Mechanics Cycle 1 Chapter 5. Chapter 5

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1 Chpter 5 Contct orces: ree Body Digrms nd Idel Ropes Pushes nd Pulls in 1D, nd Newton s Second Lw Neglecting riction ree Body Digrms Tension Along Idel Ropes (i.e., Mssless Ropes) Newton s Third Lw Bodies in Contct s Blocks or... Points BODIES in CONTACT: Consider bodies in contct with ech other nd the outside world. CONTACTS men interctions nd interctions men forces. The BODIES re books, ir molecules, horses, trins, etc., nd we focus on their 1D motion in this chpter. We won t worry bout their being twisted, rotted, squished or stretched yet. BLOCKS: So we will simply view these bodies s BLOCKS. Blocks hve mss nd the rigid shpe of boxes. Anlyzing blocks is esy but still very prcticl nd relevnt. POINTS: Insted of horse, we hve rgued tht we should use block picture (which is certinly esier to drw!). Well, for present purposes, we could just s well represent horse by POINT (which is relly esier to drw) nd forget the block ltogether! Single Block Sliding on Tble (riction Neglected) Consider severl exmples of different combintions of forces on single block, their ssocited Newton second-lw equtions, nd their fncy-shmncy free-body digrms. All forces re prllel or nti-prllel to the direction of the ccelertion, so this is truly 1D. Our convention often is to write equtions in terms of the force mgnitudes > 0 rther thn the force components x which cn be either positive or negtive. So if force is in the negtive direction, we put the minus sign in explicitly in front of the mgnitude. Plin Vnill Digrms (only horizontl forces shown) Newton s Second Lw In terms of force mgnitudes ree-body Digrms* push: m m ( x > 0) pull: m m ( x > 0) 1 m m ( 1x 1 > 0, 2x 2 > 0) 1 m m ( 1x 1 > 0, 2x 2 < 0) (to the right is the positive x-direction) *BD convention: Plce tils of ll force rrows t the point used to represent the block. While the verticl forces don t come into the horizontl force equtions, we show them nywy. The support force on the block due to the tble is N (perpendiculr or norml to the tble the tble or ny surfce cnnot exert horizontl forces until we include friction). In this simple cse, the weight Wmg of the block must blnce N by the second lw becuse the block is not even moving, let lone ccelerting, verticlly. Remember, we cn isolte ny prt of system nd drw its BD. 5-1

2 Two Blocks in Contct nd Sliding on rictionless Tble Due to Single Externl Push 1 2 externl force: A) A constrint: the blocks move together which mens their ccelertions re the sme. We define this common ccelertion to be simply (we re using mgnitudes for ccelertion, too, so, for exmple, 1x + 1 since block 1 is ccelerting to the right, etc.) 1 2 B) The system s common ccelertion is determined with the second lw pplied to the two blocks considered together s the overll system. Since we know the externl force: ( totl mss ) ( + ) externl + Agin, we use right is positive once gin. Agin, we think of the horizontl direction s the x-xis, with the positive x-xis pointing to the right. Agin, x + > 0 where insted of components, we use plus or minus mgnitudes if the direction is right or left, respectively. C) Thus we know the forces on ech individul block. Why? Since we know ech block s common ccelertion (from bove) long with its mss, then we know the force on ech from the second lw: force on : must 1 1 force on : must D) Next we cn dig inside nd clculte the contct force between the two blocks becuse, s we will show on the next pge, we now know both the force on ech individul block nd their common ccelertion. Preliminry remrks: We will drw both plin vnill digrms nd BD s for ech block. In the ll the drwings, we just guess the rrow mgnitudes (nd sometimes even the directions!). The clcultions tell us how big they re, nd confirm their directions, fterwrds. We don t dre drw ny contct force into ny picture unless it relly IS in contct with the body we re nlyzing. 5-2

3 Clculte the contct force on the first block due to the second block: irst Block (isolted) on 1 by 2 The force on block 1 is: 1 - on 1 by 2 rom the previous pge, we know: 1 + Putting the bove together: 1 - on 1 by 2 + Solve for on 1 by 2 : on 1 by Combine terms: on 1 by m m on 1 by 2 + Clculte the contct force on the second block due to the first block: Second Block (isolted) on 2 by 1 N.B. There is no here. It does not ct directly on The force on block 2 is: 2 on 2 by 1 rom the previous pge, we know: 2 + Putting the bove together: 2 on 2 by

4 Shedding light on Newton s third lw through this exmple: rom the results on the previous pge, we find the contct forces on 1 by 2 nd on 2 by 1 hve the sme mgnitudes: on 1 by 2 + on 2 by 1 While their mgnitudes re the sme, notice in the figures tht the directions of the two forces re drwn in opposite directions. It is lso importnt to note tht the two contct forces ct on two different bodies: one on block 1 nd the other on block 2. This is ll consistent with the third lw, which sys tht the contct force on block 1 due to block 2 must be equl nd opposite to the contct force on block 2 due to block 1. The third lw comes out utomticlly becuse we ssume the only contribution to externl which crries the hidden ssumption tht internl forces cncel. is, Shortcut nottion: rom now on, utomticlly put the contct forces between blocks equl nd opposite, such s shown in the following plin vnill digrm, nd mke your life esier: f f m 1 As in the bove exmple, is NOT directly pplied to the second block, so PLEASE PLEASE PLEASE drw f but NOT on in the bove digrm. If cuses f nd if the whole system ccelertes, is prtly eten up by on its wy to INDIRECTLY ccelerting. **************************************************************************************************** Problem 5-1 Two blocks with msses s shown nd on frictionless horizontl surfce re in contct with ech other. If n externl force (mgnitude) is pplied s shown on ' the left side of block 1, nd nother externl force (mgnitude ') pushes bck in the opposite direction on the right side of block 2, ) Drw the BD* with both horizontl nd verticl forces shown (lwys show ll forces in BD s) for the whole system s point nd use its horizontl forces to determine the ccelertion of the system in terms of,,, nd b) Determine the force on ech individul block in terms of,,, nd c) Drw the BD for block 2, nd, from its horizontl forces nd (b), determine the mgnitude of the force tht block 1 exerts on block 2 in terms of,,, nd d) Drw the BD for block 1, nd, from its horizontl forces nd (b), determine the mgnitude of the force tht block 2 exerts on block 1 in terms of,,, nd Of course, the nswers (c) nd (d) should gree! * We will often be sking for BD s becuse 1) teching reserch hs shown tht when students drw them, they tend to do better job on writing down nd understnding the second lw, nd 2) experts need them s much s beginners! The BD s help us blend our physicl mentl spce better with our mth mentl spce! **************************************************************************************************** 5-4

5 Pulling With Ropes: The orce is Clled the Tension T Consider two blocks tied together by rope. Also, consider n externl force pulling one of the blocks (perhps through nother rope). Becuse of this externl force, the first mss pulls on the in-between rope nd thus the in-between rope pulls on the second mss. The in-between rope hs become tense which leds to nturl nme, tension, for the force it exerts on both blocks:? In this chpter, we use light ropes, which mens tht we re using mssless ropes, which we refer to s idel ropes. A nice connection to the previous pge: Consider rope whose mss is not negligible compred with the other msses in the system. If we pull on one end of this rope nd there re no other externl forces, the rope ccelertes such tht the rope s mss ets up the force, nd the tension force decreses long the length of the rope s we go frther wy from our hnds. By the wy, if the externl force is zero on mssive rope, the tension is constnt long the rope see below nd Ch. 5+. Bck to idel ropes nd their feelings: We go bck now to idel ropes where the tension is the sme ll long n idel rope, whether the rope ccelertes or not! Let s prove this. irst, wht forces does the in-between rope in the bove picture feel? The only externl forces cting directly on this rope re the pulls from the two blocks (but nothing directly from, remember!). Thus, drw the following BD nd note the opposite directions (rrows) for these forces. Newton s 2 nd lw pplied to this BD in terms of the force mgnitudes nd with minus sign for the rrow pointing to the left: on rope on rope by m1 on - rope by m2 m rope 0 I the rope is mssless (OR not ccelerted) on rope Hence, for mssless rope, by m1 on rope on m by m2, nd by Newton s third lw, 1 on m by rope 2 by rope Therefore, the in-between rope in the picture bove pulls forwrd on block 2 with the sme on m mgnitude of force tht it pulls bckwrd on block 1: 1 on m by rope 2 by rope Second, using the sme rgument on ny little piece of the rope, the force must be zero on tht piece becuse it s mssless. Thus tht piece is pulled left nd right by forces with identicl mgnitudes, nd thus tht little piece pulls bck on its two pullers with identicl forces by Newton s third lw! So every lousy little piece of the rope feels the sme force pulling on it left nd right. This is wht we men by constnt tension long rope. Usul nottion: We will lwys cll the force long the rope, the tension T. It might be less confusing if we clled it T but why strt now in trying to be cler! 5-5

6 Anlysis of the two blocks with rope between them nd n externl pull: Like the contct force f in the push contct cse erlier, we re interested in the tension T tht rises in the two-block system, given nd the msses nd. Putting double rrows on the tense 1 rope in figures like this one reminds us tht the rope pulls left nd right with equl force. 2 T We strt gin with the constrint tht the blocks move together so tht they hve common ccelertion, 1 2. Now, we utomticlly stisfy Newton s third lw by tking the sme tension force T to be pulling on both msses. Thus, if we wrote down three second-lw equtions (one for the whole system, one for the first block nd one for the second block), one would be redundnt. This is consistent with the fct tht there re only two unknowns nd T so we only need two independent equtions. This is why the dditionl eqution is redundnt. Our procedure: irst, we will drw the BD s, guessing tht T <, which will be confirmed by the clcultion. (You won t get thrown in jil if you misguess the length of your rrows.) Second, for ech BD, write the two corresponding second-lw equtions for the horizontl forces. Third, solve them for T nd. Our detils: Plin vnill nd BD for s the system: T The second lw for the horizontl motion of : - T Plin vnill nd BD for s the system: T The second lw for the horizontl motion of : T The two boxed equtions re system of equtions to be solved for the two unknowns, nd T, on the next pge. 5-6

7 Solution by substitution (just like on Sesme Street): Plug T into - T : - + ( + ) Solve for : + Solve for T by substituting bck in : T + We see indeed tht nonzero gives rise to nonzero tension T, but T < becuse ets up prt of the force for this ccelerted system. Solution by combintion: By combintion we men tht we will dd or subtrct the two equtions. The ide is to try to cncel out get rid of one of the unknowns. Add: T + - T ( + ) Once we hve ( + ), solve for nd substitute this into T to get the sme nswers s bove: /( + ) nd T /( + ). Notice how T cnceled in the bove ddition. This mth corresponds to the cncelltion of internl forces, ccording to Newton s third lw. We lredy built Newton s third lw into our system, when we used equl nd opposite tension forces in the two individul block BD s. Comments: Verify the BD for the whole system gives nothing new: The second lw for the horizontl motion of + : ( + ) Clerly, this is the sme s wht we hd. It is short-cut you could use in plce of either BD of the individul blocks. Notice there s no T here, consistent with the third lw. In the ltest exmple, the forces hve been referred to s forces of tension. In the exmple t the beginning of the chpter, we pressed one block into nother. The forces there could thus be clled forces of compression. 5-7

8 **************************************************************************************************** Problem 5-2 Two blocks with msses s shown nd on frictionless horizontl surfce re connected by light rope. Assume the force mgnitudes 1, 2 nd msses, re known, long with the directions of the forces s shown. 1 2 T ) Before you strt, wht would you guess the tension T would be if 1 2? Don't clculte, just guess. You will get full credit for ny honest guess. You cn check this below when you re done clculting. Now consider ny two of the following three systems: the whole system, the isolted system, or the isolted system. b) Drw BD s for the two systems you consider. Remember, ech BD involves ONLY those forces DIRECTLY on the considered body. We tke this opportunity to remind you tht you don t hve to know which force rrows re longer. We hven t told you yet who is bigger 1 or 2. c) Derive two equtions from considertion of the horizontl forces in your BD s using the convention tht the common ccelertion is defined to the right.* If the system turns out to be ccelerting to the left, will be negtive. d) Solve your two equtions for nd T in terms of 1, 2,, nd. Check your formul for : must be positive if 1 < 2 (the system would ccelerte to the right) nd must be negtive if 1 > 2 (ccelertion is to the left). e) Now ssume 1 2 in your nswer in (d) nd see if your guess in () ws right. If it ws wrong, try hrd to nlyze why, so your physics intuition cn be strengthened. Guessing is lwys good self-disciplinry step. We hve our own preconceptions nd, if they fil to mtch up with the world round us, we need to work relly hrd to reshpe them. (Of course, if your own privte world is relly much nicer thn the one described in these notes, frnkly, you might like to keep it just s it is.) f) inlly, independent of whether 1 2, wht is the force on ny prt of the in-between rope with tension T in it? This nswer should lso help you in building n intuition bout the mening of tension nd questions like ()! **************************************************************************************************** * Here we define s the x-component of the ccelertion vector with the convention of positive x-xis pointing to the right: x. Thus if we find < 0, it s ccelerted left. 5-8

9 Shedding More Light on Newton s Third Lw In both exmples in this chpter, good old Newton s third lw tells us tht the force block 1 exerts on block 2 must be equl nd opposite to the force block 2 exerts on block 1. Let s bet it to deth by prphrsing wht we sid in Chpter 0: If the force on body 1 is due to n interction with body 2, then body 2 feels n equl nd opposite force due to body 1. on 1 by 2 - on 2 by 1 Comments: ) This mens the two forces hve the sme mgnitude nd the opposite direction. b) Thus, "single isolted force" is impossible! We need two or more bodies for interctions/forces! A force mens n interction!! If there s n ction, there must be rection. We must hve ction-rection. Stop us before we try to sy it yet nother wy! c) Here is net thing. An ction-rection pir of forces lwys involves two forces tht re ech cting on two different bodies!!!!!!!! They my be equl nd opposite, but they ct on different bodies nmely, the two bodies intercting with ech other. d) Newton s third lw sys tht, for n BD of whole system mde up of bunch of bodies, we cn ignore internl forces (like f nd T in the two exmples). So the force on the whole system is just the sum of ll externl forces. See lter chpters for lots of exmples. ******************************************************************************************** Problem 5-3 Reltively quick nd esy, but... importnt! ) If you push ginst tree with some force of mgnitude, wht is the force mgnitude nd reltive direction with which the tree pushes bck on you? b) Suppose person replces the tree nd you gin push ginst her with the sme force of mgnitude s in (). If the two of you re t stlemte, wht is the force mgnitude nd direction with which this person pushes bck on you? c) You re tlking bout n ction-rection force pir in ech of the cses () nd (b). Notice how ech force in given pir cts on different body! Explin why the support force on your shoes is NOT n ction-rection pir with your weight. Wht is the other member of the ction-rection pir involving your weight? Wht is the other member of the ction rection pir involving the support force on your shoes? continued 5-9

10 d) Bsed on wht we hve lerned, we cn quickly figure out nswers to the next two questions before we get to them in the second cycle. Are the contct forces between the blocks in the following string of identicl blocks (mss m) the sme, lrger, or smller s we go from left to right? Plese think bout this by drwing n BD for the lst block N isolted, from which the second lw for the horizontl force tells us tht the contct force on N is simply equl to m. Now drw n BD for the two blocks, N-1 nd N, together. Wht must the contct force due to N-2 on the two-block system equl now? Do you see pttern emerging? In greement with being eten up s we go long the row of blocks? e) Wht bout this string? Tht is, re the contct forces between the blocks in the following string of blocks the sme, lrger, or smller s we go from left to right? Plese use BD s nd the second lw gin to be sure of your nswer. ******************************************************************************************** 5-10

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