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Physics 110 1435-1436 H Instructor: Dr. Alaa Imam E-mail: [email protected] Chapter 5 FORCE AND MOTION -I Sections 5-2, 5-3, 5-4, 5-5, 5-6 Newtonian Mechanics Newton’s First Law Force Mass Newton’s Second Law Important skills from this lecture: 1. 2. 3. 4. 5. 6. Explain Newton’s first law Define the force and its unit Resolve forces and find the resultant along x & y axes Define the mass and its relation to force Explain newton second law Relate the force component along an axis to its acceleration 7. Define Newton unit 8. Draw free-body diagram 9. Apply Newton 2nd law in one & two dimensions Newtonian Mechanics The study of the relation between force & acceleration is called Newtonian mechanics Newtonian mechanics does not apply to all situations; If the speeds of the interacting bodies are very large (near the speed of light) à Einstein’s special theory of relativity applied If the interacting bodies are on the scale of atomic structure, e.g. electrons à quantum mechanics is applied Newtonian mechanics is applied to the motion of objects ranging in size from the very small to astronomical objects Send a puck sliding over the ice of a skating rink, however, and it goes a lot farther. You can imagine longer and more slippery surfaces, over which the puck would slide farther and farther. In the limit you can think of a long, extremely slippery surface (said to be a frictionless surface), over which the puck would hardly slow. (We can in fact come close to this situation by sending a puck sliding over a horizontal air table, across which it moves on a film When there is no force acting on a body: of air.) From these we can conclude that a body will keep moving with If theobservations, body is at rest, it stays at rest constantvelocity no force acts on itit.continues That leads to us move to the first Newton’s three If the if body is moving, withofthe laws of motion: same velocity (same magnitude & direction) Newton’s First Law Newton’s First Law: If no force acts on a body, the body’s velocity cannot change; that is, the body cannot accelerate. In other words, if the body is at rest, it stays at rest. If it is moving, it continues to move with the same velocity (same magnitude and same direction). 5-4 Force We now wish to define the unit of force. We know that a force can cause the the vector rules of Chapter 3. d so on. Thus in magnitude general, if our antity, with both and standard direction.body Is force also a vector This means thatdirection when two magnitude know that force F must be the easily assigna,awe direction to aaforce (just assign of or more forces act on a body, w force, or resultant force, bythat adding the individual forces vector of force (in sufficient. newtons) is equal the magbutthe that is not We musttoprove by experiment that has forces the magnitude direction of the net force has the s per second per second). quantities. Actually, that has been done: are indeedand vector body asacceleraallcombine the individual forces e acceleration produces. However, ave magnitudesitand directions, and they according to together. This fact is called the position forces. The world would be quite strange if, for e magnitude also afor vector Chapter 3.and direction. Is force rection a force (just assign the friend werewe toofcan pullfind on their the standard body in the same direction hat whento two or more forces act on direction a body, net the A force isbymeasured by the acceleration produces ufficient. must prove experiment that of 1 N, and yet somehow netitpull was 14 N. force, by We adding individual forces vectorially. A singlethe force y, thatand has direction been done: forces indeed vector this book, forces most of Acceleration isInahas vector quantity àare force is aoften vectorrepresented quantity with a vec nitude the netare force the same effect on the : : nd directions, and they combine according to is represented ( ) This F, is and a net ividual forces together. fact called theforce principle of super- with the vector symbol Fnet. As aisforce orfor a (N) net force can s. The world would quite strange if, example, you have and acomponents along coordinate ax be Force unit Newton more act on body a body, find theiranet ll on forces the standard in we thecan same direction, each with forceare single-component forces. The only along single axis, athey the individual forces vectorially. Aprinciple single force Superposition for forces: mehow the net pull was 14 N. on of the force thetwo same the act forces are net most often represented with aon vector symbol astheir net force or has When or effect more forces on asuch body, : ether. This factwith is called the principle superresultant force ( Fof are theother vector addition of the is represented the vector symbol with vectors, net.)As ld be if, foralong example, a When forces act individual forcesyou and rce canquite have strange components coordinate axes. theare same eachforces. with force ebody axis, in they single-component Then can drop the direction, A single force that ahas thewe magnitude & direction of the net ll was 14 N. force has the same effect on the body as all the individual ften represented with a vector symbol such as forces together : h the vector symbol other vectors, Fnet. As canwith have many components along coordinate axes ponents along coordinate When act a single axis, they are single Whenaxes. forces act forces only along ngle-component forces. Then weforces can drop thearrows could be replaced by signs component à the to indicate the forces directions Force of the forces along that axis. Instead of the wording used in Section 5-3, the more proper statement of 5 FORCENewton’s AND MOTION—I Law is in terms of a net on force: If First many forces are acting a body, and their net force is zeroà the body cannot accelerate are made in, say, an elevator that is accelerating relative to the ground, then : and the results can be the measurements are being made in a noninertial frame Newton’s First Law: If no net force acts on a body (F net ! 0), the body’s velocity surprising. cannot change; that is, the body cannot accelerate. CHECKPOINT 1 : Which of the figure’s six arrangements correctly show the vector addition of forces F1 : : yield the thirdforces vector, which is meant Fnet?net force and F may be acting ontoarepresent body,their butnetifforce their 2 tomultiple There body cannot accelerate. F1 Inertial Reference Frames F F1 is zero, the F1 ü F F 2 2 2 (b) (c) Newton’s(a) first law is not true in all reference frames, but we can always find reference framesF 2 in which it (as well as the rest of Newtonian mechanics) is true. Such special frames are referred to as inertial reference frames, or simply F1 F1 F1 inertial frames. ü (d) ü (e) F2 (f ) F2 An inertial reference frame is one in which Newton’s laws hold. Mass Force produces different magnitudes of acceleration for different bodies e.g., if a baseball & a bowling ball are given the same kick à baseball acceleration > bowling ball? Because the mass of the baseball differs from the mass of the bowling A less massive baseball receives a greater acceleration than a more massive bowling ball when the same force is applied to both The mass of a body is the characteristic that relates a force on the body to the resulting acceleration Mass is: an intrinsic characteristic of a body a scalar quantity E All the definitions, experiments, and observations we have discussed so far can be This equation is statement: simple, but we must use it cautiously. First, we must b summarized in one neat : 5-6 Newton’s Second Law certain about which body we are applying it to. Then Fnet must be the vector sum of all the act body. Only forces that act on that All theforces definitions, we discussed sobody’s farbody can beare to b Newton’s Secondthat Law:experiments, Theon net that forceand on aobservations body is equal to have the product of the summarized one neatsum, statement: included inacceleration. thein vector not forces acting on other bodies that might b mass and its involved in the given situation. For example, if you are in a rugby scrum, the ne force on you isSecond the vector of all and pulls on your body. It doe Newton’s Law: Thesum net force on athe bodypushes is equal to the product of the body’s In equation form, mass andany its acceleration. not include push or pull on another player from you or from anyone els : Every time you work F: a force problem, your firstlaw). step is to clearly state the body t (Newton’s second (5-1) ! ma net which you are applying Newton’s law. In equation form, Likeequation equations, is component equivalent toFirst, three This is simple, but weEq. must use it cautiously. wecomponent must be equa other This vector equation is equivalent to5-1 three equations, one : : :coordinate for eachbody axis of an xyz coordinate system tions, for each axis an xyz system: certainone about which we are applying it to. Then F must be the vector(5-1) sum (Newton’s second net law). Fnet ! ma Newton’s Second Law of all the forces that act on that body. Only forces that act on that body are to be Fnet,isxsum, ! ma Fnet,y ! mause and Fnet,z ! ma . might includedThis in the vector not forces acting other bodies thatzwe equation simple, we must cautiously. First, must be be (5-2) x,but y,onit : involved in about the given situation. if to. you are Fin rugby the sum net certain which body weFor areexample, applying it Then bescrum, the vector netamust forceofon the vector of all theOnly pushes andthat pulls body.are Itan does allyou the is forces that actsum on that body. acton onyour that along body toaxis be to th Each of these equations relates the net forces force component AND MOTION—I not include push or pull on not another player from you or from anyone else. included inalong the vector sum, forces acting on other bodies that might be us tha any Each of these equations relates the net force component along acceleration that same axis. For example, the first equation tells an to acceleration along that same axis Every time work a the force problem, first step isxare to clearly state thexbody to involved in axis the situation. For your example, if you in acauses rugby scrum, the net the sum ofyou all thegiven force components along the axis the component a force on you is the component vector sum of all the no pushes and pulls body. It zdoes which you areacceleration applying Newton’s law. The along a given axisacceleration is caused only byon the sum force of the body’s acceleration, but causes inyour theof y the and direction not include any push pullaxis, on another player from orcomponent from anyone else. Like other vector equations, Eq. 5-1 to you three equaalong thatorsame and notisbyequivalent force components along any other Turnedcomponents around, the acceleration component ax is caused only byaxis. the sum of th time youaxis work force your first step is to clearly state the body to tions,Every one for each ofaan xyzproblem, coordinate system: forcewhich components alongNewton’s the x axis. you are applying law.In general, Fnet, ma Fnet,yifEq. !the ma ma . component (5-2) x, that y,is and net,za ! zis Equation 5-1x ! tells us force Fon body zero, the body’s Like other vector equations, 5-1net equivalent to three equa- OTION—I From Newton’s law, if Fnet = 0 ⇒ a = 0 The acceleration a given by theat sum of the force Ifcomponent the body along is at rest (v axis = 0 is&caused a = 0),only it stays rest components along that same axis, and not by force components along any other axis. If the body is moving (v ≠ 0 & a = 0), it continues to move at constant velocity on such balance another, andthe body’s Equation 5-1Any tellsforces us that if thebody net force on one a body is zero, thebody forces and the body at are said in acceleration : a ! both is at rest, it stays rest; if ittoisbe moving, it continues 0. If the equilibrium to move at constant velocity. In such cases, any forces on the body balance one Thethe forces could one another, another, and both forces andcancel the body are said to be in equilibrium. but still the Commonly, the forces are act alsoon said tobody cancel one another, but the term “cancel” is 2nd tricky. It does not mean that the forces cease to exist (canceling forces is not like canceling dinner reservations). The forces still act on the body. For SI units, tells us that InEq. SI 5-1 units 1 N ! (1 kg)(1 m/s2) ! 1 kg " m/s2. (5-3) Some force units in other systems of units are given in Table 5-1 and Appendix D. Table 5-1 For SI units, Eq. 5-1 tells us that 1 N ! (1 kg)(1 m/s2) ! 1 kg " m/s2. (5-3) Some forceunits in other units are given of in Table Appendix D. Some forcesystems units inofother systems units 5-1 areand given in Table 5-1 Table 5-1 Units in Newton’s Second Law (Eqs. 5-1 and 5-2) System Force Mass Acceleration SI CGSa Britishb newton (N) dyne pound (lb) kilogram (kg) gram (g) slug m/s2 cm/s2 ft/s2 1 dyne ! 1 g " cm/s2. a 1 lb ! 1 slug " ft/s2. b To solve problems with Newton’s second law, we often draw a free-body diagram in which the only body shown is the one for which we are summing forces. A sketch of the body itself is preferred by some teachers but, to save space accelerationEveryday for different bodies. Put a baseball and a bowling ball on the floor experience tells us that a given force produces different magnitudes of and give both the same kick. EvenPut if you don’tand actually do this, youtheknow acceleration forsharp different bodies. a baseball a bowling ball on floor the result: The noticeably larger acceleration the you bowling and baseball give both receives the same a sharp kick. Even if you don’t actuallythan do this, know ball. The two differ because the mass of acceleration the baseball differs from the accelerations result: The baseball receives a noticeably larger than the bowling The two accelerations differ exactly, because is the mass of the the mass ofball. the bowling ball — butunknown what, mass? Calculate an mass bybaseball differs from mass ofhow the bowling ball —mass but what, exactly, is mass? We canthe explain to measure by imagining a series of experiments in knowing another mass and their We can to measure mass by imagining series of experiments in an inertial frame. In explain the firsthow experiment we exert a forcea on a standard body, an inertial frame. In the first experiment we exert a force on a standard body, whose masswhose m0 ismass defined to be 1.0 kg. Suppose that that the the standard body acceleraccelerations m is defined to be 1.0 kg. Suppose standard body acceler0 2 ates at 1.0 m/s . We can then say the force on that body is 1.0 N. ates at 1.0 m/s2. We can then say the force on that body is 1.0 N. We next apply thatapply same force would needneed some wayway of of being We next that same(we force (we would some beingcertain certain itit is the sameisforce) to aforce) second whose massmass is not known. the same to a body, secondbody body,X, body X, whose is not known.Suppose Suppose If afind = 1body N isXapplied on at 2 0.25 bodies, standard body, 2 m/s2.aWe know that a2 lessmassive massive we that Fthis accelerates . We know that a less we find that thisforce body X accelerates at 0.25 m/s whose mass m = 1.0 kg, and acceleration a = 1.0 m/s , and ogreater acceleration than a more massive o baseballareceives aacceleration bowlingball ballwhen when baseball receives greater than a more massive bowling nd thesame 2 force body X whose mass (mX)Let is not known, and its the (kick) is applied to both. us then make the following conjecthe same force (kick) is applied to both. Let us then make the following conjec2 acceleration = masses 0.25 of m/s To findis m X : to the inverse of the ratio of ture: The ratio ofaX the two. bodies equal ture: The ratio of the masses of two bodies is equal to the inverse of the ratio of their accelerations when the same force is applied to both. For body X and the their accelerations whenthis thetells same force is applied to both. For body X and the standard body, us that Fo = Fx standard body, this tells us that mX a0 mX ma0 0 ! aX . ! . Solving for mX yields m0 aX Solving for mX yields a0 1.0 m/s2 mX ! m0 aX ! (1.0 kg) 2 m/s2 0.25 1.0 m/s ! 4.0 kg. a0 ! (1.0 kg) ! 4.0 kg. mX ! m0 2 Our conjecture be useful, of course, only if it continues to hold when awill 0.25 m/s X we change the applied force to other values. For example, if we apply an 8.0 N force Free-body Diagram To solve problems with Newton’s second law, a freebody diagram is drawn A free-body diagram is a diagram that contains the only body under the forces In the free-body diagram, the body is represented with a dot, and each force on the body is drawn as a vector arrow with its tail on the body tion is if a vector quantity, with both magnitude and direction. Is force also a ve , and so on. Thus in general, our standard body quantity? canFeasily a direction to a force (just assign the directio n of magnitude a, we know that aWe force must assign be the acceleration), but magthat is not sufficient. We must prove by experiment ude of the force (in newtons) is equal to the forces are vector quantities. Actually, that has been done: forces are indeed v eters per second per second). quantities; they have magnitudes directions, and they combine accordin Ait system consists ofacceleraone or moreand bodies y the acceleration produces. However, the vector rulesalso of Chapter th magnitude and direction. Is force a vector3. means that whenof two or more forces act on a body, we can find thei Types ofThis forces that affect any system: a direction to aforce (just assign the direction force, resultant force,that by adding the individual forces vectorially. A single ot sufficient. We must proveorby experiment External force: any force on the bodies that comes from that has the magnitude and ually, that has been done: forces indeed vector direction of the net force has the same effect on outside theare system as all the according individual to forces together. This fact is called the principle of su es and directions, and body they combine Internal forces two bodies inside the if, for example, you a positionforces: for forces. Thebetween world would be quite strange system were to pull on the standard body in the same direction, each with a f or more forces act on afriend body, we can find their net 1 N, and yet net pull 14 N.connected to If of the bodies making athe system arewas rigidly ng the individual forces vectorially. Asomehow singleup force In book, forces are most often represented with a vector symbol su one ection of the net force hasanother, thethis same effect on the : : F , and a net force is represented with the vector symbol F à the system is treated as one composite body, and net. As with other vec together. This fact is called the principle of supera if, force or a netisforce can components alongforces coordinate axes. When force on the the of all external would be quite strange forsystem example, youvector andhave a sum onlya along a single axis, they are single-component ard body in the same direction, each a force e.g., system ofwith a connected railroad engine & car forces. Then we can drop pull was 14 N. If a tow line pulls on the front of the engine t often represented with a vector symbol as à tow force acts on such the whole engine – car system : with the vector symbol other vectors, à Fnet. As on with the system can be related to its omponents along coordinate axes. When act 2nd law acceleration usingforces Newton’s e single-component forces. Then drop theof the system (m) will be we thecan total mass law, Fnet ! ma , where m is the total mass of the system. CHECKPOINT 2 The figure here shows two horizontal forces acting on a block on a frictionless floor. If a : third horizontal force F3 also acts on the block, : 3N 5N what are the magnitude and direction of F3 when the block is (a) stationary and (b) moving to the left with a constant speed of 5 m/s? F1 = 5N, F2 = −3N From Newton’s 2nd law, Fnet = ma (a) The block is stationary à v = 0, a = 0, à Fnet = 0 F3 + F1 + F2 = 0 à F3 + 5 −3 =0 à F3 = −2N (b) The body is moving with a constant velocity à a = 0 From Newton’s 2nd law, Fnet = 0 à F3 = −2N Sample Problem acceleration. Situation A: For Fig. 5-3b, where only on 93 5-6 KNEWTON’S SECOND LAW (a) acts, Eq. 5-4 Sample Problem gives us EY IDEA One- and two-dimensional forces, puck This is a free-body F1 ! max, In each situation we can relate the acceleration : a to the net Puck F1 x diagram. : axis, we with can simplify situation writing the second Parts A, B, and C of Fig. 5-3 show three force situations which acting on the puck Newton’seach second law, by which, Fnet in with given data, yields : : law forthe x components only:only the x one or two forces act on a puckProblem that moves frictionless Sample Fnetover . However, because motion is along ! ma (b) F1 4.0 N ice along an x axis, in one-dimensional motion. The puck’s : : Fnet, x ! main a(5-4) ! can simpli ! 20 m x. which x ! we axis, Parts A, B, and C of Fig. 5-3 show three situations mass is m ! 0.20 kg. Forces F1 and F2 are directed alongAthe m 0.20 kg B The free-body diagrams for the three situations are alsofor x componen axis and have magnitudes N and F2 ! 2.0 N. Force ! 4.0 law oneF1 or two forces act on a puck that moves over frictionless : The positive answer indicates The horizontal force These forces compete.that the ac by F3 is directed at angle u ! 30° and has magnitude F3 ! 1.0 F1 given in Fig. 5-3, with the puck represented F2 F1 a dot. positive direction thecauses x axis. causes a horizontal iceis the along an x ofaxis, in one-dimensional motion. The puck’s Their net of force x In each situation, acceleration the puck? axis, we can simplifyxSituation situation by writing the second hreeN.situations in what which :each : acceleration. a horizontal 5-3b, wherealong only onethe horizontal force acceleration. F mass is m ! 0.20 kg. Forces and FA:2 For areFig.directed 1 (a) Situation B: In Fig. 5-3d, two horizonta (c) law for x componentsacts, only: moves over frictionless Eq. 5-4 gives us : KEY IDEA puck, F1 in the positive direction of xdia and The free-body axis and have magnitudes F ! 4.0 N and F2 ! 2.0 N. Force Puck F1 1 This is a free-body onal Inmotion. The puck’s F F : : F ! ma , 2x 1 direction. NowisEq. 5-4 gives us This a free-body 1 each situation we can relate the acceleration a to the net x diagram. x (5-4) : Fand ! ma given in Fig. 5-3, wi is directed at angle u ! 30° magnitude F ! 1.0 F : net, x has x. 3 3 diagram. directed along F2 are force on thethe puck with Newton’s second law, Fnet acting which, with given data, yields F1 " F2 ! max, : : N. In each situation, what Fnet ! ma . However, because the motion is along only the x is the acceleration of the puck? (b) (d) The free-body diagrams for the three situations are N and F2 ! 2.0 N. Force F 4.0 N which, withalso given data, yields ax ! 1 ! ! 20 m/s2. (Answer) Situation A: For F m 0.20 kg given in Fig. 5-3,B with the puck represented by a Cdot. has magnitudeAF3 ! 1.0 F1 " F2 4.0 N " 2.0 N Eq. a x ! acts, ! 5-4 gives u positive indicates that the acceleration is in The horizontal force K EFYTheIThese D E forces A answer compete. mhorizontal 0.20 kg eration of the puck? Only thethe F1 F2 F2 1 positive direction of the x axis. causes a horizontal Their net force causes component of F3 x x x force Situation A: For Fig. 5-3b, where only one horizontal : θ acceleration. horizontal acceleration. F3 Thus, with F2. net force accelerates the puck in In each situation we can relatea the acceleration netthe competes a to the : acts, Eq. 5-4 gives (a) forces act on the (c) us Situation B: In Fig. 5-3d, two horizontal (e) tion of the x axis. : : force acting on the puck with Newton’s second law, F which, with given d F F puck, in the positive direction of x and net 1 2 in the negative : Puck F1 : This is a free-body : F2 F1 direction. F ThisFismotion aNow free-body This free-body Eq. 5-4 gives us 2only the Situation C: is InaFig. 5-3f, force F3 is not : x netdiagram. ! ma , F . However, because is along x ! ma x the x 1 x θ cceleration a to the net diagram. diagram. the puck’s acceleration; only F1 " F2 ! maxF,3 direction of : a the!m Newton’s second law, (b) (d) data, yields ( f ) is. (Force F3 is two-dimensional but x which, with given which, with given data, yields A dimensional.) Thus, we write Eq. 5-4 as One- and two-dimensional forces, puc One- and two-dimensional forces, puck otion is along only the x B ontal force horizontal on. F2 F1 x (c) F2 F1 (d) x Fig. 5-3 In three situations, forces act on a puck that moves F1 " F2 4.0 N " 2.0 N F3,shown. F1 horizontal x " F2 ! max. along are also a x4.0 ! N ! an x axis. ! 10diagrams m/s2. The 2 Free-body positive answe The force . ! (Answer) These forces compete. F F1a x ! Only the horizontal m ! 20 m/s 0.20 kg 2 From the figure, we see that F3,x ! F3 cos m 0.20 Their net force causes component ofkg F3 positive direction o (Answer) a horizontal x x θ causes acceleration and substituting for F3,x yield F3 a horizontal acceleration. competes with Faccelerates Thus, the net force the puck in the positiveAdditional direc2. examples, vid acceleration. (e) The positive answer indicates that the accelerationa is! in F3,xthe F3 cos $ " F2 " F2 tion of the x axis. ! x (a) Situation m mB: In Fi of the x axis. : F2 : This is apositive free-body direction This is a free-body along the 30#) " 2.0 N xSituation C: In Fig. 5-3f, force F3 is not directed (1.0 N)(cos F1 in the!pos puck, θ diagram. diagram. "5.7 ! This is aof free-body direction the puck’s acceleration; only x component F3, x Puck F1F3 : 0.20 kg direction. Now Eq. F3 is two-dimensional is.5-3d, (Force two but the motion is only (f x ) Fig. Situation B: In horizontal forces act on theonediagram. C : dimensional.) Thus, we write Eq. :5-4 as Kthe E Yacceleration IDEA N. In each situation, what is of the puck? : Situation A: For Fig eachsituation situationwe wecan canrelate relatethe theacceleration acceleration: thenet net InIneach aatotothe : : acts, Eq. 5-4 giveswith us forceFFnetacting actingon onK the puck withNewton’s Newton’ssecond secondlaw, law, E Ypuck IDEA which, force the with which, with net : :: : . However, because the motion is Fnet!!ma alongonly only the x ma FIn . However, the motion is: neteach situation we canbecause relate the acceleration to the net the x aalong o force F F acting on the puck with Newton’s second law, m = 0.20kg, F1 = 4 N, F2 = 2N, = 1N, θ = 30 3 AA : Fnet : net : ! ma . However, because the motion is along only the x Fnet ,x = max 1 (a) (a) F1 = max F 4N ax = 1 = = 20m / s 2 m 0.20Kg F1 Puck Puck x causes a horizontal acceleration. This is a free-body (a)Puck F1 F1F1 x This is a free-body diagram. x This diagram. is a free-body diagram. x (b) (b) Theseforces forcescompete. compete. F1 These These forces compete. F F1 1 x Their net force causes Their causes net net forceforce causes x x Their F2 FF22 a horizontal acceleration. a horizontal acceleration. a horizontal acceleration. (c) (c) (c) F2 F2 F1 F1 This This is a free-body is a free-body xF1 x This is a free-body F2 x diagram. diagram. diagram. (d)(d) (d) F2 F3,x − F2 F cosθ − F2 = 3 = m m (1)cos(−30 o ) − 2 ax = = −5.7m / s 2 0.20kg F F2 2 x θ F3 (e)F3 θ ax = F3 (e) (e) F2 F2 F2 F3 θ (f ) θ F3 ax ! F1 a"ax xF ! m Thus, the n Thus, the ne Thus, the net force ac tion of the tion of the x axis. tion of the x Situation Situation C: In Fig.C Situation directiondirection of the puck o : direction of: F3(Force is. (Forceis. is two-d :F is. (Force 3 dimensional.) Thus,Fw dimensiona F C C F3,x − F2 = max direction. Now Eq. 5- dimensiona C c) Fnet ,x = max Situation : B :F in puck, 1in t F1 Fig. puck, Situation B: In : direction. N N puck, F1direction. in the positi which, with given dat B B B The positive answer positive direction of t Situation which,with with which, (b) b) Fnet ,x = max F1 − F2 = max F − F2 4 N − 2N ax = 1 = = 10m / s 2 m 0.20Kg F ax ! Thepositive positiv The m positivedire dir positive F F1 A a) Thehorizontal horizontalforce force The causes a horizontal x x causes a horizontal The acceleration. horizontal force acceleration. which, with given dat θ θ x Only the horizontal Only the Fhorizontal component Only theofhorizontal 3 component of F xcompetes with F 2. F 3 component of competes with3F . competes with F2.2 This is a free-body diagram. This is a free-body x x x This is a free-body diagram. diagram. From the figure, we s From the fi acceleration Fromand thesubs fig acceleratio acceleration F3,x " F2 ax ! F!3 m a x ! F3,x ! a x N)(cos (1.0 3 ! 0.20 (1 ! (1.0 ! F3 Thus, the net force a f) In three (situations, forces act on a puck that moves (f ) rection of the x axis. along an x axis. Free-body diagrams are also shown. Thus, the n Fig. 5-3 In three situations, forces act on a puck that moves Thus, the ne Fig. 5-3 In three situations, forces act on a puck that moves rection of t Fig. 5-3 along an x axis. Free-body diagrams are also shown. along an x axis. Free-body diagrams are also shown. rection of th : F1 # : F2 # : F3 : ! ma , Sample 93 Problem 5-6 (5-6) NEWTON’S" SECOND LAW (20 N) sin 90° ! (2.0 kg)(3.0 m/s ) sin 50° " (10 N) sin("150°) which gives us ! "10.4 N. : F3 : ! ma " : F1 " : F 2. Sample Problem (5-7) Two-dimensional forces, cook Vector: In unit-vector notation, we can write : F3 ! F3, x î # F3, y ĵ ! (12.5 N)î " (10.4 N) ĵ ! (13 N)î " (Answer) :(10 N) ĵ. In the overhead view of Fig. 5-4a, a 2.0 kg cookie tin is accelerated at 3.0 m/s in the direction shown by a , over a frictionless horizontal surface. The acceleration is caused by three horizontal forces, only two of which are shown: F of magnitude 10 N and F of magnitude 20 N. What is the third force F in unit-vector notation and in magnitude-angle notation? x compone Calculations: Because this is a two-dimensional problem, : One-weand two-dimensional forces, cannot find F 3 merely by substituting magnitudes 2 the puck for the vector quantities on the right side of Eq. 5-7. Instead, We can now use a vector-capable calculator to get the mag-3,x : : : : we must vectorially add ma , "F 1 (the reverse of F 1), and nitude and the angle of F 3 . We can also use Eq. 3-6 to obtain : situation by writing the second ree situations in which: axis, we can simplify eachthe "F 2 (the reverse of F 2), as shown in Fig. 5-4b. This addition magnitude and the angle (from the: positive direction of law for x components only:the x axis) as moves over can befrictionless done directly on a vector-capable calculator because 1 : 2 2 we know both magnitude and angle for all three vectors. F3 ! 2F 3,x # F 3,y ! 16 N nal motion. The puck’s 2 Eq. 5-7 inF However, here we shall evaluate the right side of (5-4) net, x ! max. F3,y "1 are directed the : terms ofalong components, first along the x axis and then along $ ! tan ! "40%. and (Answer) 3 F3, x 3,x the y axis. The free-body diagrams for the three situations are also N and F ! 2.0 N. Force 2 as magnitude F3 ! 1.0 eration of the puck? given in Fig. 5-3, with the puck represented by a dot. y We draw the product F2 y This is the resulting TheseSituation are two A: ForKFig. oneand horizontal of mass acceleration E Y5-3b, I Dwhere E A only horizontal acceleration of the three as a vector. acts, Eq. vector. us horizontal force : 5-4 gives – F1 vectors. net a F1 ! : max, – F2 net force The net force F on the tin is the sum of the three forces celeration : a to the 50° a via Newton’s and is related to the acceleration second law ma : x : Newton’s second law,! ma which, with givenx data, yields (Fnet ).30°Thus, tion is along only the x ntal force orizontal on. F F1 4.0 N F3 : a ! : 1 ! : (b) : ! 20 m/s2. x F 1 # Fm ma ,Then we can add the(Answer) three(5-6) 2 # F0.20 3 !kg (a) which gives Theuspositive answer indicates that vectors to find the missing third force vector. the acceleration is in positive direction of the : x axis. : Fig. 5-4 (a) An overhead forces :view of two of three horizontal : that act on a cookie tin, : the F ! ma ! m( Then, subst F ! (2. " ! 12 y compone F3, y ! may ! m(a ! (2.0 "( ! "10 (5-7) Vector: In Situation B: In Fig. 5-3d, two horizontal forces act on the : F ! ma " F " F . : : resulting in acceleration : , "F1, and a . F3 is not shown. (b) An arrangement of vectors ma 3 2 1 : : "F2 to find force F3. : : F m = 2kg, a F = ma F1 + F2 + F3 = ma for the vector quantities on the right side of Eq. 5-7. Instead, : : : we must vectorially add ma , "F 1 (the reverse of F 1), and : : "F 2 (the reverse of F 2), as shown in Fig. 5-4b. This addition can be done directly on a vector-capable calculator because know magnitude all three vectors. =we3m / s 2both , F = 10Nand , angle F2 =for20N 1 However, here we shall evaluate the right side of Eq. 5-7 in terms of components, first along the x axis and then along the y axis. F3 = ma − F1 − F2 For x component F3,x = max − F1,x − F2,x F3,x = m(a cos50) − (F1 cos(30 −180)) − (F2 cos90) and y These are two of the three horizontal force vectors. F2 y This is the resulting horizontal acceleration vector. a 50° F3,x = 2(3cos50) −10cos(−150) − 20cos90 30° F3,x = 12.5N For y component We can now u nitude and th the magnitud the x axis) as x F1 (a) (b) F3,y = may − F1,y − F2,y F3,y = m(a sin 50) − (F1 sin(−150)) − (F2 sin 90) Fig. 5-4 F3,y = 2(3sin 50) − 10 sin(−150) − 20 sin 90 F3,y = −10.4 N (a) An overhead view of two of three horizontal force : resulting in acceleration : a . F3 is not shown. (b) An arrangement : : "F2 to find force F3. Additional examples, video, and practice F3 in unit vector F3 = F3,x iˆ + F3,y ĵ F3 = 12.5iˆ − 10.4 ĵ The magnitude F3 = (12.502 + (−10.4) 2 F3 = 16N The angle θ = tan −1 θ = −40o F3. y F3,x = tan −1 −10.4 12.5 Examples: Q.1: One Newton equals: (a) Kg.m (b) kg.m/s2 (c) kg/s2 (d) m/s2 (c) kg/s2 (d) m/s2 F = ma à N = kg.m/s2 Q.2: The basic SI unit of the force is: (a) Kg.m (b) kg.m/s2 Q.3: A box is moving with a constant speed of 24.7 m/s. The net force on the box is: (a) Zero (b) 4N If ΣF = 0 à a = 0 (c) 5N (d) 45N Q.4: Three forces act on a particle of mass m, F1 = 80i + 60j, F2 = 40i + 100j. If the particle moves with a constant speed of 4m/s, then F3 is: (a) 80i + 60j (b) 80i − 60j (c) −80i + 60j (d) −120i−160j v constant à a = 0 àΣF = 0 F1 + F2 + F3 = 0 à F3 = − (F1 + F2 ) = −120i − 160j Q.5: Two forces act on a particle that moves with constant velocity. If F1 = 6i − 2j, then F2 is: (a) 6i − 2k (b) − 2i + 6k (c) − 6i + 2j (d) − 2i + 6j v constant à a = 0 à ΣF = 0 F1 + F2 = 0 à F2 = − F1 = − 6i + 2j Chapter 5 FORCE AND MOTION -I Sections 5-7 Some Particular Forces Important skills from this lecture: 1. Define the gravitational force & write it in unit vector notation and magnitude and angle notation 2. Define the weight and differentiate between mass and weight 3. Define the normal force and calculate it 4. Define the frictional force 5. Calculate the tension force these early chapters, we do not nature of this force a second body. In the these early chapters, we do not discuss the nature th und. We shall assume that the discuss ground is an inertial frame. SOM y axis along the body’s path, with the positive direction upward. Forof th and usually consider situations in which the second body is Earth. Thus, when5-7 we and usuallybody consider situations in which thewe second body is Earth. Thus, w : is Earth. der situations inofwhich the second Thus, when : usually Suppose a body of mass m is in free fall with the free-fall acceleration o ! ma which, Newton’s second law can be written in the form Fanet,y F speak the gravitational force on a body, we mean force thaty ,pulls : g F speak of the gravitational force on a body, we usually mean a force th g Fdirectly tational g. force onwe a body, we usually mean a the force that pulls g if on it5-7 toward the center of Earth — that is, directly down toward the situation, becomes nitude Then, neglect the effects of air, the only force acting ontowa th on it directly toward the center of Earth — that is, directly down Some Particular Forces : ward the center of Earth — thatthat is,shall directly down toward the ground. We shall assume the ground is anthe inertial frame. ground. We assume that ground is andownward inertial frame. force and F y is the gravitational force . We can relate this g "F !mm("g) Suppose a is body of mass frame. m is in of free with acceleration of gfall assume that the ground an inertial The Gravitational Force Suppose a body mass is : inthe freefree-fall fall : with the free-fall accelera (F ! ).force nward acceleration with second law We place a the vertica magnitude g.free Then, if Newton’s we neglect the effects of the air, thema only acting on : g. Then, magnitude if we neglect the effects of the air, the only force ody of mass m is in fall with the free-fall acceleration of : A gravitational force force Fgon on aa body: body isaaforce certain type of pull that is directed towardacting : pulls body A gravitational that on the body or body’s mg. Fpositive is the gravitational force can relate thiscan downward force and axis g! Fg.discuss body is the gravitational force We relate this downward for g. We F the path, with the direction upward. For this a second body. In these early chapters, we do not the nature of this force : n,isifalong we neglect the effects of the air, the only force acting on the : directly toward the center of Earth (downward ê) : : (F ! ma ) downward with Newton’s second law . We place a vertical : acceleration (F ! ma ). We we downward acceleration with Newton’s second law place a and usually consider situations in which the second body is Earth. Thus, when ! ma , which, inForou wton’s second law can be written in the form F Fg. the itational yIn force We can relate this downward force and net,y yto : words, magnitude of the gravitational force is equal the product y axis along the body’s path, with the positive direction upward. th axis along the body’s path, with the positive direction upward. For this axis, F:g onof of the gravitational force a body, usually mean a force that pulls m Inspeak free fall motion, when : the effects air is we neglected, the only force ation, (F !written ma ).gravitational ration becomes withNewton’s Newton’s second law We place a force vertical ! in ma which, Newton’s second law can bethe indirectly the form Fnet,y , which, our second law bem in the form Fnet,y y ,the ydown This same gravitational force, with same magnitude, still acts on th on it on directly toward the center of Earth —written that is, toward acting a body of can mass is the F! ma The Gravitational Force g situation, becomes situation, becomes body’s path, with the positive direction upward. For this axis, ground. We shall assume thatin the ground isbut an inertial frame. even when the body is not free fall is, say, at rest on a pool table or m "F ! m("g) g m is nd form Suppose body of à mass in free fall with the free-fall acceleration of ma , which, in our law can across beFrom written in athe Fgravitational Newton’s 2(For law , or "F ! m("g) net,y y g "F! ! m("g) the table. the force to disappear, Earth would h g magnitude g. Then, if we neglect the effects of the air, the only force acting on the s : disappear.) or Fg ! this mg. downward force F body is the gravitational force and Fg !Fmg. (5-8) g. We can relate or (5-8) : g ! mg. : We write Newton’s second law forlaw the(Fgravitational force in these ! ma ). We place downward acceleration with Newton’s second a vertical "Fcan g ! m("g) In words, the magnitude the gravitational force is equal the product m y axisthe along the body’s path, with theofpositive direction upward. Fortomg. this axis, forms: In words, magnitude of the gravitational force is equal to the product ords, the magnitude of theThis gravitational force isform equal to!the product mg. same gravitational force, with the same magnitude, still acts ma , which, in our on th Newton’s second law can be written in the F net,y y This same gravitational force, with the same magnitude, still acts on the body : Fg ! mg. (5-8) : at rest on a pool table or even whenwith the body is"F not in free fall but is,mg say, even Using unit vector notation: situation, becomes This same gravitational force, the same magnitude, still actsoron the bod F , table ! ĵ ! "mg ĵ ! g say, at rest on a pool when the body is not in free gfall but is, moving across the table. (For the gravitational force to disappear, Earth would h across the table. (For theisgravitational to disappear, would have to !product m("g) nnitude whenofthe body is not in free fall but is, say, at rest ontheaaEarth pool or away movin gforce gravitational force equal to"F the mg. the The gravitational force acts on the body even when body not in disappear.) where ĵ is the unit vector that points upward along y axis,istable directly fro disappear.) : theor free fall situation We can write Newton’s second law for thea gravitational force in dow these ss the table. (For the gravitational force to disappear, Earth would have to vitational force, with the same magnitude, still acts on the body F ! mg. (5-8) g ground, and is the free-fall acceleration (written as vector), directed g We can write forms: Newton’s second law for the gravitational force in these vector ppear.) dy is not in free fallgravitational but is, say, at resttoon a pool table or moving forms: For the force disappear, Earth has to disappear : In words, the magnitude of the gravitational force is equal to the product mg. : F mg , ! "F ĵ ! "mgĵ ! g g For the gravitational force to disappear, Earth would have to We can write Newton’s secondF:law for the force vecto : This same gravitational force, with thegravitational same magnitude, still actsinonthese the(5-9) body mg , ! "F ĵ ! "mg ĵ ! g g where ĵ is the unit vector thatis,points a ytable axis,or directly away fr even when the body is not in free fall but say, atupward rest onalong a pool moving ms: Weight where ĵ is the unit vector that points upward along a y axis, directly away from the : The weight W of a body is the magnitude of the net force required to preven The weight W of a body is the magnitude of the net forcebyrequired theFor examp body from falling freely, as measured someoneto onprevent the ground. body from falling freely, as measured theyou ground. example, keep a ball at restbyinsomeone your handon while stand For on the ground, to you must pro an upward forceyou to stand balanceonthe the ball from E keep a ball at rest in your hand while thegravitational ground, youforce muston provide Suppose the of theforce gravitational 2.0 N. Then the magnitu an upward force to balance the magnitude gravitational on theforce ballisfrom Earth. your upward force mustforce be 2.0isN, and thus thethe weight W of theofball is 2.0 N Supposethe magnitude gravitational 2.0 N.net Then magnitude The weight (W)ofofthe a body: is the magnitude of the force required also say that the ball weighs 2.0 N and speak about the ball weighing 2.0 N. your upward force must be 2.0 N, falling and thus theasweight W of ball ison 2.0 N. We to prevent the body from freely, measured bythe someone A ball with a weight of 3.0 N would require a greater force from y TERalso 5 FORCE AND MOTION—I the ground say that the ball weighs 2.0 N and speak about the ball weighing 2.0 N. namely, a 3.0 N force — to keep it at rest. The reason is that the gravitational A ball with a weight of 3.0 N would a greater—force from you —say that this you must has a require greater magnitude e.g., if the magnitude ofbalance the gravitational force on a ball is namely, 2.0 N. 3.0 N. We W namely, a 3.0 N force — to keep it at rest. The reason is that the gravitational force ond ball is heavier than the first ball. To keep a ball magnitude of 2.0 N or at rest, an upward W !force Fg with (weight, with ground as inertial frame). 5-7 you must balance has a greater magnitude — namely, 3.0 N. We sayathat Now let us generalize the situation. Consider bodythis thatsechas an acceler has to be applied to balance the gravitational one : equation tells us (assuming the ground is an inertial frame) that This athe relative to N the ground, which we again assume to be an inertial fr ond ball is heavier than first à the weight Wofofzero the ball. ball is 2.0 : F Two forces act on the body: a downward gravitational force Now let us generalize the situation. Consider a body that has an acceleration g and a balan : relative If another ball requires a greater force toW. keep at write rest Newton’s secondFglaw for a verti upward force of magnitude We itcan a of zero to the ground, which we again assume to be an inertial frame. nd ball is heavier than the 1st one : à the 2 axis, theW positive direction as F Thewith weight of a body is equalupward, to theforce magnitude Fg of the gravitational force mR Force E AND MOTION—I Two forces act onThe theGravitational body: a downward gravitational g and a balancing body. : ! type malaw Fbalancing upward force of forces magnitude We can write Newton’s second vertical y If two act onW. a body; net,y y.upward Fg(downward) A gravitational force on a body is&a acertain offor pulla that is directed towa force ofa magnitude W upward, axis, with the positive direction asbecomes second body. In these early chapters, we do not discuss the nature of this for In our situation, this or W ! F ( weight, with ground as inertial frame). (5-11) nd g APTER 5 FORCE AND MOTION—I à fromand Newton’s 2 law: usually consider situations inEq. which the second body is Earth. Thus, when w Substituting mg for F from 5-8, we find : g ! ma . F à net,y àan W Fgframe) !we m(0) y is tellsof usthe (assuming the ground inertial that mean a force that pu( F speak gravitational force a" body, usually g on FgR =This mRg equation on it directly toward the center ofWEarth — that is, directly down toward th In our situation, this becomes mgground (weight), or W ! Fg à (weight,!with as inertial frame). (5 à l-arm balance. When ground. We shall assume that the ground is an inertial frame. Watells " Fus ! m(0) equation (assuming ground is an inertial frame) that nce, the gravitational body m is to in free with theforce free-fall The weight W Suppose ofThis a body isaequal toof the magnitude Fthe of thefall gravitational on(5-10) theacceleration gmass gits which relates body’s weight mass. y being weighed (on body. magnitude Then,aifbody we neglect effects of the air, the only force acting on th Tog.weigh meansthe to : measure its weight. One way to do this is e total gravitational body is the gravitational force Fg. We can relate this downward force an : the body on one of the pans of an equal-arm balance (Fig. 5-5) and then p : rence bodies m (on the (F ! ma ). We placeforce downward acceleration with Newton’s second law a vertic The weight W of a body is equal to the magnitude F of the gravitational on t g R Weight 5-7 Some Particular Forces : force FgL on the body being weighed (on the left pan) and the total gravitational : force FgR on the reference bodies (on the right pan) are equal. Thus, the mass mL of the body being weighed is equal to the total mass mR of the reference bodies. How to weight a body CHAPTER 5 FORCE AND 96 Two ways for weighting a body: 1. Equal-arm balance: the body is placed on one of the pans, a reference bodies (with known masses) is placed on the other pan when a balance is obtained: à the gravitational forces on the two sides match à the masses on the pans match à we know the mass of the body mL mR FgL = mLg FgR = mRg An equal-arm balance. When the device is in balance, the gravitational : force FgL on the body being weighed (on the left pan) and the total gravitational : force FgR on the reference Scalebodies marked(on the in either right pan) are equal. Thus, the mass mL of or to the total the body being weighedweight is equal mass units mass mR of the reference bodies. Fig. 5-5 1. Spring scale: the body stretches a spring, which causes a movement of a pointer along a scale. The scale has been calibrated & marked in either mass or weight units It is accurate only when the value of g is the same as when the scale was calibrated Fg = mg Fig. 5-6 A spring scale. The reading is proportional to the weight of the object on which r To the bod erence ance (s pans th the loca We a sprin MOTION either m way an or un mass was cal This Th vertica scale Th in mentbod w weight weight. Subs Ca and is r of g is weight whic on EarT butthe theb eren ance The No pans the lo If you s The reaW spry upaon eithe buckled wayd floor mass floor co was Thc comes T The weight of a body must be measured when the body is not accelerating vertically relative to the ground A body’s weight is not its mass Weight is the magnitude of a force Weight is related to mass by Newton’s 2nd law If a body is moved to a point where the value of g is different: The body’s mass will not differ The body’s weight will differ e.g., for a bowling ball of mass 0.3 kg On Earth On Moon M = 0.3 Kg M = 0.3 Kg g = 9.8 m/s2 g = 1.6 m/s2 W= (0.3) (9.8) = 2.9 N W= (0.3) (1.6) = 0.49 N is due to reason Earth'sis that the mattress, because The it deforms downward due to you, pushes he gravitational The forces x Fg x a floor, on you. it deforms (it is compressed, bent, or downward pull. Similarly,Fgif you orce on theupblock balance. (a) stand on (b) e gravitational The forces Fg up on you. buckled ever so slightly) and pushes The gravitational TheEven forcesa seemingly rigid concrete F due Earth's Fgbalance. : g ce ontothe block ravitational The forces FFg gsitting directly onbalance. floor does this (iftable it isexperiences not theFground, enoughto people on the force on the block 7 (a) A block resting on a a normal force perpendicular F ownward pull. (a) (b) g N on the block balance. due to Earth's floor could break it). is due to Earth's etop. (b) The free-body for the block. : to Earth's wnward pull. The pushdiagram (a) (b) :or floor is a normal force FN . The name on you from the mattress is block downward A one.g., a table experiences a normal FN(b) perpendicular to ward pull.restingpull. (a) (b) orforce if you stand(a) on a floor a mattress The Normal Force comes from the mathematical term normal, meaning perpendicular: The force on t on : buckled slightly) b)block The free-body diagram for the block. à they deform (compressed, bent, or you from, say, the floor is perpendicular to the floor. downtoon a table, : : resting on a table experiences a normal force F perpendicular ture if resting 5-7a shows an example. A block of mass m presses N ock on a table experiences a normal force F perpendicular (a) A block resting on pulls a table experiences normal force FNto perpendicular to Earth you down, andathey pushes you up N : The free-body diagram for the block. e free-body diagram theremain block. ing it somewhat because ofstationary the gravitational force Fg on the block. The àfor you op. e (b) The free-body diagram for the block. : 7a shows anthe example. A block of mass m Fpresses down on a diagram table, for the ushes block with normal force . The free-body N the up onWhen : a body presses : against:a surface, the surface (even a seemingly rigid : The somewhat because of the gravitational force on block. Fonly s gaon FN body sahows given Fig. 5-7b. Forces are the two forces on the block Fg and aninexample. A block ofpushes mass m: presses down athe table, one) deforms and on the with normal force F that is perpendicular to shows an example. A block of mass m presses down on a Ntable, : re 5-7a shows an example. A block of mass m presses down on a table, : on up on the block normal force free-body diagram Fblock mewhat because ofwith the Thus, gravitational on can the write block. The for the F we the surface. N . The : Newton’s y are both vertical. for theforce second law g omewhat because of:theofgravitational force block. The The Fg onFthe : : g it somewhat because the gravitational force on the block. : g theFig. block with diagram foron the nnin 5-7b. Forces and FF areyfree-body only two forces the block F(F NN. The gforce : free-body ositive-upward ynormal ! ma )the as :axis :net, y force pFig. on the block with normal . The diagram for the F N two hes up onForces theThus, block with normal force . The free-body diagram Fwrite are theonly only forces on Newton’s the block the on the block Fg and Nforces :F : N are both5-7b. vertical. for the block we can second lawfor the : : Fand Fig.in5-7b. Forces are the only two second forces on the block Fblock they are both vertical g and N" hin vertical. Thus, for the we can write Newton’s law F given Fig. 5-7b. Forces are the only two forces on the block F F F ! ma . g N N g y -upward y axis (F ! ma ) as net, y y nd à Newton’s 2thelaw in ycan axis: The force oth Thus, fory) the we Newton’s second law ward y axis vertical. (Fnet, asforblock Normal forceFFlaw Thenormal normal force are vertical. both Thus, block wewrite can write Newton’s second y ! ma N Normal force (from the table)N q. 5-8, we substitute mg for F , finding is the force on is the force on F! " F! . upward y axisy(F ma asgma Nnet, yas net, y(F yg)! tive-upward ma ) y y Faxis " F ! ma . N g y the theblock blockfrom fromthe the mg. ! masupporting . N" ,substitute we substitute mg finding table. FNF ! ma g, F mg for Fgfor ,F finding F NF" g" y may. ysupportingtable. Block Block g! he magnitude of normal mg may.is Fmg " ma FNàthe we substitute mg for , finding NF" . 5-8, we substitute mg for Fgy! ,.force finding g! ude of the force is !mg mg ma #gravitational ay) à The gnitude ofnormal the normal force is!# N y ! m(g ma FF mg ! N "F y. may. The gravitational N" y y FNFN Block Block x x F (5-13)Fg forceononthe theblock block Fgg force F ! mg # ma ! m(g # a ) (5-13) (duebe to Earth's pull) accelN y# y # and offorce table (they in an vertical acceleration ayma duetotoEarth's Earth's might FNthe ! normal mgforce ! m(g ay)is isblock (5-13) nitude of the isythe due magnitude ofnormal is downward pull.accel- relative (a) elevator).ayIfof the tableand and block notbeaccelerating to the the table block (theyare might inpull. an cceleration downward (a) cal acceleration a of the table and block (they might be in an accel- The forces The forces balance. balance. Fg (b) (b) Fg g and F are the downward on the block balance. (a) (b)only : ockforce is given inpull. Fig. 5-7b. Forces two forces on the block F F g N g force onbecause the block of the gravitational balance. deforming it somewhat force on the block. The F xg is due to Earth's : write nd they are both vertical. for the block we can Newton’s second law : is dueThus, to Earth's table pushes up on the block with normal force . The free-body diagram forLAR theFORCE S F A block resting a table experiences a normal force to N FN perpendicular 5-7forces SOM E PARTICU downward pull. on (a) (b) pull. : : The gravitational The r a(a) positive-upward ydownward axis (Fnet, ! ma ) as (a) (b) 9 y y Fg 5-7forces SOME PARTICULAR FORCES top. (b) The free-body diagram for the block. F block is given in Fig. 5-7b. Forces and are the only two on the block F FNg g force on the block balance. : : y F " F ! ma . Fig. 5-7 A block resting on a table experiences a normal force F to and are both(a) vertical. Thus, for the block we can write Newton’s law y (a) Athey block resting on a table experiences a normal force F perpendicular to second is due to Earth'sN g y N perpendicular N The normal force the tabletop. (b)The The free-body diagram the block. Normal force FNfor normal force downward pull. (a) (b) for a positive-upward y axis (F ! ma top. (b) The free-body diagram for the block. Normal force F) as net, y y ure 5-7a shows an example. A block of mass m presses down on a table, is the force on om Eq. 5-8, we substitute mg onfor Fg, finding is the force : the blockthe from thefrom block ng it somewhat because of thethegravitational force theF: block. The to Fg on F " F ! ma . Fig. 5-7 (a) A block resting on a table experiences a normal force perpendicular N g y N : supporting table. supporting table. Figure 5-7a shows an A ma block Fof mass m presses down on a table, " mg F ! .Nfree-body FNexample. Block yF Block : Block ushes up on the block with normal force . The diagram for the the tabletop. (b) The free-body diagram for the block. Block N ure 5-7a shows an example. m presses down on a table, deforming it somewhat because the gravitational force on the block. The F :A block : ofofmass g : x From Eq. 5-8, we substitute mg for F , finding : x forces on the block g the only Fwith given inmagnitude Fig. 5-7b. Forces and gthe N are ng itthe somewhat because ofF gravitational forcetwo the block. Thefor the table pushes up on the block force . The free-body diagram FNF hen of the normal force isnormal g on : : : The gravitational The forces yshes areup both vertical. Thus, for the block we can write Newton’s second law F mass 5-7a an example. A block of m presses down on a table, F is given inshows Fig. 5-7b. Forces are the only two forces on the block F The gravitational The forces on block theFigure block with normal . The free-body diagram for the F FNand g N " mg ! ma . F force on the block force balance. F NF y g : : : gforthe Fblock mg # ! #two abalance. ) forces (5-13) force the sitive-upward ythey axis (F !! ma asma deforming iton somewhat because of gravitational force theblock block. The and are both vertical. Thus, them(g block we write F Newton’s second law isnet, due to Earth's y the ycan yN y) g on F given in Fig. 5-7b. Forces and are only on the F : g N is due toup Earth's downward (a) table onthe thepull. block with FN . The free-body diagram for the for a pushes positive-upward ynormal axis (F ! ma as (b) Then the magnitude of force is net, ynormal y)force :. can :write yr are vertical. Thus,pull. for the block we second downward (a) (b) Newton’s F " F ! ma of the table and block might beon inlaw anblock accelanyboth vertical acceleration a gForces y block Fig. is given Fig.Nresting 5-7b. theF(they only two forces the Fyg anda F 5-7 (a)in A block on a table experiences normal force perpendicular to N are Fthe " Fg! !not may.accelerating Nma : sitive-upward y(a) axis (F ! )Thus, asblock the tabletop. (b) The free-body for block. Fma ! mg # m(g # awrite (5-13) ating elevator). If the table and are relative tolaw the ± net, y vertical. ydiagram Ntable y y) are both for the block we can Newton’s second Fig.and 5-7 they A block resting on a experiences a normal force F − N perpendicular to q. 5-8, we substitute mg for Fg, finding the tabletop. (b) The diagram for the block. From Eq. 5-8,free-body we substitute mgnet, for Fgma , finding for positive-upward y axis (F ound, then aya! 0 Figure and Eq. 5-13 yields y! y) as F " F ! ma . 5-7a shows an a example. A block of mass m presses down on a table, of the table and block (they might be in an accelfor any 1. vertical acceleration N g y y !not If the table and block are accelerating relative to the ground, " mg ma . F deforming it somewhat because of the gravitational force the block. The N y" F mg ! ma . F onaccelerating FNblock ma g! y.ynot àtable ay =pushes 0Ifà erating elevator). the table and are relative to the F ! mg. (5-14) Figure 5-7a shows an example. A block of mass m presses down on afortable, up on F the,block withN normal force F . The free-body diagram the q. 5-8, we substitute mg for finding : g isnormal given in because Fig.force 5-7b. Forces andforce are theforce only two the block Fgravitational deforming it somewhat ofis the on theonblock. The Fg forces e magnitude ofblock the Then the magnitude of the normal From Eq. 5-8, we substitute mg for FgF, finding ground, then a ! 0 and Eq. 5-13 yields : is y andup they both vertical. Thus, for the block can write Newton’s second table onare the block withare normal force The free-body diagram forlaw the ground, FN .we 2. pushes If the table and block accelerating up-ward relative to the : : " mg ! ma . F for a positive-upward y axis (F ! ma ) as N y FN block is à given theay! only two FFgyNand !F mg #! ma m(g # ay) on the block (5-13)(5-13) " mg FFig. !5-7b. mg # ma ! m(g # )ma ay in is +ve à Forces N are y. forces N y! F mg. (5-14) N F block " F !we macan . write Newton’s second law and they are both vertical. Thus, for the CHECKPOINT 3 evertical magnitude ofany thevertical normal force isnormal table and block (they in an accelfor acceleration a y)ofasthe the magnitude of the force is: (they for aThen positive-upward axis (Fnet, the table block might bemight in anbeaccelacceleration a yof y ! mayand The Normal Force N N g g : N : g net, y : N : N : g y N g y 3. the If the table and are accelerating relative to the ground, y g FNdown-ward n Fig. 5-7, is magnitude ofblock the normal force greater than, less than, or equal to erating elevator). If the table and block are not accelerating relative to the ay F is −ve à elevator). If à the table and block are not accelerating relative to the F " F ! ma . " mg ! ma . F −mayayy! F !Ngmg #y# m(g # ay) (5-13) Ny mg #an ma m(g ) upward (5-13) N! N! mg if the block and table are in elevator moving (a) at constant speed and ground, then a ! 0 and Eq. 5-13 yields y 3 Then the magnitude of the normal force is thenCHECKPOINT ay From ! for 0 Eq. and Eq. 5-13 yields 5-8, we substitute mg for Fg, finding of the table(they and (they might beaccelin an accelany vertical acceleration ayand b) at increasing speed? : block of the table block might be in an vertical acceleration a F ! mg. (5-14) y F ! mg # ma ! m(g # a ) (5-13) In Fig. 5-7, is the magnitude ofFthe normal forceare N " mg y N yFN greater than, less than, or equal to ! ma erating elevator). If the table and block not accelerating relative to the N y. From Eq. 5-8, we substitute mg for F , finding FN ! mg. (5-14) (5-14) FN ! mg. CHECKPOINT 3 : In Fig. 5-7, is the magnitude of the normal force FN greater than, less than, or equal to 5-7 SOM E PARTICU LAR FORCE S mg if the block and table are in an elevator moving upward (a) at constant speed and 5-7 SOME PARTICULAR FORCES (b) at increasing speed? P 9 97 y y normal force Normal force FN The normal force (from theNormal table)force F e force on N FN = mg + ma y a) is the force on block from the the block from the to slide a body over a surface, the motion is resisted porting table.If we either slide or attempt F a =0 Friction N Block supportingby table. y F Block a bonding between the body and the surface. (We discuss this bonding more in Block N Block : the next chapter.) The resistance isx considered to be a F single force f , called either = mg x N the frictional force or simply friction. This force is directed along the surface, opgravitational TheThe forces The gravitational forces (Fig. 5-8). Sometimes, to simplify a sitposite the direction of the Fintended motion g Fg F F = ma y + mg g e onforce the on block Fg the block(due balance. (the surface Nis frictionless). to Earth's pull) uation, friction is assumed to be balance. negligible b) ue to is Earth's due to Earth's nward pull. pull. downward Tension (a) (a) (b) (b) : : a y = +ve F > mg N (a) Aresting block resting on a experiences table experiences a normal force FN perpendicular block on a table a normal force FN perpendicular toto op. (b) The free-body diagram forblock. the block. The free-body diagram the When a for cord (or a rope, cable, or other such object) is attached to a body and : pulled taut, the cord pulls on the body with a force T directed away from the ure 5-7a shows an example. A block of mass m presses down on a table, shows an example. A block of mass m presses :down on a table, : F on the block. The ng it somewhat because the gravitational force g mewhat because of theofgravitational force F : g on the block. The : FN . The free-body diagram for the shes up on the block with normal force : force:FN . The free-body diagram for the on the block with normal : : FN are the only two forces on the block given in Fig. 5-7b. Forces Fg and Fig. 5 attem upward (a) at constant speed and CHECKPOINT 3 : In Fig. 5-7, is the magnitude of the normal force FN greater than, less than, or equal to mg if the block and table are in an elevator moving upward (a) at constant speed and (b) at increasing speed? Friction rriction a surface, the motion is resisted If a body is slid over a surface, the motion is resisted by a (We discuss this bonding more in : to slide we either slide or attempt a body over a surface, motion is resisted bonding between the body the and the surface be a single force called either f , y a bonding between the body and the surface. (We discuss this bonding more in : isnext directed the resistance surface, ophe chapter.)along The is considered a single force f, ,called called either resistance This is to a be single force frictional force or he frictional force or simply friction. This force is directed along the surface, op5-8). Sometimes, to simplify a sitsimply friction osite the direction of the intended motion (Fig. 5-8). Sometimes, to simplify a siturface is frictionless). Direction of ation, friction is assumed to be negligible (the surface is frictionless). attempted slide ension Direction of attempted slide f f : When a cord (or a rope, cable, or other such object) is attached to a body and Fig. 5-8 A frictional force f opposes the : Frictional force is directed along the surface, opposite the : bject) is attached to a body and Fig. 5-8 A frictional f opposes the slide of a body over a surface. ulled taut, the cord pulls on the body with a force T directed away from theforceattempted : direction of the motion attempted slide of a body over a surface. force T directed away from the To simplify a situation, sometimes a friction is assumed to be negligible (the surface is frictionless) friction. This force is directed along the surface, op: dmply to be negligible (the surface is frictionless). the next chapter.) The resistance is considered to be a single force f , called eitherDirection of he intended motion (Fig. 5-8). Sometimes, to simplify a sitthe frictional force or simply friction. This98force CHAPTE is directed along surface, op-attempted R 5 FORCE AN Dthe MOTION—I slide ed beanegligible (the surface pt toto slide body over a surface, the motionisisfrictionless). resisted Direction of posite direction of the motion (Fig. 5-8). Sometimes, to simplify a sitbody and the the surface. (We discuss thisintended bonding more in attempted f : stance is considered a single force either , called uation, frictiontoisbeassumed tof be negligible (the surface is frictionless). slide 98 : CHAPTER 5 FORCE ply friction. This force is directed along the surface, op- AND MOTION—I T T cable, or other such object) is attached to a body and Fig. 5-8 A frictional force f opposes the f intended motion (Fig. 5-8). Sometimes, to :simplify a sit(theTension If a cordaway is attached to a body and attempted slide of a body over a surface. ls onnegligible the body with ais force force( T):directed from the to be surface frictionless). Direction of : pulled, theobject) cord pulls on the body tension force attempted e,Tension cable, or other such is98attached to a5with body and Fig. 5-8 A frictional force f opposes CHAPTER FORCE AND MOTION—I : slide forces at the two ends of 5-9 away (a) The cord, pulled taut, is The along theFig. cord, the body attempted slide TofThe a body over a surface Tfrom T ulls on the body withdirection a forceof T isdirected away from the the cord are equal in magnitude under tension. If its mass is negligible, f When a cord (or a rope, cable, or other such object) is and attached to a body and Fig. 5-8 A the cord pulls on the body the hand : : such Tension is on the magnitude of force with force , even if the runs f opposes attempted sl cable, or other object) attached to athe body and with pulled taut, the cord:is (T): pulls body a Tforce Tcorddirected away Fig.the 5-8 A frictional force the from the Tension : a massless, frictionless pulley as T attempted slide of a body over a surface. if T has magnitude = 50around N on the body with e.g., a force directed away fromTthe in (b) and (c). T T T (a) The forces at the two ends of T Fig. 5-9 (a) The cord, pulled taut, is à the tension in the cord is 50 N the cord are equal in magnitude. under tension. If its mass is negligible, the cord pulls on the body and the hand body and along the cord (Fig. 5-9a). The : T with force , even if the cord runs because the cord is said to be in a state of The cord properties: The forces at the two ends of T Fig. 5-9pulley (a) The around a massless, frictionless as cord, pulled taut, is means that it is being pulled taut. The tensi the cord are equal (b ) in magnitude. (c ) under tension. If its mass is negligible,(a) in (b) and (c). Unstretchable force on the body. For example, if the force the cord pulls on the body and the hand tude T ! 50 N, the tension in the cord is 50 : T , even if the cord force runs 98 CHAPTE Massless (itsAN mass is negligible compared to the body’s mass) R 5 FORCE D with MOTION—I cordforce is often said tocalled be massless (m body andasalong the cord (Fig. 5-9a).AThe is often a tension around a massless, frictionless pulley the of body’s mass) unstretchable. The because cord is said to be(a) in a to state tension (or and to be under (b ) tension), Exists only as a inconnection between 2 the bodies (b) and (c). between two on both bod means that it is being pulled taut. The tension in bodies. the cordItispulls the magnitude T even if the bodies and the cord are acceler Pulls on both bodies with the same force magnitude T force on the body. For example, if the force on the body from the cord has m body and along the cordfrictionless (Fig. 5-9a). The (Figs. force 5-9b is ofte a massless, pulley an tude T ! 50 N, the tension in the cord is 50 N. T T T because the cord is said totobe a stateand of tension (orfric to compared theinbodies negligible A cord is often said to be massless (meaning its mass is negligible com If the cord wraps halfway around a pulley, the netmeans force on cord wrapstaut. halfway aroundin a pulley, as that it isthe being pulled The tension the cord to the body’s mass) and unstretchable. The cord then exists only as a conn from theexample, cord hasifthe magnitude 2T.body f the pulley from the cord has the magnitude 2T force on the body. For force on the Tthe between two bodies. It pulls on both bodies with the same force magnit tude T ! 50 N, the tension T in the cord is 50 N. The forces at the two bodies ends of and the cordT are accelerating even if the and even if the cord runs a Fig. 5-9 (a) The cord, pulled taut, is A cord is often said to be massless (meaning its mas the cord aare equal in magnitude. CHECKPOINT under tension. If its mass is negligible, massless, frictionless pulley (Figs. 5-9b and c). Such a4pulley has negligibl to the body’s mass) and unstretchable. The cord then ex the cord pulls on the body and the hand N. The friction suspended in Fig. 5-9c weighs compared to the bodies and negligible onbody its axle opposing its 75 rota : between two bodies. It pulls on both bodies with the with force T , even if the cord runs N when body5-9c, is moving upward cos the cord wraps halfway around a pulley, as the in Fig. the net force(a) onatthe around a massless, frictionless pulley as even if the bodies the cord speed? are accelerating and eve (c)and at decreasing from the cord has the magnitude 2T. (a) (b ) (c ) in (b) and (c). a massless, frictionless pulley (Figs. 5-9b and c). Such a p the cord wraps halfway around a pulley, as in Fig. 5-9c, the net force on the pulley from the cord has the magnitude 2T. CHECKPOINT 4 The suspended body in Fig. 5-9c weighs 75 N. Is T equal to, greater than, or less than 75 N when the body is moving upward (a) at constant speed, (b) at increasing speed, and (c) at decreasing speed? T 5-8 Newton’s Third Law T Two bodies are said to interact when they T push or pull on each other — that is, when a force acts on each body due to the other body. For example, suppose you T The forces at the two ends of T position a book B so it leans against a crate C (Fig. 5-10a). Then the book and the cord are equal in magnitude. : crate interact: There is a horizontal force FBC on the book from the crate (or due : to the crate) and a horizontal force FCB on the crate from the book (or due to the book). This pair 5-10b. Newton’s third law states that (a) (b ) of forces is shown in(cFig. ) F = ma a) Fy = ma y c) Fy = may b) y y T − W = ma T − W = m(−a) − W =force ma along the cord (Fig. 5-9a). The force is often called a Ttension a a=state 0 ofThird he cord is said to be in tensionLaw: (or to be under which Newton’s When twotension), bodies from each T =W − ma T = W interact, + ma the forces on the bodies at it is being pulled taut. tension cordinismagnitude the magnitude of the T The −W =always 0m in the other are equal andTopposite in direction. T = 75 − ma nheBbody. For example, T = 75 + ma if the force on the body from the cord has magniT =W s50the same T <W N, the tension in the cord is 50 N. T > W T = 75 N is often said to be massless (meaning its mass is negligible compared sd the For the book and crate, we can write this law as the scalar relation dy’s to mass) due B. and unstretchable. The cord then exists only as a connection two bodies. It pulls on both bodies with the same force T, magnitudes) FBC ! Fmagnitude (equal CB e bodies and the cord are accelerating and even if the cord runs around ainst crate or(Figs. as the vector s,efrictionless pulley 5-9b and c).relation Such a pulley has negligible mass book Examples: Q.1: In which figure of the following the y-component of the net force is zero? (a) (b) 6N 3N 2N (c) 6N 3N 2N 3N 2N 3N 2N 7N 3N 2N 2N 2N 4N 4N (d) 5N 4N 4N Q.2: In which figure of the following the particle moves with a constant velocity? 6N 6N (a) (b) 3N 2N (c) 3N 2N 3N 2N 4N ΣFx = 0, ΣFy = 0 7N 5N (d) 3N 3N 2N 1N 4N 3N 2N 4N 4N Q.3: A particle of mass 2kg at a point where g = 9.8m/s2, the weight of this particle at point where g = 0 is: (a) 49N (b) 98N (c) zero (d) 9.8N W = mg = 2 X 0 = 0 Q.4: The direction of the acceleration of a body is: (a) Opposite to the net force (b) The same direction of the net force (c) Perpendicular to the direction of the net force (d) The same of the initial velocity Q.5: In which figure of the following the particle moves up if it starts from rest? (a) (b) 6N 3N 6N 5N 5N 3N 2N 4N ΣFx = 0, ΣFy=+ve (c) 8N 3N 3N 7N 3N 2N 2N 1N 4N (d) 5N 4N 4N Q.6: In which figure of the following the acceleration of the particle moves to right? 6N (a) (b) 3N 2N 6N (c) 2N 2N 3N 2N 4N 5N (d) 3N 3N 3N 5N 2N 1N 4N 7N 4N 4N (a) ΣFx = 1, ΣFy = 0 (b) ΣFx = 0, ΣFy = −1 (c) ΣFx = 0, ΣFy = 0 (d) ΣFx = −2, ΣFy = 1 the correct answer is (a) because the direction of a has to be as the same direction as Fnet Q.7: In the figure, the net force on the block is: (a) 1N right (b) 6N up (c) 2N left (d) 4N down 6N ΣFx = 3 − 2 = 1N, ΣFy = 6 − 2 − 4 = 0N 3N 2N 2N 4N Q.8: When a force of 10N is applied to a body, the body accelerates with 2m/s2. The mass of the body is: (a) 20kg (b) 10kg (c) 0.5kg (d) 5kg m = F/a = 10/2 = 5kg Q.9: From the figure, the acceleration of the block of mass = 0.5kg moving along the x-axis on a horizontal frictionless table is: (a) 10m/s2 (b) − 10m/s2 (c) − 6.3m/s2 (d) − 8.3m/s2 Motion direction a = F/m =− 5/0.5 = −10m/s2 F=5N Q.10: A force of 7N is applied to a mass of 7kg, the resulting acceleration is: (a) 3m/s2 (b) 1m/s2 (c) 2m/s2 (d) 4m/s2 a = F/m =7/7 = 1m/s2 Q.11: A force accelerate a 5kg particle from rest to a speed of 12m/s in 4s. The magnitude of this force is: (a) 10N (b) zero (c) 20N (d) 15N M = 5kg, vo = 0, v = 12m / s, t = 4s v = vo + at ⇒ a = 3m / s 2 F = Ma = 3(5) = 15N Q.12: A body of mass 1kg is accelerating by a = 3i + 4jm/s2, the magnitude of the acting force F on the body is: (a) 2.5N (b) 7.5N (c) 12N (d) 5N F = ma = (1)(3i + 4 j) = 3i + 4 j F = 9 + 16 = 5N Q.13: A net force of 15N acts on a body of weight 29.4N. The acceleration of the body is: (a) 9.8m/s2 (b) 5m/s2 (c) 6.5m/s2 (d) 2.4m/s2 W = mg ⇒ m = a= W 29.4 = = 3kg g 9.8 F 15 = = 5m / s 2 m 3 Q.14: Only two forces are acing on a particle of mass 2kg that moves with an acceleration of 3m/s2 in the positive direction of y axis. If F1 = 8i N, the magnitude of F2 is: (a) 12N (b) 10N (c) 17N (d) 15N F1 + F2 = ma 8i + F2 = 2(3 j) ⇒ F2 = −8i + 6 j F2 = 64 + 36 = 10N Chapter 5 FORCE AND MOTION -I Sections 5-8, 5-9 Newton’s Third Law Applying Newton’s Laws Important skills from this lecture: 1. Explain Newton’s third law and apply it to different cases 2. Apply Newton’s laws to solve problems for one and two body system when a force on has each due2T. to the other body. For example, suppose you from acts the cord thebody magnitude : ts on each body due toLaw the other body. Forthe example, suppose you from the position a book B so it leans against a crate C (Fig. 5-10a). Then the book and nteract: There is a horizontal force on book from the crate (or due F 5-8 Newton’s Third BC ewton’s Third Law CHECKPOINT 4a crate:C (Fig. : B so it leans against 5-10a). Then the book anddue crate interact: There is a horizontal force on the book from thetocrate F rate) and a horizontal force F on the crate from the book (or the (or dueCHEC BC : CHECKPOINT 4 CB N. IsorT: equalon to,each greater than,—orthat less is, than 75 The suspended body in Fig.when 5-9c weighs Two bodies are said to interact they 75 push pull other here aforce horizontal force on the book from the crate (or due Fupward dies areissaid to interact when they push orFig. pull on each other — that is,to,from the crate) and a horizontal force FNewton’s on the the BC when theon body is moving at constant speed, increasing speed, andbook N.(b) Is at T crate equal greater than, orthat less(or thandue 75 to the The suspended body in(a) 5-9c weighs 75 CB This pair of forces is shown in 5-10b. third law states when atoN acts each body due toFig. the other body. For example, suppose you : The susp acts book). on each body due toof the other body. For example, suppose you due N when the body is moving upward (a) at constant speed, (b) at increasing speed, and (c) at decreasing speed? dorce aposition horizontal force F on the crate from the book (or to the This pair forces is shown in Fig. 5-10b. Newton’s third law states that a book B so it leans N when CB against a crate C (Fig. 5-10a). Then the book and Newton’s Third Law : 5-10a). Then the book and (c) at decreasing a book B so it leans against a crate speed? C (Fig. crate interact: There is a horizontal force from thestates crate (orthat due F : BC on the book of forces isashown in force Fig. 5-10b. Newton’s third law : the eract: There is horizontal on book from the crate (or due F BC F to the crate) and a horizontal force : CB on the crate from the book (or due to the ate) and aThis horizontal force Fistwo onbodies the fromNewton’s thethe book (or due to thebodies CB book). pair ofWhen forces shown in crate Fig.interact, 5-10b. third law states that ton’s Third Law: forces on the from eachfrom each Newton’s Third Law: When two bodies interact, the forces on the bodies his pair of forces is shown in Fig. 5-10b. Newton’s third law states that are alwaysother equal magnitude and oppositeand in direction. areinalways equal in magnitude opposite in direction. Two bodies are said to interact when they push or pull on each other — that is, d Law: When two Two bodies interact, the forces on the bodies from each bodies are said to interact when they push or pull on each other — that is, Newton’s Third Law: When two bodies interact, the other forces body. on theFor bodies from each ame when a force acts on each body due to the example, suppose youB suppose you Book Crate C sCrate equal in magnitude and opposite in direction. when a force acts on each body due to the other body. For example, other are always equal in magnitude and opposite in direction. on’s Third Law: When two bodies interact, the forces on the bodies from each C Both book & crate in the figure interact: Book B Crate C position a book B soaitbook leansBagainst a crate C (Fig. 5-10a). Then the book and the book and position so it law leans against alaw crate Crelation (Fig. 5-10a). Then :as For the book and crate, we can write this as the scalar relation ebook always and equal in magnitude and opposite in direction. crate, we can write this the scalar : There is a horizontal thebook book from crate interact: There is a horizontal force on force the from the crate (or due FBC on crate interact: There isforce a horizontal F B. BC on the book from the crate (a) (or due : : the crate) and horizontal force FCB! the crate from the book (or due the (or due to the (a) to For thetobook and crate, can write law ason the scalar relation to theawe crate) athis horizontal force FCB on the crate from the book the crate and aand horizontal force on the crate F F (equal magnitudes) BC CB F !pair F (equal magnitudes) d crate, we can write this law as the scalar relation BC CB book).This pair of forces is shown in Fig. 5-10b. Newton’s third law states thatlaw states that book). This of forces is shown in Fig. 5-10b. Newton’s third ate book and crate, we can write this law as the scalar relation from the book FCB FBC FBC ! FCB (equal magnitudes) 5-8 Newton’s Third LawThird Law 5-8 Newton’s or as the vector relation eFCBvector or as the relation vector relation ! F F! FF (c) at dec 5-8 Ne Two bodi when a fo position a crate inte to the cra book). Th FCB FBC magnitudes) (equal magnitudes) B C : CB : (equal BC BC CB B C e "F (equal magnitudes and opposite directions), (5-15) Newto : :FBC ! :Newton’s :Newton’s (b) each CB When Third Law: two bodies interact, the forces on the bodies from Third Law: When two bodies interact, the forces on the bodies from each vector relation The force on B other are ! "F F (equal magnitudes and opposite directions), (5-15) (b) BC CB ! "F F (equal magnitudes and opposite directions), (5-15) gnirelation BC CB The force on B other are always equal in magnitude and opposite in direction. force on: B other:are always equal in magnitude and opposite in direction. due to C has the same where the minus sign means that these two forces are in opposite directions. due to C has the same We where the minus sign means that these two forces are in opposite directions. We ! "F F (equal magnitudes and opposite directions), (5-15) These two forces are called action & reaction forces :to C hasBC the:sameCB magnitude as the same magnitude asWe the the minus sign means that these forces are in opposite directions. ! "F (equal magnitudes andtwo opposite (5-15) can call the forces between two interacting bodies a When third-law force pair. When the forces between two interacting bodies a directions), third-law force pair. nitude ascall the BC can CB the bo force on C dueFor to B. For the book and crate, we can write this law as the scalar relation e he minus sign means that these two forces are in opposite directions. We force on C due to B. For the book and crate, we can write this law as the scalar relation el on C due toB. A the forces between two interacting bodies a pair. third-law pair. When third-law force pair: athe forces between two force othe B. forces between two interacting bodies third-law force When Fig. 5-10 (a) Book B leans against crate F ! F (equal magnitudes) s sign means that these two forces are in opposite directions. We BC CB : against crate Fig. 5-10 (a) Book B leans FBC ! FCB (equal magnitudes) C. leans against crateinteracting bodies : (b) Forces FBC (the force on the book or as the v C. (b) Forces F (the force on the:book or as the vector relation crate ces bodies a third-law force pair.from When the crate) and F (the force on the rce onbetween the book or as thetwo vectorinteracting relation from the crate) and F (the force on the BC ok force on the (the thethe same magnive magnidirection. : CB CB When any two bodies interact(equal in any situationopposite (stationary, crate from the book) (5-15) have the same magni! magnitudes and directions), : : crate from the book) have the same magnitude(5-15) and are opposite in direction. "FCB FBC !accelerating), (equal a magnitudes and force oppositepair directions), moving, third-law is present where the tude and are opposite in direction. : FBC : "FCB where the minus sign means that these two forces are in opposite directions. We where the minus signthe means thatbetween these two forces are in opposite can call forces two interacting bodies adirections. third-law We force pair. When can call the forces between two interacting bodies a third-law force pair. When can call t eand interaction. focus on the forces acting on the cantaloupeCantaloupe (Fig. Force THIRD LAW 5-8 NEWTON’S : : ( b ) 5-11b). forceLet’s is the F first :the cantaloupe : interaction. The third-law force pair for the cantaloupe – table interaction. CE ( afor ) the cantaloupe are (d) – Earth FCT and CE. (c) The third-law force pair is the normal force Fon the cantaloupe from the table, and force FCT FCE is the interaction.force (d) The5-11 third-law pair for thelies cantaloupe – table hey a third-law (a)force A cantaloupe onEarth. a table thatinteraction. stands on Earth. (b)force The forces on Cantaloupe gravitationalFig. force on the cantaloupe due to Are they a Fthird-law : : CE ( cCantaloupe ) C the cantaloupe are FCT and FCE. (c) The third-law force pair for the cantaloupe – Earth FFCT from tab ntaloupe, and not on EC (normal force These forces pair? No, because they are forces on a single body, the cantaloupe, and not on F CT rce pair isany present. The interact two bodies in any situation, third-law force is present. The interaction. (d) The third-law force pair forathe cantaloupe – tablepair interaction. F just happen CE These are two interacting bodies. FECpresent. The Cantaloupe C any two bodies interact in any situation, a third-law force pair is force table) F (gravitational force) lawfrom would still hold if Table T CE book and crate in Fig. 5-10a are stationary, but tothe third law would still So on are the these. be balanced. third-law force hold if ntaloupe but on the To find a third-law pair, we must focus not cantaloupe but on the book and crate in Fig. 5-10a are stationary, but the third law would still holdpairs. if F TC (b) Earth they were moving and even if they were accelerating. These are F interaction between the onesituation, other body. InEC the cantaloupe –is present. T ional In force) the cantaloupe – two they were moving and even if they wereand accelerating. Table T cantaloupe any bodies interact in any a third-law force pair third-law force Fig. 5-11 A cantaloupe lies onpairs a table that stands on the Earth.can(b) T Asthe another example, let(the us find the(a) third-law force involving pairs involving can: pairs : Earth interaction (Fig. 5-11c), Earth pulls on the cantaloupe with a gravitational ) d ) Earth E As another example, let us find third-law force involving the canEarth e with a gravitational pairs. book and crate in Fig. 5-10a are stationary, third law force would hold the cantaloupe are FCT andbut FCE.the (c) The third-law pairstill for the can : : taloupe in Fig. 5-11a, which lies on a table that stands on Earth. The cantaloupe : Earth. The cantaloupe force the cantaloupe pulls Earth with aThe gravitational Are FCE taloupe inand Fig. 5-11a, which lies on aon table that stands on Earth.force Theforce cantaloupe interaction. (d) third-law pair forFthe – table inte EC.cantaloupe n Earth. (b) The forces on they were moving and even if they were accelerating. tional force . Are F EC ) the ( con ) Earth E ( atable interacts with table and with Earth (this time, there are whose three bodies whose these forces a third-law force pair? Yes, because they are two interactinteracts with the and with Earth (this time, there areforces three bodies ir for the cantaloupe – Earth are three bodies whose As another example, let us find the third-law force pairs involving F CT the ca rces on two interacte – table interaction. interactions we must sortto out). ing bodies, the force on each due the other. Thus, by Newton’s third law, interactions we must sort out). taloupe in Fig. 5-11a, which lies on a table that stands on Earth. The cantalou F (normal force from table) any two bodies interact in any situation, a third-law force pS ( a ) ( c ) CT on’s third law, These forces Let’s first focus on the forces acting on the cantaloupe (Fig. 5-11b). Force Let’s first focus on the forces acting on the cantaloupe (Fig. 5-11b). Force : : Theinteracts cantaloupe interacts with the table & with Earth Fthird with the table and with Earth (this time, there are three bodies who F book and crate in Fig. 5-10a are stationary, but the law TC : : CT upe (Fig. 5-11b). Force : : (cantaloupe ! Earth interaction). F " !F just happen CE EC isCT the force on the cantaloupe from the table, and isaccelerating. the FCT Fwere F CE (gravitational force) is:normal the normal force onsort the cantaloupe from theforce table, force FCE is the F CEand they were moving and even if they F (normal force from table) interactions we must out). aw force pair is present. The to be balanced. CT n). le, andgravitational force is the F So These forces (Earth. b) (d) CEcantaloupe–table force on the cantaloupe due to Are they a third-law force The interaction: As another example, let us find the third-law force pairs gravitational force on the cantaloupe due to Earth. Are they a third-law force F Next, in the cantaloupe – table interaction, the force on the cantaloupe from e third law would still hold ifjustfirst TC Let’s focus on the forces acting on cantaloupe (Fig. 5-11b). For happen : : : : they a third-law force pair? No, because they are forces on single body, the cantaloupe, onthat F CEa(gravitational force) Fig. 5-11 (a) A cantaloupe lies on a table that stands on Earth. (b) The forces on taloupe in Fig. 5-11a, which lies and on a not table stands on Ear g. the table is and, conversely, the force on the from thetable) cantaloupe is F FTCforce pair? No, because they are forces on atable single body, the cantaloupe, and Fnot ont to balanced. (normal force on the cantaloupe from the : be : is the normal force on the cantaloupe from the table, and is F CT the cantaloupe from CT CE the cantaloupe are and F . (c) The third-law force pair for the cantaloupe – Earth F (b) ( d ) there are t CT CE interacts withpair, the and tableso and with Earth (this time, two5-11d). interacting bodies. :the force (Fig. pairs involving cancantaloupe, and not on These forces are also a third-law force two interacting bodies. gravitational force on the cantaloupe due to Earth. Are they a third-law for interaction. (d) The third-law force pair for the cantaloupe – table interaction. (force on the table from the cantaloupe) the cantaloupe is F interactions we must sort out). TC To find a third-law pair, we must focus not on the cantaloupe but on the Fig. 5-11 (a) A cantaloupe lies on a table that stands on Earth. (b) The forces on nds on Earth. The cantaloupe : No, : : third-law : To find a pair, we pair must focus not on thethe cantaloupe onnot the pair? because they are forces on a single body, cantaloupe, and Let’s first focus on the forces acting on thebut cantaloupe the cantaloupe are and F The third-law force for the cantaloupe –In Earth F CT CE so interaction between cantaloupe and one other body. the cantaloupe – there are three bodies whose (cantaloupe – table interaction). FCTthe ". (c) !F : TC interaction. (d) The third-law forceinpair forsituation, the cantaloupe –the table interaction. two interacting bodies. cantaloupe but on the interaction between the cantaloupe and one other cantaloupe – is the normal force on the cantaloupe from the table, an F any two bodies interact any aon third-law force pair iswith present. The In the Earth interaction (Fig. 5-11c), Earth pullsCT cantaloupe abody. gravitational book and crate in Fig.(Fig. areinteraction: stationary, but the third law would still holdthe if: cantaloupe gravitational force on the cantaloupe due to Areon they Tocantaloupe find a5-10a third-law pair, we must focus not on but t Earth interaction 5-11c), Earth pulls on the cantaloupe with a Earth. gravitational F: The cantaloupe–Earth dy. the cantaloupe – antaloupe (Fig. 5-11b). Force ). In force and the pulls on Earth with a gravitational force . Are F CE EC : :the canta : they were moving and between even if they were accelerating. pair? No,force because they are forces on a single body, interaction the cantaloupe and one other body. In the cantaloup he table, and force is the F any two bodies interact in any situation, a third-law pair is present. The force and the cantaloupe pulls on Earth with a gravitational force F F CE these third-law force pair? Yes, because they are forces on upe with a forces gravitational EC. Are AsaCE (Earth pulls the cantaloupe with a involving gravitational force) another example, leton us find the third-law force pairs thetwo can-interactCHECKPOINT 5 two interacting bodies. : and crate in Fig. 5-10a are lies stationary, butthat theEarth third law would still hold if h. Are they a third-law force Earth interaction (Fig. pulls onThe the cantaloupe with a gravitation taloupe inforce Fig. 5-11a, which onthe a 5-11c), table stands on Earth. cantaloupe these forces a third-law force pair? Yes, because they are forces on two ingbook bodies, the on each due to other. Thus, by Newton’s third law, : : cant itational force . Are F To find a third-law pair, we must focus not oninteractthe (cantaloupe pulls on Earth with a gravitational force) they were moving and even if they were accelerating. EC Suppose that the cantaloupe and table of Fig. 5-11 are in an elevator cab that begins to y, the cantaloupe, and not on force and the cantaloupe pulls on Earth with a gravitational force F interacts with F the table and with Earth (this time, there are three bodies whose EC. A : the other. : ing bodies, the force on each due to Thus, by Newton’s third law, :CE :find interaction between the cantaloupe and one other body. In As another example, let us the third-law force pairs involving the canforcesaccelerate on two interact(a) Do the magnitudes of FTCpair? increase, decrease, or stay the on two intera upward. FYes, interactions we must sort out). CT (cantaloupe !and Earth interaction). F " !F these forces a third-law force because they are forces CE EC Earth interaction 5-11c), Earth pulls on the cantaloupe w taloupe inLet’s Fig. 5-11a, which lies a table stands on Earth. (Fig. The cantaloupe : :that firstthe focus on theon forces acting on the cantaloupe (Fig. 5-11b).in Force same? (b) Are those two forces still equal in magnitude and opposite direction? (c) on the cantaloupe but on : wton’s third law, : with : by Newton’s third law, bodies, the force on(this each due toand the other. Thus, : and : Earth interacts the table with time, there are three bodies whose (cantaloupe ! Earth interaction). F "increase, !F the cantaloupe pulls Earth with a gravitatio FCE CE ECforce is ing the normal force on the cantaloupe from the table, and force the on FCTcantaloupe F(d) ator cabDo that begins to CE isAre Next, in the cantaloupe – table interaction, the force on the cantaloupe from the magnitudes of and F decrease, or stay the same? those two F er body. In the – CE EC interactions we must sort out). : : these forcesAre a third-law force pair? they are forc gravitational force on the cantaloupe due to they a third-law force Yes, because : : Earth. Example: Q.1: A book rests on a table, exerting a downward force on it. The reaction to this force is: (a) Force from the Earth on the table (b) Force from the book on Earth (c) Force from the Earth on the book (d) Force from the the table on the book Applying Newton’s Laws tal of five forces act on the blocks, as shown in Fig. 5-13: the cord does not stretch certainblock time, block S mov table, hanging time. This means that th accelerations have the sam 1. The cord pulls to the right on sliding block S with a forceon Block of magnitude T. Sample Problem 2. The pulls upward on S hanging block H with a force Figure 5-12cord shows a block (the sliding block) with mass of the same This to upward force keepsablock Q How do I classify this Block on block hanging M table, ! 3.3 kg. Themagnitude block isT.free move along horizontal ticular law of physics H from falling freely. frictionless surface and connected, by a cord that wraps over Yes. Forces, masses, a 3. Earth pulls down on block S with the gravitational force g block) witha mass : frictionless pulley, to a second block H (the hanging they should suggest Newt FgS, which has a magnitude equal to Mg. FN Block S along a horizontal : block), with mass m ! 2.1 kg. The cord and pulley have neg. That is our starting Ke ma 4. Earth pulls down on block H with the gravitational force T ord that wraps overmasses : ligible compared to the blocks (they are “massless”). FgH, which has a magnitude equal to mg. M Q If I apply Newton’s se k H (the hanging : body should I apply it The hanging block as the sliding block 5. The table pushesHupfalls on block S with a normal force S FNaccelerates . nd pulley have to negthe right. Find (a) the acceleration of block S, (b) the ac- We focus on two bodie block. Although they ar ey are “massless”). FgScord. celeration of block H, and (c) the tension in the points), we T can still treat block S accelerates Sliding every part of it5-13 moves Th in e Fig. block S block S, (b) the acQ What is this problem all about? macting Idea is to apply Newton’s Block onHthe tws in the cord. M You are given two bodies — sliding block and hanging Q Whatblocks about the pulley of Fig. 5block — butFig. must5-13 also consider Earth, which pulls on both We Fcannot represent The forces gH different parts of it move bodies. (Without nothing would happen here.) A toactingEarth, on Frictionless the two There is a rotation, we shall deal wit surface block and hanging tal of five forces actofon the blocks, as shown in Fig. 5-13: blocks Fig. 5-12. does eliminatethe thecord pulley fro Hanging m hich pulls on both certain time, b H S with a force mass to be negligible com 1. The cord pulls to the right on sliding block block appen here.) A toIts only This function There is another thing you should note.blocks. We assume thatmei time. of magnitude T. wn in Fig. 5-13: the cord does not stretch, so that if block H falls accelerations 1 mm in a 2. The cord pulls upward on hanging block H with a forceQ OK. Now how do I ap certain block S moves 1 mmofto the rightRepresent in that block sameS as block time, S of mass is connected to a block mass ock S with a force ofFig. the5-12 sameAmagnitude T.MThis upward forceHkeeps block Q How do I Block on table, block hanging s a block S (the sliding block) with mass block is free to move along a horizontal by Wea cord needthat to find Fnetover in x & y directions for both e and connected, wraps blocks of masses M & m lley, to a second block H (the hanging m ! 2.1 kg. The cord and pulley have neg For block S of mass M: mpared to the blocks (they are “massless”). Fnet ,x S= accelerates Max ⇒ T = Max = Ma (1) k H falls as the sliding block (a) the acceleration of block S, Ma (b) the acFnet ,y = y ⇒ FN − FgS = M (0) = 0 k H, and (c) the tension in the cord. ⇒ FN = FgS = Mg (2) Fig. 5-13 The forces FN Block S T M FgS T m Block H F gH For block H of mass m: roblem all about? acting on the two n two bodies — sliding block and hanging Fnet ,x = 0 blocks of Fig. 5-12. also consider Earth, which pulls on both Fnet ,y here.) = mayA⇒to-T − FgS = may y Earth, nothing would happen There is another thing you should note. We assume that ct on the blocks, as shown in Fig. 5-13: ⇒ T − mg y 1 mm in a the=cord −madoes (3)not stretch, so that if block H falls certainintime, S moves 1 mm to the right in that same −a block because block H accelerates the block negative to the right on The sliding S with a force a direction of the y axis FN time. This means that the blocks move together and their T. T accelerations have the same a. T M magnitude m upward on hanging block H with a force x x Both blocks M & m accelerate with the same gnitude T. Thismagnitude upward force Q How do I classify this problem? Should it suggestHanging a para keeps block Sliding FgH block H a ticular law of physics to Fme? freely. gS block S By substituting 1 into 3: Yes. Forces, masses, and accelerations are involved, and wn on block S with the gravitational force : m they should suggest Newton’s second law of motion, F a magnitude equal to Mg.Ma − mg = −ma ⇒ a = net ! g (4) : (a) (b) M .+That m is our starting Key Idea. ma wn on block H with the gravitational force a magnitude equal to mg. : es up on block S with a normal force FN. 5-14 (a) A free-body diagram for block S of Fig. 5-12. Q If I applyFig. Newton’s second law to this problem, to which (b) A free-body diagram for block H of Fig. 5-12. body should I apply it? Now note that Eqs. 5-18 and 5-20 are simultaneous equations with the same two unknowns, T and a. Subtracting these equations eliminates T. Then solving for a yields By substituting 4 into 1: m (5-21) a" g. M$m Mm T =5-18 yieldsg Substituting this result into Eq. M +m 12. Mm T" g. (5-22) M $ m Putting in the values of M, m, and g from the problem, we obtain: Putting in the numbers gives, for these two quantities, m 2.1 kg g" (9.8 m/s2) M$m 3.3 kg $ 2.1 kg " 3.8 m/s2 (Answer) a" es sense ction in and T " o apply s explain ea: The an write (5-16) f the net mponent block S Mm (3.3 kg)(2.1 kg) g" (9.8 m/s2) M$m 3.3 kg $ 2.1 kg " 13 N. (Answer) Q The problem is now solved, right? That’s a fair question, but the problem is not really finished until we have examined the results to see whether they make sense. (If you made these calculations on the job, wouldn’t you want to see whether they made sense before you turned them in?) Look first at Eq. 5-21. Note that it is dimensionally correct and that the acceleration a will always be less than g. This is as it must be, because the hanging block is not in free tem and a free-body diagra not determine acceleration along the plane. The acceleration along the plane is set by the force compo: Sample Problem 102 Cord accelerates blockFrom up aFig. ramp positive direction CH A P of Tthe ExR F 5-15h, we write Newton’s second law ( net " nents along the plane (not by force components perpendicular : from the cord is up the plan a ) for motion along the x axis as m : to expressed by aNewton’s second lawup (Eq. 5-1).a In the Fig.plane), 5-15a, as a cord pulls on box of sea biscuits along component along the plane is down the plane and has mag- yforce F N. The gravitational g Cord accelerates block up aFig. ramp 5-15 (a) A box is pull 2 Sample Problem Tindicated ! mg sin in u "Fig. ma.5-15g.tude (5-24) (To see why that nitude mg sin u as frictionless plane inclined at u " 30°. The box has mass m " mg " (5.00 Cord Calculation: For convenience, we draw a coordinate sysFig. 5-15 (a)(b) A box isthree pulled upkg)(9.8 a acting planem/s by The forces on trig function is involved, we go through the steps of Figs. 5.00 kg, and the force from the cord has magnitude T " 25.0 Norm : the In Fig. a cord p Substituting andasolving for three a,5-15a, we find tem and a free-body shown in Fig. In Fig. diagram 5-15a, aascord pulls on a5-15b. box The of:sea biscuits updata along along plane is down (b)component The forces acting on the box: the cor force forc T , the gravitational Fig. 5-15 (a) A box is pulled up a plane by a cord. 5-15c to h to relate the given the force compo: : angle to frictionless plane inc N. What is the box’s acceleration component a along the inpositive direction of the x plane axis isinclined up the plane. Force force gravitational force andin the nof T , the FFinding 2 mg nitude sinAcan uF:as indicated Fig. frictionless at u " 30°.TThe box hasTo mass m" 5.00 kg, and the The box accelerates. g,force FNthe , (Answer) a " 0.100 m/s force nents.) indicate the direction, we write the clined plane? : N. (c)–(i) (b) three forces acting box: the N. What is the box’s fromThe the cord is up the plane and on has the magnitude T "cord’s 25.0 force (c)–(i) Finding the component .plane? F!mg Nfunction Cordforce :the force from trig is involved, we go th : : 5.00 kg, and the cord has magnitude T " 25.0 the plane and perpendicular down-the-plane component as sin u. The normal force clined where the positive result indicates that the box accelerates N. The T gravitational force Fg is downward and has magni: force force and the normal , the gravitational , F the plane plane (Fig. and perpendicular to does it. g KisEthe Y I D E A acceleration T FN the isaperpendicular and thus 5-15c to h5-15i) to relate the given ang N. kg)(9.8 What component along the in-to the tude mg: (5.00 m/s2) "box’s " with 49.0 N. More important, its up plane. block) mass force Finding the force components along FN. (c)–(i) 5-15 (a) A boxacceleration is pulledhe upBlock a plane bynents.) a cord. notFig. determine along the acceleration alon θ boxplane. To indicate the directio clined The acceleration alongplane? the plane is set by the force compo: FT S Fig. 5-15 (a) A box N The accel e rates. nents along the plane along a horizontal From Fig. 5-15h, we write Newton’s second law (Fnet " is pulle the and perpendicular to it. perpendicular nentsplane along the plane (not by force components (b) The three forces acting on the box:the cord’ s down-the-plane component as !mg to the plane), as expr (b) The three forces acting onC : : a m ) for motion along the x axis as : : : y Athe to plane), as expressed lawI(Eq. gravitational force , the TF Thisforce is aT:right Calculation: co rd that wraps overby Newton’s second KEY D E 5-1). A is perpendicular toFor the plane (F force T, the gravitational force and the normal , F N This is a right g tem and a free-body Normal force force (c)–(i) Finding the fo . F M T ! mg sin u " ma. (5-24) N : (a) acceleration 90°along −θ triangle. positive direction o Calculation: For convenience, we draw a coordinate sysnot determine th H (the The hanging F the perpendicular − θplane and This is galsto triangle. forceofforce Finding theand force components along FN.x(c)–(i)axes x is acceleration along the plane set bySubstituting the compoThe box is moving in the +veisdirection from theFN 90° cord up θ θ data solving for a, we find tem and a free-body diagram as shown in Fig. 5-15b. The Cord N. The gravitational From Fig. 5-15h, we write NewG : dpositive pulley have nentsofnegalong by force components perpendicular tude mg "T (5.00 kg) the plane and perpendicular to it. direction the xthe axisplane is up (not the plane. Force T : 2 Perpendicular Fig. 5-15 (a) A box is pulled up a plane by a cord. m am/s ) for along the x axis as , motion (Answer) a " 0.100 This is a right The acceleration component along the inclined plain The box accelerates. to the plane), as expressed by Newton’s second law (Eq. 5-1). from the cord is up the plane and has magnitude T 25.0 " fo θ ey are Cord's pull (b) The“massless”). three forces acting on:the box: the cord’s F A θ gS : : component of where boxisaccelerates N.force The Tgravitational force and hasismagniright is required − θthe positive result indicates that the This 90°force −Fθg Fisg, downward This also. 90° Ta ! mg sin u " m triangle. and the normal , the gravitational : 2 (a) T Calculation: For convenience, we draw a coordinate syslock S accelerates tude mg (5.00 kg)(9.8 m/s " ) " 49.0 N. More important, its 90° − θ (b) F up the plane. triangle. force FN. (c)–(i) Finding theneed force components Fgg à We to find Falong θ net in the x directionθ Gravitational A This is a right triangle. 90° −θ This Block on table, block hanging θ Substituting data and(a) solving for a, w θ (c) A bo 5-15 force m (c) (b) The three forces Block H Hy 2 force yT the gravitati , , a " 0.100 m/s (b) Perpendi c ul a r θ F force F . (c)–(i) Find and a free-body diagram as shown in Fig. 5-15b. The the plane andtem perpendicular to it. θ lock S, (b) the ac: positive direction of the x axis is up the plane. Force T Fnet ,x = max nA the cord. This is a right(a)T " 25.0 θ from the cord is up the plane and has magnitude Fig. : g : N (c) ⇒ T − mgsin component of indicatesA where result N. The gravitational forceθF=g ma is downward and has Perpendicular 90° − θ theFNpositive 90°magni−θ Thi s i s al s o. tri a ngl e . x This is a right F Fig. 5-13 The forces gH T 2 Cord Adjacent leg(c) component T tude kg)(9.8 " 49.0 N. More important, itsof up the plane.Parallel 90°)− θ 90° mg − θ " (5.00This is also.m/s triangle. (u F T : a = plane − gsin θ the two Fg (a) A box is pulled up aacting by on a cord. m (c) (d) The box accelerates. θ θ (b) The three forces acting on the box: the cord’s : blocks of Fig. 5-12. : 25the normal force T, the gravitational force Fg, and F A : θ = (9.8)sin 30 = 0.1m / s 2 g θ force FN. (c)–(i) Finding the force components along θ the plane and perpendicular to it. 5 Fig. 5-15 block and hanging hich pulls on both appen here.) A to(c) n in Fig. 5-13: Normal force the plane and perpen This is a triangle. (e) θ Hypotenuse θ g right (use cos θ ) mg Gravitational (c ) Opposite leg force y θ mg cos θ (f ) (b) mg(use sin θsin θ )mg mg sin θ Cord mg fo − th θ Cord's pull of θ component θ mg cos θ Hypotenuse θFgmg cos θ θ y FF θ 90° g g Parallel There(d)is another you should note. We assume that The net thing of(e)these component of (a) the cord does not stretch, so 1 lemm in a O l Fg that if block H xfalls Paral forces determines (a) certain A box is pulled up ablock plane bySa cord. Perpendicular (c) 1 mm (d) right (e)in component (f)θ (u mgFsin time, moves to the thatof same y accelerates. N The box T The net of these the acceleration. θ These forces ( g) Adjacent leg component of 5-15 a righta Fig. ockThis S iswith force (b) The three forces acting on the box: the cord’s us, F2 ! 2.00 N and e x axis. Eq. 5-25 leads to ax ! #2.00 m/s2. Sample Problem (Answer) Forces within an elevator c Sample Problem In Fig. 5-17a, a passenger of mass m ! 72.2 kg stands on Block on table, block hanging a platform scale in an elevator cab. We are concerned with Sample Problem the scale readings when the cab is stationary and when it is g block) with mass movingan upelevator or down.cab Forces within FN Block S along a horizontal y T whatever ord that wraps over (a) Find a general solution for the scale reading, ! 72.2 kg stands on M k H (the hanging vertical motion of the cab. are concernedthe with FN nd pulley have it negnary and when is ey are “massless”). FgS KEY IDEAS T block S accelerates eblock reading, whatever (1) The S, (b) the ac- reading is equal to the magnitude of the normal force m : Block H other force actin the cord. FN on the passenger from the scale. The only assenger Passenger : ing on the passenger is the gravitational force Fg, as shown in F Fig. 5-13 The forces These the free-body diagram of Fig. 5-17b. (2) We can forces relate the gH acting on the two : compete. forces on the passenger to his acceleration a by using of the normal force block and hanging Fig. 5-17 blocks of Fig.: 5-12. F g : Their net force Newton’s second law (F net ! ma ). However, recall that we only force actdicates eit hichother pulls on both : causes a vertical can use this lawThere only in an inertial frame. If the cab accelerorce as shown Fg, here.) body diag appen A in tois another thing you should note. We assume that acceleration. ( a )frame. So( bwe ) choose We can relate the on him fro ates, then it is not an inertial the ground wn in Fig. 5-13: the cord does not stretch, so that if block H falls 1 mm in a : ration a by using Fig. 5-17 (a)block A passenger stands a platform scale that in- same certain time, S moves 1 on mm to the right in that the readings (F when the cab is stationary and when it is for yscale components net, y " may) to get moving up or down. FN ! Fg " ma (a) Find a general solution for the scale reading, whatever or F " F # ma. (5-27) the vertical motion of theNcab. g This tells us that the scale reading, which is equal to FN, depends on the vertical acceleration. Substituting mg for Fg Fnet ,y = ma ⇒ F − F = ma ⇒ F = mg + ma K E Y I D E A S y N g N gives us ⇒ FN = m(a g)normal force (1) The reading is equal to m(g the magnitude of + the # a) (Answer) (5-28) FN " : FN on the passenger from the scale. The only other force act: for on anythe choice of acceleration a. ing passenger is the gravitational force Fg, as shown in the diagram of Fig. 5-17b. (2)cab We is canstationary relate theor (b) free-body What does the scale read if the : forces the passenger to 0.50 his acceleration a by using movingon upward at a constant m/s? : : Newton’s second law (F net ! ma ). However, recall that we When the cab is stationary or moving with a constant speed, a = 0 can use this law only in an inertial frame. If the cab accelerKEY IDEA ates, then it is not an inertial frame. So we choose the ground For any constant velocity (zero or otherwise), the accelera= m(a + g) → isFNzero. = mg = (72.2)(9.8) = 708N tion a F ofN the passenger Calculation: Substituting this and other known values into Eq. 5-28, we find FN " (72.2 kg)(9.8 m/s2 # 0) " 708 N. (Answer) This is the weight of the passenger and is equal to the mag- FN " (7 " 47 For an upward acce is increasing or its d reading is greater th is a measurement o in a noninertial fram decreasing upward the scale reading is l (a) (d) During the upw Fig. 5-17 (a) pass th magnitude FnetAof dicates either his weigh the magnitude ap,cab body diagram theDp frame of the for cab? on him from the scale a Calculation: The m on the passenger do senger or the cab; so the magnitude FN of the upward accelerat the net force on the p Fnet " FN ! Fg " during the upward passenger ong the y w written FN " (72.2 kg)(9.8 m/s2 # 3.20 m/s2) This is the weight of the passenger and is equal to the mag" 939 N, (Answer) nitude Fg of the gravitational force on him. and for a " !3.20 m/s2, it gives (c) What does the scale read if the cab accelerates upward 2 2 " (72.2 kg)(9.8 at m/s3.20 ! 3.20 at 3.20 m/sF2 Nand downward m/s2m/s ? ) " 477 N. (5-27) ual to FN, mg for Fg (5-28) ionary or (Answer) alues into (Answer) during the upwar ap,cab relative to th inertial frame of map,cab, and Newto For an upward acceleration (either the cab’s upward speed isAccelerates increasing upward or its downward the scale FN =speed m(a +isg)decreasing), → FN = 72.2(3.2 + 9.8) = 939N reading is greater than the passenger’s weight. That reading is a measurement of an apparent weight, because it is made Sample Problem in a noninertial frame. For (either + 9.8) = 476.52N Accelerates downward FNa =downward m(−a + g)acceleration → FN = 72.2(−3.2 decreasing upward speed or increasing downward speed), of block pushing on bloc Acceleration the scale reading is less than the passenger’s weight. : In Fig. 5-18a, constant horizontal force (d) During the aupward acceleration in part (c),Fwhat the app ofismagnitude " 4.0 20 N is applied A the of passenger, mass mAand force on whatkg, is which magnitude Fnet of thetonetblock the magnitude ap,cab of hisBacceleration asB measured the blocks pushes against block of :mass m " 6.0 kg.inThe : " ma p,cab frame of theacab? Does Fnetsurface, ? slide over frictionless along an x axis. accelera- Fnet " FN ! F (a) What is the Calculation: Theacceleration magnitude Fgofofthe theblocks? gravitational force F = F − F net doesNnot depend g : motion of the pason the passenger on the Serious Error: Because force Fapp is applied directly senger or theFcab; so, from part (b), Fg .is8708 ) N. From part (c), net = 939 − (72 .2)(9 to block A, we use Newton’s second law toduring relate that the magnitude FN of the normal force on the passenger Facceleration = 231N : net a ofNblock force to the the motion the upward acceleration is the 939 readingA. onBecause the scale.Thus, is along x axis, we isuse that law for x components the net forcethe on the passenger (Fnet, x " max), writing it as Fnet " FN ! Fg " 939 N ! 708 N " 231 N, (Answer) only horizontal fo : force FAB from blo Dead-End Solut ing, again for the x (We use the min Because FAB is a equation for a. Successful Solu Sample Problem Acceleration of block pushing o Sample Problem : only horiz In Fig. 5-18a, a constant horizontal force Fapp of magnitude : Block on20table, force FAB fr N is block appliedhanging to block A of mass mA " 4.0 kg, which pushes against block B of mass mB " 6.0 kg. The blocks g block) with mass Dead-End slide over a frictionless surface, along anFNx axis. Block S along a horizontal ing, again f " total mass m A B What is the acceleration ofThis theforce blocks? ord that wraps(a) over causes the T values, we find M Fapp A : acceleration of the full k H (the hanging Fapp system. Serious Error: Becausex force is applied directly Fa two-block nd pulley havetonega !use t (We block A, we use (a) Newton’s second law to relate that mA " : ey are “massless”). F Because F gS force to the acceleration a of block A. Because the motion T equation fo block S accelerates two x forces is along the x axis, we use These that are lawthefor components Thus, the acceler Fapp FAB block S, (b) the acacting on just block A. A m (Fnet, x " max), writing it asx in the positive Block H dir Their net force causes in the cord. Successfu 2.0 m/s2. : (b) Fapp " its macceleration. Aa. force Fapp i F : (b)gH What is the Fig. 5-13 The forces system. We However, this is seriously wrong because Fapp is not the block A (Fig. 5-18 acting on the two B block and hanging This is the only force blocks of Fig. 5-12. FBA causing the acceleration hich pulls on both x of block B. appen here.) A toThere that the (c) is another thing you should note. We assume We can relate wn in Fig. 5-13: not stretch, soF:that if block H fallseration 1 mmwith in New a (a) Adoes constant horizontal force Fig. the 5-18cord app is applied to block A, which pushes againstblock block B. Two horizontal act on certain time, S(b) moves 1 mm forces to the right in that same to block s act on re, once pushes against block B of mass mB " 6.0 kg. The blocks PA R T 1 slide over a frictionless surface, along an x axis. 105 REVIEW & SUMMARY B (a) What is the acceleration of the blocks? F A : app F = 20N, m = 4Kg, m = 6Kg app A B Serious Error: Because force Fapp is applied directly total mass m " m . Solving for a and substituting known A Dead-End ing, again x B (a) that to block A, we use Newton’s second law to relate values, we find : Fnetto (mAacceleration + mB )a ,x =the force the motion Fapp 20 a N of block A. Because !(m we ! 2.0law m/s2. for x components is along the x axis, use that Fappa +!Fm − F = + m )a 4.0 BAA " m ABB A kg "B 6.0 kg Fapp FAB A ),6)a writing it as (Fnet, (Answer) 20x+" 0 =ma (4x+ Thus, the acceleration of the system and ofa.each block is F " m 2 app A a =positive 2m / sdirection of the x axis and has the magnitude: in the 2.0 m/s2. However, this is seriously wrong because Fapp (b) What is the (horizontal) force block A (Fig. 5-18c)? KEY IDEA : FBA FBA = mB a = 6 × 2 = 12N on block B from (b) is not the B FBA x (c) This force ca acceleration two-block sys (We use t Because F These are the equation f x acting on jus Their net forc Successf : its acceleratio force Fapp system. We This is the on causing the a of block B. : Fig. 5-18 (a) A constant horizontal force F We can relate the net force on block B to the block’s accelA, which pushes against block B. (b) Two hori eration with Newton’s second law. block A. (c) Only one horizontal force acts on Calculation: Here we can write that law, still for components along the x axis, as eration of the system with Newton’s seco FBA ! mBa, again for the x axis, we can write that law a which, with known values, gives Examples: Q.1: A constant force of 46N is applied at an angle of 60o to a block A of a mass 10kg as shown in the F figure. Block A pushes another block B of mass 36kg. Assuming a frictionless surface, the total acceleration of the blocks along the x-axes is: (a) 1.5m/s2 (b) 0.25m/s2 (c) 0.5m/s2 (d) 2m/s2 A B FNA Fnet ,x = (mA + mB )a a= θ F cosθ 46 cos 60 = = 0.5m / s 2 mA + mB 10 + 36 FNB o F Fcosθ Fsinθ A B mAg mBg Motion direction Q.2: A 3kg box is placed in the top of a 10kg box. The bottom box is pushed with a force F. The two boxes moves together with an acceleration of 2m/s2. The horizontal force F is: (a) 3N (b) 26N (c) 1N (d) 5N A F Motion direction B Fnet ,x = (mA + mB )a = (10 + 3)2 = 26N Q.3 In the figure, two blocks are connected and pulled on a horizontal table by a force with a magnitude of 20N. If the m1 =3kg, m2 = 2kg, then T and a are: (a) 5N, 4m/s2 (b) 8N, 4m/s2 (c) 5N, 4m/s2 (d) 10N, 3m/s2 Fnet ,x = (m1 + m2 )a F 20 ⇒a= = = 4m / s 2 m1 + m2 5 To find T we apply Newton's 2nd law on one of the mass: F − T = m1a ⇒ T = F − m1a = 20 − 12 = 8N or T = m2 a = 2(4) = 8N Motion direction m2 T T m1 F Q.4: An elevator of total mass 2000kg moves upward. The tension in the cable pulling it is 24000N. The acceleration of the elevator is: T (a) 2.2m/s2 (b) 9.8m/s2 (c) 12m/s2 (d) 3.6m/s2 Fnet ,y = ma T − mg = ma a ⇒a= T − mg 24000 − 2000(9.8) = = 2.2m / s 2 m 2000 mg Q.5: A 70kg man stands on a spring scale in an elevator that has a downward acceleration of 2.8m/s2. The scale will read: (a) 980N (b) 680N (c) 490N (d) 343N T a Fnet ,y = ma T − mg = −ma ⇒ T = m(g − a) = 70(9.8 − 2.8) = 490N mg Q.6: An elevator has a body of 10kg. The tension in the cable when the elevator is moving upward at a constant speed of 10m/s is: (a) zero (b) 98N (c) 1.5N (d) 7.3N T Fnet ,y = ma T − mg = 0 v ⇒ T = ma = 10(9.8) = 98N mg Q.7: Two masses m1 = 4kg and m2 = 6kg are connected to a rope of a negligible mass. An upward force of 198N is applied as shown. The magnitude of the acceleration of the system is: (a) 10m/s2 (b) 40.2m/s2 (c) 50.2m/s2 (d) 70.2/s2 T Fnet,y = Ma T − (m1 + m2 )g = (m1 + m2 )a = 198 − (10)9.8 = 10m / s 2 10 T − (m1 + m2 )g ⇒a= m1 + m2 a Mg Q.8: A block slides down a frictionless inclined plane with an acceleration of 4.9m/s2. The angle between the plane and the horizontal is: (a) 30o (b) 26o (c) 21.55o (d)14.32o FN Fnet ,x = max ⇒ −mgsin θ = −ma ⇒ a = gsin θ a ⇒ θ = sin −1 = 30 o g mg sinθ θ mg cosθ Q.9: A 40kg crate is held at rest on a frictionless incline by a force parallel to the incline. If the incline is 30o above the horizontal, the magnitude of the applied force is: FN F (a) 20N (b) 40N (c) 23.5N (d) 10N Fnet,x = max mg sinθ at rest à a = 0 30o ⇒ F − mgsin θ = 0 ⇒ F = mgsin θ = 40(9.8)sin 30 = 20N o mg cosθ Q.10: A block of mass 4kg is pushed up a smooth 30o inclined plane by a constant force of magnitude 40N and parallel to the incline. The magnitude of the acceleration of the block is: (a) zero (b) 9.8m/s2 (c) 1.2m/s2 (d) 5.1m/s2 Fnet ,x = max ⇒ F − mgsin θ = ma F − mgsin θ 40 − 4(9.8)sin 30 o ⇒a= = = 5.1m / s 2 m 4 FN mg sinθ 30o Q.11: If the mass of the block is 5kg. Find T if the block moves with a constant velocity upward the smooth inclined plane (or at rest). (a) 45N (b) 24.5N (c) 42N (d) 25N Fnet ,x = max ⇒ T − mgsin θ = 0 ⇒ T = mgsin θ = 5(9.8)sin 30 = 24.5N o F FN mg cosθ F mg sinθ 30o mg cosθ -15) conyou able Eqs. x! ese Acceleration Equation Missing Number Equation Quantity Q.12: A 5kg block is pushed upward 30o inclined plane with initial velocity of 14m/s. The distance 2-11 ! v0 is: " at x # x0 that the blockvgoes 1 2 ! v t " x # x 0 0 (a) 20m (b) 10m (c) 18m(d) 24m 2 at 2-15 v 2-16 t v 2 ! v 20 " 2a(x # x0) 1 2-17 x # x ! v)t θ a 0 0 " 2 (v Fnet ,x = max ⇒ ma = −mgsin 1 2 x # x ! vt # 2-18 v0 0 2 at / s 2 ⇒ a = −gsin θ = −4.4m FN mg sinθ 30o mg cosθ vo =the 14macceleration / s, v = 0 is indeed Make sure that 2 constant before using in this table. vo2 the14equations Δx = − = = 20m a 4.4 a Q.13: From the figure, the normal force FN on a block of weight 60N sliding down a frictionless plane is: (a) 50N (b) 30N (c) 25N (d) 40N Fnet ,y = may ⇒ FN − mg cosθ = 0 ⇒ FN = mg cosθ = 60 cos 60 = 30N o FN mg sinθ 60o mg cosθ Q.14: Show the correct direction of the tension force T: B X B A (a) B X A X (c) Q.15: A block of mass m is connected to a block of mass M as shown. The normal force on the block m is: (a) mg − T (b) mg (c) Mg − T (d) Mg A (d) m X M FN T X T mg X A (b) m B M Mg Q.16: Referring to the last example, if the block M is moving downward, the net force acting on it is: (a) Ma − T = Mg (b) T = Ma (c)T = Mg (d) T − Mg = − Ma Fnet ,y = Ma ⇒ T − Mg = −Ma FN m T X T mg M Mg Motion direction !!"#$%&"'((")*(+,$*-."/%0%"!! !!"#$%&"'((")*(+,$*-."/%0%"!! •1 Only two horizontal forces act on a 3.0 kg body that can move sec. 5-6 Newton’s Second Law over a!!"#$%&"'((")*(+,$*-."/%0%"!! frictionless floor. One force is 9.0 N, acting due east, and the •1 Only two forces act on a 3.0 kg body can move sec. 5-6 Newton’s Second other is 8.0 N,horizontal acting 62° northLaw of west. What is thethat magnitude of SSM Worked-out solution available in Student Solutions Manual WWW Worked-out solution is at http://w Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com – Number of dots indicates level of problem difficulty ILW Interactive solution is at • ••• sec. 5-6 Newton’s Second Law ••8 A 2.00 kg object is subjec Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com a! eration : over a frictionless floor. One force is 9.0 N, acting due east, and the Problems: the acceleration? •1 body’s Only two horizontal forces act on a 3.0 kg body that can move other is 8.0 N, acting 62° north of west. What is the magnitude of overTwo a frictionless floor. One acting due east,that andcan the •2 horizontal forces actforce on a is 2.09.0kgN,chopping block the body’s acceleration? otherover is 8.0 N, acting 62° north counter, of west. What theinmagnitude of slide a frictionless kitchen which islies an xy plane. : •2 forces on aN)ĵ. 2.0 Find kg chopping block that the Two body’s One forcehorizontal isacceleration? the acceleration of can the F1 ! (3.0 N)î act # (4.0 slide over a frictionless kitchen counter, which lies in an xy plane. chopping block the other force is can (a) •2 Two horizontal forces actnotation on a 2.0 when kg chopping block that : in unit-vector : : One force is Find the acceleration of the F ! (3.0 N)î # (4.0 N)ĵ. 1 # ("4.0 kitchen and Fslide N)ĵ, (b)counter, F2 ! ("3.0 over aN)î frictionless whichN)î lies#in(4.0 an N)ĵ, xy plane. 2 ! ("3.0 : : in unit-vector notation when the other force is (a) chopping block (c) ! (3.0 F2 force ("4.0 One is N)î F1 !#(3.0 N)îN)ĵ # .(4.0 : N)ĵ. Find the acceleration of the : Fchopping N)î #in("4.0 N)ĵ, (b) F2 ! ("3.0 # (4.0force N)ĵ, isand 2 ! ("3.0block unit-vector notation when N)î the other : 2 (a) •3 If the 1 kg standard body has an acceleration of 2.00 m/s at : : (c) (3.0N)î N)î # # ("4.0 ("4.0N)ĵ, N)ĵ . (b) F ! ("3.0 N)î # (4.0 N)ĵ, and F2 F!2 ! ("3.0 2 20.0°:to the positive direction of an x axis, what are (a) the x com(c) F ! (3.0 N)î # ("4.0body N)ĵ . has an acceleration of 2.00 m/s2 at •3 If2 the ponent and1 kg (b)standard the y component of the net force acting on the 20.0° to the positive direction of an axis, what notation? are (a) the xm/s com2 •3 Ifand the(c) 1 kg standard body has acceleration of 2.00 at body, what is the net force in xan unit-vector ponent and (b) the y component of the net force acting on the 20.0° to the positive direction of an x axis, what are (a) the x com••4 While forces actnet on force it, a parbody, andand (c)two what is the in of unit-vector notation? y acting on the ponent (b) the y component the net force ticle is to move at the constant velocbody,While and (c) is the on net force in unit-vector notation? ••4 twowhat forces a par" (4 act ity : One of m/s)ĵ. it, v ! (3 m/s)î y : at the constant velocticle is to move ••4:forces While forces act#on("6 it, aN)ĵ. parthe is two F1 ! (2 N)î y " ity One of ! (3 m/s)î (4 m/s)ĵ. v ticle isisto at the constant velocWhat themove other force? : F1 : the forces is F ! (2 N)î # ("6 N)ĵ. 1 ity v ! (3 m/s)î " (4 m/s)ĵ. One of θ1 : astronauts, ••5 Three propelled x What is the other force? F1 the forces is F1 ! (2 N)î # ("6 N)ĵ. θ3 by jet backpacks, push and guide a θ1 F2 What is the other force? ••5 Three astronauts, propelled x F1 120 kg asteroid toward a processing θ3 θ1 F2 by backpacks, push andshown guide in a ••5jet exerting Threethe astronauts, propelled F3 dock, forces x 120 kg asteroid toward a processing θ3 by jet backpacks, guide F2 Fig. 5-29, with F1 !push 32 N,and F2 ! 55 N,a F dock, exerting the forces shown in 3 asteroid toward a processing F120 41 N, u1 ! 30°, and u3 ! 60°. 3 !kg Fig. 5-29 Problem 5. Fig. 5-29, with F ! 32 N, F ! 55 N, 1 2 F dock,isexerting the forces shown(a)in What the asteroid’s acceleration "(8.00 m/s2)î # ( : 2.00 kg object is subjec ••8 are FA N 1 ! : (30.0 N)î # (16.0 2 a ! "(8.00 eration m/s )î # ( find force. ••8 the 2.00 kg object is subjec :A third are F1 !: (30.0 N)î # (16.0 N a0.340 eration ! "(8.00 m/s2)î #m ••9 A kg particle : third force. find the are F (30.0 N)î # (16.0 N to x(t)1!!"15.00 # 2.00t " 4.0 ••9 0.340 kg particle m find the third force. with xAand y in meters and t in to x(t) ! "15.00 # 2.00t " an 4.0 the (b) the ••9 magnitude A 0.340 and kg particle m with x and y in meters and t in of axis) of the force tothe x(t)x! "15.00 # net 2.00t " 4.o the magnitude and direction (b) the ano gle of particle’s with xthe and y in meters and t i of the x axis) of the net the magnitude and (b) force the ano ••10 A 0.150 kg particle m gle of the particle’s direction o of x(t) the x!axis) of the net force to "13.00 # 2.00t # 4.0 ••10 kg direction particle m gle of A theIn0.150 particle’s seconds. unit-vector notatio to x(t) ! # s? 2.00t # 4.0 particle t ! 3.40 ••10 Aat"13.00 0.150 kg particle seconds. unit-vector notatio to x(t)A!In "13.00 # 2.00t # 4. ••11 2.0 kg particle moves particle 3.40 s? notati seconds.atforce Int ! unit-vector variable directed along particle at t ! 3.40 s?ct 2moves ••11 A (4.0 2.0 kg particle 3.0 m # m/s)t # " (2. variable force ••11 AThe 2.0 factor kgdirected particle move seconds. c is 2aalong const 3.0 mhas # (4.0 m/s)t # ct " (2. variable directed ticle aforce magnitude of along 36 N seconds. The factor c is a 2 const 3.0 m # (4.0 m/s)t # ct " (2 axis. What is c? ticle has a magnitude ofa36 N seconds. The factor c is cons •••12What Two forc axis. is c? horizontal ticle has a magnitude of 36 slides over frictionless ice,No axis. What is c? •••12 Two horizontalsyste forc which an xy coordinate : slides frictionless horizontal is•••12 laid over out.Two Force inforto F1 is ice, which xy frictionless coordinate syste slides an over positive direction of x ax : the ice, is laid out. Force in F1 is which coordinate and hasan a xy magnitude of syste 7.0t : : direction of the x ax positive is laidF2out. F1 is in Force has Force a magnitude of 9t e cab and its en that occuis 8.00 m/s2 from a height of 10.0 m by holding onto a rope that runs over a frictionless pulley to a 65 kg sandbag. With what speed does the man hit the ground if he started from rest? heir humanom a cannon ersion of the els to land in non and at a or 5.2 m and e underwent magnitude of ough it were ••53 In Fig. 5-48, three connected blocks are pulled to the right on a horizontal frictionless table by a force of magnitude T3 ! 65.0 N. If m1 ! 12.0 kg, m2 ! 24.0 kg, and m3 ! 31.0 kg, calculate (a) the magnitude of the system’s acceleration, (b) the tension T1, and (c) the tension T2. onnected by by the cable m1 T1 m2 Fig. 5-48 '((")*(+,$*-."/%0%"!! T2 m3 Problem 53. T3 and 65. are F1 ! (30.0 N)î # (16.0 N over a frictionless floor. One force is 9.0 N, acting due east, and the find the third force. other is 8.0 N, acting 62° north of west. What is the magnitude of 1. We are only concerned with horizontal forces in this problem (gravity plays no direct the body’s acceleration? ••9 A 0.340 kg particle role). We take East as the +x direction and North as +y. This calculation is efficiently on forces a vector-capable using magnitude-angle notation •2 Twoimplemented horizontal act on acalculator, 2.0 kg chopping block that can units understood). slide over a frictionless kitchen counter, which lies in an xy plane. : One force is F1 ! (3.0 N)î F# (4.0 0 g Find 118 acceleration b9.0 N)ĵ. b8.0 the g b2.9 53 g of the a m notation3.0when the other force is (a) chopping block in unit-vector : : F2 ! ("3.0 N)î # ("4.0 N)ĵ, (b) F2 ! ("3.0 N)î # (4.0 N)ĵ, and Therefore, the acceleration has a magnitude of 2.9 m/s2. : (c) F2 ! (3.0 N)î # ("4.0 N)ĵ . (withtoSIx(t) ! "15.00 # 2.00t " 4 with x and y in meters and t i the magnitude and (b) the an of the x axis) of the net force gle of the particle’s direction ••10 A 0.150 kg particle 2. We apply Newton’s second law (Eq. 5-1 or, equivalently, Eq. 5-2).2The net force to x(t) ! "13.00 # 2.00t # 4. •3 If the 1 kg has 2.00addition m/s isatdone using applied on standard the choppingbody block is FnetanF1acceleration F2 , where theof vector seconds. In unit-vector notat 20.0° to unit-vector the positive direction of an of x axis, what are by (a)a the x comnotation. The acceleration the block is given F F / m . d 1 2i particle at t ! 3.40 s? ponent and (b) the y component of the net force acting on the body, and what is the net force in unit-vector notation? (a) (c) In the first case ••4 While two forces on it, aˆi parF1 Fact 3.0N 4.0N 2 ticle is to move at the constant velocity : (3am/s)î v ! so 0 . " (4 m/s)ĵ. One of : the forces is F1 ! (2 N)î # ("6 N)ĵ. (b) In the second case, the acceleration a What is the other force? ˆj equals 3.0N ˆi y 4.0N ˆj F1 0 ˆi 3.0Npropelled 4.0N ˆj 3.0N ˆi 4.0N ˆj θ1 ••5 ThreeFastronauts, x F 1 2 (4.0m/s 2 )ˆj. θ3 by jet backpacks,m push and guide a 2.0kg F2 120 kg asteroid toward a processing (c) In this final situation, a is F3 dock, exerting the forces shown in Fig. 5-29, with F1 ! 32 N, F ! 55 N, 3.0N 2 ˆi 4.0N ˆj 3.0N ˆi 4.0N ˆj F1 F2 ˆ (3.05. m/s 2 )i. F3 ! 41 N, u1 ! 30°, and u3 ! 60°. Fig. 5-29 Problem 2.0 kg m What is the asteroid’s acceleration (a) in unit-vector notation and as law (b)(specifically, a 3. We apply Newton’s second Eq. 5-2). Alex magnitude and (c) a direction relative (a) We find the x component of the force is Charles ••11 A 2.0 kg particle move variable force directed along 3.0 m # (4.0 m/s)t # ct 2 " (2 seconds. The factor c is a cons ticle has a magnitude of 36 N axis. What is c? •••12 Two horizontal for slides over frictionless ice, which an xy coordinate syst : is laid out. Force F1 is in positive direction of the x a and has a magnitude of 7.0 : Force F2 has a magnitude of N. Figure 5-32 gives the x co ponent vx of the velocity of disk as a function of time t d One force is F1 ! (3.0 N)i # (4.0 N)j. Find the acceleration of the ˆi ˆj 3.0N ˆi notation 4.0N ˆj when3.0N 4.0N of2 the F1 F2 in unit-vector chopping block the other force is (a) ˆj. x axis) of the net force o (4.0m/s ) : : gle of the particle’s direction o F2 ! ("3.0 N)î ("3.0 N)î # (4.0 N)ĵ, and m # ("4.0 N)ĵ, (b) F2 ! 2.0kg : (c) F2 ! (3.0 N)î # ("4.0 N)ĵ . ••10 A 0.150 kg particle m to x(t) ! "13.00 # 2.00t # 4.0 is has an acceleration of 2.00 m/s2 at In the this final •3(c) If 1 kg situation, standard abody seconds. In unit-vector notatio 20.0° to the positive direction of an x axis, what are (a) the x comˆi ˆi particle at t ! 3.40 s? 4.0Nofˆj the net 3.0N 4.0Nonˆj the ponent and y component force acting F1 (b) F2 the 3.0N 2 ˆ (3.0 m/s )i. A 2.0 kg particle moves ••11 body, and (c)mwhat is the net force in unit-vector notation? 2.0 kg variable force directed along ••4 While two forces act on it, a pary 3.0 m # (4.0 m/s)t # ct 2 " (2. 3. We apply Newton’s second law (specifically, Eq. 5-2). ticle is to move at the constant velocseconds. The factor c is a const ity : v ! (3 m/s)î " (4 m/s)ĵ. One of : x component of the force is ticle has a magnitude of 36 N (a)forces We find the is the F1 ! (2 N)î # ("6 N)ĵ. axis. What is c? What is the other force? F2 1 F ma ma cos 20.0 1.00kg 2.00m/s cos 20.0 1.88N. •••12 x x 172 CHAPTER 5 Two horizontal forc θ1 ••5 Three astronauts, propelled x slides over frictionless ice, o θ3 by jet backpacks, push and guide a F2 which an xy coordinate syste (b)kg Theasteroid y component of the force is : 120 toward a processing is laid out. Force F1 is in t F3 dock, exerting the forces shown in 2 171 Fy F ma ma sin 20.0 1.0kg 2.00m/s sin 20.0 0.684N.positive direction of the x ax y Fig. 5-29, with ! 32 N, F ! 55 N, 1 2 and has a magnitude of 7.0 : F3 ! 41 N, u1 ! 30°, and u3 ! 60°. Fig. 5-29 Problem 5. Force F2 has a magnitude of 9 (c) Inisunit-vector notation, the force(a) vector is What the asteroid’s acceleration N. Figure 5-32 gives the x com in unit-vector notation and as (b) a ponent vx of the velocity of t ˆ F ˆj (1.88 N)i ˆ ˆ F Frelative Alex (0.684 N) j . magnitude and (c) a direction xi y disk as a function of time t du Charles to the positive direction of the x axis? ing the sliding. What is the a 4. Since v = constant, we have a = 0, which implies gle between the constant dire ••6 In a two-dimensional tug-of: : tions of forces F1 and F2? war, Alex, Betty, and Charles Fpull F F ma 0 . net 1 2 horizontally on an automobile tire sec. 5-7 Some Particular F atThus, the angles shown in the overhead the other force must be •13 Figure 5-33 shows an arr view of Fig. 5-30. The tire remains 20.0° to the positive direction of an x axis, what are (a) the x com172 (b) the y component of the net force acting on the ponent and body, and (c) what is the net force in unit-vector notation? (b) The component ofit, thea force ••4 While twoyforces act on par- is y ticle is to move at the constant velocFy ma y ma sin 20.0 1.0kg 2.00m/s 2 sin 20.0 : ity v ! (3 m/s)î " (4 m/s)ĵ. One of : the forces is F1 ! (2 N)î # ("6 N)ĵ. (c) In unit-vector notation, the force vector is What is the other force? F 1 particle at t5! 3.4 CHAPTER ••11 A 2.0 kg p variable force di 3.0 m # (4.0 m/s seconds. The fact 0.684N. ticle has a magni axis. What is c? •••12 Two h slides over frict which an xy coo is laid out. Forc positive directio and has a magn : Force F2 has a m N. Figure 5-32 g ponent vx of the disk as a functio the vector 5. The net force applied on the chopping block is Fnet F1 F2 F3 , where ing the sliding. W is done using unit-vector given bythe gleisbetween ••6 In addition a two-dimensional tug-of- notation. The acceleration of the block : a Betty, F1 F2andF3Charles / m. tions of forces F1 war, Alex, pull horizontally on an automobile tire sec. 5-7 Some at the angles shown the overhead (a) The forcesin exerted by the three astronauts can be expressed in unit-vector notation as •13 Figure 5-33 view of follows: Fig. 5-30. The tire remains Betty 137° pended by cords stationary in spite of the three pulls. θ1 N)ˆj . F Fx ˆi Fy ˆj (1.88 N)iˆ (0.684 ••5 Three astronauts, propelled x θ3 by jet backpacks, push and guide a F2 = constant, we have a = 0, which implies 4. Since vtoward 120 kg asteroid a processing F3 dock, exerting the forces shown in Fig. 5-29, with F1 ! 32 N, F2 ! 55 N,Fnet F1 F2 ma 0 . F3 ! 41 N, u1 ! 30°, and u3 ! 60°. Fig. 5-29 Problem 5. Thus, the other force must be What is the asteroid’s acceleration (a) in unit-vector notation and as (b) a F2 F1 ( 2 N) ˆi ( 6 N) ˆj . Alex magnitude and (c) a direction relative Charles to the positive direction of the x axis? d i en that occuis 8.00 m/s2 heir humanom a cannon ersion of the els to land in non and at a or 5.2 m and e underwent magnitude of ough it were onnected by by the cable With what speed does the man hit the ground if he started from rest? ••53 In Fig. 5-48, three connected blocks are pulled to the right on a horizontal frictionless table by a force of magnitude T3 ! 65.0 N. If m1 ! 12.0 kg, m2 ! 24.0 kg, and m3 ! 31.0 kg, calculate (a) the magnitude of the system’s acceleration, (b) the tension T1, and (c) the tension T2. m1 T1 m2 Fig. 5-48 '((")*(+,$*-."/%0%"!! T2 m3 Problem 53. T3 v 2(1.3 m/s 2 )(10 m) 2a y 5.1 m/s. 53. We apply Newton’s second law first to the three blocks as a single system and then to the individual blocks. The +x direction is to the right in Fig. 5-48. (a) With msys = m1 + m2 + m3 = 67.0 kg, we apply Eq. 5-2 to the x motion of the system, in which case, there is only one force T3 T3 i . Therefore, T3 msys a 65.0 N (67.0 kg)a which yields a = 0.970 m/s2 for the system (and for each of the blocks individually). (b) Applying Eq. 5-2 to block 1, we find T1 m1a 12.0kg 0.970m/s 2 11.6N. (c) In order to find T2, we can either analyze the forces on block 3 or we can treat blocks 1 and 2 as a system and examine its forces. We choose the latter. T2 m1 m2 a 12.0 kg 24.0 kg 0.970 m/s 2 34.9 N . 54. First, we consider all the penguins (1 through 4, counting left to right) as one system, to which we apply Newton’s second law: