10.0-m wire making an angle of 5.04" with the vertical (Fig.
`P5.19). Determine (a) the horizontal and vertical c