PHYS 218sec. 517-520ReviewChap. 5Applying Newton’s lawsThe purpose of this chapter is to make you be familiar with application of Newton’s laws of motion.Therefore, this chapter deals with various examples and problems. There are lots of different examples in end-of-chapter problems and you should solve as many problems as possible. Several important examples, such as the Atwood’s machine, are not covered by the text but are covered by end-of-chapter problems. Pay special attention to these problems.First draw all forces acting on the system. Then draw free body diagrams for each object. Decompose forces into x and y directions, if needed. The sum of these forces should give ma according to Newton’s second law.Particles in Equilibrium=∑F0Ex 5.1We are neglecting the mass of the rope. on on on on on on on on For Gymnast, For Rope, ( action-reaction pair)Therefor0e,0RG GCR GRRG GRRG G CR GRFT mgFT TTTTmgTT= −= −=== ===∑∑∵Since this system is in equilibrium.Ex 5.2Same as Ex 5.1, but with a massive ropeon RGTGymnastRopeGymnast + Ropeas a composite bodyon GRTGwon CRTon CRTGRww+RwAction-reaction pair on on on on on on on on on For Gymnast, For Rope, ( action-reaction pair)Therefor()0e0,RG GCR GR RRG GRRG GR GCR GR R R GFT mgFT T mgTTTTmgTTmgmmg= −= −−======+=+∑∑∵ on on on You can have the same answer by considering the gymnast and rope as one composite body() (),which gives(0)CR G R CR G RCR G RFT w w T m mgTmmg= − += − +=+=∑Ex 5.3Two-dimensional equilibriumθEngine Ring OOxyw1T1T2T3Tθθ3cosT θ3sinT θ132 31Engine: Ring: cos , sinyxyFTmgFT T FT Tθθ= −= − = −∑∑∑132 31113 2300This system is in equilibrium, so . This givesEngine: Ring: cos , sinBy solving the coupled equations, we get,,coscossin si0nsinyxyFTmgFT T FT TTmg mgTmgT TTθθθθθθ θ=== −= − = −=== ====∑∑∑∑F0.tanmgθ=Ex 5.4An inclined planeα90 α−ααTnmgcosmgαsinmg αxy-direction: sin-direction: cosso that we havesin ,0cos0xyxFTmgyFnmgTmg nmgαααα= −= −====∑∑Tension over a frictionless pulleyEx 5.5θ1m2mxy2for myx1for mTT2mg1mg1cosmg θ1sinmg θθCoordinate systems may differ for m1and for m2n221121For ,For , sinBy solving the above equations, 00sinyxmFTmgmFTmgmmθθ====− =−∑∑Dynamics of Particlesm=∑FaEx 5.8Tension in an elevator cable: Obtain the tension when the elevator slows to a stop with constant acceleration in a distance of d. Its initial velocity is v0.Moving downwardwith decreasingspeedxyTwMg=ya()()22 200000By Newton's second law,.Now you need to know , which can be obtained from the distance and velocities:(0)2:initial position, :final position, :yyyyy yyyyFTMgTMgaavv vadyyyy dyyvMa= −⇒ =+−==== − >−∑∵20final velocity (0)yvTMgd=⎛⎞⎟⎜⎟⎜∴ =+⎟⎜⎟⎟⎜⎝⎠Ex 5.9Apparent weight in an accelerating elevator: A woman is standing on a scale while riding the elevator in Ex 5.8. What’s the reading on the scale?xynwmg=yaFree body diagram forwomanReading on the scale = force exerted by the woman on the scale= normal force exerted by the scale on the woman (by Newton's 3rd law) The question is asking the magnitude of the normal force .n∴()yyyF manmgnmga= −∴ =+=∑Apparent weight, if 0 (i.e., accelerating upward), if 0 (i.e., accelerating downward)yynw anw a>><<extreme case: when (i.e., free falling)0yagn= −=Apparent weightlessAcceleration down a hill: What’s the acceleration?Ex 5.10αyxαwmg=nsinw αcosw αsincos (no motion in direction)This givess, cos0inxyxxymFwFnw yag nmamagαααα======−=∑∑Two bodies with the same magnitude of acceleration: Find the acceleration of each body and the tension in the stringEx 5.121m2mfrictionlessMassless, inelastic stringxy+n1mg1xaTxT2 ya2mgy+Coordinate systems may differ for m1and for m2The two masses are connected, so theiraccelerations are equal12xyaaa==1111121111220For ,For ,xy yxyyma mammFTFnmgmF aT mgaamm== −====−===∑∑∑m1has no motion in its y-directionEx 5.12 (cont’d)The equation of motion for m1in y-direction gives no information on a or T1122 221212 12From , ,From ,two equations and two unknowns ( , ),mT mammgT maaTmmmagTgmm mm=− =⇒ ==++The final answers should be written in terms of quantities given in the problem, i.e., m1, m2and g in this caseFriction forcesKinetic friction• The kind of friction that acts when a body slides over a surface• Contact force (other contact force is normal force)• Friction and normal forces are always perpendicular to each other• Microscopically, these forces arise from interactions between molecules.Therefore, frictionless surface is an idealization. • Direction of frictions force: always to oppose relative motion of the two surfaces• Empirically, it is known that the magnitude of friction force is proportional tothe magnitude of normal force,fr(relation between magnitudes)fn∝, :coefficient of kinetic frictionkk kfnµµ=This symbol means ‘proportional to’.Depends on many conditions such as velocity, etcStatic friction• The kind of friction that acts when there is no relative motion.See Fig. 5.19 in p.151 of the textbook.• Static friction is always less than or equal to its maximum value.(because there is a critical value where the object starts to move)(E.g, if there is no applied force, the static friction is zero.)Rolling frictionThe friction force when an object is rolling on a surface,max, :coefficient of static frictionss s sff nµµ≤ =, :coefficient of rolling friction or tractive resistancerr rfnµµ=Minimizing the kinetic friction: What is the force to keep the crate moving with constant velocity?Ex 5.15θnwTkf()cos 0 cos ( )sin 0 sincos sincos sinxk kkkykkkFTfTnfnFT nw nwTwTwTTθθµµθθµθ µ θθ µ θ= − = ⇒ ===+− = ⇒ = −∴ = −⇒=+∑∑∵The value of θwhich givesthe minimum value of Twhen µk=11cos sinxx+Ex 5.16 & 5.17 Toboggan ride with frictionαyxαwmg=nsinw αcosw αkf()sincosThe second equation givescos , and cos .Then from the first equation,sin cos sin cos .0xkykkxkxk kmFmg fFnmgnmg f n mgaggagααα µµ αα µ ααµ α= −= −==== − = −==∑∑Some special casesIf the hill is vertical 90 If there is no friction 0 sin If the object is moving with a constant speed 0 (Ex 5.10) sin cos tan (Ex xkxxkkagagaαµ αα µ α µ α⇒ = ⇒ =⇒ = ⇒ =⇒ =⇒ = ⇒ =5.16)Fluid resistance & terminal
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