Slide 1 / 139 Slide 2 / 139 P Physics C - Mechnics Dynmics - pplictions of Newtons Lws 2015-12-03 www.njctl.org Tble of Contents Slide 3 / 139 Click on the topic to go to tht section Introduction Sliding locks Fixed xis Pulley Suspended Pulleys Plumb bob in cr The nked Curve Inclined Plne nd Pulley Flling objects with ir Resistnce Non Uniform Circulr Motion
Slide 4 / 139 Introduction Return to Tble of Contents Introduction Slide 5 / 139 This is not typicl chpter presenttion. There re no multiple choice or free response questions to test student knowledge. Rther, eight comprehensive Dynmics questions re solved, step by step. Questions re built into the mteril - you should try to nswer them before you move onto the next slide. These cn be done in clss, led by the techer, or they cn be done by the students outside of clss. Slide 6 / 139 Sliding locks Return to Tble of Contents
Sliding locks Slide 7 / 139 Ground lock, of mss 36 kg, is bout to be pulled cross the ground by rope with force of F pp. The coefficient of kinetic friction between block nd the ground is 0.27. lock, of mss 42 kg is resting on lock with coefficient of sttic friction of 0.71 between the two blocks. Sliding locks Slide 8 / 139 Ground We're going to use dynmics concepts to find the mximum force tht my be pplied by the rope before block strts sliding on block. Sliding locks Slide 9 / 139 Ground Once someone strts pulling on the rope, wht do we expect will hppen?
Sliding locks Slide 10 / 139 Ground lock will strt ccelerting to the right, s it now s net pplied force to the right. The friction between the ground nd block will oppose the pplied force, so tht will reduce the ccelertion. Wht bout block? Sliding locks Slide 11 / 139 Ground The most importnt thing to first notice bout block is tht the pplied force is NOT cting on block, nd when the free body digrms re drwn, this force will not be shown on block! Intuitively, we know tht something will be cting in the x direction on block. lock will either sty right where it is, reltive to block, nd move to the right, or it will strt sliding to the left. Reltive to the ground, it will be moving to the right in both cses. If not the force due to the rope, wht is this force? Sliding locks Slide 12 / 139 Ground The sttic friction force. Let's tke two limiting cses to explin this. If μ s = 0, then tht implies frictionless surfce between blocks nd. When is pulled to the right, block would just sty where it is reltive to the ground, nd would slide to the left on block. If μ s were infinitely lrge, then it would just stick to block nd slide with it to the right.
Sliding locks Slide 13 / 139 Ground Let's formlly strt the problem: Given: where μ s is between blocks nd nd μ k is between block nd the ground. Find: the mximum F pp before lock strts sliding on lock. Sliding locks Slide 14 / 139 Ground s in ll dynmics problems, the first step is to drw the free body digrm. In this cse, there re two objects of interest, blocks nd, so there should be two free body digrms (FD). ut, they will not be the ones you'd expect! Here's clue - there will not be FD for block nd FD for block. We will define different systems - two of them. nd drw FDs for ech one. Tke few minutes with your group nd propose two systems. Sliding locks Slide 15 / 139 Ground System 1 will be lock. System 2 will be locks nd. Given this strting point, try drwing free body digrms for both systems. Remember - when working with system, only the externl forces on the system re shown. System 1 System 2
Sliding locks Slide 16 / 139 System 1 System 2 Here they re. Don't worry if you didn't get these exctly right. Ech system will be now be nlyzed in detil. System 1 System 2 f s f k (+) F pp m g (m +m )g Sliding locks Slide 17 / 139 Let's look t System 1 first. The Norml force nd the force due to grvity re pretty stright forwrd. Grvity pulls down on block (m g), nd block exerts n upwrds Norml force () on block. Note the dditionl subscript on the Norml force. Tht is to distinguish it from the Norml force cting on System 2. m g f s ut wht of the sttic friction force? Why is it pointed to the right? Doesn't the sttic friction force oppose motion? Isn't block moving to the right? Sliding locks Slide 18 / 139 Ground f s m g Sttic friction seeks to mintin the reltive position of the objects tht re initilly t rest. In most cses, this mens if force is pplied to n object in one direction, then the sttic friction force is in the opposite direction. ut, here, the pplied force is not on block, it is on block. nd for block to mintin its position reltive to block, s block strts moving to the right, the sttic friction force cts to the right. This results in block stying on the right edge of block - their reltive positions don't chnge.
Sliding locks Slide 19 / 139 Ground f s m g nother wy to look t this problem is using your intuition gin. gin, tht doesn't lwys work in physics, but it helps here. If you pull block, your experience tells you tht block will move to the right. Mybe not s fst s block is ccelerting, but it will move to the right. Tht requires Force (Newton's Second Lw). The only force vilble in the x direction is the sttic friction force between block nd. So, f s goes on the free body digrm pointing to the right! Sliding locks Slide 20 / 139 Time for System 2 now. Grvity is cting on the combintion of blocks nd, so it is equl to (m + m )g. The ground exerts n upwrd Norml force on the blocks, nd is represented s (+). Note the dditionl subscript on the Norml force. Tht is to distinguish it from the Norml force in System 1. f k (+) F pp (m +m )g Next, we'll look t the forces in the x direction. Sliding locks Slide 21 / 139 F pp is cting to the right, nd f k cts between the ground nd block nd is directed to the left. The sttic friction force between blocks nd is not drwn here. Why? f k (+) F pp (m +m )g
Sliding locks Slide 22 / 139 System 2 ws defined s the combintion of blocks nd. Therefore, the sttic friction force between blocks nd is n internl force. (+) Internl forces re not shown on FDs (Discuss in your groups how Newton's Third Lw comes into ply here). f k F pp (m +m )g Sliding locks Slide 23 / 139 System 1 f s Time to pply Newton's Second Lw to the FDs. We'll strt with System 1. m g x direction y direction Took little shortcut on the nottion - since the objects re moving in the x direction, the subscript, x, will be omitted on the ccelertion vector. Slide 24 / 139
Slide 25 / 139 Sliding locks Slide 26 / 139 FN FN(+) fs fk Fpp mg System 1 System 2 (m+m)g System 2 equtions: This gives us n expression for the ccelertion of the totl system (blocks nd ). If we wnt to find the point t which block strts to slide, wht ssumption cn we mke bout the ccelertion of block? Sliding locks Slide 27 / 139 FN FN(+) fs fk Fpp mg System 1 System 2 (m+m)g Switching to the System 1 equtions, for block not to slide, its ccelertion must be equl to the ccelertion of the totl system, so = +. Why? If is greter thn +, then block would slide forwrd on block. If is less thn + then it would slide bckwrds. So, since we're looking for the exct point where the pplied force cuses to slide, we hve:
Sliding locks Slide 28 / 139 FN FN(+) fs fk Fpp mg System 1 System 2 (m+m)g Clculting f s from the Norml force: Putting these two eqution together nd cnceling m : Slide 29 / 139 Slide 30 / 139
Slide 31 / 139 Slide 32 / 139 Fixed xis Pulley Return to Tble of Contents Fixed xis Pulley Slide 33 / 139 So fr, pulleys hve been used in constructions like twood's Mchine nd to llow for the horizontl nd verticl motion of two objects in system (the force is redirected) s shown below. oth of these systems used "fixed xis pulleys" s the pulley's xis of rottion does not chnge reltive to reference frme ttched to the ground or tble or ceiling.
Fixed xis Pulley Slide 34 / 139 ssume mssless pully nd string, nd tht there is no friction between the pulley nd its xle nd the string. This llows us to ignore the effects of rottionl motion in the pulley. In both cses below, the tension force on either side of the pulley is the sme. nd, for the twood's mchine - if the block on the left weighed 100 N, you would need force on the right side of 100 N to keep it sttionry. Fixed xis Pulley Slide 35 / 139 Let's use fixed xis pulley, but try something different. In the below picture, two window cleners re sitting in hrnesses. The picture doesn't show it, but the two lines from ech hrness loop over fixed pulley on top of the building. ssuming the window clener on the left hs mss of 82.0 kg nd the hrness is 21.0 kg wht force does he need to exert to sty sttionry? (Of course, there re sfety fetures ssocited with the hrness, but to simplify the problem, we'll ssume tht the clener needs to exert force to sty where he is). Ie:Settle_Public_Librry_window_wshers_06.jpg Fixed xis Pulley Slide 36 / 139 s in ll dynmics problems, we'll strt with free body digrms. Wht forces re cting on the window clener nd the hrness? The rope tht he is holding with his hnds is pulling him up. The hrness set is pushing him up with Norml force (nd the window clener is pushing down on the hrness with norml force). The rope tht is ttched to the hrness is pulling up on the hrness nd the window clener. Grvity is pulling down on the window clener nd the hrness. Now, drw FDs for the window clener, the hrness nd the combined system of the two.
Fixed xis Pulley Slide 37 / 139 In order from left to right, the FDs for: hrness, window clener, combined system of hrness plus window clener. m hg m wcg m wc+hg The Norml forces between the window clener nd the hrness cncel out in the combined system - they re internl forces nd do not ffect the motion of the system. The tension force is the force the window clener needs to exert to sty motionless - wht do you notice bout this force? Fixed xis Pulley Slide 38 / 139 There re two Tension forces cting on the system! The window clener is holding on to the right rope with force tht, by Newton's Third Lw, is blnced by the Tension force in the rope. The two Tension forces re equl to ech other, since the rope is ssumed to be mssless nd is not stretching or compressing - so the force tht the window clener is pulling down on the right trnsmits itself throughout the rope. Wht's the next step? Think Newton. m wc+hg Fixed xis Pulley Slide 39 / 139 pply Newton's Second Lw to the system. m wc+hg Pretty strightforwrd. ut wht's very interesting bout this result? Wht's the weight of the window clener nd the hrness?
Fixed xis Pulley Slide 40 / 139 The weight of the system is 1009 N (W = ). ut, becuse of the pulley configurtion, the window clener only needs to exert hlf of the force to overcome the grvittionl force. Why? Doesn't this look like violtion of the conservtion of energy? The work required to lift n object distnce Δy is: ut with this configurtion, the window clener is exerting hlf of the required force - F/2. How is the window clener still rising distnce of Δy? Think bout how fr the rope is trvelling, d (s tht is the distnce the force is cting over). Slide 41 / 139 Fixed xis Pulley Slide 42 / 139 Now tht we've shown tht the reduced force obtined by this rrngement requires tht it be exerted over greter distnce (nd energy is conserved), let's move on to nother question: m wc+hg The window clener pulls down on the rope, nd long with the hrness, he ccelertes upwrd t rte of 0.330 m/s 2. Wht force must he pull with, nd wht is the force tht he now exerts on the hrness?
Fixed xis Pulley Slide 43 / 139 Let's strt with the system FD to find the Tension in the rope (which is equl to the window clener's pulling force) nd then pick either of the individul FDs to clculte the Norml force (the force tht the window clener exerts on the hrness). m hg m wcg m wc+hg Slide 44 / 139 Fixed xis Pulley Slide 45 / 139 We'll now solve both the hrness nd the window clener FD to show we get the sme result for the norml force. First, the hrness system. FN FT mhg FN FT mwcg FT FT mwc+hg
Fixed xis Pulley Slide 46 / 139 Now, the window clener system. FT FN FT FT FT FN mhg mwcg mwc+hg Tht's s we expected, = 309 N s clculted for both systems. Fixed xis Pulley Slide 47 / 139 Where the pulley merely redirected the force, there ws no benefit, in terms of reduced force required, obtined. The benefit ws tht the person lifting the object could stnd in the best plce to move the object - he didn't hve to stnd under it. ut, n dvntge ws gined in the second cse - the force required to lift the object ws reduced - but hd to be exerted over greter distnce. We're now going to move on to the cse of the suspended pulley where the pulley's xis is not fixed nd see if it cn reduce the lifter's required force even more. Slide 48 / 139 Suspended Pulleys Return to Tble of Contents
Suspended Pulleys Slide 49 / 139 First, let's define wht Suspended Pulley is! The fixed xis pulley's xis of rottion (xle) did not chnge reltive to the ground or other sttionry reference frme. In other words, the pulley didn't move. The rope moved, the lod moved, but not the pulley. So, wht could suspended pully do? Suspended Pulleys Slide 50 / 139 It cn move. It will move up or down nd tke the lod with it. There re two systems of multiple pulleys here. Picture crte ttched to the ring on the bottom of the bottom pulley. Which is the fixed pulley? Which is the supended pulley? Suspended Pulleys Slide 51 / 139 It cn move. It will move up or down nd tke the lod with it. The top system is ttched to the boom overhed nd is fixed pulley. The lower one is the one with suspended pulleys.
Suspended Pulleys Slide 52 / 139 elow is schemtic of fixed nd suspended pulleys tht will be used to lift the mss t the bottom by pplying the force to the right. C F pp Identify the fixed nd the suspended pulleys: M Pulleys nd C re fixed. Pully is suspended, s it will trvel with the mss. Suspended Pulleys Slide 53 / 139 The problem t hnd is to find the required pplied force to lift the mss M. C M F pp This my look tricky, but we'll be using free body digrms to figure out the required force. First, lbel this picture with ll the forces you cn think of. Remember, we're deling with mssless pulleys nd strings. Suspended Pulleys Slide 54 / 139 6 2 5 C 7 3 M 4 1 F pp Here they re - the lines do not represent the nitudes of the vrious forces - they're just schemtic digrm - we don't hve free body digrm yet. The next slide will show the FDs. Give them try before the slide chnges. Mg
Suspended Pulleys Slide 55 / 139 5 C 6 7 3 M 4 Mg 2 1 F pp C 5 Here they re! Next, Newton's Second nd Third Lws nd the mssless string nd frictionless pulley ssumptions will be used to relte the forces to ech other. 6 7 3 M 4 Mg 3 5 7 4 2 1 Suspended Pulleys Slide 56 / 139 C 6 7 2 1 The mssless string nd frictionless pulley ssumptions llow us to sy: F pp = 1 = 3 = 5 = 7 5 3 F pp Why? 4 M Mg ecuse the string is not stretching or compressing, it hs no mss to consider in the FD, nd the pulleys hve no mss, nor do they use ny energy in their rottion. So, the tension is the sme everywhere in the string. Suspended Pulleys Slide 57 / 139 C 5 6 7 3 3 5 7 4 2 1 ssume the mss is sttionry - it is not ccelerting. Write Newton's Second Lw equtions for ech FD. Note how Newton's Second Lw tells us which ction-rection forces re equl. M 4 Mg The equtions re on the next slide. Try them yourself first.
Suspended Pulleys Slide 58 / 139 6 C 6-5 - 7 = m y = 0 5 7 5 7 3 2 1 2-3 - 1 = m y = 0 3 3 + 5 + 7-4 = m y = 0 4 M 4 Mg 4 - Mg= m y = 0 6 Suspended Pulleys 2 We ctully only need three of these equtions: Slide 59 / 139 5 C 7 3 1 F pp F pp = 1 = 3 = 5 = 7 4 - Mg = m y = 0 3 + 5 + 7-4 = m y = 0 Combine the first nd third to get: 4 4 = 3F pp M Substitute this into the second eqution: Mg 3F pp = Mg Suspended Pulleys Slide 60 / 139 6 2 C 1 One more step: 5 7 3 4 F pp F pp = Mg/3 Thus it only tkes one third of the force to hold mss M sttionry thn if it ws being held up just by rope. M Mg
Slide 61 / 139 Plumb bob in cr Return to Tble of Contents Plumb bob in cr Slide 62 / 139 First, we need to describe Plumb bob. It is weight (normlly pointed t the bottom) tht is ttched to string tht is then ffixed to horizontl support. Wht would be the purpose of such n object? free body digrm (FD) is gret wy to figure tht out. Wht re the forces cting on the Plumb bob? Plumb bob in cr Slide 63 / 139 The only forces cting on the weight (bob) re the grvittionl force,, nd the Tension in the string,. oth forces ct in the y direction, nd there re no forces in the x direction. So, the point of the bob points down, in perfectly verticl direction, which gives perfect reference line. This is helpful to people who wnt to build house tht is stright or ny other constructions tht need to be ligned with the y xis (wllpper in room, skyscrper, etc.).
Plumb bob in cr Slide 64 / 139 ssume the bob is not moving, nd is redy to be used s reference line. Let's pply Newton's Second Lw. Plumb bob in cr Slide 65 / 139 For physics purposes, let's now ttch this plumb bob to the overhed light in cr nd ccelerte the cr forwrd in sfe mnner. Will the plumb bob sty pointing in verticl direction? Think of wht hppens to n object on the center console of cr when the cr ccelertes forwrd. Plumb bob in cr Slide 66 / 139 From the driver's perspective, n object on center console will slide bckwrds. Since the plumb bob is ttched to the overhed light, it will swing bckwrds, s seen by the driver. n inertil observer, stnding outside on the rod, will see the plumb bob swing bckwrds. The problem will now be solved from the inertil observer's point reference frme, where the cr is ccelerting. Wht does the FD look like now?
Plumb bob in cr Slide 67 / 139 The grvittionl force is still the sme. hs x nd y components. The bob is no longer pointed verticlly down. Wht is the source of this force in the x direction? When the cr ws not moving, the bob pointed stright down. Now tht the cr is ccelerting to the left, the bob hs moved. So, the force is relted to the cr's ccelertion nd the originl Tension force. Since there re now non perpendiculr forces in two dimensions, it's time for vector resolution. Plumb bob in cr Slide 68 / 139 y x We're going to find nother use for the plumb bob - once the forces re resolved, the ccelertion of the cr will be found without using ny other cr guges! The ngle, which is the ngle between the verticl nd the string of the Plumb bob is mesured. Wht re the x nd y components of the Tension? ssume the ccelertion is constnt, so the bob is sttionry in this new position. Plumb bob in cr Slide 69 / 139 y direction F F T Ty x x direction Let's stck these two equtions on top of ech other: Wht cn you sy bout these two equtions, nd does nything bout the trigonometry jump out t you?
Plumb bob in cr Slide 70 / 139 y x These equtions define the motion of the plumb bob in two dimensions, t the sme point of time, so they must both be true. Hence, they re simultneous equtions. Mny times, you dd/subtrct simultneous equtions to get the solution. ut, in this cse - wht bout dividing? This will consolidte the trigonometric functions, nd eliminte the mss of the bob from the solution. Plumb bob in cr Slide 71 / 139 y x Whenever you see sine nd cosine function - think of dividing the equtions. This is very hndy trick. Plumb bob in cr Slide 72 / 139 y x Just by mesuring the ngle tht the string mkes with the verticl, the ccelertion of the cr cn be clculted. Note how the mss of the plumb bob does not mtter. If = 11 0, wht is the ccelertion of the cr?
Plumb bob in cr Slide 73 / 139 y x This is bout s fst s four cylinder compct cr cn ccelerte when it is trying s hrd s it cn. Hopefully, you'll never push the cr to this ccelertion, so if you hve plumb bob ttched to the interior overhed light, it should never exceed n ngle of 11 0. Slide 74 / 139 The nked Curve Return to Tble of Contents The nked Curve Slide 75 / 139 n interesting problem tht involves centripetl ccelertion, non rotted coordinte system tht looks like it should be rotted nd sttic nd kinetic friction, is the cse of cr on bnked, curved rodwy. http://commons.wikimedi.org/wiki/file: 1968_MC_mbssdor_DPL_sttion_w gon_fl-r1.jpg
The nked Curve Slide 76 / 139 First thing tht we'll do is to prepre for the FD. The photogrph of the 1968 sttion wgon (sk your prents wht those were) will be replced by shpe. The cr hs just entered turn - which cn be modeled s circulr orbit when seen from bove. It is lso on bnked curve - nd the bnk mkes n ngle with the flt ground. Cr mking turn s seen from bove v Cr seen from behind on bnked curve Drw the cclertion vector on ech sketch. The nked Curve Slide 77 / 139 Cr mking turn s seen from bove v Cr seen from behind on bnked curve First, ssume there is no sttic friction between the cr nd the rod nd drw the FD for the sketch on the right. This is done not just to mke the mth esier, but it ctully is the most conservtive cse to ensure the cr does not slide up or down the incline (both cses involve leving the rodwy, which is clerly not sfe). The nked Curve Slide 78 / 139 y x The first surprise, is even though we're deling with n inclined plne, we're not going to rotte the coordinte system! Plese discuss this within your group nd suggest n explntion. lso, wht kind of ccelertion is represented by the vector,, bove?
The nked Curve Slide 79 / 139 y x Certinly, one of the dvntges of rotted coordinte system is to reduce the mount of vectors tht need to be resolved into component prts. ut, even more importnt, is to line up the coordinte system so the ccelertion is long one of its xes. nd tht's wht's done bove. Since the cr is going in circle - it's Centripetl ccelertion. Wht force is cusing the centripetl ccelertion? The nked Curve FNsin y FNcos Slide 80 / 139 x y resolving the Norml force into its x nd y components, it cn be seen tht Nsin points towrds the center of the circle tht the cr is driving in, nd is cusing the centripetl ccelertion. Tke minute to verify tht the ngle in the tringle mde by the Norml force is the sme s the ngle of inclintion of the bnked rod. Next step? Newton's Second Lw. In the x nd the y directions. The nked Curve Slide 81 / 139 First we'll write Newton's Second Lw for the y direction, nd ssume the cr stys on the rod surfce nd isn't bouncing up nd down so y = 0. FNsin y FNcos x Why ws the y direction solved first?
The nked Curve ecuse we'll need the vlue of for Newton's Second Lw in the x direction. FNsin y FNcos x Slide 82 / 139 Wht is the significnce of this result? nd wht property of the cr is missing? Slide 83 / 139 The nked Curve Slide 84 / 139 Now, let's bring in friction. When cr negotites turn, it's best tht it stys in the middle of the rod. If it moves up the rmp - well, then it cn fil to mke the turn nd it will go flying off where it could hit trees or other obstructions or be stopped by brrier of some sort. If it moves down the rmp, then it could hit whtever problems wit it on the inner curve - such s brriers, prked crs or the like. Wht type of friction is relevnt here?
The nked Curve Slide 85 / 139 Sttic Friction. This is covered in the Rottionl Motion unit of this course. ut here's quick explntion. tire rottes s it moves the cr forwrd. When the tire comes in contct with the rod, the liner velocity of the tire is pointed towrds the rer of the cr. It is exctly equl to the liner velocity of the entire cr. These velocities dd to zero - so the tire is t rest with the rodwy t the contct point. Hence - no reltive motion - sttic friction! The nked Curve Slide 86 / 139 First, find the mximum speed tht cr cn trvel t without sliding up the bnk s it goes round bnked curve. fter this is worked out, the problem of wht minimum speed cr will need to prevent it from slipping down the bnk will be presented. Let's build on the previous work tht left out friction. Here it is: Is this still relevnt? Does nything need to be dded or subtrcted? FNsin y FNcos x The nked Curve Slide 87 / 139 We don't need to subtrct nything. The ccelertion vector still points towrds the center of the curve. The non rotted coordinte system will still work. The Norml vector is still vlid. ut, the frictionl force needs to be dded. The cr needs to be prevented from sliding up the bnk. Wht direction is the force of friction? FNsin y FNcos x
The nked Curve Slide 88 / 139 Sttic friction works to mintin the reltive position of the two objects. So, to prevent the cr from moving up the bnk, the friction force must point down the bnk s shown below. Wht ngle does the sttic friction force mke with the negtive x xis? Once you figure this out, resolve the force into its x nd y components. sin y f sf FNcos x The nked Curve Slide 89 / 139 The ngle is. Here's the completed FD. Next, work out Newton's Second Lw equtions, strting, s lwys, with the y direction. sin y f sf cos f sf sin f sf cos x sin y FNcos f sfcos f x sfsin f sf The nked Curve y direction: ssume the cr does not bounce up nd down, nd f sf = μ sf. The inequlity is removed here, since the mximum vlue of the sttic friction force is the point t which the cr will strt slipping. Slide 90 / 139
FNsin y The nked Curve Slide 91 / 139 FNcos f sfcos f x sfsin f sf x direction: work out Newton's Second Lw, nd then substitute in the vlue of obtined from the y direction eqution. The nked Curve Slide 92 / 139 Now, let's consider the minimum speed the cr needs to trvel t to prevent it from sliding down the bnk nd running off the rod there. s in the previous cse, its the force of sttic friction. ut, which wy is the sttic friction force this time? FNsin y FNcos x The nked Curve Slide 93 / 139 Sttic friction works to mintin the reltive position of the two objects. So, to prevent the cr from moving down the bnk, the friction force must point up the bnk s shown below. It' now time to resolve the friction force into its x nd y components. sin y cos f sf x
The nked Curve Slide 94 / 139 Here's the completed FD. Next, work out Newton's Second Lw equtions, strting, s lwys, with the y direction. sin y cos Cn you see why the ngle between f sf nd the x xis is? f sf f sf sin f sf cos x The nked Curve Slide 95 / 139 FNsin y FNcos f sf f sfsin f sfcos x y direction: ssume the cr does not bounce up nd down, nd f sf = μ sf. The inequlity is removed here, since the mximum vlue of the sttic friction force is the point t which the cr will strt slipping. The nked Curve Slide 96 / 139 FNsin y FNcos f sf f sfsin f sfcos x x direction: work out Newton's Second Lw, nd then substitute in the vlue of obtined from the y direction eqution.
Slide 97 / 139 Inclined Plne nd Pulley Return to Tble of Contents Inclined Plne nd Pulley Slide 98 / 139 Let's put together two concepts tht hve been covered in the Dynmics Chpter notebook (inclined plne) nd in this pplictions of Newton's Lws notebook (the pulley). Inclined Plne nd Pulley Slide 99 / 139 Find the ccelertion of the two msses on the incline, where there is kinetic friction between the box nd the incline. ssume the two msses re connected by string of negligible mss nd does not stretch or compress. lso ssume frictionless nd mssless pulley, so tht rottionl effects need not be considered.
Inclined Plne nd Pulley Slide 100 / 139 Drw ll the forces you cn think of on the two boxes nd show the ccelertion vectors for ech box. Wht if you choose the wrong direction for the ccelertion? Inclined Plne nd Pulley Slide 101 / 139 m 2g f k Next step - drw the free body digrms for ech box. m 1g Here it is. nd if you chose different direction for the ccelertion thn the ctul ccelertion - no problem - you will get negtive vlue for the ccelertion fter the equtions re solved, which tells you it's opposite to your choice. Inclined Plne nd Pulley System 1 Slide 102 / 139 System 2 There is lot going on here. The next few slides will explin the free body digrms in more detil. Ech mss will be treted s its own system, nd then it will be shown how they cn be considered s one system of two msses. F y=m 2gcos f k F x=m 2gsin
Inclined Plne nd Pulley Slide 103 / 139 System 1 System 1 is the m 1 block. We're ssuming tht the block is flling, hence the direction of is down. Once we solve the Newton's Second Lw equtions, if we get negtive vlue for, tht just mens it's ctully ccelerting in the opposite direction (up). This one is firly strightforwrd - grvity is pulling down on the mss nd the tension in the string is pulling it up. F y=m 2gcos F x=m 2gsin Inclined Plne nd Pulley f k System 2 This system is bit more complex. It involves rotted coordinte system nd the resolution of the grvittionl force long the new rotted xes. For review of this lgebr, plese refer to the Dynmics Chpter Notebook presenttion on the Inclined Plne. ccelertion is chosen in the up direction on the plne. While it isn't importnt which direction is chosen, it is importnt to choose the directions for both boxes tht re consistent! If box m 1 is ccelerting downwrds, then box m 2 must be ccelerting up the plne s the boxes re connected. Slide 104 / 139 F y=m 2gcos F x=m 2gsin Inclined Plne nd Pulley f k System 2 The friction force is shown opposite the direction of ccelertion - of the reltive motion of the box nd the plne. The tension force is shown cting upwrds. Wht is the reltionship of this tension force with the tension force shown for System 1? Slide 105 / 139 Since the string is mssless nd it is ssumed not to stretch, the two tension forces re equl.
Inclined Plne nd Pulley System 1 Slide 106 / 139 System 2 F y=m 2gcos f k Time for the ppliction of Newton's Second Lw nd writing the equtions for these free body digrms. Try System 1 first. F x=m 2gsin Inclined Plne nd Pulley Slide 107 / 139 System 1 Tht's ll for this system. The only thing to wtch out for is tht this is non rotted coordinte system - nd y for this system will ctully equl x for the box in System 2! Flip bck to the previous slide to see this. Inclined Plne nd Pulley y-xis: Slide 108 / 139 F y=m 2gcos F x=m 2gsin f k System 2 This will be used to clculte f k: x-xis: We now hve two simultneous equtions for nd. Wht re they?
Inclined Plne nd Pulley Slide 109 / 139 System 1 FT FT Fy=m2gcos fk System 2 Fx=m 2gsin Here they re. Two equtions. Erlier, it ws stted tht the ccelertion of both boxes ws the sme - even though they re nmed differently due to the coordinte system being rotted in System 2. So, x = y =. Wht's next? Inclined Plne nd Pulley Slide 110 / 139 System 1 FT FT Fy=m2gcos fk System 2 Substitute in for x nd y nd dd the equtions together. Fx=m 2gsin cncels out! Inclined Plne nd Pulley Slide 111 / 139 We've now found the ccelertion of the two box system - nd without using single number! Tht's the wy to solve physics problems - work out the lgebr first, nd then substitute in the numbers. This hs few benefits - one of which is your techer is more pt to give you prtil credit if the work is shown. In clss, the work is more importnt thn the numericl nswer. It's only when rockets nd bridges re built tht numbers re very importnt!
Inclined Plne nd Pulley Slide 112 / 139 nother benefit is tht limiting cses cn be tken to both vlidte the lgebr nd to observe interesting phenomen. Wht if the ngle ws set equl to 0 0 or 90 0 (the two limiting cses)? Wht would the system look like? Iine rotting the incline clockwise until the incline is verticl for the 90 0 cse or rotting it counter-clockwise until the incline is horizontl for the 0 0 cse. Strt with the 0 0 system. Inclined Plne nd Pulley Slide 113 / 139 m2 m1 Did you come up with the physicl picture for the 0 0 cse? If not, strt with the lgebr nd set = 0 0 : Is this fmilir eqution? Inclined Plne nd Pulley Slide 114 / 139 m2 Set = 0 0 m2 m1 m1 We've got the cse of block pulling nother block off of tble where there is friction between the block nd the surfce.
Inclined Plne nd Pulley Slide 115 / 139 m2 m1 Time for the 90 0 system. ny luck? If not, strt with the lgebr nd set = 90 0 : The coefficient of kinetic friction is gone - becuse m 2 is brely in contct with the rotted incline. Cn you visulize it now? Inclined Plne nd Pulley Slide 116 / 139 m2 Set = 90 0 m1 1 = 2 The inclined plne becomes n twood Mchine! m 1 m 2 twood Mchine Inclined Plne nd Pulley Slide 117 / 139 System 1 m1 m2 FT FT Fy=m2gcos Fx=m 2gsin fk System 2 One more thing to tlk bout for this problem. Look t ll the work bove. You cn see the forces cting on the two msses in the ccelertion eqution. ut which force is conspicuously missing? Does tht give you n ide bout different wy to solve the problem?
Inclined Plne nd Pulley Slide 118 / 139 System 1 m1 m2 FT FT Fy=m2gcos Fx=m 2gsin fk System 2 The Tension force, doesn't pper! Wht if we considered just ONE system - the system of both msses? is now n internl force - nd it is left out of the FD becuse it consists of n infinite number of ction-rection forces within the string tht cncel out. Only externl forces contribute to the motion of system. Inclined Plne nd Pulley Slide 119 / 139 System 1 m1 m2 FT FT Fy=m2gcos Fx=m 2gsin fk System 2 Look t the bove two FDs nd the sketch. The pully's only purpose is to redirect m 1g so it cn pull m 2 up the incline. Or, m 1 pulls on m 2 with, nd m 2 pulls on m 1 with. So wht cn we cll tht? Inclined Plne nd Pulley Slide 120 / 139 m1 m2 m1g Fy=m2gcos fk System of both msses Fx=m 2gsin Tht's n ction-rection pir, nd is n internl force to the system nd will not tke prt in the FD. Since the pulley is merely redirecting the force, m 1g, we drw it on the new "System of both msses" FD, s shown bove.
Inclined Plne nd Pulley Slide 121 / 139 y-xis: m2 m1 This will be used to clculte f k: m1g Fy=m2gcos fk Fx=m 2gsin System of both msses x-xis: We did this before - the only chnge is replcing with m 1g, nd not hving to solve simultneous equtions. Inclined Plne nd Pulley Slide 122 / 139 m1 m2 One lst step is expressing x s, since tht is the only motion we're observing, nd we hve: m1g Fy=m2gcos fk System of both msses Fx=m 2gsin Slide 123 / 139 Flling Objects with ir Resistnce Return to Tble of Contents
Flling Objects with ir Resistnce Slide 124 / 139 The Kinemtics unit covered objects in free fll - where the only force cting ws the grvittionl force. The objects fll with constnt ccelertion, nd the velocity increses s liner function of time, or s the squre root of the distnce fllen. Ner the surfce of the erth, there is n opposing force to grvity for flling objects - the ir. ir is fluid, nd it provides resistnce force (lso clled drg) tht opposes the object's downwrd motion. This force increses s fctor of the object's velocity squred. t the point where this force equls the grvittionl force, the net force will be zero, nd the object will fll with constnt velocity - terminl velocity. Flling Objects with ir Resistnce Slide 125 / 139 The resistnce force cn lso be incresed by incresing the surfce re of the object flling (try this yourself with sheet of pper, nd then crumple it up nd let it fll). This is prt of the theory behind prchutes - which reduce the terminl velocity of the prchutist - good thing. Newton's Second Lw for n object with terminl velocity (=0): f = kv 2 terminl velocity Flling Objects with ir Resistnce Slide 126 / 139 more interesting problem is finding the velocity t ny point s the object flls. This requires the use of integrl clculus, specificlly, seprtion of vribles nd chnge of vribles. We're going to chnge the frictionl force to liner dependence on velocity(f = kv), so tht the integrtion is simpler. Motion in y direction is understood - drop subscript Differentil Eqution f = kv Seprtion of vribles - v on the left, nd t on the right Integrting from time, 0, to the velocity t specific time, t
Flling Objects with ir Resistnce Slide 127 / 139 Integrting from time, 0, to the velocity t specific time, t. f = kv Use chnge of vribles, to turn the integrl into something we cn work with Not mthemticlly rigorous here - we should've chnged the limits of integrtion for u, but we'll fix tht lter on ( shortcut) Flling Objects with ir Resistnce Slide 128 / 139 Go bck pge, nd see wht would hve hppened if the friction force ws kv 2. This integrl would hve required tble of integrls or solving by trigonometric substitution. Now, we substitute u=-kv bck, nd we use the originl limits of integrtion (this w shortcut) f = kv Tke the exponentil of ech side Flling Objects with ir Resistnce Slide 129 / 139 f = kv Clculte the terminl velocity for this liner frictionl force (drg): Do you see this term to the left? Mke the substitution, vt = /k This is good - s t pproches, v pproches vt.
Grphicl Representtion of Flling Objects with ir resistnce (drg) Slide 130 / 139 v -y liner g v = vt vt qudrtic t t h t = 0 v = vt Object strts t height, h, nd flls - s f(t 2 ) until it reches vt; then f(t). Slide 131 / 139 Non Uniform Circulr Motion Return to Tble of Contents Non-Uniform Circulr Motion Verticl Loop Slide 132 / 139 tn rd rd rd rd rd tn UCM ssumes constnt speed motion - which is chieved by hving only one force ct on the object - for exmple, horizontl tension force, which provides the centripetl motion. rd tn tn This works well for horizontl rottion. ut wht if you re swinging bll in verticl circle?
Non-Uniform Circulr Motion Verticl Loop Slide 133 / 139 tn rd rd rd rd rd tn Grvity provides downwrd force t ll times. This force resolves into tngentil nd rdil component. The tngentil force chnges the speed, nd the rdil force combines with the tension force to chnge the centripetl ccelertion. tn rd tn The non zero tngentil force is dded to the rdil force to crete n ccelertion vector tht points inside the circle - but not t the center. This is not UCM. Non-Uniform Circulr Motion Verticl Loop v Ttop Tbottom r T sin cos v Here is the most generl free body digrm. We re not ssuming tht the tension force stys constnt - it certinly cn if mchine is providing the force, but if it's person just swinging bll, it probbly won't be. Since this is rdil motion, the most nturl coordinte system is polr - the xes re lined up long the rdil vector nd tngent to the circle. Slide 134 / 139 Non-Uniform Circulr Motion Verticl Loop v Ttop Newton's Second Lw: Rdil: Slide 135 / 139 r Tbottom T sin cos Tngentil: v
Non-Uniform Circulr Motion Verticl Loop Slide 136 / 139 rd tn rd rd tn rd rd rd tn tn little more complex thn UCM, s the ccelertion nd velocity is different t every point on the circle, depending on nd T. Sttic Equilibrium Slide 137 / 139 There is whole field of problems in engineering nd physics clled "Sttics" tht hs to do with cses where no ccelertion occurs nd objects remin t rest. nytime we construct bridges, buildings or houses, we wnt them to remin sttionry, which is only possible if there is no ccelertion or no net force. There re two types of motion tht we need to consider (nd in both cses, motion is to be prevented!). Wht re they? Torque nd Rottionl Equilibrium Slide 138 / 139 good exmple is opening door, mking door rotte. The door does not ccelerte in stright line, it rottes round its hinges. Think of the best direction nd loction to push on hevy door to get it to rotte nd you'll hve good sense of how torque works.
Tension Force Slide 139 / 139 T y T x T x T y F x = m x = 0 T 1x - T 2x = 0 x - xis Tsin = Tsin which just confirms tht if the ngles re equl, the tensions re equl y - xis F y = m y = 0 T 1y + T 2y - = 0 Tcos + Tcos = 2Tcos = T = / (2cos) Note tht the tension rises s cos becomes smller...which occurs s pproches 90 o. It goes to infinity t 90 o, which shows tht the ropes cn never be perfectly horizontl.